%%latex file
%\documentstyle[12pt,amssymb]{amsart}
\documentclass[12pt]{amsart} \usepackage{amssymb} \usepackage{mathabx} \usepackage{bm} \usepackage{wasysym} \usepackage{graphicx} \usepackage{stmaryrd}
\usepackage{imakeidx}
\usepackage{color,hyperref}
\hypersetup{colorlinks,breaklinks,
             linkcolor=blue,urlcolor=blue,
           anchorcolor=blue,citecolor=blue}
\newenvironment{pf}{\proof[\proofname]}{\endproof}

\makeindex[name=terms,title=Index of terms, columns=2]

\renewcommand\baselinestretch{1.1}
\renewcommand\theequation{\arabic{section}.\arabic{equation}}

\newcommand{\wrlab}[1]{\label{#1}}
\renewcommand{\arraystretch}{1}

\newtheorem{thm}{THEOREM}[section]
\numberwithin{thm}{subsection}
\newtheorem{axiom}[thm]{AXIOM}
\newtheorem{defn}[thm]{DEFINITION}

\newcommand{\secref}[1]{\S\ref{#1}}

\newcommand{\tref}[1]{Theorem~\ref{#1}}
\newcommand{\aref}[1]{Axiom~\ref{#1}}
\newcommand{\dref}[1]{Definition~\ref{#1}}

% BLACK BOARD BOLD

\def\Bbb{\mathbb}  %needed in Nancy

\def\A{{\mathbb A}} \def\B{{\mathbb B}} \def\C{{\mathbb C}} \def\D{{\mathbb D}}
\def\N{{\mathbb N}} \def\I{{\mathbb I}} \def\Q{{\mathbb Q}} \def\R{{\mathbb R}}
\def\Z{{\mathbb Z}} 

\def\T{{\mathbb T}}

\def\P{{\mathbb P}} 

% BOLD
\def\bfzero{{\bf 0}}
\def\bfP{{\bf P}}
\def\bfQ{{\bf Q}}
\def\bfS{{\bf S}}
\def\bfT{{\bf T}}

% CAL 
\def\Cal{\mathcal} \def\scra{{\mathcal A}} \def\scrb{{\mathcal B}}
\def\scrd{{\mathcal D}} \def\scrf{{\mathcal F}} \def\scrg{{\mathcal G}}
\def\scri{{\mathcal I}} \def\scrk{{\mathcal K}} \def\scrl{{\mathcal L}}
\def\scrm{{\mathcal M}} \def\scrp{{\mathcal P}} \def\scrq{{\mathcal Q}}
\def\scrr{{\mathcal R}} \def\scrs{{\mathcal S}} \def\scrt{{\mathcal T}}
\def\scru{{\mathcal U}} \def\scrv{{\mathcal V}} \def\scrw{{\mathcal W}}
\def\scre{{\mathcal E}} \def\scrx{{\mathcal X}} \def\scry{{\mathcal Y}}
\def\scrz{{\mathcal Z}} \def\scrh{{\mathcal H}} \def\scrc{{\mathcal C}}
\def\scrn{{\mathcal N}} \def\scro{{\mathcal O}}
\def\scrpf{{\scrp\scrf}}
\def\scral{{\scra\scrl}}
\def\scrsb{{\scrs\scrb}}

% HATTED ROMAN
\def\hatd{{\widehat{d}}}
\def\hatX{{\widehat{X}}}

% HATTEDCAL
\def\hatscrp{{\widehat{\scrp}}}
\def\hatscrp{{\widehat{\scrp}}}
\def\hatpi{{\widehat{\pi}}}

% HATTED BB BOLD
\def\hatbbR{{\widehat{\R}}}

\newcommand\scrlito{
  \mathchoice
    {{\scriptstyle\mathcal{O}}}% \displaystyle
    {{\scriptstyle\mathcal{O}}}% \textstyle
    {{\scriptscriptstyle\mathcal{O}}}% \scriptstyle
    {\scalebox{.7}{$\scriptscriptstyle\mathcal{O}$}}%\scriptscriptstyle
  }

\def\scrbigo{{\widehat{\scro}}}

% LIE ALGEBRAS

\font\tenolde=eufm10 at 10pt
\font\sevenolde=eufm7
\font\fiveolde=eufm5
\newfam\oldefam
\textfont\oldefam=\tenolde
\scriptfont\oldefam=\sevenolde
\scriptscriptfont\oldefam=\fiveolde
\def\fraktwo{\fam\oldefam\tenolde}

\newfam\cyrfam
\font\tencyr=wncyr10
\def\cyr{\fam\cyrfam\tencyr}

\newcommand\sbullet[1][.75]{\mathbin{\vcenter{\hbox{\scalebox{#1}{$\bullet$}}}}}

\def\aff{{\mathfrak{aff}(\R)}} \def\La{{\mathfrak a}}
\def\Lb{{\mathfrak b}} \def\Lc{{\mathfrak c}} \def\Lg{{\mathfrak g}}
\def\Lh{{\mathfrak h}} \def\Lk{{\mathfrak k}} \def\Ll{{\mathfrak l}}
\def\Lm{{\mathfrak m}} \def\Ln{{\mathfrak n}} \def\Lr{{\mathfrak r}}
\def\Ls{{\mathfrak s}} \def\Lt{{\mathfrak t}} \def\Lw{{\mathfrak w}}
\def\Lq{{\mathfrak q}} \def\Lu{{\mathfrak u}} \def\Lv{{\mathfrak v}}
\def\Lz{{\mathfrak z}} \def\Lsl{\mathfrak{sl}} \def\Lgl{\mathfrak{gl}}
\def\Lp{{\mathfrak p}} \def\Lf{{\mathfrak f}} \def\Lx{{\mathfrak x}}
\def\Lpsl{\mathfrak{psl}}
\def\Lso{{\mathfrak{so}}} 
\def\Lsu{{\mathfrak{su}}} 
\def\Lsp{{\mathfrak{sp}}} 
\def\orthog{{\mathfrak{o}}} 
\def\tilso{{\mathfrak{so}}}
\def\tilco{{\mathfrak{co}}} 
\def\co{\mathfrak{co}}

% OVERLINE
\def\barscru{{\overline{\scru}}}
\def\barPsi{{\overline{\Psi}}}
\def\barA{{\overline{A}}}
\def\barB{{\overline{B}}}
\def\barC{{\overline{C}}}
\def\barD{{\overline{D}}}
\def\barE{{\overline{E}}}
\def\barF{{\overline{F}}}
\def\barbarF{{\overline{\overline{F}}}}
\def\barG{{\overline{G}}}
\def\barH{{\overline{H}}}
\def\barI{{\overline{I}}}
\def\barJ{{\overline{J}}}
\def\barK{{\overline{K}}}
\def\barL{{\overline{L}}}
\def\barM{{\overline{M}}}
\def\barN{{\overline{N}}}
\def\barP{{\overline{P}}}
\def\barbarP{{\overline{\overline{P}}}}
\def\barQ{{\overline{Q}}}
\def\barR{{\overline{R}}}
\def\barS{{\overline{S}}}
\def\barT{{\overline{T}}}
\def\barU{{\overline{U}}}
\def\barV{{\overline{V}}}
\def\barW{{\overline{W}}}
\def\barbarW{{\overline{\overline{W}}}}
\def\barX{{\overline{X}}}
\def\barY{{\overline{Y}}}
\def\barZ{{\overline{Z}}}
\def\barLr{{\overline{\Lr}}}
\def\barLg{{\overline{\Lg}}}
\def\barLh{{\overline{\Lh}}}
\def\barLp{{\overline{\Lp}}}
\def\barLn{{\overline{\Ln}}}
\def\barLm{{\overline{\Lm}}}
\def\barLl{{\overline{\Ll}}}
\def\bara{{\overline{a}}}
\def\bare{{\overline{e}}}
\def\barf{{\overline{f}}}
\def\barg{{\overline{g}}}
\def\bark{{\overline{k}}}
\def\baru{{\overline{U}}}
\def\barv{{\overline{v}}}
\def\barw{{\overline{w}}}
\def\barz{{\overline{z}}}
\def\barbeta{{\overline{\beta}}}
\def\bargamma{{\overline{\gamma}}}
\def\barI{{\overline{I}}}

% TILDE
\def\tilvec{\widetilde{\vec}}

% HAT
\def\hatA{{\widehat{A}}}
\def\hatC{{\widehat{C}}}
\def\hatH{{\widehat{H}}}
\def\hatI{{\widehat{I}}}
\def\hatN{{\widehat{N}}}
\def\hatQ{{\widehat{Q}}}
\def\hatR{{\widehat{R}}}
\def\hatS{{\widehat{S}}}
\def\hatV{{\widehat{V}}}
\def\hatt{{\widehat{t}}}
\def\hatu{{\widehat{u}}}
\def\hatv{{\widehat{v}}}
\def\hatw{{\widehat{w}}}
\def\hatx{{\widehat{x}}}
\def\haty{{\widehat{y}}}
\def\hatz{{\widehat{z}}}
\def\hatalpha{{\widehat{\alpha}}}
\def\hatrho{{\widehat{\rho}}}
\def\hatkappa{{\widehat{\kappa}}}
\def\hatlambda{{\widehat{\lambda}}}
\def\hatsigma{{\widehat{\sigma}}}
\def\hatmu{{\widehat{\mu}}}
\def\hatnu{{\widehat{\nu}}}
\def\hattau{{\widehat{\tau}}}

% CHECK
\def\ckt{{\widecheck{t}}}
\def\ckw{{\widecheck{w}}}
\def\ckx{{\widecheck{x}}}
\def\cky{{\widecheck{y}}}
\def\ckz{{\widecheck{z}}}
\def\cksigma{{\widecheck{\sigma}}}
\def\cktau{{\widecheck{\tau}}}
\def\ckH{{\widecheck{H}}}
\def\ckN{{\widecheck{N}}}
\def\ckS{{\widecheck{S}}}

\def\ds{\displaystyle}
\def\bfindex[#1]{{\bf#1}\index[terms]{#1}}

\def\timestimes{{\times\!\!\times}}

%operators
\def\Qd{{\rm Qd}}
\def\PM{{\rm PM}}
\def\RC{{\rm RC}}
\def\IC{{\rm IC}}
\def\TS{{\rm TS}}
\def\ES{{\rm ES}}
\def\sgn{{\rm sgn}}
\def\PF{{\rm PF}}
\def\set{{\rm set}}
\def\bi{{\rm bi}}
\def\fn{{\rm fn}}
\def\ss{{\rm setset}}
\def\sf{{\rm setfn}}
\def\fs{{\rm fnset}}
\def\seq{{\rm seq}}
\def\EUCL{{\rm EUCL}}
\def\TNSR{{\rm TNSR}}
\def\starstar{{**}}
\def\fincap{{\hbox{{\rm fin}}\,\cap}}
\def\fincup{{\hbox{{\rm fin}}\,\cup}}
\def\ctblcap{{\hbox{{\rm ctbl}}\,\cap}}
\def\ctblcup{{\hbox{{\rm ctbl}}\,\cup}}
\def\finamalg{{\hbox{{\rm fin}}\,\amalg}}
\def\two{{\rm two}}
\def\three{{\rm three}}
\def\ten{{\rm ten}}
\def\Lin{{\rm Lin}}
\def\Bilin{{\rm Bilin}}
\def\Trilin{{\rm Trilin}}
\def\PS{{\rm PS}}
\def\UE{{\rm UE}}
\def\FV{{\rm FV}}
\def\CH{{\rm CH}}
\def\DQ{{\rm DQ}}
\def\SS{{\rm SS}}
\def\SUMS{{\rm SUMS}}
\def\res{{\rm res}}
\def\add{{\rm add\,}}
\def\FS{{\rm FS}}
\def\LIMS{{\rm LIMS}}
\def\NOT{{\rm NOT}}
\def\ORB{{\rm ORB}}
\def\dlin{{\rm dlin}}
\def\LINS{{\rm LINS}}
\def\opLIMS{\mathop{\LIMS}}
\def\IDS{{\rm IDS}}
\def\UL{{\rm UL}}
\def\UB{{\rm UB}}
\def\LB{{\rm LB}}
\def\LP{{\rm LP}}
\def\LPD{{\rm LPD}}
\def\Isol{{\rm Isol}}
\def\Cl{{\rm Cl}}
\def\sCl{{\rm sCl}}
\def\id{{\rm id}}
\def\dim{{\rm dim\,}}
\def\DIMS{{\rm DIMS\,}}
\def\Bas{{\rm Bas\,}}
\def\Int{{\rm Int}}
\def\IntD{{\rm IntD}}
\def\det{{\rm det\,}}
\def\exp{{\rm exp}}
\def\End{{\rm End}}
\def\SO{{\rm SO}}
\def\CO{{\rm CO}}
\def\O{{\rm O}}
\def\Stab{{\rm Stab}}
\def\Fix{{\rm Fix}}
\def\GL{{\rm GL}}
\def\Ad{{\rm Ad}}
\def\Aut{{\rm Aut}}
\def\der{{\rm der}}
\def\ad{{\rm ad}}
\def\diag{{\rm diag}}
\def\Mink{{\rm Mink}}
\def\Tay{{\rm Tay}}
\def\lin{{\rm lin}}
\def\switch{{\rm switch}}
\def\SOtwortwo{{\SO(2)\ltimes\R^2}}
\def\O{{\rm O}}
\def\OB{{\rm OB}}
\def\SO{{\rm SO}}
\def\SL{{\rm SL}}
\def\PSL{{\rm PSL}}
\def\SBF{{\rm SBF}}
\def\STF{{\rm STF}}
\def\SM{{\rm SM}}
\def\QF{{\rm QF}}
\def\Id{{\rm Id}}
\def\dom{{\rm dom}}
\def\im{{\rm im}}
\def\dd{{\rm dd}}
\def\dct{{\rm dct}}
\def\sym{{\rm sym}}
\def\Sym{{\rm Sym}}
\def\as{{\rm as}}
\def\mm{{\rm mm}}
\def\cof{{\rm cof}}
\def\adj{{\rm adj}}
\def\det{{\rm det}}
\def\DET{{\rm DET}}
\def\ADJ{{\rm ADJ}}
\def\DNZ{{\hbox{\rm DNZ}}}
\def\CVZ{{\hbox{\rm CVZ}}}
\def\BNZ{{\hbox{\rm BNZ}}}
\def\ev{{\rm ev}}
\def\sl{{\rm s\ell}}

%misc
\def\vep{{\varepsilon}}
\def\bksl{{\backslash}}
\def\sigmaalg{{\rm s.alg}}
\def\drta{\dashrightarrow}
\def\ltlt{{<\!\!<}}
\def\phx{\phantom{x}}

\def\Xint#1{\mathchoice
{\XXint\displaystyle\textstyle{#1}}%
{\XXint\textstyle\scriptstyle{#1}}%
{\XXint\scriptstyle\scriptscriptstyle{#1}}%
{\XXint\scriptscriptstyle\scriptscriptstyle{#1}}%
\!\int}
\def\XXint#1#2#3{{\setbox0=\hbox{$#1{#2#3}{\int}$ }
\vcenter{\hbox{$#2#3$ }}\kern-.6\wd0}}
\def\ddashint{\Xint=}
\def\dashint{\Xint-}

\begin{document}
\bibliographystyle{alpha}

%TITLE: 

\title[Class Notes]
{Class Notes}

\author{Scot Adams}

\date{Printout date: \today}

\maketitle

\tableofcontents

\vfil\eject
 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Foundations}

\subsection{Abbreviations}
\quad
\vskip.1in\noindent
We will be using the following abbreviations:
\begin{center}
\begin{tabular}{|c|l|}
\hline
 $\forall$ & forall \qquad (or, sometimes, for any) \\ 
\hline
 $\exists$ & there exists \qquad (or, sometimes, there exist) \\ 
\hline
 s.t. & such that \\ 
\hline
 $\neg$ & not \\ 
\hline
 \& & and \\ 
\hline
 $\vee$ & or \\ 
\hline
 $\Rightarrow$ & implies \\ 
\hline
\end{tabular}
\end{center}

\vskip.1in\noindent
For any text strings $A$ and $B$,\\
\phx\qquad $A\,\,\Leftrightarrow\,\,B$ \qquad\qquad
means \qquad\qquad $[\,\,A\,\Rightarrow\,B\,\,]\,\,\&\,\,[\,\,B\,\Rightarrow\,A\,\,]$.
\vskip.1in\noindent
For any text string $A$,\\
\phx\qquad\quad $\not\exists\,A$ \qquad\qquad
$\,\,\,\,\,\,$ means $\,\,\,$ \qquad\quad\qquad\qquad $\neg\,(\,\exists\,A)$.

\subsection{The logic purist}
\quad
\vskip.1in\noindent
Mathematics is Truth.

We won't be purists about this, but we do describe briefly what how the purist would like
mathematics to be organized, beginning with a finite alphabet which would include:
\begin{itemize}
\item[]lowercase Roman letters: \quad a,b,c,\ldots,z \qquad ;
\item[]uppercase Roman letters: \quad A,B,C,\ldots,Z \qquad ;
\item[]lowercase italic Roman letters: \quad $a$,$b$,$c$,\ldots,$z$ \qquad ;
\item[]uppercase italic Roman letters: \quad $A$,$B$,$C$,\ldots,$Z$ \qquad ;
\item[]uppercase script letters: \quad $\scra$,$\scrb$,$\scrc$,\ldots,$\scrz$ \qquad ;
\item[]uppercase blackboard bold letters: \quad $\A$,$\B$,$\C$,\ldots,$\Z$ \qquad ;
\item[]lowercase Greek letters: \quad $\alpha$,$\beta$,$\gamma$,\ldots,$\omega$ \qquad ;
\item[]some uppercase Greek letters: \quad $\Gamma$,$\Delta$,$\Theta$,$\Lambda$,$\Xi$,$\Pi$,$\Sigma$,$\Phi$,$\Psi$,$\Omega$ \qquad ;
\item[]digits: \quad $0$,\ldots,$9$ \qquad ; \qquad\qquad the symbol $\infty$ \qquad ;
\item[]the abbreviations from the preceding section: \quad $\forall$,\ldots,$\not\exists$ \qquad ;
\item[]more special characters: \quad $\in$,$=$,$<$,$($,$)$ \qquad ;
\item[]a blank space to separate words \qquad ;
\item[]the symbol $\frownie$ \qquad .
\end{itemize}
According to the purist, the entire alphabet should be explained at the start and no characters added later.

We are not purists; later, we'll, in fact, add many characters, {\it e.g.}\\
\phx\qquad\qquad\qquad $>$,$\le$,$\ge$,$\emptyset$,$\notin$,$\bigcup$,$\bigcap$,$\subseteq$, \qquad {\it etc.}\\
\vskip.1in\noindent
Note that $\in$ is not the same as the Greek letter $\vep$.\\
Note that $\emptyset$ is not the same as the Greek letter $\phi$.\\

The purist would have us give a list of strings called \bfindex[axioms].
There should be one string on each line.
The list may be infinite, but there should be an algorithm (the \bfindex[axiom algorithm]) that prints it out.
For example, the first two axioms might be:\\
\phx\qquad $\forall x,\,\,x=x$\\
\phx\qquad $\forall x,\,\,x/0=\frownie$

The purist would ask for an algorithm (the \bfindex[inference algorithm])
that would take, as input, a FINITE list of strings (one per line),
and then produce, as ouput, a FINITE list of strings (one per line).

Typically, for example, if the two axioms above were input into a typical inference algorithm,
then, somewhere in the output, we would find a line that reads:\\
\phx\qquad $(\,\,\forall x,\,\,x=x\,\,)\,\,\&\,\,(\,\,\forall x,\,\,x/0=\frownie\,\,)$\\
The typical inference algorithm would take each pair of input lines and, somewhere in the output,
produce a line obtained by surrounding each of the two by parentheses and then concatenating them with an
ampersand, $\&$, in between.

This carries the idea that, if you know two things separately, then you know both of them are true together.

The purist also asks that every line of input to the inference algorithm is one of the lines of output.

This carries the idea that, if you know something, then you know it.

We won't go into details about all the requirements of the inference algorithm, and it might vary depending on
the exact nature of the mathematics you're trying to develop.
The basic idea is that the purist wants complete
clarity on what kinds of statements can be inferred when other statements are assumed.

Once the axiom algorithm and inference algorithm are written, we write a \bfindex[theorem algorithm] that does the following:\\
\phx\qquad create the first 100 axioms,\\
\phx\qquad\qquad\qquad and call them {\bf the first batch of axioms},\\
\phx\qquad input the first batch of axioms into the inference algorithm,\\
\phx\qquad\qquad\qquad and call the output the {\bf first batch of theorems},\\
\phx\qquad print out the first batch of theorems,
\\
\phx\qquad create the next 100 axioms,\\
\phx\qquad\qquad\qquad and call them {\bf the second batch of axioms},\\
\phx\qquad append the second batch of axioms\\
\phx\qquad\qquad\qquad to the end of the first batch of theorems,\\
\phx\qquad\qquad\qquad and call the result the {\bf first appended list},\\
\phx\qquad input the first appended list into the inference algorithm,\\
\phx\qquad\qquad\qquad and call the output the {\bf second batch of theorems},\\
\phx\qquad print out the second batch of theorems,
\\
\phx\qquad create the next 100 axioms,\\
\phx\qquad\qquad\qquad and call them {\bf the third batch of axioms},\\
\phx\qquad append the third batch of axioms\\
\phx\qquad\qquad\qquad to the end of the second batch of theorems,\\
\phx\qquad\qquad\qquad and call the result the {\bf second appended list},\\
\phx\qquad input the second appended list into the inference algorithm,\\
\phx\qquad\qquad\qquad and call the output the {\bf third batch of theorems},\\
\phx\qquad print out the third batch of theorems,
\\
\phx\qquad create the next 100 axioms,\\
\phx\qquad\qquad\qquad and call them {\bf the fourth batch of axioms},\\
\phx\qquad append the fourth batch of axioms\\
\phx\qquad\qquad\qquad to the end of the third batch of theorems,\\
\phx\qquad\qquad\qquad and call the result the {\bf third appended list},\\
\phx\qquad input the third appended list into the inference algorithm,\\
\phx\qquad\qquad\qquad and call the output the {\bf fourth batch of theorems},\\
\phx\qquad print out the fourth batch of theorems,
\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad {\it etc.}

\vskip.2in

The theorem algorithm produces, as output, an infinite list of strings (one per line).
These strings are called \bfindex[theorems].\\

The logic purist says, if you think some string is true, you just have to wait to see if it appears in the list of theorems.

If it's NOT true, you have to wait forever to find that out.

We are not logic purists, but the possibility of such a pure system underlies everything that follows.

We start over.

\subsection{Some basic set theory}
\quad
\vskip.1in\noindent
\begin{defn}
By an \bfindex[object], we mean:\\
\phx\qquad a set \quad or \quad a real number \quad or \quad $\infty$ \quad or \quad $-\infty$ \quad or \quad $\frownie$.
\end{defn}

\begin{defn}
Let $x$ be an object and let $S$ be a set.\\
By $x\notin S$, we mean: $\neg\,(\,x\in S\,)$.
\end{defn}

\begin{defn}
Let $S$ and $T$ be sets. \quad
Then:\\
\phx\qquad $S\bigcup T:=\{x\,|\,(x\in S)\,\vee\,(x\in T)\}$,\\
\phx\qquad $S\bigcap T:=\{x\,|\,(x\in S)\,\&\,(x\in T)\}$ \quad and\\
\phx\qquad $S\bksl T:=\{x\,|\,(x\in S)\,\&\,(x\notin T)\}$.\\
\end{defn}

\begin{thm}
Let $S:=\{1,2,3\}$, $T:=\{3,4,5\}$. \quad
Then:\\
\phx\qquad $S\bigcup T=\{1,2,3,3,4,5\}=\{1,2,3,4,5\}$ \quad $\&$\\
\phx\qquad $S\bigcap T=\{3\}$ \quad $\&$ \quad $S\bksl T=\{1,2\}$ \quad $\&$ \quad $T\bksl S=\{4,5\}$.
\end{thm}

\begin{proof}
Omitted.
\end{proof}

\begin{defn}
We define:\\
\phx\qquad $\emptyset:=\{\,\,\}$,\\
\phx\qquad $\N:=\{1,2,3,\ldots\}$,\\
\phx\qquad $\N_0:=\{0,1,2,3,\ldots\}$,\\
\phx\qquad $\Z:=\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}$,\\
\phx\qquad $\Q:=\{k/\ell\,|\,k\in\Z,\,\ell\in\N\}$ \qquad and\\
\phx\qquad $\R:=\{\,\hbox{real numbers}\,\}$.
\end{defn}

It is a nontrivial theorem that there are elements of $\R$ that are not elements of $\Q$.
In particular, $\sqrt{2}\in\R\bksl\Q$.
Proving this would require us to delve more deeply into the axioms of $\R$ than we have time for.

\subsection{Some axioms}
\quad
\vskip.1in\noindent

\begin{defn}
Let $S$ and $T$ be sets.\\
\phx\qquad\qquad\qquad\qquad Then $S\subseteq T$ means: $\forall x\in S$, $x\in T$.\\
\phx\qquad\qquad\qquad\qquad Also, $T\supseteq S$ means: $S\subseteq T$.\\
\phx\qquad\qquad\qquad\qquad Also, $S\not\subseteq T$ means: $\neg(S\subseteq T)$.\\
\phx\qquad\qquad\qquad\qquad Also, $T\not\supseteq S$ means: $S\not\subseteq T$.\\
\phx\qquad\qquad\qquad\qquad Also, $S\subsetneq T$ means: $(\,S\subseteq T\,)\,\&\,(\,S\ne T\,)$.\\
\phx\qquad\qquad\qquad\qquad Also, $T\supsetneq S$ means: $S\subsetneq T$.
\end{defn}

The next result is an axiom, called the \bfindex[Axiom of Extensionality]:

\begin{axiom}
Let $S$ and $T$ be sets.\\
\phx\qquad Then: \qquad
$[\,\,S=T\,\,]\,\,\Leftrightarrow\,\,[\,\,(\,S\subseteq T\,)\,\,\&\,\,(\,T\subseteq S\,)\,\,]$.
\end{axiom}

\begin{thm}
Let $A:=\{1,2,3\}$, $B:=\{1,2,3,4\}$, $C:=\{2,3,4\}$.\\
Then: \quad $(\,\,A\subsetneq B\,\,)\,\,\&\,\,(\,\,B\not\subseteq A\,\,)$\\
\phx\quad\quad $\,\,\&\,\,(\,\,B\not\subseteq C\,\,)\,\,\&\,\,(\,\,C\subsetneq B\,\,)$\\
\phx\quad\quad $\,\,\&\,\,(\,\,A\not\subseteq C\,\,)\,\,\&\,\,(\,\,C\not\subseteq A\,\,)$.
\end{thm}

In the preceding theorem, the sets $A$ and $C$ are said to be \bfindex[incomparable] exactly because: \qquad
$(\,\,A\not\subseteq C\,\,)\,\,\&\,\,(\,\,C\not\subseteq A\,\,).$

\begin{thm}
$\N\subsetneq\N_0\subsetneq\Z\subsetneq\Q\subsetneq\R$.
\end{thm}

\begin{defn}
We define:\\
\phx\qquad\qquad\qquad $\R^*\,\,:=\,\,\,\,\R\,\bigcup\,\{\,\infty\,,\,-\infty\,\}$ \qquad and\\
\phx\qquad\qquad\qquad $\Z^*\,\,:=\,\,\,\,\Z\,\bigcup\,\{\,\infty\,,\,-\infty\,\}$ \qquad and\\
\phx\qquad\qquad\qquad $\N^*\,\,:=\,\,\,\,\N\,\bigcup\,\{\,\infty\,\}$ \qquad and\\
\phx\qquad\qquad\qquad $\N_0^*\,\,:=\,\,\N_0\,\bigcup\,\{\,\infty\,\}$.
\end{defn}

\begin{axiom}
\qquad $\forall x$, \quad $x=x$.
\end{axiom}

\begin{axiom}
\qquad $\forall x$, \quad $x/0=\frownie$.
\end{axiom}

\begin{axiom}
\qquad $\forall$set $S$, \quad $S\ne\frownie\notin S$.
\end{axiom}

In the preceding axiom, ``$\,\,S\ne\frownie\notin S\,\,$'' is an abbreviation of:\\
\phx\qquad\qquad\qquad\qquad $\,\,$ ``$\,\,(\,\,S\ne\frownie\,\,)\,\,\&\,\,(\,\,\frownie\notin S\,\,)\,\,$''.\\
Generally, we will make these kinds of abbreviations without comment.

\begin{axiom}
Let $a,b,c\in\R^*$.\\
\phx\qquad Assume $a<b<c$. \qquad
Then $a<c$.
\end{axiom}

\begin{axiom}
Let $a,b\in\R^*$.\\
\phx\qquad Assume $a<b$. \qquad
Then $a\ne b$.
\end{axiom}

If we wish to develop set theory and the theory of the real number system in parallel, also making room for $\frownie$,
it requires a number of axioms, which you can read about in the notes from last year's course.

A more standard approach involves developing set theory first, then defining each real number as a specific set,
then proving basic results about the real number system, then moving on to real analysis. In this standard approach,
the symbol $\frownie$ is never used.

This year, we will not develop the foundations so carefully, and will instead rely on the reader's sense
(based on previous learning) of how sets and real numbers work.
So, for example, the last two axioms might simply have been omitted,
since most readers know that a double strict inequality contracts to a strict inequality,
and, also, most readers know that less than implies not equal to.

\subsection{The Axiom of Choice}
\quad
\vskip.1in\noindent

A set $S$ is said to be \bfindex[nonempty] if $S\ne\emptyset$.

The next axiom is called \bfindex[the Axiom of Choice]:

\begin{axiom}
Let $S$ be a nonempty set.\qquad
Then $\CH_S\in S$.
\end{axiom}

We will use $\CH(S)$ as an alternate notation for $\CH_S$.

\begin{axiom}
$\CH_\emptyset=\frownie$.
\end{axiom}

\begin{thm}
Let $S:=\{1,2,3\}$, $x:=\CH_S$.\\
\phx\qquad\qquad\qquad Then: \qquad
$(\,\,x=1\,\,)\,\,\vee\,\,(\,\,x=2\,\,)\,\,\vee\,\,(\,\,x=3\,\,)$.
\end{thm}

\begin{thm}
Let $S:=\{\{1,2,3\}\}$, $x:=\CH_S$.\\
\phx\qquad\qquad\qquad Then: \qquad
$x=\{1,2,3\}$.
\end{thm}

\subsection{Singleton sets and the Unique Element operator}
\quad
\vskip.1in\noindent

A singleton set is a set with exactly one element:

\begin{defn}
Let $S$ be an object.\\
By $S$ is a \bfindex[singleton set], we mean:\\
\phx\qquad $(\,\,S\hbox{ is a set}\,\,)\quad\&\quad(\,\,\forall x,y\in S,\,\,x=y\,\,)$.
\end{defn}

\begin{defn}
Let $S$ be a set. \quad
Then:\\
\phx\qquad\qquad\qquad\qquad $\ds{\UE_S:=\begin{cases}\CH_S,&\hbox{if }S\hbox{ is a singleton set} \\ \frownie,&\hbox{if }S\hbox{ is not a singleton set.} \end{cases}}$
\end{defn}

We sometimes use $\UE(S)$ to denote $\UE_S$, and sometimes even leave out the parentheses in $\UE(S)$:

\begin{thm}
We have: \quad $\UE\{5\}=5$ $\,$ and $\,$ $\UE\{\{1,2\}\}=\{1,2\}$\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,\,$ and $\,$ $\UE\{1,2\}=\frownie$ $\,$ and $\,$ $\UE\emptyset=\frownie$.
\end{thm}

\subsection{Functions}

\begin{thm}
We have: \quad $\{5,8\}=\{8,5\}$.\\
\phx\qquad\qquad\quad $\,\,\,\,\,\,$ Also, we have: \quad $\{5,5\}=\{5\}$.\\
\phx\qquad\qquad $\,\,$ Also, we have: \quad $\{\{5\},\{5\}\}=\{\{5\}\}$.
\end{thm}

\begin{defn}
Let $a$ and $b$ be objects.\\
\phx\qquad\qquad Then: \quad $\langle\langle a,b\rangle\rangle\,\,\,:=\,\,\,\{\,\,\{a\}\,\,,\,\,\{a,b\}\,\,\}$
\end{defn}

\begin{thm}
We have:\\
\phx\qquad\qquad $\langle\langle5,8\rangle\rangle=\{\{5\},\{5,8\}\}$,\\
\phx\qquad\qquad $\langle\langle8,5\rangle\rangle=\{\{8\},\{8,5\}\}=\{\{8\},\{5,8\}\}$,\\
\phx\qquad\qquad $\langle\langle5,8\rangle\rangle=\{\{5\},\{5,8\}\}\ne\{\{8\},\{5,8\}\}=\langle\langle8,5\rangle\rangle$,\\
\phx\qquad\qquad $\langle\langle5,5\rangle\rangle=\{\{5\},\{5,5\}\}=\{\{5\},\{5\}\}=\{\{5\}\}$ \quad and\\
\phx\qquad $\,\,\,\langle\langle\,1\,,\,\{3\}\,\rangle\rangle=\{\,\,\,\{1\}\,\,\,,\,\,\,\{\,1\,,\,\{3\}\,\}\,\,\,\}$.\\
\end{thm}

\begin{thm}
Let $a,b,c,d$ be objects.\\
\phx $\,$ Then: \qquad\quad $\,\,\,\,[\quad\langle\langle a,b\rangle\rangle\quad]\qquad\Leftrightarrow\qquad[\quad\,\,\,\,\,(a=c)\,\&\,(b=d)\,\,\,\,\,\quad]$.\\
\phx $\,$ Also: \quad $\,[\quad\{a,b\}=\{c,d\}\quad]\qquad\Leftrightarrow\qquad[\quad(\,\,(a=c)\,\&\,(b=d)\,\,)$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\,\,\,\,\,\vee\,(\,\,(a=d)\,\&\,(b=c)\,\,)\quad]$.\\
\end{thm}

\begin{defn}
Let $R$ be an object.\\
\phx\qquad By $R$ is a \bfindex[relation], we mean:\\
\phx\qquad\qquad $(\,R\hbox{ is a set}\,)\quad\&\quad(\,\forall c\in\R,\,\exists a,b~\hbox{s.t.}~c=\langle\langle a,b\rangle\rangle\,)$.
\end{defn}

\begin{thm}\ \\
\phx\quad The set $\{\langle\langle1,2\rangle\rangle,\langle\langle1,3\rangle\rangle,\langle\langle2,3\rangle\rangle\}$ is a relation.\\
\phx\quad The set $\{\langle\langle1,2\rangle\rangle,\langle\langle2,3\rangle\rangle\}$ is a relation.
\end{thm}

We graphed the two relations in the preceding theorem, noting the vertical line test failing for the frist, but not the second.

\begin{defn}\wrlab{defn-fn}
Let $f$ be an object.\\
By $f$ is a \bfindex[function], we mean:\\
\phx\qquad (1) $f$ is a relation \qquad\qquad\qquad\qquad and\\
\phx\qquad (2) $\forall a,b,c,\,\,\,[\,\,\,(\,\,\langle\langle a,b\rangle\rangle\in f\,\,)\,\,\&\,\,(\,\,\langle\langle a,b\rangle\rangle\in f\,\,)\,\,\,]\,\,\,\Rightarrow\,\,\,[\,\,\,b=c\,\,\,]$.
\end{defn}

Condition (2) in the \dref{defn-fn} is called the \bfindex[vertical line test].

\begin{thm}
Let $R\,\,\,:=\,\,\,\{\,\,\langle\langle1,2\rangle\rangle,\,\,\langle\langle1,3\rangle\rangle,\,\,\langle\langle2,3\rangle\rangle\,\,\}$\\
\phx\qquad\qquad\qquad\quad and let $f\,\,\,\,:=\,\,\,\{\,\,\langle\langle1,2\rangle\rangle,\,\,\langle\langle2,3\rangle\rangle\,\,\}$.\\
\phx\quad Then: \qquad $R$ is NOT a function \qquad and \qquad $f$ IS a function.
\end{thm}

\begin{defn}
Let $f$ be a function and let $x$ be an object.\\
\phx\qquad\qquad\qquad Then $f(x)\,\,\,:=\,\,\,\UE\,\{\,y\,|\,\langle\langle x,y\rangle\rangle\,\in\,f\,\}$.\\
\phx\qquad\qquad\qquad Also, $\,\,\,f_x\,\,\,\,\,\,:=\,\,\,f(x)$.
\end{defn}

\begin{thm}
Let $f\,\,:=\,\,\{\,\langle\langle1,2\rangle\rangle\,,\,\langle\langle2,3\rangle\rangle\}$.\\
\phx\qquad\qquad Then: \quad $f$ is a function \quad and \quad $f(1)=2$\\
\phx\qquad\qquad\quad and \qquad\quad $f_2=3$ \qquad $\,\,$ and \qquad $f_3=\frownie$.
\end{thm}

The function in the preceding theorem is denoted
$\ds{\left(\begin{tabular}{c}$1\mapsto2$\\$2\mapsto3$\end{tabular}\right)}$.
From this point forward, instead of writing
$$f\,\,:=\,\,\{\,\langle\langle1,2\rangle\rangle\,,\,\langle\langle2,3\rangle\rangle\},$$
we strongly prefer
$$f\,\,:=\,\,\left(\begin{tabular}{c}$1\mapsto2$\\$2\mapsto3$\end{tabular}\right).$$

\begin{defn}
Let $f$ be a function. Then:\\
\phx\qquad\qquad\qquad $\D_f\,\,:=\,\,\{\,x\,|\,f_x\ne\frownie\,\}$\\
\phx\qquad and \qquad $\,\,\,\I_f\,\,:=\,\,\{\,y\,|\,\exists x\in\D_f~\hbox{s.t.}~f_x=y\,\}$.
\end{defn}

In the preceding definition, $\D_f$ is called the \bfindex[domain] of $f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad and $\,\I_f$ is called the \bfindex[image] of $f$.

\begin{thm}
Let $\ds{f\,\,:=\,\,\left(\begin{tabular}{c}$1\mapsto4$\\$3\mapsto2$\\$4\mapsto2$\end{tabular}\right)}$.\\
\phx\quad Then: \quad $\D_f\,\,=\,\,\{1,3,4\}$ \quad and \quad $\I_f\,\,=\,\,\{4,2,2\}\,\,=\,\,\{2,4\}$.
\end{thm}

\begin{defn}
Let $X$ and $Y$ be sets, and let $f$ be an object.\\
Then $f:X\drta Y\,$ means: $(f\hbox{ is a function})\,\&\,(\D_f\subseteq X)\,\&\,(\I_f\subseteq Y)$.\\
Also, $f:X\to Y$ $\,\,$ means: $(f\hbox{ is a function})\,\&\,(\D_f=X)\,\&\,(\I_f\subseteq Y)$.\\
Also, $f:X\to>Y$ means: $(f\hbox{ is a function})\,\&\,(\D_f=X)\,\&\,(\I_f=Y)$.\\
\end{defn}

\begin{defn}
Let $a,b\in\R^*$.\\
\phx\qquad Then \qquad $[a;b]\,:=\,\{\,x\in\R^*\,|\,a\le x\le b\,\}$\\
\phx\qquad $\,$ and \qquad $\,[a;b)\,:=\,\{\,x\in\R^*\,|\,a\le x<b\,\}$\\
\phx\qquad $\,$ and \qquad $\,(a;b]\,:=\,\{\,x\in\R^*\,|\,a<x\le b\,\}$\\
\phx\qquad $\,$ and \qquad $\,(a;b)\,:=\,\{\,x\in\R^*\,|\,a<x<b\,\}$\\
\phx\qquad $\,$ and \qquad $\,[a..b]\,:=\,\{\,x\in\Z^*\,|\,a\le x\le b\,\}$\\
\phx\qquad $\,$ and \qquad $\,[a..b)\,:=\,\{\,x\in\Z^*\,|\,a\le x<b\,\}$\\
\phx\qquad $\,$ and \qquad $\,(a..b]\,:=\,\{\,x\in\Z^*\,|\,a<x\le b\,\}$\\
\phx\qquad $\,$ and \qquad $\,(a..b)\,:=\,\{\,x\in\Z^*\,|\,a<x<b\,\}$.\\
\end{defn}

\begin{thm}
Let $f\,\,:=\,\,\{\,\langle\langle x,y\rangle\rangle\,|\,(\,x,y\in\R\,)\,\&\,(\,y=x^2\,)\,\}$.\\
\phx\qquad\quad Then: \quad $f$ is a function \quad $\,$ and $\,$ \quad $f_2=4$ \quad and \quad $f_{-2}=4$\\
\phx\qquad\quad\quad and \qquad\quad $\D_f=\R$ \qquad $\,\,$ and \qquad $\I_f=[0;\infty)$\\
\phx\qquad\quad\quad and \quad\quad $f:\R\to\R$ \qquad $\,$ and \qquad $f:\R\to>[0;\infty)$.
\end{thm}

From this point forward, instead of writing\\
\phx\qquad ``Let $f\,\,:=\,\,\{\,\langle\langle x,y\rangle\rangle\,|\,(\,x,y\in\R\,)\,\&\,(\,y=x^2\,)\,\}$'',\\
we strongly prefer\\
\phx\qquad ``Define $f:\R\to\R$ by: $\forall x\in\R$, $f_x=x^2$''.\\
We graphed this function $f$; the graph is a parabola.

Define $g:\R\drta\R$ by: $\forall x\in\R$, $g_x=(1/x)+5$.\\
We graphed $g$; the graph is a hyperbola.

\begin{defn}
Let $S$ be a set and $q$ and object.\\
\phx\qquad Then \quad $S_q^+:=S\bigcup\{q\}$ \quad and \quad $S_q^\times:=S\bksl\{q\}$.
\end{defn}

\begin{thm}
Let \quad $S:=\{1,2,3\}$, \quad  $y:=3$, \quad $z:=4$.\\
\phx\qquad Then \quad $S_y^+=\{1,2,3\}$ \quad and \quad $S_z^+=\{1,2,3,4\}$\\
\phx\qquad $\,$ and \quad $\,S_y^\times=\{1,2\}\!$ \qquad and \quad $S_z^\times=\{1,2,3\}$.
\end{thm}

\begin{thm}
Define $g:\R\drta\R$ by: $\forall x\in\R$, $g_x=(1/x)+5$.\\
\phx\qquad\qquad Then \qquad $\D_g=\R_0^\times$ \qquad and \qquad $\I_g=\R_5^\times$\\
\phx\qquad\qquad $\,$ and \qquad $\!\!g:\R_0^\times\to\R\,\,$ \quad and \quad $g:\R_0^\times\to>\R_5^\times$.
\end{thm}

\begin{thm}
Let $f$ and $g$ be functions.\\
Then: $(\,f=g\,)\,\Leftrightarrow(\,\forall x\,f_x=g_x\,)$.
\end{thm}

\begin{thm}
Let $f$ and $g$ be functions and let $S$ be a set.\\
Assume $\D_f\subseteq S$ and $\D_g\subseteq S$.
Then: $(\,f=g\,)\,\Leftrightarrow(\,\forall x\in S\,f_x=g_x\,)$.
\end{thm}

\begin{defn}
Let $f$ and $g$ be functions.\\
Then $g\circ f$ is the function defined by:
$\forall x$, $(g\circ f)_x=g_{f_x}$.
\end{defn}

Frownie is infective:

\begin{defn}
$\forall f$, \quad $(\,\frownie\circ f=\frownie\,)\,\&\,(f\circ\frownie=\frownie\,)$.
\end{defn}

\begin{thm}
Define $f,g:\R\to\R$ by:\\
\phx\qquad\qquad\qquad\qquad\qquad $\forall x\in\R$, \qquad $f_x=x+1$ \quad and \quad $g_x=x^2$.\\
Then: $\forall x\in\R$, $(g\circ f)_x=(x+1)^2=x^2+2x+1$ and $(f\circ g)_x=x^2+1$.\\
\phx\qquad\qquad\qquad Also, we have: $(g\circ f)_1=4\ne2=(f\circ g)_1$.\\
\phx\qquad\qquad\qquad Also, we have: \qquad $g\circ f\quad\ne\quad f\circ g$.
\end{thm}

\begin{thm}
Let $f$, $g$ and $h$ be functions.\\
\phx\qquad\qquad Then $(h\circ g)\circ f=h\circ(g\circ f)$.
\end{thm}

\begin{proof}
Want: $\forall w,\,\,((h\circ g)\circ f)_w=(h\circ(g\circ f))_w$.\\
Given $w$. \qquad
Want: $((h\circ g)\circ f)_w=(h\circ(g\circ f))_w$.\\
Let $x:=f_w$. \qquad Let $y:=g_x$. \qquad Let $z:=h_y$.\\
Then $((h\circ g)\circ f)_w=(h\circ g)_{f_w}=(h\circ g)_x=h_{g_x}=h_y=z$.\\
Want: $(h\circ(g\circ f))_w=z$. \qquad
We have $(g\circ f)_w=g_{f_w}=g_x=y$.\\
Then $(h\circ(g\circ f))_w=h_{(g\circ f)_w}=h_y=z$, as desired.
\end{proof}

\subsection{Injective functions and inverse functions}

\begin{defn}
Let $f$ be a function.\\
By $f$ is \bfindex[one-to-one], we mean:\\
\phx\qquad\qquad\qquad $\forall a,b\in\D_f$, \quad $(\,f_a=f_b\,)\,\Rightarrow\,(\,a=b\,)$.\\
Also, by $f$ is \bfindex[injective], we mean: \quad $f$ is one-to-one.
\end{defn}

\begin{defn}
Let $X$ and $Y$ be sets, and let $f$ be an object.
Then $f:X\hookrightarrow Y\,\,\,\,$ means: $\,\,(\,f:X\to Y\,)\,\,\,\,\&\,(\,f\hbox{ is one-to-one}\,)$.\\
Also, $f:X\hookrightarrow>Y$ means: $(\,f:X\to>Y\,)\,\&\,(\,f\hbox{ is one-to-one}\,)$.
\end{defn}

\begin{thm}
Let $\ds{f\,\,:=\,\,\left(\begin{tabular}{c}$1\mapsto4$\\$2\mapsto5$\\$3\mapsto6$\end{tabular}\right)}$.\\
\phx Then: \quad $f:\{1,2,3\}\hookrightarrow>\{4,5,6\}$ \quad and \quad $f:\{1,2,3\}\hookrightarrow\{3,4,5,6,7\}$.
\end{thm}

\begin{thm}
$\emptyset:\emptyset\hookrightarrow>\emptyset$
and $\emptyset:\emptyset\hookrightarrow\{7,8\}$.
\end{thm}

\begin{defn}
Let $R$ be a relation.\\
\phx\qquad\qquad\qquad Then: \qquad $R^\wedge\quad:=\quad\{\,\,\langle\langle y,x\rangle\rangle\,\,|\,\,\langle\langle x,y\rangle\rangle\in R\,\,\}$.
\end{defn}

\begin{thm}
Let $\,\,R\,\,:=\,\,\{\,\langle\langle 1,2\rangle\rangle,\,\langle\langle 2,2\rangle\rangle,\,\langle\langle 2,4\rangle\rangle\,\}$.\\
\phx\qquad\qquad\qquad\qquad Then $R^\wedge\,=\,\{\,\langle\langle 2,1\rangle\rangle,\,\langle\langle 2,2\rangle\rangle,\,\langle\langle 4,2\rangle\rangle\,\}$.
\end{thm}

\noindent
Let $R\,:=\,\{\,\langle\langle 1,2\rangle\rangle,\,\langle\langle 2,2\rangle\rangle,\,\langle\langle 2,4\rangle\rangle\,\}$.\\
We graphed $R$ and $R^\wedge$ and observed:\\
\phx\qquad $R^\wedge$ is obtained from $R$ by reflection through the $45$-degree line.

Frownie is infective:

\begin{defn}
\quad $\forall x$, $\frownie_x:=\frownie$. \quad
Also, $\forall x$, $\frownie(x)=\frownie$.
\end{defn}

\begin{defn}
Let $f$ be a function.\\
\phx\qquad\qquad\qquad Then $\ds{f^{-1}\,\,:=\,\,\begin{cases}f^\wedge,&\hbox{ if }f\hbox{ is one-to-one}\\ \frownie,&\hbox{ if }f\hbox{ is not one-to-one}.\end{cases}}$
\end{defn}

\begin{thm}
Let $\ds{f\,\,:=\,\,\left(\begin{tabular}{c}$1\mapsto5$\\$2\mapsto7$\\$3\mapsto4$\end{tabular}\right)}$,
$\ds{\,\,g:=\,\,\left(\begin{tabular}{c}$7\mapsto1$\\$8\mapsto2$\\$9\mapsto1$\end{tabular}\right)}$.\\
\phx\qquad Then \qquad $f^{-1}=\left(\begin{tabular}{c}$5\mapsto1$\\$7\mapsto2$\\$4\mapsto3$\end{tabular}\right)$ \quad and \quad $g^{-1}=\frownie$\\
\phx\qquad and \quad $f^{-1}$ is a function \quad and \quad $g^{-1}$ is not a function\\
\phx\qquad and \quad $f_7^{-1}=2$ \quad and \quad $f_1^{-1}=\frownie$ \quad and \quad $g_2^{-1}=\frownie$\\
\phx\qquad and \quad $\forall x,\,g_x^{-1}=\frownie$ \quad and \quad $\forall x,\,(g^{-1}\circ g)_x=\frownie=(g\circ g^{-1})_x$\\
\phx\qquad and \quad $(f^{-1}\circ f)_3=3$ \quad and \quad $(f^{-1}\circ f)_4=\frownie$\\
\phx\qquad and \quad $(f\circ f^{-1})_3=\frownie$ \quad and \quad $(f\circ f^{-1})_4=4$.
\end{thm}

\begin{defn}
Let $S$ be a set.\\
\phx\qquad Then $\id^S:S\to S$ is defined by:
\quad $\forall x\in S$, $\,\,\,\id_x^S=x$.
\end{defn}

\begin{thm}
Let $\ds{f\,\,:=\,\,\left(\begin{tabular}{c}$1\mapsto5$\\$2\mapsto7$\\$3\mapsto4$\end{tabular}\right)}$.\\
\phx\qquad\qquad Then: \quad $f^{-1}\circ f=\id^{\D_f}$ \quad and \quad $f\circ f^{-1}=\id^{\I_f}$.
\end{thm}

\begin{thm}
Let $f$ be a one-to-one function.\\
Then: \quad $f^{-1}$ is a function\quad and\quad $\D_{f^{-1}}=\I_f$\quad and\quad $\I_{f^{-1}}=\D_f$\\
\phx\qquad\qquad and \quad $f^{-1}\circ f=\id^{\D_f}$ \quad and \quad $f\circ f^{-1}=\id^{\I_f}$.
\end{thm}

\begin{thm}
Let $\ds{\,\,g:=\,\,\left(\begin{tabular}{c}$7\mapsto1$\\$8\mapsto2$\\$9\mapsto1$\end{tabular}\right)}$.\\
Then: \quad $g^{-1}=\frownie$ \quad and \quad $g\circ g^{-1}=\frownie$ \quad and \quad $g^{-1}\circ g=\frownie$.\\
\end{thm}

\begin{thm}
Let $g$ be a function.\\
\phx\qquad\qquad\qquad\qquad Assume $g$ is not one-to-one.\\
Then: \quad $g^{-1}=\frownie$ \quad and \quad $g\circ g^{-1}=\frownie$ \quad and \quad $g^{-1}\circ g=\frownie$.\\
\end{thm}

\begin{defn}
Let $f$ be a function and $S$ a set.\\
\phx\qquad Then \quad $f_*(S)\,\,:=\,\,\{f_x\,|\,x\in S\cap\D_f\}$\\
\phx\qquad and \quad $\,\,\,f^*(S)\,\,:=\,\,\{x\in\D_f\,|\,f_x\in S\}$.
\end{defn}

We drew a Venn diagram with a downward function to illustrate the preceding definition.

\begin{thm}
Let $\ds{f\,\,:=\,\,\left(\begin{tabular}{c}$1\mapsto7$\\$2\mapsto8$\\$3\mapsto9$\end{tabular}\right)}$.\\
Then $f^*\{6,7,8\}=\{1,2\}$ and $f_*\{2,3,4\}=\{8,9\}$.
\end{thm}

\begin{defn}
Let $S$ be a set and let $a$ be an object.\\
\phx\qquad\qquad Then $C_S^a:S\to\{a\}$ is defined by: \quad $\forall x\in S,\,\,\,C_S^a(x)=a$.
\end{defn}

\noindent
Let $S$ be a set and let $a$ be an object.\\
Then $C_S^a$ is called the \bfindex[constant function] on $S$ with value $a$.

We graphed $C_{(0;\infty)}^2$; it is a horizontal ray to the right out of the point $(0,2)$.
It does not include the point $(0,2)$.

\begin{defn}
Let $T$ be a set and let $S\subseteq T$.\\
\phx\qquad\qquad\qquad Then $\chi_S^T:T\to\{0,1\}$ is defined by:\\
\phx\qquad\qquad\qquad\qquad $\forall q\in T$, \quad $\chi_S^T(q)\,\,=\,\,\begin{cases}1,&\hbox{if }q\in S\\0,&\hbox{if }q\notin S.\end{cases}$
\end{defn}

\noindent
Let $T$ be a set and let $S\subseteq T$.\\
Then $\chi_S^T$ is called the \bfindex[characteristic function] of $S$ in $T$.

\begin{defn}
Let $f$ be a function, $A\subseteq\D_f$.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,\,$ Then $f|A$ is the function defined by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\forall x$, \quad $(f|A)_x\,\,=\,\,\begin{cases}f_x,&\hbox{if }x\in A\\\frownie,&\hbox{if }x\notin A.\end{cases}$
\end{defn}

In the preceding definition, $f|A$ is called the \bfindex[restriction] of $f$ to $A$.

\begin{thm}
Let $f:=\left(\begin{tabular}{c}$1\mapsto2$\\$3\mapsto4$\\$5\mapsto6$\end{tabular}\right)$ and let $A:=\{1,5\}$.\\
\phx Then $\,\,$ $f|A=\left(\begin{tabular}{c}$1\mapsto2$\\$5\mapsto6$\end{tabular}\right)$
$\,\,$ and $\,\,$ $f_1=2$ $\,\,$ and $\,\,$ $f_3=4$ $\,\,$ and $\,\,$ $f_5=6$\\
\phx\qquad\qquad and $\,\,$ $(f|A)_1=2$ $\,\,$ and $\,\,$ $(f|A)_3=\frownie$ $\,\,$ and $\,\,$ $(f|A)_5=6$.
\end{thm}

\subsection{Max, min, inf and sup}

\begin{defn}
Let $a\in\R^*$, $S\subseteq\R^*$.\\
\phx\qquad Then $a<S$ means: $\forall x\in S$, $a<x$.\\
\phx\qquad Also, $a\le S$ means: $\forall x\in S$, $a\le x$.\\
\phx\qquad Also, $a>S$ means: $\forall x\in S$, $a>x$.\\
\phx\qquad Also, $a\ge S$ means: $\forall x\in S$, $a\ge x$.\\
\phx\qquad Also, $S>a$ means: $a<S$.\\
\phx\qquad Also, $S\ge a$ means: $a\le S$.\\
\phx\qquad Also, $S<a$ means: $a>S$.\\
\phx\qquad Also, $S\le a$ means: $a\ge S$.
\end{defn}

\begin{defn}
Let $S\subseteq\R^*$.\\
\phx\qquad Then $\UB_S\,\,:=\,\,\{a\in\R^*\,|\,S\le a\}$.\\
\phx\qquad Also, $\LB_S\,\,:=\,\,\{a\in\R^*\,|\,a\le S\}$.\\
\phx\qquad Also, $\UB(S)\,\,:=\,\,UB_S$.\\
\phx\qquad Also, $\LB(S)\,\,:=\,\,LB_S$.
\end{defn}

\noindent
The set $\LB_S$ is the set of \bfindex[lower bounds] of $S$.\\
The set $\UB_S$ is the set of \bfindex[upper bounds] of $S$.\\

\vskip.1in
We are sometimes sloppy and omit $()$ in $\UB(S)$ and $\LB(S)$.\\
So, for example, $\UB\{1,3\}$ means $\UB(\{1,3\})$.

\begin{thm}
$\UB\{1,3\}=[3;\infty]$ and $\LB\{1,3\}=[-\infty;1]$.
\end{thm}

\begin{defn}
Let $S\subseteq\R^*$.\\
\phx\qquad Then $\max_S:=\UE(S\cap\UB_S)$.\\
\phx\qquad Also, $\min_S:=\UE(S\cap\LB_S)$.\\
\phx\qquad Also, $\max(S):=\max_S$.\\
\phx\qquad Also, $\min(S):=\min_S$.
\end{defn}

\noindent
The object $\min_S$ is \bfindex[minimum] of $S$.\\
The object $\max_S$ is the set of \bfindex[maximum] of $S$.\\

\vskip.1in
We are sometimes sloppy and omit $()$ in $\max(S)$ and $\min(S)$.\\
So, for example, $\min(1;3]$ means $\min(\,\,(1;3]\,\,)$.

\begin{thm}
We have:\\
\phx $(\max\{1,2\}\!=\!2)\,\&\,(\min\{1,2\}\!=\!1)\,\&\,(\max(1;2]\!=\!2)\,\&\,(\min(1;2]\!=\!\frownie)$.
\end{thm}

The following two theorems will be used repeatedly without comment:

\begin{thm}
Let $A\subseteq\R^*$, $x:=\min A$.\\
Assume $x\ne\frownie$. \qquad
Then: $(\,x\in A\,)\,\&\,(\,x\le A\,)$.
\end{thm}

\begin{thm}
Let $A\subseteq\R^*$, $x:=\max A$.\\
Assume $x\ne\frownie$. \qquad
Then: $(\,x\in A\,)\,\&\,(\,A\le x\,)$.
\end{thm}

\begin{defn}
Let $S\subseteq\R^*$.\\
\phx\qquad Then $\sup_S:=\min(\UB_S)$.\\
\phx\qquad Also, $\inf_S:=\max(\LB_S)$.\\
\phx\qquad Also, $\sup(S):=\sup_S$.\\
\phx\qquad Also, $\inf(S):=\inf_S$.\\
\end{defn}

\noindent
The object $\inf_S$ is \bfindex[infimum] of $S$.\\
The object $\sup_S$ is the set of \bfindex[supremum] of $S$.\\

\vskip.1in
We are sometimes sloppy and omit $()$ in $\inf(S)$ and $\sup(S)$.\\
So, for example, $\inf(1;3]$ means $\inf(\,\,(1;3]\,\,)$.

\begin{thm}\ \\
\phx\qquad We have:
$(\,\,\UB(1;2]=[2;\infty]\,\,)\quad\,\,\,\,\&\quad(\,\,\sup(1;2]=2\,\,)$.\\
\phx\qquad\qquad Also: $(\,\,\LB(1;2]=[-\infty;1]\,\,)\quad\&\quad(\,\,\inf(1;2]=1\,\,)$.
\end{thm}
Some examples:

\begin{center}
\begin{tabular}{|c||c|c|c|c|c|c|}
\hline
 $S$ & $\LB$ & $\UB$ & $\min$ & $\max$ & $\inf$ & $\sup$ \\ 
\hline\hline
$\{5\}$ & $[-\infty;5]$ & $[5;\infty]$ & $5$ & $5$ & $5$ & $5$ \\  \hline
$[0;1]$ & $[\infty;0]$ & $[1;\infty]$ & $0$ & $1$ & $0$ & $1$ \\  \hline
$(0;1)$ & $[\infty;0]$ & $[1;\infty]$ & \frownie & \frownie & $0$ & $1$ \\  \hline
$[0;1)$ & $[\infty;0]$ & $[1;\infty]$ & $0$ & \frownie & $0$ & $1$ \\  \hline
$(0;1]$ & $[\infty;0]$ & $[1;\infty]$ & \frownie & $1$ & $0$ & $1$ \\  \hline
$\{0,1\}$ & $[\infty;0]$ & $[1;\infty]$ & $0$ & $1$ & $0$ & $1$ \\ \hline
$\R^*$ & $\{-\infty\}$ & $\{\infty\}$ & $-\infty$ & $\infty$ & $-\infty$ & $\infty$ \\  \hline
$\R$ & $\{-\infty\}$ & $\{\infty\}$ & \frownie & \frownie & $-\infty$ & $\infty$ \\  \hline
$\emptyset$ & $\R^*$ & $\R^*$ & \frownie & \frownie & $\infty$ & $-\infty$ \\  \hline
\end{tabular}
\end{center}

The next result is called the \bfindex[Completeness Axiom]:

\begin{axiom}
Let $S\subseteq\R^*$.
Then $\sup_S\ne\frownie\ne\inf_S$.
\end{axiom}

The following theorem will be used repeatedly without comment:

\begin{thm}
All of the following are true:\\
\phx\qquad (1) $\forall S\subseteq\R^*$, \qquad\qquad\qquad\quad $\inf S\le S\le\sup S$.\\
\phx\qquad (2) $\forall S\subseteq\R^*$, $\forall x\in\R$, \qquad $(\,S\le x\,)\,\Rightarrow\,(\,\sup S\le x\,)$.\\
\phx\qquad (3) $\forall S\subseteq\R^*$, $\forall x\in\R$, \qquad $(\,x\le S\,)\,\Rightarrow\,(\,x\le\inf S\,)$.\\
\end{thm}

Later, we will need:

\begin{thm}\wrlab{thm-below-min-implies-notinset}
Let $A\subseteq\R$ and let $x:=\min A$.\\
Assume $x\ne\frownie$. \qquad
Then: $(\,x\in A\,)\,\&\,(\,x-1\notin A\,)$.
\end{thm}

Idea of proof: From the definition of $\min$, we see
that the minimum of any set is either frownie or is an element both of the set and of its set of lower bounds.
In particular, it is an element of the set, and so we see that $x\in A$.
To show $x-1\notin A$, note that, were $x-1$ to be an element of $A$,
then we'd have $\min A\le x-1$, which would yield $x\le x-1$, which would yield $0\le -1$,
which is not true. QED

\subsection{The Archimedean Principle}
\quad
\vskip.1in\noindent

We have many unstated axioms describing the real numbers.
We also have many theorems about the real numbers that will be assumed
without proof, and used repeatedly without comment. For example:

\begin{thm}
$\forall a,b,c\in\R$, $(a<b<c)\Rightarrow(a<c)$.
\end{thm}

Note: The compound inequality $a<b<c$ means: $(a<b)\&(b<c)$.\\
We will often use these kinds of compounds without comment.

Another basic fact about $\R$ that is used repeately without comment
is the statement: \quad\qquad\quad $\,$ $-\infty<\R<\infty$.\\
In other words: $\forall x\in\R$, \quad $\,$ $-\infty<x<\infty$.\\
In other words: $\forall x\in\R$, $(-\infty<x)\&(x<\infty)$.

The following axiom is also basic:

\begin{axiom}\wrlab{axiom-supN-is-infty}
$\sup\N=\infty$.
\end{axiom}

From the preceding axiom, we get the \bfindex[Archimedean Principle]:

\begin{thm}\wrlab{thm-arch-princ}
$\forall x\in\R$, $\exists k\in\N$ s.t.~$k>x$.
\end{thm}

\begin{proof}
Given $x\in\R$. \qquad
Want: $\exists k\in\N$ s.t.~$k>x$.\\
Assume: $\neg(\,\exists k\in\N~\hbox{s.t.}~k>x\,)$. \qquad
Want: Contradiction.\\
We have: $\forall k\in\N$, $k\le x$. \qquad
Then $\N\le x$. \qquad
Then $\sup\N\le x$.\\
Since \quad $\sup\N\le x\in\R<\infty$, \quad we get \quad $\sup\N<\infty$.\\
Then $\sup\N\ne\infty$. $\,$
By \aref{axiom-supN-is-infty}, $\sup\N=\infty$. $\,$
Contradiction.
\end{proof}

\tref{thm-arch-princ} asserts:\\
\phx\qquad\qquad any real number admits a positive integer that is greater.\\
This may seem obvious; for us, this assertion is important enough
that it is named; it is called the {\bf Archimedean Principle}.

\subsection{The Principle of Mathematical Induction}
\quad
\vskip.1in\noindent

The next axiom is called the \bfindex[Well-ordering axiom]:

\begin{axiom}\wrlab{axiom-completeness}
$\forall$nonempty $T\subseteq\N$, \quad $\min T\ne\frownie$.
\end{axiom}

From this we prove the \bfindex[Principle of Mathematical Induction]:

\begin{thm}
Let $S\subseteq\N$.\\
\phx\qquad Assume: \quad $(\,1\in S\,)\quad\&\quad(\,\forall j\in S,\,j+1\in S\,)$.\\
\phx\qquad $\,\,\,\,\,$ Then: \qquad\qquad $\,\,S=\N$.
\end{thm}

\begin{proof}
Assume $S\ne\N$. \qquad Want: Contradiction.\\
Since $S\subseteq\N$ and $S\ne\N$, we have $\N\bksl S\ne\emptyset$.\\
Let $i:=\min(\N\bksl S)$. \qquad
Since $N\bksl S\ne\emptyset$, by \aref{axiom-completeness}, $i\ne\frownie$.\\
Then, by \tref{thm-below-min-implies-notinset}, we get: \qquad
$(\,i\in\N\bksl S\,)\,\&\,(\,i-1\notin\N\bksl S\,)$.\\
By hypothesis, $1\in S$. \qquad
Since $i\in\N\bksl S$, we get: $i\notin S$.\\
Since $1\in S$ and $i\notin S$, we see that $i\ne1$.\\
We have $i\in\N\bksl S\subseteq\N$, so $i\in\N$.\\
Since $i\in\N$ and $i\ne1$, we conclude that $i-1\in\N$.\\
Since $i-1\notin\N\bksl S$ and $i-1\in\N$, it follows that $i-1\in S$.\\
By assumption $\forall j\in S,\,j+1\in S$.\\
So, since $i-1\in S$, we see that $(i-1)+1\in S$.\\
Then $i\in S$. \qquad
Recall: $i\notin S$. \qquad
Contradiction.
\end{proof}

\begin{thm}\wrlab{thm-eg-use-PMI-redux}
$\forall j\in\N$, \quad $1+\cdots+j=j(j+1)/2$.
\end{thm}

\begin{proof}
Let $S:=\{j\in\N\,|\,1+\cdots+j=j(j+1)/2\}$.\\
Want: $S=\N$.\\
Since $1=1\cdot(1+1)/2$, it follows that $1\in S$.\\
So, by the PMI, it suffices to show: $\forall j\in S$, $j+1\in S$.\\
Given $j\in S$. \qquad
Want: $j+1\in S$.\\
Know: $1+\cdots+j=j(j+1)/2$.\\
Want: $1+\cdots+j+(j+1)=(j+1)((j+1)+1)/2$.
\\
We have:
\begin{eqnarray*}
1+\cdots+j+(j+1)&=&(j(j+1)/2)+(j+1)\\
&=&((j^2+j)/2)+((2j+2)/2)\\
&=&(j^2+3j+2)/2\,\,=\,\,(j+1)(j+2)/2\\
&=&(j+1)((j+1)+1)/2,
\end{eqnarray*}
as desired.
\end{proof}

\begin{thm}
$\forall j\in\N$, $1^2+\cdots+j^2=j(j+1)(2j+1)/6$.
\end{thm}

\begin{proof}
Let $S:=\{j\in\N\,|\,1^2+\cdots+j^2=j(j+1)(2j+1)/6\}$.\\
Want: $S=\N$. \quad
$1^2=1\cdot2\cdot3/6=1\cdot(1+1)\cdot(2\cdot1+1)/6$, so $1\in S$.\\
By the PMI, \quad Want: $S$ is successor closed.\\
Want: $\forall j\in S$, $j+1\in S$.\\
Given $j\in S$. \quad Want: $j+1\in S$.\\
Since $j\in S$, we know $1^2+\cdots+j^2=j(j+1)(2j+1)/6$.\\
Want: $1^2+\cdots+j^2+(j+1)^2=(j+1)((j+1)+1)(2\cdot(j+1)+1)/6$.\\
We have
\begin{eqnarray*}
1^2+\cdots+j^2+(j+1)^2&=&[1^2+\cdots+j^2]\,+\,[(j+1)^2]\\
&=&\left[\frac{j(j+1)(2j+1)}{6}\right]\,+\,\left[\frac{6(j+1)^2}{6}\right]\\
&=&\left[j+1\right]\,\left[\frac{[j(2j+1)]\,+\,[6(j+1)]}{6}\right]\\
&=&\left[j+1\right]\,\left[\frac{[2j^2+j]\,+\,[6j+6]}{6}\right]\\
&=&\left[j+1\right]\,\left[\frac{2j^2+7j+6}{6}\right]\\
&=&\left[j+1\right]\,\left[\frac{(j+2)(2j+3)}{6}\right]\\
&=&(j+1)((j+1)+1)(2\cdot(j+1)+1)/6,
\end{eqnarray*}
as desired.
\end{proof}

\subsection{Comparing sets by injections and surjections}

\begin{defn}
Let $X$ and $Y$ be sets.\\
\phx\qquad Then $\exists X\hookrightarrow Y\,$ $\,$ means: $\,\,\exists f$ s.t.~$f:X\hookrightarrow Y$.\\
\phx\qquad Also, $\exists X\to>Y\,$ means: $\,\,\exists f$ s.t.~$f:X\to>Y$.\\
\phx\qquad Also, $\exists X\hookrightarrow>Y$ means: $\,\,\exists f$ s.t.~$f:X\hookrightarrow>Y$.
\end{defn}

\begin{thm}
$\,\,\,\exists\{1,2,3\}\hookrightarrow>\{4,5,6\}$\\
\phx\qquad and \qquad\quad\qquad $\exists\{1,2,3\}\hookrightarrow\,\,\,\,\,\{3,4,5,6,7\}$\\
\phx\qquad and \qquad\quad $\,\not\exists\{3,4,5,6,7\}\hookrightarrow\,\,\,\,\,\{1,2,3\}$\\
\phx\qquad and \qquad\quad $\,\exists\{3,4,5,6,7\}\to>\,\{1,2,3\}$\\
\phx\qquad and \qquad\quad\qquad $\,\not\exists\{1,2,3\}\to>\{3,4,5,6,7\}$\\
\phx\qquad and \qquad\quad\qquad\qquad $\,\,\,\exists\emptyset\to>\{7,8\}$\\
\phx\qquad and \qquad\quad\qquad\quad $\,\not\exists\{7,8\}\to>\emptyset$.
\end{thm}

\noindent
Let $A$ and $B$ be sets.\\
Then $\exists A\hookrightarrow B$ indicates that $A$ is ``smaller'' than $B$.\\
Also, $\exists B\to>A$ indicates that $B$ is ``larger'' than $A$,\\
\phx\qquad\qquad\qquad\qquad EXCEPT when $A=\emptyset$.\\
So, $\hookrightarrow$ works slightly better than $\to>$ for comparing sets.

\vskip.1in
The next four theorems will be used, without comment in the proof of
\tref{thm-exists-surj-implies-exists-inj}, as well as in future proofs.
The proofs of these four theorems are left as unassigned HW.

\begin{thm}
Let $X,Y$ be sets, $f:X\to Y$.
Let $A$ be a set.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,\,\,\,$ Then: \quad $(x\in f^*(A))\Leftrightarrow(f(x)\in A)$.
\end{thm}

\begin{thm}
Let $a,b$ be objects. \quad
Then: $(a\in\{b\})\Leftrightarrow(a=b)$.
\end{thm}

\begin{thm}
Let $X,Y$ be sets, let $f:X\to Y$ and let $w\in\I_f$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,\,\,\,$ Then: \quad $f^*(\{w\})\ne\emptyset$.
\end{thm}

\begin{thm}
Let $X,Y$ be sets, $f:X\to Y$.
Let $p,q$ be objects.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,\,\,\,$ Assume: \quad $p\in f^*(\{q\})$.
\qquad Then: \quad $f(p)=q$.
\end{thm}

\begin{thm}\wrlab{thm-exists-surj-implies-exists-inj}
Let $S$ and $T$ be sets.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,\,\,\,$ Assume: \quad $\exists T\to>S$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,\,\,\,$ Then: \quad\quad $\,\exists S\hookrightarrow T$.
\end{thm}

\begin{proof}
We know: $\exists f$ s.t.~$f:T\to>S$. \qquad
We want: $\exists g$ s.t.~$g:S\hookrightarrow T$.\\
\phx\qquad\qquad Choose $f$ s.t.~$f:T\to>S$.\\
Define $g:S\to T$ by: $\forall x\in S$, $g_x=\CH(f^*(\{x\}))$. \quad
Want: $g:S\hookrightarrow T$.\\
Since $g:S\to T$, we need only show: $g$ is one-to-one.\\
Want: $\forall w,x\in S$, $(g_w=g_x)\Rightarrow(w=x)$.\\
Given $w,x\in S$. \qquad
Want: $(g_w=g_x)\Rightarrow(w=x)$.\\
Assume $g_w=g_x$. \qquad
Want $w=x$.\\
Since $f:T\to>S$, we have $\I_f=S$.\\
Then $w,x\in S=\I_f$, so $f^*(\{w\})\ne\emptyset\ne f^*(\{x\})$.\\
Then $\CH(f^*(\{w\}))\in f^*(\{w\})$ and $\CH(f^*(\{x\}))\in f^*(\{x\})$.\\
So, since $g_w=\CH(f^*(\{w\}))$ and $g_x=\CH(f^*(\{x\}))$,\\
\phx\qquad we get: $g_w\in f^*(\{w\})$ and $g_x\in f^*(\{x\})$.\\
It follows that: $\,\,f(g_w)=w$ and $f(g_x)=x$.\\
By assumption, $g_w=g_x$. \qquad Then $f(g_w)=f(g_x)$.\\
Then $w=f(g_w)=f(g_x)=x$, as desired.
\end{proof}

For nonempty sets, $\to>$ gives the same comparison as $\hookrightarrow$:

\begin{thm}\wrlab{thm-inj-and-surj-compare-same-on-nonempty}
Let $S$ and $T$ be nonemptysets.\\
\phx\qquad\qquad\qquad\qquad $\,$ Then: \quad $(\,\,\exists S\hookrightarrow T\,\,)\,\,\Leftrightarrow\,\,(\,\,\,\exists T\to>S\,\,)$.
\end{thm}

\begin{proof}
Omitted.
\end{proof}

$\exists\hookrightarrow>$ is reflexive:

\begin{thm}
Let $S$ be a set.
Then $\exists S\hookrightarrow>S$.
\end{thm}

Idea of proof: Show that $\id^S:S\hookrightarrow>S$. QED

\vskip.1in
The preding theorem is also true for $\exists\hookrightarrow$
and for $\exists\to>$.
That is:

\begin{thm}
Let $S$ be a set.
Then $\exists S\hookrightarrow S$ and $\exists S\to>S$.
\end{thm}

$\exists\hookrightarrow>$ is symmetric:

\begin{thm}
Let $S$ and $T$ be sets.\\
\phx\qquad\qquad\qquad\qquad $\,$ Then: \quad $(\,\,\exists S\hookrightarrow>T\,\,)\,\,\Leftrightarrow\,\,(\,\,\,\exists T\hookrightarrow>S\,\,)$.
\end{thm}

Idea of proof: Show $\forall f:S\hookrightarrow>T$, $f^{-1}:T\hookrightarrow>S$. QED

\vskip.1in
The preceeding theorem is untrue for $\exists\hookrightarrow$,
and is also untrue for $\exists\to>$.
It only works for $\exists\hookrightarrow>$.

\vskip.1in
$\exists\hookrightarrow>$ is transitive:

\begin{thm}
Let $S$, $T$ and $U$ be sets.\\
\phx\qquad\qquad Then: \quad $[\,(\,\exists S\hookrightarrow>T\,)\,\&\,(\,\exists T\hookrightarrow>U\,)\,]\,\Rightarrow\,[\,\exists S\hookrightarrow>U\,]$.
\end{thm}

Idea of proof: Show $\forall f:S\hookrightarrow>T$, $\forall g:T\hookrightarrow>U$, $g\circ f:S\hookrightarrow>U$. QED

\vskip.1in
The preceeding theorem is true for $\exists\hookrightarrow$,
and is also true for $\exists\to>$.
That is, the following two theorems are both true:

\begin{thm}
Let $S$, $T$ and $U$ be sets.\\
\phx\qquad\qquad Then: \quad $[\,(\,\exists S\hookrightarrow T\,)\,\&\,(\,\exists T\hookrightarrow U\,)\,]\,\Rightarrow\,[\,\exists S\hookrightarrow U\,]$.
\end{thm}

\begin{thm}
Let $S$, $T$ and $U$ be sets.\\
\phx\qquad\qquad Then: \quad $[\,(\,\exists S\to>T\,)\,\&\,(\,\exists T\to>U\,)\,]\,\Rightarrow\,[\,\exists S\to>U\,]$.
\end{thm}

\begin{defn}
Let $S$ be a set. \qquad
Then:\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\#S\quad:=\quad\sup\,\{\,k\in\N_0\,|\,\exists[1..k]\hookrightarrow S\,\}$.
\end{defn}

We have $[1..0]=\emptyset$, and so: \quad $\forall$ set $S$, $\exists[1..0]\hookrightarrow S$.\\
Thus we have: \qquad $\forall$set $S$, \quad $\,\,0\,\quad\in\,\quad\{\,k\in\N_0\,|\,\exists[1..k]\hookrightarrow S\,\}$,\\
\phx\qquad $\,\,$ and so: \qquad $\forall$set $S$, \quad $\emptyset\quad\ne\quad\{\,k\in\N_0\,|\,\exists[1..k]\hookrightarrow S\,\}$.

\begin{thm}
We have:\\
\phx\qquad $(\,\#\{5,6,9\}=3\,)\,\&\,(\,\#\emptyset=0\,)\,\&\,(\,\#\N=\infty\,)\,\&\,(\,\#\R=\infty\,)$\\
\phx\qquad\qquad $\&\,(\,\#\{\,\{1,2\}\,,\,3\}=2\,)\,\&\,(\,\#\{\,\{1,2\}\,\}=1\,)\,\&\,(\,\#\{1,2\}=2\,)$.
\end{thm}

\begin{defn}
Let $S$ be a set.\\
\phx\qquad By $S$ is \bfindex[finite], \qquad we mean $\#S<\infty$.\\
\phx\qquad By $S$ is \bfindex[infinite], \quad we mean $\#S=\infty$.\\
\end{defn}

The next three theorems are important, because they clarify the sense in which $\exists\hookrightarrow$ and $\exists\to>$ compare
sets for size. The proofs are, unforunately, omitted, for lack of time.
In a course on set theory, we would give proofs of all three, but this is a course in real analysis.

\begin{thm}
Let $S$ and $T$ be finite sets.\\
\phx\qquad Then: \qquad $(\,\,\exists S\hookrightarrow T\,\,)\,\,\Leftrightarrow\,\,(\,\,\#S\le\#T\,\,)$.
\end{thm}

\begin{thm}
Let $S$ and $T$ be finite sets.
Assume $S\ne\emptyset$.\\
\phx\qquad Then: \qquad $(\,\,\exists T\to>S\,\,)\,\,\Leftrightarrow\,\,(\,\,\#T\ge\#S\,\,)$.
\end{thm}

Note that, if $T=\{3,4,5\}$ and $S=\emptyset$, then $\not\exists T\to>S$.
So, in the preceding theorem, the assumption that $S$ is nonempty is necessary.
As we remarked earlier, $\exists\to>$ is a flawed way to compare sets,
and we prefer to work with $\exists\hookrightarrow$.

\subsection{The World of Sets}

\begin{thm}
Let $S$ and $T$ be sets.\\
\phx\quad Then both of the following are true:\\
\phx\qquad (A) $\,\,\,\,(\,\exists S\hookrightarrow T\,\,)\,\vee\,(\,\,\exists T\hookrightarrow S\,\,)$.\\
\phx\qquad (B) $[\,\,(\,\,\exists S\hookrightarrow T\,\,)\,\,\&\,\,(\,\,\exists T\hookrightarrow S\,\,)\,\,]\,\,\Leftrightarrow\,\,[\,\,\exists S\hookrightarrow>T\,\,]$.
\end{thm}

In the preceding, (B) is called the \bfindex[Schroeder-Bernstein Theorem].

We drew a picture of the World of Sets, in which\\
\phx\qquad two sets are $S$ and $T$ are on the same level iff $\exists S\hookrightarrow>T$ \qquad and\\
\phx\qquad a set $S$ is below a set $T$ if $\exists S\hookrightarrow T$ and $\not\exists T\hookrightarrow S$.\\
We find showed the first few levels of finite sets, where\\
\phx\qquad $\emptyset$ is the only set at the bottom level (called the $0$th level),\\
\phx\qquad the first level consists of sets $S$ for which $\#S=1$,\\
\phx\qquad the second level consists of sets $S$ for which $\#S=2$,\\
\phx\qquad the third level consists of sets $S$ for which $\#S=3$,\\
\phx\qquad {\it etc.}\\
There's a line with the finite sets below it, and the infinite sets above it.
The next theorem asserts that the level containing $\N$ is at the bottom of the infinite sets.

\begin{thm}
Let $S$ be an infinite set.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,\,\,\,$ Then $\exists\N\hookrightarrow S$.
\end{thm}

\begin{proof}
Omitted.
\end{proof}

\begin{defn}
Let $S$ be a set.\\
\phx\qquad\qquad\qquad By $S$ is \bfindex[countable], we mean: $\exists S\hookrightarrow\N$.\\
\phx\qquad\qquad $\,\,$ By $S$ is \bfindex[uncountable], we mean: $\not\exists S\hookrightarrow\N$.\\
\phx\qquad By $S$ is \bfindex[countably infinite], we mean: $\exists S\hookrightarrow>\N$.\\
\end{defn}

The countably infinite sets therefore form the level with $\N$, which is located at the bottom of the infinite sets.
The countable sets are all the sets at or below that level.
The uncountable sets are all above it.

The next two theorems tell us that $\N_0$ and $\Z$ are all countably infinite.
That is, they are both at that level that contains $\N$.

\begin{thm}
$\exists\N\hookrightarrow>\N_0$.
\end{thm}

Idea of proof:
$\ds{\left(\begin{tabular}{c}$1\mapsto0$\\$2\mapsto1$\\$3\mapsto2$\\$\vdots$\end{tabular}\right)}$. QED

\begin{thm}
$\exists\N\hookrightarrow>\Z$.
\end{thm}

Idea of proof:
$\ds{\left(\begin{tabular}{l}$1\mapsto\,\,\,\,0$\\$2\mapsto\,\,\,\,1$\\$3\mapsto-1$\\$4\mapsto\,\,\,\,2$\\$5\mapsto-2$\\$6\mapsto\,\,\,\,3$\\$7\mapsto-3$\\$\vdots$\end{tabular}\right)}$. QED

The following table lists each integer in the first column and each positive integer in the first row.
Each entry on the inside of the table is obtained by dividing\\
\phx\quad  the integer to the left of it \quad by \quad the positive integer above it.\\
Every rational number appears in the inside of the table (infinitely many times).

\vskip.1in

\begin{center}
\begin{tabular}{|r||r|r|r|r|r|r|r|r|}
\hline
Divide   & $1$      & $2$      & $3$      & $4$      & $5$      & $6$      & $7$      & $\cdots$ \\ \hline\hline
$0$      & $0$      & $0$      & $0$      & $0$      & $0$      & $0$      & $0$      & $\cdots$ \\ \hline
$1$      & $1$      & $1/2$    & $1/3$    & $1/4$    & $1/5$    & $1/6$    & $1/7$    & $\cdots$ \\ \hline
$-1$     & $-1$     & $-1/2$   & $-1/3$   & $-1/4$   & $-1/5$   & $-1/6$   & $-1/7$   & $\cdots$ \\ \hline
$2$      & $2/1$    & $2/2$    & $2/3$    & $2/4$    & $2/5$    &$2/6$     & $2/7$    & $\cdots$ \\ \hline
$-2$     & $-2/1$   & $-2/2$   & $-2/3$   & $-2/4$   & $-2/5$   & $-2/6$   & $-2/7$   & $\cdots$ \\ \hline
$3$      & $3/1$    & $3/2$    & $3/3$    & $3/4$    & $3/5$    & $3/6$    & $3/7 $   & $\cdots$ \\ \hline
$-3$     & $-3/1$   & $-3/2$   & $-3/3$   & $-3/4$   & $-3/5$   & $-3/6$   & $-3/7$   & $\cdots$ \\ \hline
$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\ddots$ \\ \hline
\end{tabular}
\end{center}

\vskip.1in

\noindent
We can list every rational number (infinitely many times) as follows:\\
\phx $\,$ start at the upper left $0$ (on the zeroth diagonal)\\
\phx $\,$ move NE along the first diagonal $1,0$ \\
\phx $\,$ move NE along the second diagonal $-1,1/2,0$ \\
\phx $\,$ move NE along the third diagonal $2/1,-1/2,1/3,0$ \\
\phx $\,$ move NE along the fourth diagonal $-2/1,2/2,-1/3,1/4,0$ \\
\phx $\,$ move NE along the fifth diagonal $3/1,-2/2,2/3,-1/4,1/5,0$ \\
\phx $\,$ move NE along the sixth diagonal $-3/1,3/2,-2/3,2/4,-1/5,1/6,0$ \\
\phx\qquad\qquad\qquad\qquad\qquad {\it etc.}\\
Concatenating these lists of diagonals, we get a list:\\
\phx\qquad $0\quad,\quad1,0\quad,\quad-1,1/2,0\quad,\quad2/1,-1/2,1/3,0\quad,\quad\ldots$\\
Every rational number appears (infinitely many times) in this list.

\begin{thm}\wrlab{thm-NontoQ}
$\exists\N\to>\Q$.
\end{thm}

Idea of proof:
$\ds{\left(\begin{tabular}{l}$1\,\,\,\,\,\,\mapsto\,\,\,\,\,\,\,\,0$\\$2\,\,\,\,\,\,\mapsto\,\,\,\,\,\,\,\,1$\\
$3\,\,\,\,\,\,\mapsto\,\,\,\,\,\,\,\,0$\\$4\,\,\,\,\,\,\mapsto\,\,\,\,-1$\\
$5\,\,\,\,\,\,\mapsto\,\,\,\,\,1/2$\\$6\,\,\,\,\,\,\mapsto\,\,\,\,\,\,\,\,0$\\$7\,\,\,\,\,\,\mapsto\,\,\,\,\,2/1$\\$8\,\,\,\,\,\,\mapsto-1/2$\\
$9\,\,\,\,\,\,\mapsto\,\,\,\,1/3$\\$10\,\,\,\mapsto\,\,\,\,\,\,\,\,0$\\$\vdots$\end{tabular}\right)}$. QED

We can now prove that $\Q$ is countably infinite,\\
\phx\qquad {\it i.e.}, that $\Q$ is at the same level as $\N$ and $\N_0$ and $\Z$.

\begin{thm}
$\exists\N\hookrightarrow\Q$.
\end{thm}

\begin{proof}
By \tref{thm-NontoQ}, \qquad $\exists\N\to>\Q$.\\
\phx\quad $\,$ So, by \tref{thm-exists-surj-implies-exists-inj}, \qquad $\exists\Q\hookrightarrow\N$.\\
Since $\id^\N:\N\hookrightarrow\Q$, \qquad we see that \quad $\exists\N\hookrightarrow\Q$.\\
Since both \qquad $\exists\Q\hookrightarrow\N$ \qquad and \qquad $\exists\N\hookrightarrow\Q$,\\
\phx\qquad by the Schoeder-Bernstein Theorem,\\
\phx\qquad we get \qquad $\exists\N\hookrightarrow>\Q$, \qquad as desired.
\end{proof}

\subsection{Power sets}

\begin{defn}
Let $S$ be a set.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ Then $2^S\,\,:=\,\,\{\,\hbox{subsets of }S\,\}$.
\end{defn}

Let $S$ be a set.
Then the set $2^S$ is the set of all subsets of $S$; it is therefore a set of sets.
It is called the \bfindex[Power set] of $S$.

To list all the subsets of $\{7,8,9\}$, build a table of YESs and NOs:
\begin{center}
\begin{tabular}{|c||c|c|c|c|c|c|c|c|}
\hline
$7$ &  NO &  NO &  NO &  NO & YES & YES & YES & YES \\ \hline
$8$ &  NO &  NO & YES & YES &  NO &  NO & YES & YES \\ \hline
$9$ &  NO & YES &  NO & YES &  NO & YES &  NO & YES \\ \hline
\end{tabular}
\end{center}
Each column gives us a particular subset, by telling us whehter or not a given element of $\{7,8,9\}$
should be in the subset or not.

We therefore arrive at eight subsets:

\begin{thm}\wrlab{thm-subsets-of-789}
$2^{\{7,8,9\}}\,\,\,=\,\,\,\{\,\,\,\emptyset\,\,\,,\,\,\,\{9\}\,\,\,,$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $\{8\}\,\,\,,\,\,\,\{8,9\}\,\,\,,\,\,\,\{7\}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $\{7,9\}\,\,\,,\,\,\,\{7,8\}\,\,\,,\,\,\,\{7,8,9\}\,\,\,\}$.\\
\end{thm}

Note that $\#(2^{\{7,8,9\}})=8=2^3=2^{\#\{7,8,9\}}$.
This is not a coincidence:

\begin{thm}
Let $S$ be a finite set. \quad
Then $\#(2^S)=2^{\#S}$.
\end{thm}

\subsection{Sets of functions}

\begin{defn}
Let $S$ and $T$ be sets.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ Then $T^S\,\,:=\,\,\{\,\hbox{functions }S\to T\,\}$.
\end{defn}

To list all the functions $\{7,8,9\}\to\{0,1\}$, build a table of $1$s and $0$s:
\begin{center}
\begin{tabular}{|c||c|c|c|c|c|c|c|c|}
\hline
$7$ & $0$ & $0$ & $0$ & $0$ & $1$ & $1$ & $1$ & $1$ \\ \hline
$8$ & $0$ & $0$ & $1$ & $1$ & $0$ & $0$ & $1$ & $1$ \\ \hline
$9$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ \\ \hline
\end{tabular}
\end{center}
Each column gives us a particular function, by telling us whehter or not a given element of $\{7,8,9\}$
should map to $1$ or $0$.

We therefore arrive at eight functions:

\begin{thm}
We have:\\
\phx\qquad\qquad\qquad $\ds{\{0,1\}^{\{7,8,9\}}=\bigg\{\left(\begin{tabular}{c}$7\mapsto0$\\$8\mapsto0$\\$9\mapsto0$\end{tabular}\right),
\left(\begin{tabular}{c}$7\mapsto0$\\$8\mapsto0$\\$9\mapsto1$\end{tabular}\right),}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\ds{\,\,\,\,\left(\begin{tabular}{c}$7\mapsto0$\\$8\mapsto1$\\$9\mapsto0$\end{tabular}\right),
\left(\begin{tabular}{c}$7\mapsto0$\\$8\mapsto1$\\$9\mapsto1$\end{tabular}\right),
\left(\begin{tabular}{c}$7\mapsto1$\\$8\mapsto0$\\$9\mapsto0$\end{tabular}\right)}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\ds{\,\,\,\,\left(\begin{tabular}{c}$7\mapsto1$\\$8\mapsto0$\\$9\mapsto1$\end{tabular}\right),
\left(\begin{tabular}{c}$7\mapsto1$\\$8\mapsto1$\\$9\mapsto0$\end{tabular}\right),
\left(\begin{tabular}{c}$7\mapsto1$\\$8\mapsto1$\\$9\mapsto1$\end{tabular}\right)\bigg\}}$.\\
\end{thm}

Note that $\#(\{0,1\}^{\{7,8,9\}})=8=2^3=(\#\{0,1\})^{\#\{7,8,9\}}$.
This is not a coincidence:

\begin{thm}
Let $S$ and $T$ be finite sets.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then $\#(T^S)=(\#T)^{\#S}$.
\end{thm}

I asked one of you to pick an element of $\{0,1\}^{\{7,8,9\}}$ and you picked the fifth element in the list above.
That is, you picked $\ds{\left(\begin{tabular}{c}$7\mapsto1$\\$8\mapsto0$\\$9\mapsto0$\end{tabular}\right)}$.

We then calculated $f^*(\{1\})$ and obtained $\{7\}$.
The fifth element of the list in \tref{thm-subsets-of-789} is $\{7\}$.
Thus we can build a bijection\\
\phx\qquad\qquad $\{0,1\}^{\{7,8,9\}}\quad\hookrightarrow>\quad2^{\{7,8,9\}}$\\
by mapping a function $f\in\{0,1\}^{\{7,8,9\}}$ to $f^*(\{1\})\in2^{\{7,8,9\}}$.

This works for any set $S$, even infinite sets:

\begin{thm}\wrlab{thm-2toS-equinumerous-01toS}
Let $S$ be a set.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then $\exists\{0,1\}^S\,\,\hookrightarrow\,\,2^S$.
\end{thm}

Idea of proof: Define $\Phi:\{0,1\}^S\to2^S$ by:\\
\phx\qquad\qquad\qquad $\forall f\in\{0,1\}^S,\quad\Phi_f=f^*(\{1\})$.\\
Show that: \qquad\qquad\quad $\Phi:\{0,1\}^S\hookrightarrow>2^S$.\\
That is, show that $\Phi$ is one-to-one, and that $\I_\Phi=2^S$. QED

\subsection{The World of Sets has no top}
\quad
\vskip.1in\noindent

The next result shows that $\N$ is not at the top of the World of Sets,
because $\{0,1\}^\N$ is higher:

\begin{thm}\wrlab{thm-2N-above-N}
$\not\exists2^\N\hookrightarrow\N$.
\end{thm}

\begin{proof}
Assume $\exists2^\N\hookrightarrow\N$. \quad
Want: Contradiction.\\
By \tref{thm-2toS-equinumerous-01toS}, $\exists\{0,1\}^\N\hookrightarrow>2^\N$, so $\exists\{0,1\}^\N\hookrightarrow2^\N$.\\
Since \quad $\exists\{0,1\}^\N\hookrightarrow2^\N$ \quad and \quad $\exists2^\N\hookrightarrow\N$,\\
\phx\qquad we see that: \quad $\exists\{0,1\}^\N\hookrightarrow\N$.\\
Then, by \tref{thm-inj-and-surj-compare-same-on-nonempty}, $\exists\N\to>\{0,1\}^\N$.\\
Choose $\Phi$ s.t.~$\Phi:\N\to>\{0,1\}^\N$.\\
Define $f:\N\to\{0,1\}$ by: $\forall j\in\N$, $f(j)=1-[\Phi_j(j)]$.\\
Since $\Phi:\N\to>\{0,1\}^\N$, we get: $\D_\Phi=\N$ and $\I_\Phi=\{0,1\}^\N$.\\
Since $f\in\{0,1\}^\N=\I_\Phi$, choose $j\in\D_\Phi$ s.t.~$f=\Phi_j$.\\
Then $f(j)=\Phi_j(j)$. \qquad Also, $j\in\D_\Phi=\N$.\\
By definition of $f$, we have: $f(j)=1-[\Phi_j(j)]$.\\
Let $x=\Phi_j(j)$. \qquad
Then $x=\Phi_j(j)=f(j)=1-[\Phi_j(j)]=1-x$.\\
Then $x=1-x$, so $2x=1$, so $x=1/2$.\\
Since $j\in\N$ and $f:\N\to\{0,1\}$, it follows that $f(j)\in\{0,1\}$.\\
Then $1/2=x=\Phi_j(j)=f(j)\in\{0,1\}$. \qquad
Then $1/2\in\{0,1\}$.\\
Since $1/2\ne0$ and $1/2\ne1$, we get $1/2\notin\{0,1\}$. \qquad
Contradiction.
\end{proof}

Idea of the preceding proof:\\
Each element of $\{0,1\}^\N$ is a sequence of bits.\\
In the proof of \tref{thm-2N-above-N}, the function $\Phi$ gives rise to
an infinite matrix $M$ of bits, by the rule:\\
\phx\qquad $\forall j,k\in\N$, \quad the $(j,k)$-entry of $M$ \quad is \quad $\Phi_j(k)$.\\
In order to get $\I_\Phi=\{0,1\}^\N$,\\
\phx\qquad we'd need that every element of $\{0,1\}^\N$ is a row of $M$.\\
However, we can make an element $f\in\{0,1\}^\N$ by the rule:\\
\phx\qquad $\forall j\in\N$, \quad the $j$-entry of $f$ \quad is \quad $1-[\Phi_j(j)]$.\\
Then $f$ cannot be the first row of $M$ because\\
\phx\qquad the first entry of $f$ \quad is not equal to\\
\phx\qquad the first entry of the first row of $M$.\\
Also, $f$ cannot be the second row of $M$ because\\
\phx\qquad the second entry of $f$ \quad is not equal to\\
\phx\qquad the second entry of the second row of $M$.\\
Also, $f$ cannot be the third row of $M$ because\\
\phx\qquad the third entry of $f$ \quad is not equal to\\
\phx\qquad the third entry of the second row of $M$.\\
Continuing, we see that, in fact $f$ cannot be any row of $M$. QED

\vskip.1in
In the preceding argument, great focus is paid to\\
\phx\qquad $\Phi_1(1)$ and $\Phi_2(2)$ and $\Phi_3(3)$ and $\cdots$.\\
These are the entries along the diagonal of $M$.
Moreover, the argument in the proof of \tref{thm-2N-above-N}
was evidently discovered by G.~Cantor.
For this reason that argument in the proof of \tref{thm-2N-above-N}
is sometimes called a \bfindex[Cantor diagonalization argument].

\vskip.1in\noindent
The World of Sets has an element at the bottom, namely $\emptyset$.\\
The next result says that there is no set $S$ that is at the top:

\begin{thm}
$\forall$set $S$, $\not\exists2^S\hookrightarrow S$.
\end{thm}

Idea of proof: If $S=\emptyset$, then $\#2^S=\#\{\emptyset\}=1$
and $\#S=\#\emptyset=0$, so $\not\exists2^S\hookrightarrow S$.
We therefore only need consider the case where $S\ne\emptyset$.
In that case, since $S\ne\emptyset\ne2^S$, we can simply change $\N$ to $S$
throughout the proof of \tref{thm-2N-above-N}, and prove
$\not\exists2^S\hookrightarrow S$, as desired. QED

\subsection{Placement of $\R$ in the World of Sets}

\begin{thm}\wrlab{thm-basetwo-map}
$\exists\{0,1\}^\N\to>[0;1]$.
\end{thm}

Idea of proof:
Define $F:\{0,1\}^\N\to[0;1]$ by:\\
\phx\qquad $\forall a\in\{0,1\}^\N$, $F(a)=0.a_1a_2a_3\cdots$ (base two).\\
Now, $0.0111111\ldots=0.1000000\ldots$ (base two),\\
\phx\qquad so this function $F$ is not one-to-one.\\
However, $F$ can be shown to be surjective onto $[0;1]$.\\
Note that: $F(1,1,1,1,1,1,\ldots)=0.111111\ldots=1$ (base two).\\
Then $F:\{0,1\}^\N\to>[0;1]$. \quad
QED

\begin{thm}\wrlab{thm-baseten-map}
$\exists\{0,1\}^\N\hookrightarrow\R$.
\end{thm}

Idea of proof:
Define $G:\{0,1\}^\N\to[0;1]$ by:\\
\phx\qquad $\forall a\in\{0,1\}^\N$, $G(a)=0.a_1a_2a_3\cdots$ (base ten).\\
Note that $0.0999999\ldots=0.1000000\ldots$ (base ten),\\
\phx\qquad but we are not using the digit nine, only $0$ and $1$.\\
This map $G$ can be shown to be injective.\\
Also, $\I_G\subseteq[0;1]\subseteq\R$. \quad
Then $G:\{0,1\}^\N\hookrightarrow\R$. \quad
QED

\begin{thm}\wrlab{R-injects-into-01toN}
$\exists\R\hookrightarrow>\{0,1\}^\N$.
\end{thm}

\begin{proof}
Define $A:\R\to(-1;1)$ by $A_t=t/\sqrt{t^2+1}$.\\
Then $A:\R\hookrightarrow(-1;1)$.\\
Define $B:(-1;1)\to[0;1]$ by $B_t=(t+1)/2$.\\
Then $B:(-1;1)\hookrightarrow[0;1]$.\\
By \tref{thm-basetwo-map}, $\exists\{0,1\}^\N\to>[0;1]$.\\
So, by \tref{thm-exists-surj-implies-exists-inj}, $\exists[0;1]\hookrightarrow\{0,1\}^\N$.\\
Choose $C$ s.t.~$C:[0;1]\hookrightarrow\{0,1\}^\N$.\\
Then $C\circ B\circ A:\R\hookrightarrow\{0,1\}^\N$,
so $\exists\R\hookrightarrow\{0,1\}^\N$.\\
By \tref{thm-baseten-map}, $\exists\{0,1\}^\N\hookrightarrow\R$.\\
Since both \quad $\exists\R\hookrightarrow\{0,1\}^\N$ \quad and \quad $\exists\{0,1\}^\N\hookrightarrow\R$,\\
\phx\qquad by the Schroeder-Bernstein Theorem,\\
\phx\qquad we see that $\exists\R\hookrightarrow>\{0,1\}^\N$, as desired.
\end{proof}

This shows that $\R$ belongs on the same level as $\{0,1\}^\N$.
By \tref{thm-2toS-equinumerous-01toS}, this is the same level as $2^\N$,
and, by \tref{thm-2N-above-N}, this is above $\N$.

\begin{defn}
Let $S$ be a set.\\
\phx\qquad\qquad By $S$ has \bfindex[continuum cardinality], we mean $\exists\R\hookrightarrow>S$.
\end{defn}

We have now shown that the continuum cardinality level is above the countably infinite level.
This level contains $\R$ and $\{0,1\}^\N$ and $2^\N$.

Using the tools we have been developing, it is not hard to prove:\\
\phx\qquad For any $a\in\R$, for any $b>a$,\\
\phx\qquad\qquad\qquad the sets $(a;b)$, $[a;b)$, $(a;b]$ and $[a;b]$\\
\phx\qquad all have continuum cardinality.

An interesting question:\\
\phx\qquad Are there any sets strictly between\\
\phx\qquad\qquad the continuum cardinality level \quad and\\
\phx\qquad\qquad the countably infinite level?\\
The standard axioms of set theory, used by most mathematicians, are together sometimes called ``ZFC'',
for Zermelo-Fraenkel with Choice.

The assertion that\\
\phx\qquad there are NO sets strictly between\\
\phx\qquad\qquad the continuum cardinality level \quad and\\
\phx\qquad\qquad the countably infinite level\\
is sometimes called ``CH'', for the Continuum Hypothesis.

It turns out that there are models of ZFC for which CH is true,
and models of ZFC for which CH is false.
One therefore has a choice:\\
\phx\qquad You can work in ZFC+CH or ZFC+($\neg$CH) or plain old ZFC.\\
In this class,\\
\phx\qquad we choose to work in plain old ZFC,\\
and the Continuum Hypothesis will be treated as undecidable, and, to us, uninteresting.
It turns out that the basics of Real Analysis don't requre that we consider
any sets that might be located strictly between $\N$ and $\R$ in the World of Sets.

\subsection{Scalars, vectors, matrices and tensors}
\quad
\vskip.1in\noindent

\noindent
For all $a$, we define $(a):=(1\mapsto a)$.

\noindent
For all $a,b$, we define $(a,b):=\left(\begin{tabular}{c}$1\mapsto a$\\$2\mapsto b$\end{tabular}\right)$.

\noindent
For all $a,b,c$, we define $(a,b,c):=\left(\begin{tabular}{c}$1\mapsto a$\\$2\mapsto b$\\$3\mapsto c$\end{tabular}\right)$.

\noindent
For all $a,b,c,d$, we define $(a,b,c,d):=\left(\begin{tabular}{c}$1\mapsto a$\\$2\mapsto b$\\$3\mapsto c$\\$4\mapsto d$\end{tabular}\right)$.\\
\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\vdots$\\
\\
For all $a,\ldots,z$, we define $(a,\ldots,z):=\left(\begin{tabular}{c}$1\mapsto a$\\ \vdots \\$26\mapsto z$\end{tabular}\right)$.

\vskip.2in\noindent
For all sets $A,B$, we define\\
\phx\qquad\qquad $A\times B:=\{(a,b)\,|\,a\in A,b\in B\}$.\\
For all sets $A,B,C$, we define\\
\phx\qquad\qquad $A\times B\times C:=\{(a,b,c)\,|\,a\in A,b\in B,c\in C\}$.\\
For all sets $A,B,C,D$, we define\\
\phx\qquad\qquad $A\times B\times C\times D:=\{(a,b,c,d)\,|\,a\in A,b\in B,c\in C,d\in D\}$.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\vdots$\\
For all sets $A,\ldots,Z$, we define\\
\phx\qquad\qquad $A\times\cdots\times Z:=\{(a,\ldots,z)\,|\,a\in A,\ldots,z\in Z\}$.

\begin{defn}
$\forall$set $S$, $\forall k\in\N$, \quad $S^k:=S^{[1..k]}$.
\end{defn}

Since $[1..k]=\{1,\ldots,k\}$,\\
\phx\qquad the set $S^k$ is the set of all functions $\{1,\ldots,k\}\to S$.

\begin{thm}
$[1..4]=\{1,2,3,4\}$ \qquad and\\
\phx\qquad $\R^4=\R^{[1..4]}=\{\hbox{functions }[1..4]\to\R\}$ \qquad and\\
\phx\qquad $\ds{(8,9,6,6)=\left(\begin{tabular}{c}$1\mapsto8$\\$2\mapsto9$\\$3\mapsto6$\\$4\mapsto6$\end{tabular}\right)\in\R^{[1..4]}=\R^4}$.
\end{thm}

\begin{thm}
Let $v:=(8,9,6,6)$.\\
\phx\quad Then $v\in\R^4$ and $v_1=8$ and $v_2=9$ and $v_3=6$ and $v_4=6$.
\end{thm}

\begin{thm}
For any set $S$,\\
\phx\qquad $\exists S^1\hookrightarrow>S$ \qquad and\\
\phx\qquad $S^2=S\times S$ \qquad and\\
\phx\qquad $S^3=S\times S\times S$ \qquad and\\
\phx\qquad $S^4=S\times S\times S\times S$.
\end{thm}

If you wish, you can continue the preceding theorem out to\\
\phx\quad $S^{26}=S\times S\times S\times S\times S\times S\times S\times S\times S\times S\times S\times S\times S\times S\times S\times S\times$\\
\phx\quad\qquad\qquad\qquad\qquad $S\times S\times S\times S\times S\times S\times S\times S\times S\times S$,

\vskip.2in

\noindent
For all $a$, we define $[a]:=((1,1)\mapsto a)$.

\noindent
For all $a,b$, we define $\ds{\left[\begin{array}{c}a\\b\end{array}\right]:=\left(\begin{tabular}{c}$(1,1)\mapsto a$\\$(2,1)\mapsto b$\end{tabular}\right)}$.

\noindent
For all $a,b$, we define $\ds{\left[\begin{array}{cc}a&b\end{array}\right]:=\left(\begin{tabular}{c}$(1,1)\mapsto a$\\$(1,2)\mapsto b$\end{tabular}\right)}$.

\begin{thm}
All of the following are true:\\
\phx\qquad\qquad\qquad $(5;9)=\{x\in\R\,|\,5<x<9\}$ \qquad and\\
\phx\qquad\qquad\qquad $\ds{(5,9)=\left(\begin{tabular}{c}$1\mapsto a$\\$2\mapsto b$\end{tabular}\right)}$ \qquad\qquad\quad and\\
\phx\qquad\qquad\qquad $\ds{\left[\begin{array}{cc}5&9\end{array}\right]=\left(\begin{tabular}{c}$(1,1)\mapsto5$\\$(1,2)\mapsto9$\end{tabular}\right)}$ \qquad and\\
\phx\qquad\qquad\qquad $\ds{\left[\begin{array}{c}5\\9\end{array}\right]=\left(\begin{tabular}{c}$(1,1)\mapsto5$\\$(2,1)\mapsto9$\end{tabular}\right)}$.\\
\end{thm}

\noindent
For all $a,b,c,d$, we define $\ds{\left[\begin{array}{cc}a&b\\c&d\end{array}\right]:=
\left(\begin{tabular}{c}$(1,1)\mapsto a$\\$(1,2)\mapsto b$\\$(2,1)\mapsto c$\\$(2,2)\mapsto d$\end{tabular}\right)}$.

\begin{thm}
We have:\\
\phx\qquad\qquad\qquad\qquad $\ds{\left[\begin{array}{cc}(5;9)&(5,9)\\\left[\begin{array}{cc}5&9\end{array}\right]&\left[\begin{array}{c}5\\9\end{array}\right]\end{array}\right]=
\left(\begin{tabular}{c}$(1,1)\mapsto(5;9)$\\$(1,2)\mapsto(5,9)$\\$(2,1)\mapsto\left[\begin{array}{cc}5&9\end{array}\right]$\\$(2,2)\mapsto\left[\begin{array}{c}5\\9\end{array}\right]$\end{tabular}\right)}$.
\end{thm}

We can continue these definitions to $2\times3$ and $3\times2$ matrices:\\
\phx\qquad\qquad For all $a,b,c,d,e,f$, we define:\\
\phx\qquad\qquad\qquad\qquad $\ds{\left[\begin{array}{ccc}a&b&c\\d&e&f\end{array}\right]:=
\left(\begin{tabular}{c}$(1,1)\mapsto a$\\$(1,2)\mapsto b$\\$(1,3)\mapsto c$\\$(2,1)\mapsto d$\\$(2,2)\mapsto e$\\$(2,3)\mapsto f$\end{tabular}\right)}$ \qquad\qquad and\\
\phx\qquad\qquad\qquad\qquad $\ds{\,\,\,\,\,\left[\begin{array}{cc}a&d\\b&e\\c&f\end{array}\right]:=
\left(\begin{tabular}{c}$(1,1)\mapsto a$\\$(2,1)\mapsto b$\\$(3,1)\mapsto c$\\$(1,2)\mapsto d$\\$(2,2)\mapsto e$\\$(3,2)\mapsto f$\end{tabular}\right)}$.

\begin{thm}
Both of the following are true:\\
\phx\qquad\qquad\qquad\qquad $\ds{\left[\begin{array}{ccc}4&5&6\\9&8&7\end{array}\right]:=
\left(\begin{tabular}{c}$(1,1)\mapsto4$\\$(1,2)\mapsto5$\\$(1,3)\mapsto6$\\$(2,1)\mapsto9$\\$(2,2)\mapsto8$\\$(2,3)\mapsto7$\end{tabular}\right)}$ \qquad\qquad and\\
\phx\qquad\qquad\qquad\qquad $\ds{\,\,\,\,\,\left[\begin{array}{cc}4&9\\5&8\\6&7\end{array}\right]:=
\left(\begin{tabular}{c}$(1,1)\mapsto4$\\$(2,1)\mapsto5$\\$(3,1)\mapsto6$\\$(1,2)\mapsto9$\\$(2,2)\mapsto8$\\$(3,2)\mapsto7$\end{tabular}\right)}$.
\end{thm}

We can continue these definitions to $2\times4$, $4\times2$, $3\times3$ matrices, {\it etc.},\\
but we leave it to the reader to do this work.

\begin{defn}
$\forall i,j\in\N$, $i\times j:=[1..i]\times[1..j]$ \quad and\\
\phx\qquad\qquad\qquad\qquad $\forall i,j,k\in\N$, $i\times j\times k:=[1..i]\times[1..j]\times[1..k]$
\end{defn}

The reader can continue these definitions to include\\
\phx\qquad\qquad $i\times j\times k\times\ell$ \quad and \quad $i\times j\times k\times\ell\times m$ \quad and so on.

\begin{thm}
$2\times2=[1..2]\times[1..2]=\{1,2\}\times\{1,2\}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $=\{(1,1),(1,2),(2,1),(2,2)\}$.\\
\phx\qquad Also, $\R^{2\times2}\,\,=\,\,\{\,\,\hbox{functions }\{(1,1),(1,2),(2,1),(2,2)\}\to\R\,\,\}$.\\
\phx\qquad Also, $\ds{\left[\begin{array}{cc}6&7\\8&9\end{array}\right]=\left(\begin{tabular}{c}$(1,1)\mapsto6$\\$(1,2)\mapsto7$\\$(2,1)\mapsto8$\\$(2,2)\mapsto9$\end{tabular}\right)\in\R^{2\times2}}$.
\end{thm}

\begin{thm}\ \\
\phx\qquad We have $2\times2\times2=[1..2]\times[1..2]\times[1..2]$\\
\phx\qquad\qquad\qquad\qquad\qquad $\,=\{1,2\}\times\{1,2\}\times\{1,2\}$\\
\phx\qquad\qquad\qquad\qquad\qquad $\,=\{(1,1,1),(1,1,2),(1,2,1),(1,2,2),$\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\,\,\,\,(2,1,1),(2,1,2),(2,2,1),(2,2,2)\}$.
\end{thm}

\begin{thm}\wrlab{thm-eg-three-tensor}
Let\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\ds{T:=\left(\begin{tabular}{l}
$(1,1,1)\mapsto\,\,\,\,\,4$\\
$(1,1,2)\mapsto\,\,\,\,\,3$\\
$(1,2,1)\mapsto\,\,\,\,\,9$\\
$(1,2,2)\mapsto-6$\\
$(2,1,1)\mapsto\,\,\,\,\,6$\\
$(2,1,2)\mapsto-9$\\
$(2,2,1)\mapsto\,\,15$\\
$(2,2,2)\mapsto\,\,\,\,\,8$
\end{tabular}\right)}$.\\
\phx\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ Then $T\in\R^{2\times2\times2}$.
\end{thm}


\noindent
A real number like $\sqrt{2}$ is sometimes called\\
\phx\qquad\qquad a \bfindex[scalar] \quad or \quad a $0$-\bfindex[tensor].\\
A one-dimensional array like $(1,2,3)$ is sometimes called\\
\phx\qquad\qquad a \bfindex[vector] \quad or \quad a $1$-{\bf tensor}.\\
A two-dimensional array like $\ds{\left[\begin{array}{ccc}4&5&6\\9&8&7\end{array}\right]}$ is sometimes called\\
\phx\qquad\qquad a \bfindex[matrix] \quad or \quad a $2$-{\bf tensor}.\\
The object $T$ from \tref{thm-eg-three-tensor} is hard to visualize on a page, but can be thought of as a three-dimesional array of real numbers (with shape $2\times2\times2$).
This kind of object is sometimes called\\
\phx\qquad\qquad a $3$-{\bf tensor}.\\
Even harder to visualize would be an element of, say $\R^{5\times7\times2\times3}$,
which can be thought of as a four-dimensional array of real numbers (with shape $5\times7\times2\times3$),
and is sometimes called\\
\phx\qquad\qquad a $4$-{\bf tensor}.\\
Continuing, we have $5$-{\bf tensors} and $6$-{\bf tensors} and so on.

\subsection{Functionals}
\quad
\vskip.1in\noindent

Frownie is infective:

\begin{defn}
$\forall a$, $a+\frownie:=\frownie$ and $\frownie+a=\frownie$\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ and $a-\frownie=\frownie$ and $\frownie-a=\frownie$\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ and $\,a\cdot\frownie=\frownie\,$ and $\,\frownie\cdot a=\frownie$\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ and $\,\,a/\frownie=\frownie\,\,$ and $\,\,\frownie/a=\frownie$.\\
\phx\qquad\qquad Also, $\sqrt{\frownie}:=\frownie$.\\
\phx\qquad\qquad Also, $\CH_{\frownie}:=\frownie$ and $\CH(\frownie):=\frownie$.\\
\phx\qquad\qquad Also, $\UE_{\frownie}:=\frownie$ and $\UE(\frownie):=\frownie$.
\end{defn}

\begin{defn}
Let $f$ be an object.\\
\phx\qquad By $f$ is a \bfindex[functional], we mean:\\
\phx\qquad\qquad\qquad $f$ is a function \quad and \quad $\I_f\subseteq\R$.
\end{defn}

\begin{defn}
Let $f$ be a functional and let $a\in\R$.\\
\phx\qquad Then $af$ is the functional defined by:\\
\phx\qquad\qquad\qquad $\forall x$, \quad $(af)_x=a\cdot f_x$.
\end{defn}

\begin{defn}
Let $f$ and $g$ be functionals.\\
\phx\qquad Then $f+g$ is the functional defined by:\\
\phx\qquad\qquad\qquad $\forall x$, \quad $(f+g)_x=f_x+g_x$.\\
\phx\qquad Also, $f-g$ is the functional defined by:\\
\phx\qquad\qquad\qquad $\forall x$, \quad $(f-g)_x=f_x-g_x$.\\
\phx\qquad Also, $fg$ is the functional defined by:\\
\phx\qquad\qquad\qquad $\forall x$, \quad $(fg)_x=f_x\cdot g_x$.\\
\phx\qquad Also, $f/g$ is the functional defined by:\\
\phx\qquad\qquad\qquad $\forall x$, \quad $(f/g)_x=f_x/g_x$.\\
\end{defn}

\begin{thm}
$6\cdot(7,8,9)=(42,48,54)$.
\end{thm}

\begin{proof}
We have:\\
\phx\quad
$6\cdot(7,8,9)=\left(\begin{tabular}{c}$1\mapsto7$\\$2\mapsto8$\\$3\mapsto9$\end{tabular}\right)=\left(\begin{tabular}{c}$1\mapsto42$\\$2\mapsto48$\\$3\mapsto54$\end{tabular}\right)=(42,48,54)$.
\end{proof}

\begin{thm}
$\ds{6\cdot\left[\begin{array}{ccc}7&8&9\\1&2&3\end{array}\right]=
\left[\begin{array}{ccc}42&48&54\\5&12&18\end{array}\right]}$
\end{thm}

\begin{thm}
$(3,1)+(5,6)=(8,7)$.
\end{thm}

\begin{proof}
We have:\\
\phx\quad
$(3,1)+(5,6)=\left(\begin{tabular}{c}$1\mapsto3$\\$2\mapsto1$\end{tabular}\right)+\left(\begin{tabular}{c}$1\mapsto5$\\$2\mapsto6$\end{tabular}\right)=
\left(\begin{tabular}{c}$1\mapsto8$\\$2\mapsto7$\end{tabular}\right)=(8,7)$.
\end{proof}

\begin{proof}
We have:\\
\phx\quad
$(3,1)+(5,6)=\left(\begin{tabular}{c}$1\mapsto3$\\$2\mapsto1$\end{tabular}\right)+\left(\begin{tabular}{c}$1\mapsto5$\\$2\mapsto6$\end{tabular}\right)=
\left(\begin{tabular}{c}$1\mapsto8$\\$2\mapsto7$\end{tabular}\right)=(8,7)$.
\end{proof}

\begin{thm}\ \\
\phx\qquad\qquad $\ds{\left[\begin{array}{ccc}3&2&1\\5&-1&7\end{array}\right]+\left[\begin{array}{ccc}8&-1&4\\9&16&2\end{array}\right]=\left[\begin{array}{ccc}11&1&3\\14&15&9\end{array}\right]}$.
\end{thm}

\begin{defn}
Let $i\in\N$, $V:=\R^i$.
Then $0_V:=C_{[1..i]}^0$.
\end{defn}

\begin{thm}
Let $V:=\R^4$.\\
\phx\qquad Then $0_V=C_{[1..4]}^0=C_{\{1,2,3,4\}}^0=\left(\begin{tabular}{c}$1\mapsto0$\\$2\mapsto0$\\$3\mapsto0$\\$4\mapsto0$\end{tabular}\right)=(0,0,0,0)$.
\end{thm}

\begin{defn}
Let $i,j\in\N$, $V:=\R^{i\times j}$.
Then $0_V:=C_{i\times j}^0$.
\end{defn}

Recall: $2\times3=[1..2]\times[1..3]=\{1,2\}\times\{1,2,3\}$.

\begin{thm}
Let $V:=\R^{2\times3}$.\\
\phx Then $0_V=C_{2\times3}^0=C_{\{1,2\}\times\{1,2,3\}}^0=\left(\begin{tabular}{c}$(1,1)\mapsto0$\\$(1,2)\mapsto0$\\$(1,3)\mapsto0$\\$(2,1)\mapsto0$\\$(2,2)\mapsto0$\\$(2,3)\mapsto0$\end{tabular}\right)
=\left[\begin{array}{ccc}0&0&0\\0&0&0\end{array}\right]$.
\end{thm}

\begin{defn}
Let $i,j,k\in\N$, $V:=\R^{i\times j\times k}$.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,\,$ Then $0_V:=C_{i\times j\times k}^0$.
\end{defn}

The reader is invited to continue these definitions,\\
with $i\times j\times k\times\ell$ and $i\times j\times k\times\ell\times m$ and so on.

\subsection{Translating and dilating sets of reals}

\begin{defn}
Let $a\in\R$, $S\subseteq\R$.\\
\phx\qquad\qquad Then $a\cdot S:=\{ax\,|\,x\in S\}$.\\
\phx\qquad\qquad Also, $a+S:=\{a+x\,|\,x\in S\}$.\\
\phx\qquad\qquad Also, $S\cdot a:=\{xa\,|\,x\in S\}$.\\
\phx\qquad\qquad Also, $S+a:=\{x+a\,|\,x\in S\}$.\\
\end{defn}

\begin{thm}
$(-4)\cdot[1;2)=(-8;-4]$ \quad and \quad $4+(2;5]=(6;9]$\\
\phx\qquad\qquad and \quad $3\N_0+2=\{2,5,8,11,14,17,20,\ldots\}$.
\end{thm}

\begin{defn}
Let $s$ be an object.\\
\phx\qquad By $s$ is a \bfindex[sequence], we mean:\\
\phx\qquad\qquad\qquad $s$ is a function \quad and \quad $\D_s=\N$.
\end{defn}

Let $s_1,s_2,s_3,\ldots$ be objects.
Then, by $(s_1,s_2,s_3,\ldots)$, we mean:\\
\phx\qquad\qquad the sequence \quad $\ds{\left(\begin{tabular}{c}$1\mapsto s_1$\\$2\mapsto s_2$\\$3\mapsto s_3$\\\vdots\end{tabular}\right)}$.

\begin{thm}
Let $s:=(5,6,7,5,6,7,5,6,7,5,6,7,5,6,7,\ldots)$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,\,\,\,$ Then \quad $s$ is a sequence \quad and \quad $s_{1000}=5$.
\end{thm}

A purist is uncomfortable with ellipses ($\cdots$), and would prefer that we replace\\
\phx\qquad Let $s:=(5,6,7,5,6,7,5,6,7,5,6,7,5,6,7,\ldots)$\\
with\\
\phx\qquad Let $s:\N\to\R$ be defined by: $\forall j\in\N$,
$s_j=\begin{cases}
5,&\hbox{if }j\in3\N_0+1\\
6,&\hbox{if }j\in3\N_0+2\\
7,&\hbox{if }j\in3\N_0+3.
\end{cases}$\\

\subsection{Absolute value and norms}

\begin{defn}
$\forall x\in\R$, $|x|:=\sqrt{x^2}$.
\end{defn}

In the preceding definition, the number $|x|$ is called
the \bfindex[absolute value] of $x$.

\begin{thm}
$|-7|=\sqrt{(-7)^2}=\sqrt{49}=7$.
\end{thm}

\begin{thm}
The following are all true:\\
\phx\qquad (1) $\forall x\in\R$, \quad $(\,\,|x|=0\,\,)\,\,\Leftrightarrow\,\,(\,\,x=0\,\,)$ \qquad and\\
\phx\qquad (2) $\forall a\in\R$, \quad $\forall x\in\R$, $|ax|=|a|\cdot|x|$ \qquad and\\
\phx\qquad (3) $\forall x,y\in\R$, \quad $|x+y|\le|x|+|y|$.
\end{thm}

In the preceding theorem, (1) says absolute value ``separates $0$'' (from other real numbers),
while (2) says absolute value is ``absolute homogeneous'', while (3) says absolute value is ``subadditive''.

\begin{defn}
Let $i\in\N$, $V:=\R^i$. \quad
Then:\\
\phx\qquad\qquad\qquad\qquad $\forall x\in V$, \quad $\ds{|x|:=\sqrt{\sum_{s=1}^i\,x_s^2}}$
\end{defn}

In the preceding definition, the number $|x|$ is called the \bfindex[norm] of $x$,
or the $V$-{\bf norm} of $x$.
It is sometimes written $|x|_V$ for clarity.

\begin{thm}
$|(-4,3)|=\sqrt{(-4)^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$.
\end{thm}

\begin{thm}
Let $i\in\N$, $V:=\R^i$. \quad
Then:\\
\phx\qquad (1) $\forall x\in V$, \quad $(\,\,|x|_V=0\,\,)\,\,\Leftrightarrow\,\,(\,\,x=0_V\,\,)$ \qquad and\\
\phx\qquad (2) $\forall a\in\R$, \quad $\forall x\in V$, $|ax|=|a|\cdot|x|_V$ \qquad and\\
\phx\qquad (3) $\forall x,y\in V$, \quad $|x+y|_V\le|x|_V+|y|_V$.
\end{thm}

In the preceding theorem, (1) says the $V$-norm ``separates $0_V$'' (from other vectors),
while (2) says the $V$-norm is ``absolute homogeneous'', while (3) says  the $V$-norm is ``subadditive''.
Subadditivity is tricky to prove, and its proof will be deferred until spring semester.

\begin{defn}
Let $i,j\in\N$, $V:=\R^{i\times j}$. \quad
Then:\\
\phx\qquad\qquad\qquad\qquad $\forall x\in V$, \quad $\ds{|x|:=\sqrt{\sum_{s\in i\times j}\,x_s^2}}$
\end{defn}

In the preceding definition, the number $|x|$ is called the {\bf norm} of $x$,
or the $V$-{\bf norm} of $x$.
It is sometimes written $|x|_V$ for clarity.

\begin{thm}
$\left|\left[\begin{array}{ccc}7&8&9\\4&5&6\end{array}\right]\right|=\sqrt{7^2+8^2+9^2+4^2+5^2+6^2}$.
\end{thm}

\begin{thm}
Let $i,j\in\N$, $V:=\R^{i\times j}$. \quad
Then:\\
\phx\qquad (1) $\forall x\in V$, \quad $(\,\,|x|_V=0\,\,)\,\,\Leftrightarrow\,\,(\,\,x=0_V\,\,)$ \qquad and\\
\phx\qquad (2) $\forall a\in\R$, \quad $\forall x\in V$, $|ax|=|a|\cdot|x|_V$ \qquad and\\
\phx\qquad (3) $\forall x,y\in V$, \quad $|x+y|_V\le|x|_V+|y|_V$.
\end{thm}

In the preceding theorem, (1) says the $V$-norm ``separates $0_V$'' (from other matrices),
while (2) says the $V$-norm is ``absolute homogeneous'', while (3) says  the $V$-norm is ``subadditive''.
Subadditivity is tricky to prove, and its proof will be deferred until spring semester.

\begin{defn}
Let $i,j,k\in\N$, $V:=\R^{i\times j\times k}$. \quad
Then:\\
\phx\qquad\qquad\qquad\qquad $\forall x\in V$, \quad $\ds{|x|:=\sqrt{\sum_{s\in i\times j\times k}\,x_s^2}}$
\end{defn}

In the preceding definition, the number $|x|$ is called the {\bf norm} of $x$,
or the $V$-{\bf norm} of $x$.
It is sometimes written $|x|_V$ for clarity.

\begin{thm}
Let $i,j,k\in\N$, $V:=\R^{i\times j\times k}$. \quad
Then:\\
\phx\qquad (1) $\forall x\in V$, \quad $(\,\,|x|_V=0\,\,)\,\,\Leftrightarrow\,\,(\,\,x=0_V\,\,)$ \qquad and\\
\phx\qquad (2) $\forall a\in\R$, \quad $\forall x\in V$, $|ax|=|a|\cdot|x|_V$ \qquad and \\
\phx\qquad (3) $\forall x,y\in V$, \quad $|x+y|_V\le|x|_V+|y|_V$.
\end{thm}

In the preceding theorem, (1) says the $V$-norm ``separates $0_V$'' (from other matrices),
while (2) says the $V$-norm is ``absolute homogeneous'', while (3) says  the $V$-norm is ``subadditive''.
Subadditivity is tricky to prove, and its proof will be deferred until spring semester.

These definitions can be continued to develop norms for four-tensors, five-tensors, {\it etc}.
In all cases, we have separation of zero, absolute homogeneity and subadditivity.
In all cases subadditivity is most easily proved using the ``dot product'' and Cauchy-Schwarz,
both of which will be exposed in the spring semester.

\subsection{Metrics and metric spaces}

\begin{defn}
Let $S$ be a set, $d:S\times S\to[0;\infty)$.\\
By $d$ is a \bfindex[metric] on $S$, we mean:\\
\phx\qquad (1) $\forall x,y\in S$, \quad $[\,\,x=y\,\,]\,\,\Leftrightarrow\,\,[\,\,d(x,y)=0\,\,]$ \qquad and\\
\phx\qquad (2) $\forall x,y\in S$, \quad $d(x,y)=d(y,x)$ \qquad and\\
\phx\qquad (3) $\forall x,y,z\in S$, \quad $d(x,z)\le[d(x,y)]+[d(y,z)]$.\\
\end{defn}

In the preceding definition, (1) says that $d$ ``separates points'', while
(2) says that $d$ is ``symmetric'', while (3) says that $d$ satisfies the ``triangle inequality''.

\begin{defn}
For any set $S$, $\scrm(S):=\{\hbox{metrics on }S\}$.
\end{defn}

\begin{defn}
Define $d_0:\R\times\R\to[0;\infty)$ by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\forall x,y\in\R$, \quad $d_0(x,y)=|y-x|$.
\end{defn}

\begin{thm}
$d_0(2,7)=d_0(7,2)=5$.
\end{thm}

\begin{thm}
$d_0\in\scrm(\R)$.
\end{thm}

The metric $d_0$ is the {\bf standard metric} on $\R$.

\begin{defn}
Let $i\in\N$, $V:=\R^i$.\\
Define $d_i:V\times V\to[0;\infty)$ by: \qquad $\forall x,y\in V$, \quad $d_i(x,y)=|y-x|_V$.
\end{defn}

\begin{thm}
$d_3(\,(4,1,5)\,,\,(4,-2,9)\,)=\sqrt{0^2+(-3)^2+4^2}$.
\end{thm}

\begin{thm}
Let $i\in\N$, $V:=\R^i$.
Then $d_i\in\scrm(V)$.
\end{thm}

Let $i\in\N$.
The metric $d_i$ is the {\bf standard metric} on $\R^i$.

\begin{defn}
Let $i,j\in\N$, $V:=\R^{i\times j}$.\\
Define $d_{(i,j)}:V\times V\to[0;\infty)$ by:\\
\phx\qquad\qquad\qquad $\forall x,y\in V$, \quad $d_{(i,j)}(x,y)=|y-x|_V$.
\end{defn}

\begin{thm}\ \\
\phx\qquad $d_{(2,3)}\left(\,\left[\begin{array}{ccc}1&5&-6\\4&10&9\end{array}\right]\,,\,\left[\begin{array}{ccc}1&4&-6\\8&10&15\end{array}\right]\,\right)=\sqrt{1^2+4^2+6^2}$.
\end{thm}

\begin{thm}
Let $i,j\in\N$, $V:=\R^{i\times j}$.
Then $d_{(i,j)}\in\scrm(V)$.
\end{thm}

Let $i,j\in\N$.
The metric $d_{(i,j)}$ is the {\bf standard metric} on $\R^{i\times j}$.

\begin{defn}
Let $i,j,k\in\N$, $V:=\R^{i\times j\times k}$.\\
Define $d_{(i,j,k)}:V\times V\to[0;\infty)$ by:\\
\phx\qquad\qquad\qquad $\forall x,y\in V$, \quad $d_{(i,j,k)}(x,y)=|y-x|_V$.
\end{defn}

\begin{thm}
Let $i,j,k\in\N$, $V:=\R^{i\times j\times k}$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then $d_{(i,j,k)}\in\scrm(V)$.
\end{thm}

Let $i,j,k\in\N$.
The metric $d_{(i,j,k)}$ is the {\bf standard metric} on $\R^{i\times j\times k}$.

\vskip.1in
These definitions can be continued to develop metrics for four-tensors, five-tensors, {\it etc}.

\begin{defn}
By a \bfindex[metric space], we mean: an ordered pair $(S,d)$ s.t.: \qquad $(\,\,S\hbox{ is a set}\,\,)\,\,\&\,\,(\,\,d\in\scrm(S)\,\,)$.
\end{defn}

\begin{thm}
$(\R,d_0)$ is a metric space.\\
\phx\qquad\quad $\,\,\,\,\,$ Also, $\forall i\in\N$, $(\R^i,d_i)$ is a metric space.\\
\phx\qquad $\,\,\,\,\,\,$ Also, $\forall i,j\in\N$, $(\R^{i\times j},d_{(i,j)})$ is a metric space.\\
\phx\quad $\,\,\,\,\,\,\,$ Also, $\forall i,j,k\in\N$, $(\R^{i\times j\times k},d_{(i,j,k)})$ is a metric space.\\
\end{thm}

\begin{thm}
Let $X:=(\R^5,d_5)$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,\,$ Then $X_1=\R^5$ and $X_2=d_5$.
\end{thm}

\begin{defn}
Let $X$ be a metric space.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,\,$ Then $X_\set:=X_1$ and $d_X:=X_2$.
\end{defn}

In the preceding definition, $X_\set$ is called the \bfindex[underlying set] of $X$,
while $d_X$ is called the \bfindex[underlying metric] of $X$.

\begin{thm}
Let $X:=(\R^5,d_5)$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,\,$ Then $X_\set=\R^5$ and $d_X=d_5$.
\end{thm}

\begin{thm}
Let $X:=(\R^{9\times8\times7},d_{(9,8,7)})$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,\,$ Then $X_\set=\R^{9\times8\times7}$ and $d_X=d_{(9,8,7)}$.
\end{thm}

\noindent
Out of sloppiness, we almost always write $X$ for $X_\set$.\\

\noindent
Out of sloppiness, we sometimes write $d$ for $d_X$.\\

\noindent
When a set $S$ has a standard metric $d$, then, out of sloppiness, we frequently write
$S$ to mean $(S,d)$.
So, for example, out of sloppiness, we frequently write $\R^{9\times8\times7}$ to mean $(\R^{9\times8\times7},d_{(9,8,7)})$.

\noindent
In particular, when we write $d_\R$, we would mean $d_{(\R,d_0)}$.\\
Then $d_\R=d_{(\R,d_0)}=d_0$.\\

\begin{thm}
$d_\R(2,7)=d_\R(7,2)=5$.
\end{thm}

Let $V:=\R^{2\times3}$.
Then $d_V=d_{(\R^{2\times3},d_{(2,3)})}=d_{(2,3)}$.

\begin{thm}
Let $V:=\R^{2\times3}$. Then\\
\phx\qquad\qquad $\,\left[\begin{array}{ccc}1&5&-6\\4&10&9\end{array}\right]\,,\,\left[\begin{array}{ccc}1&4&-6\\8&10&15\end{array}\right]\in V$ \qquad\qquad and \\
\phx\qquad $d_V\left(\,\left[\begin{array}{ccc}1&5&-6\\4&10&9\end{array}\right]\,,\,\left[\begin{array}{ccc}1&4&-6\\8&10&15\end{array}\right]\,\right)=\sqrt{1^2+4^2+6^2}$.
\end{thm}

\begin{thm}
Let $X:=\R^{2\times2}$. Then\\
\phx\qquad\qquad $\,\left[\begin{array}{cc}5&9\\6&7\end{array}\right]\,,\,\left[\begin{array}{cc}8&1\\0&0\end{array}\right]\in X$ \qquad\qquad and \\
\phx\qquad $d_X\left(\,\left[\begin{array}{cc}5&9\\6&7\end{array}\right]\,,\,\left[\begin{array}{cc}8&1\\0&0\end{array}\right]\,\right)=\sqrt{3^2+8^2+6^2+7^2}$.
\end{thm}

\begin{thm}
Let $X$ be a metric space and let $S\subseteq X$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then $d_X|(S\times S)\,\,\in\,\,\scrm(S)$.
\end{thm}

\noindent
Let $X$ be a metric space and let $S\subseteq X$.\\
Then $d_X|(S\times S)$ is called the \bfindex[relative metric] on $S$ inherited from $X$.\\
It is the standard metric on $S$, and, by sloppiness,\\
\phx\qquad we sometimes write $S$ for the metric space $(S,d_X|(S\times S))$.\\
So, for example, in the next theorem, when we write\\
\phx\qquad ``Let $S:=[0;\infty)$'',\\
the meaning is\\
\phx\qquad ``Let \quad $S\,\,\,:=\,\,\,(\,\,[0;\infty)\,\,,\,\,d_\R|([0;\infty)\times[0;\infty))\,\,)$''.\\

\begin{thm}
Let $S:=[0;\infty)$. Then:\\
\phx\qquad $(\,\,d_S(2,7)=5\,\,)\,\,\&\,\,(\,\,d_S(4,-3)=\frownie\,\,)\,\,\&\,\,(\,\,d_\R(4,-3)=7\,\,)$.
\end{thm}

\begin{thm}
Let $S:=[3;4]$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: $\forall x,y\in S$, $d_S(x,y)=|y-x|$.
\end{thm}

\begin{thm}
Let $T:=[-8;-6]$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: $\forall p,q\in T$, $d_T(p,q)=|q-p|$.
\end{thm}

\begin{defn}
Let $X$ be a metric space, $q\in X$ and $t\in\R$.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,\,$ Then: $B_X(q,t):=\{p\in X\,|\,d_X(p,q)<t\}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad and $\barB_X(q,t):=\{p\in X\,|\,d_X(p,q)\le t\}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad and $\,S_X(q,t):=\{p\in X\,|\,d_X(p,q)=t\}$.
\end{defn}

\begin{defn}
Let $X$ be a metric space and let $q\in X$.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,\,$ Then: $\scrb_X(q):=\{B_X(q,t)\,|\,t>0\}$.
\end{defn}

\begin{defn}
Let $X$ be a metric space.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,\,$ Then: $\scrb_X:=\{B_X(q,t)\,|\,q\in X,\,t>0\}$.
\end{defn}

The next theorem is called the \bfindex[Recentering Theorem].

\begin{thm}
Let $X$ be a metric space, $B\in\scrb_X$ and $q\in X$.\\
\phx\qquad\qquad Then: \qquad (1) $\,$ $\exists C\in\scrb_X(q)~\hbox{s.t.}~B\subseteq C$\\
\phx\qquad\qquad $\,$ and $\,$ \qquad (2) $\,$ $(\,q\in B\,)\,\Rightarrow\,(\,\exists A\in\scrb_X(q)~\hbox{s.t.}~A\subseteq B\,)$.
\end{thm}

\begin{proof}
{\it Proof of (1):}\\
Since $B\in\scrb_X$, choose $p\in X$, $r>0$ s.t.~$B=B_X(p,r)$.\\
Let $s:=d_X(p,q)$. \quad Let $C:=B_X(q,r+s)$. \quad Then $C\in\scrb_X(q)$.\\
Want: $B\subseteq C$. \quad Want $\forall z\in B$, $z\in C$.\\
Given $z\in B$. \quad Want $z\in C$. \quad Want: $z\in B_X(q,r+s)$.\\
Since $z\in B=B_X(p,r)$, we get $d_X(z,p)<r$, so $[d_X(z,p)]+s<r+s$.\\
Then \quad $d_X(z,q)\,\,\le\,\,[d_X(z,p)]\,+\,[d_X(p,q)]\,\,=\,\,[d_X(z,p)]+s<r+s$,\\
\phx\qquad\qquad\qquad\qquad and so $z\in B_X(q,r+s)$, as desired.\\
{\it End of proof of (1).}\\
\\
{\it Proof of (2):}\\
Unassigned HW.\\
{\it End of proof of (2).}
\end{proof}

\begin{thm}
$\forall a\in\R$, $\forall\vep>0$, \quad $B(a,\vep)=(a-\vep;a+\vep)$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad and \quad $\barB(a,\vep)=[a-\vep;a+\vep]$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad and \quad $\,S(a,\vep)=\{a-\vep,a+\vep\}$.
\end{thm}

\begin{thm}
Let $a,x\in\R$ and let $\vep>0$.\\
\phx\qquad\qquad\qquad Then: \quad $(\,\,|x-a|<\vep\,\,)\,\,\Leftrightarrow\,\,(\,\,d_\R(a,x)<\vep\,\,)$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\,\,\,\,\Leftrightarrow\,\,(\,\,x\in B_\R(a,\vep)\,\,)$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\,\,\,\,\Leftrightarrow\,\,(\,\,x\in(a-\vep;a+\vep)\,\,)$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\,\,\,\,\Leftrightarrow\,\,(\,\,a-\vep<x<a+\vep\,\,)$.
\end{thm}

\begin{thm}
Let $a,x\in\R$ and let $\vep>0$.\\
\phx\qquad\qquad\qquad Then: \quad $(\,\,|x-a|<\vep\,\,)\,\,\Leftrightarrow\,\,(\,\,a-\vep<x<a+\vep\,\,)$.
\end{thm}

\begin{thm}
Let $x\in\R$ and let $\vep>0$.\\
\phx\qquad\qquad\qquad Then: \quad $(\,\,|x|<\vep\,\,)\,\,\Leftrightarrow\,\,(\,\,-\vep<x<\vep\,\,)$.
\end{thm}

\begin{thm}
Let $x\in\R$ and let $\vep>0$.\\
\phx\qquad\qquad\qquad Then: \quad $(\,\,d_\R(x,0)<\vep\,\,)\,\,\Leftrightarrow\,\,(\,\,-\vep<x<\vep\,\,)$.
\end{thm}

\begin{thm}
Let $a,x\in\R$ and let $\vep>0$.\\
\phx\qquad\qquad\qquad Then: \quad $(\,\,|x-a|\le\vep\,\,)\,\,\Leftrightarrow\,\,(\,\,a-\vep\le x\le a+\vep\,\,)$.
\end{thm}

\begin{thm}
Let $x\in\R$ and let $\vep>0$.\\
\phx\qquad\qquad\qquad Then: \quad $(\,\,|x|\le\vep\,\,)\,\,\Leftrightarrow\,\,(\,\,-\vep\le x\le \vep\,\,)$.
\end{thm}

\begin{thm}
Let $x\in\R$ and let $\vep>0$.\\
\phx\qquad\qquad\qquad Then: \quad $(\,\,d_\R(x,0)\le\vep\,\,)\,\,\Leftrightarrow\,\,(\,\,-\vep\le x\le\vep\,\,)$.
\end{thm}

\begin{thm}
$\forall z\in\R$, $(\,|z-5|<0.3\,)\,\Leftrightarrow\,(\,4.7<z<5.3\,)$.
\end{thm}

\begin{thm}
$\forall z\in\R$, $(\,|z-5|\le0.3\,)\,\Leftrightarrow\,(\,4.7\le z\le5.3\,)$.
\end{thm}

\begin{thm}
$\forall q\in\R$, $(\,|4-q|<0.01\,)\,\Leftrightarrow\,(\,3.99<q<4.01\,)$.
\end{thm}

\begin{thm}
$\forall q\in\R$, $(\,|4-q|\le0.01\,)\,\Leftrightarrow\,(\,3.99\le q\le4.01\,)$.
\end{thm}

\subsection{Lipschitz functions}

\begin{defn}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad Let $f:X\drta Y$ and let $K\ge0$.\\
\phx\qquad\qquad\qquad By $f$ is \bfindex[Lipschitz]-$K$ from $X$ to $Y$, we mean:\\
\phx\qquad\qquad\qquad\qquad\qquad $\forall p,q\in\D_f$, \quad $d_Y(f_p,f_q)\le K\cdot(d_X(p,q))$.
\end{defn}

\noindent
Lipschitz-$0$ is the same as constant.\\
Lipschitz-$1$ is sometimes called ``distance semi-decreasing'';\\
Lipschitz-$1$ means that the function\\
\phx\qquad\qquad\qquad may move two points closer together,\\
\phx\qquad\qquad\qquad but it never moves them farther apart.\\
Recall that a secant line for a function $f:\R\drta\R$ is\\
\phx\qquad a line that crosses through least two points of the graph of $f$.\\
Let $f:\R\drta\R$. \quad
Then $f$ is Lipschitz-$K$ means that\\
\phx\qquad\qquad\qquad every secant line has slope in $[-K;K]$.

\begin{thm}
Let $S:=[3;4]$ and $T:=[-7;-5]$.\\
\phx\qquad\qquad Define $f:S\to T$ by: $\forall x\in S$, $f_x=-2x+1$.\\
\phx\qquad\qquad Then $f$ is Lipschitz-$2$ from $S$ to $T$.
\end{thm}

\begin{proof}
Want: $\forall x,y\in S$, $d_T(f_x,f_y)\le2\cdot[d_S(x,y)]$.\\
Given $x,y\in S$. \qquad
Want: $d_T(f_x,f_y)\le2\cdot[d_S(x,y)]$.\\
It suffices to show: $d_T(f_x,f_y)=2\cdot[d_S(x,y)]$.\\
We have: $d_T(f_x,f_y)=|f_y-f_x|=|(-2y+1)-(-2x+1)|$\\
\phx\qquad\qquad\qquad\qquad $\,\,=|(-2y)-(-2x)|=|(-2)\cdot(y-x)|$\\
\phx\qquad\qquad\qquad\qquad $\,\,=|-2|\cdot|y-x|=2\cdot|y-x|=2\cdot(d_S(x,y))$.
\end{proof}

\begin{defn}
Define $|\bullet|:\R\to[0;\infty)$ by $\forall x\in\R$, $|\bullet|_x=|x|$.
\end{defn}

\begin{thm}
$|\bullet|$ is Lipschitz-$1$ from $\R$ to $[0;\infty)$.
\end{thm}

\begin{proof}
Want: $\forall x,y\in\R$, \quad $d(\,\,|\bullet|_x\,\,,\,\,|\bullet|_y\,\,)\,\,\,\le\,\,\,d(x,y)$.\\
Given $x,y\in\R$. \qquad\qquad\qquad\quad $\,$
Want: $d(\,\,|\bullet|_x\,\,,\,\,|\bullet|_y)\,\,\,\le\,\,\,d(x,y)$.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\,\,$ Want: $d(|x|,|y|)\,\,\,\le\,\,\,d(x,y)$.\\
Let $a:=|x|$ and $b:=|y|$. \qquad\qquad\qquad $\,\,\,\,\,\,$
Want: $d(a,b)\,\,\,\le\,\,\,d(x,y)$.\\
Want: $|b-a|\le|y-x|$. \qquad Let $t:=|y-x|$. \qquad Then $t=|x-y|$.\\
Want: $|b-a|\le t$. \qquad\qquad\qquad\qquad\qquad
Want: $a-t\le b\le a+t$.\\
Since $a=|x|=|y+(x-y)|\le|y|+|x-y|=b+t$,\\
\phx\qquad we get $a\le b+t$, \qquad
so $a-t\le b$. \qquad\quad $\,\,$
Want: $b\le a+t$.\\
We have $b=|y|=|x+(y-x)|\le|x|+|y-x|=a+t$.
\end{proof}

\begin{defn}
Let $i\in\N$ and let $V:=\R^i$.\\
\phx\qquad Define $|\bullet|_V:V\to[0;\infty)$ by: \qquad $\forall x\in V$, \quad $(|\bullet|_V)_x=|x|_V$.
\end{defn}

\begin{thm}
Let $i\in\N$ and let $V:=\R^i$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,\,\,\,$ Then $|\bullet|$ is Lipschitz-$1$ from $V$ to $[0;\infty)$.
\end{thm}

\begin{defn}
Let $i,j\in\N$ and let $V:=\R^{i\times j}$.\\
\phx\qquad Define $|\bullet|_V:V\to[0;\infty)$ by: \qquad $\forall x\in V$, \quad $(|\bullet|_V)_x=|x|_V$.
\end{defn}

\begin{thm}
Let $i,j\in\N$ and let $V:=\R^{i\times j}$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,\,\,\,$ Then $|\bullet|$ is Lipschitz-$1$ from $V$ to $[0;\infty)$.
\end{thm}

\begin{defn}
Let $i,j,k\in\N$ and let $V:=\R^{i\times j\times k}$.\\
\phx\qquad Define $|\bullet|_V:V\to[0;\infty)$ by: \qquad $\forall x\in V$, \quad $(|\bullet|_V)_x=|x|_V$.
\end{defn}

\begin{thm}
Let $i,j,k\in\N$ and let $V:=\R^{i\times j\times k}$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,\,\,\,$ Then $|\bullet|$ is Lipschitz-$1$ from $V$ to $[0;\infty)$.
\end{thm}

These definitions and theorems can be continued to four-tensors, five-tensors, {\it etc}.
Keep in mind that, in all of these theorems, Lipschitz-$1$ means distance semi-decreasing.
This concept comes up in other ways:

\begin{thm}
Define $p:\R^2\to\R$ by: \qquad $\forall v\in\R^2$, \quad $p_v=v_1$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,$ Then $p$ is Lipschitz-$1$.
\end{thm}

\begin{proof}
Let $X:=\R^2$. \qquad
Want: $\forall u,v\in X$, $\,$ $d_\R(p_u,p_v)\le d_X(u,v)$.\\
Given $u,v\in X$. \qquad
Want: $d_\R(p_u,p_v)\le d_X(u,v)$.\\
Want: $|p_v-p_u|\le|v-u|_X$. \quad $\,\,\,$
We have $p_u=u_1$ and $p_v=v_1$.\\
Want: $|v_1-u_1|\le|v-u|_X$. \qquad
We have $u=(u_1,u_2)$ and $v=(v_1,v_2)$.\\
Want: $|v_1-u_1|\le|(v_1,v_2)-(u_1,u_2)|_X$.\\
Want: $|v_1-u_1|\le|(v_1-u_1,v_2-u_2)|_X$.\\
We have $0\le(v_1-u_1)^2$ and $0\le(v_2-u_2)^2$.\\
Since $0\le(v_2-u_2)^2$, we get $(v_1-u_1)^2\le(v_1-u_1)^2+(v_2-u_2)^2$.\\
\phx\qquad\qquad\qquad\quad $\,$ Since $0\le(v_1-u_1)^2\le(v_1-u_1)^2+(v_2-u_2)^2$,\\
\phx\qquad\quad we conclude that $\,\sqrt{(v_1-u_1)^2}\le\sqrt{(v_1-u_1)^2+(v_2-u_2)^2}$.\\
That is, $|v_1-u_1|\le|(v_1-u_1,v_2-u_2)|_X$, as desired.
\end{proof}

\begin{defn}
Let $X$ and $Y$ be metric spaces, $f:X\drta Y$.\\
\phx\qquad\qquad\qquad\qquad\qquad\quad By $f$ is \bfindex[Lipschitz] from $X$ to $Y$, we mean:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\exists K\ge0$ s.t.~$f$ is Lipschitz-$K$.
\end{defn}

\section{Limits and Continuity}

\subsection{A doubly quantified statement}

\begin{thm}
$\forall\vep>0$, $\exists\delta>0$ s.t.~$\delta^2+\delta\le\vep$.
\end{thm}

\begin{proof}
Given $\vep>0$. \quad
Want: $\exists\delta>0$ s.t.~$\delta^2+\delta\le\vep$.\\
Let $\delta\,\,:=\,\,\min\{\,\,\sqrt{\vep/2}\,\,,\,\,\vep/2\,\,\}$.\quad
Then $\delta>0$.\\
Want: $\delta^2+\delta\le\vep$.\\
By the definition of $\delta$, we have\\
\phx\qquad\qquad both $\delta\le\sqrt{\vep/2}$ and $\delta\le\vep/2$.\\
Since $0<\delta\le\sqrt{\vep/2}$, we get $\delta^2\le(\sqrt{\vep/2})^2$.\\
Then $\delta^2\le\vep/2$. \quad
So, since $\delta\le\vep/2$, \quad $\delta^2+\delta\le(\vep/2)+(\vep/2)$.\\
Then $\delta^2+\delta\le(\vep/2)+(\vep/2)=\vep$, as desired.
\end{proof}

\subsection{Limits of sequences}

\begin{defn}
Let $X$ be a metric space.\\
\phx\qquad\qquad\qquad\qquad\qquad Let $s\in X^\N$ and let $z\in X$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then $s\to z$ in $X$ means:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\forall\vep>0$, $\exists K\in\N$ s.t.~$\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_j,z)<\vep\,\,)$.
\end{defn}

\begin{thm}
Define $s\in\R^\N$ by: $\forall j\in\N$, $s_j=1/j$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,$ Then $s\to0$ in $\R$.
\end{thm}

\begin{proof}
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_\R(s_j,0)<\vep\,\,)$.\\
Given $\vep>0$. \quad Want: $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_\R(s_j,0)<\vep\,\,)$.\\
By the Archimedean Principle, choose $K\in\N$ s.t.~$K>1/\vep$.\\
Want: $\forall j\in\N$, $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_\R(s_j,0)<\vep\,\,)$.\\
Since $j\ge K>1/\vep$, we get $j>1/\vep$.\\
Since $j>1/\vep>0$, we get $1/j<\vep$.\\
Since $j>0$, we get $1/j>0$, and so $|1/j|=1/j$.\\
Then $d_\R(s_j,0)=|s_j-0|=|s_j|=|1/j|=1/j<\vep$, as desired.
\end{proof}

\subsection{Some precalculus}
\quad
\vskip.1in\noindent

The results in this section will be used in the future without comment.

\begin{defn}
$\forall k\in\N$, $\forall a\in\R$, $\sqrt[k]{a}:=\max\{x\in\R\,|\,x^k=a\}$.
\end{defn}

\begin{thm}
We have:\\
\phx\qquad $(\,\,\forall a\in\R,\,\,\sqrt[2]{a}=\sqrt{a}\,\,)\,\,\&$\\
\phx\qquad $(\,\,\forall k\in\N,\forall a\ge0\,\,(\sqrt[k]{a})^k=a\,\,)\,\,\&$\\
\phx\qquad $(\,\,\forall k\in2\N_0+1,\forall a\in\R\,\,(\sqrt[k]{a})^k=a\,\,)$.
\end{thm}

\begin{thm}
$\forall\vep\in\R$, $(\vep/3)+(\vep/3)+(\vep/3)=\vep$.
\end{thm}

\begin{thm}
$\forall a,b,c,d,e,f\in\R$,\\
\phx\qquad\qquad $((a\le b)\&(c\le d)\&(e\le f))\Rightarrow(a+c+e\le b+d+f)$.
\end{thm}

\begin{thm}
Let $a,b,c\in\R^*$. \quad
Let $m:=\min\{a,b,c\}$.\\
\phx\qquad\qquad Then: $(\,m\in\{a,b,c\}\,)\,\&\,(\,m\le a\,)\,\&\,(\,m\le b\,)\,\&\,(\,m\le c\,)$.
\end{thm}

\begin{thm}
$\forall C\ge0$, $\forall\lambda>0$, \quad $\ds{\frac{C}{C+1}\cdot\lambda<\lambda}$.
\end{thm}

\subsection{The Precalculus Product Rule}

\begin{thm}
$\forall a,b,x,y\in\R$,\\
\phx\qquad\qquad $xy-ab\quad=\quad(x-a)\cdot b\,\,+\,\,a\cdot(y-b)\,\,+\,\,(x-a)\cdot(y-b)$.
\end{thm}

To prove this theorem, simply expand the right hand side and cancel.

To remember this theorem:

The theorem is true even if some of the variables are negative, and even if $x\le a$ or $y\le b$.

However, the easiest case to picture is the case when $x>a>0$ and $y>b>0$.
We imagine an $a\times b$ rectangle that grows to an $x\times y$ rectangle,
with the lower left corner staying fixed.
The change in area is $xy-ab$, and it can be viewed as the sum of the areas
of three subrectangles of the $x\times y$ rectangle.
This yields the formula in the preceding theorem.
\vskip.1in
\noindent
One can think of $x-a$ as the ``change to $a$'' and denote it by $\triangle a$.\\
One can think of $y-b$ as the ``change to $b$'' and denote it by $\triangle b$.\\
One can think of $xy-ab$ as the ``change to $ab$'' and denote it by $\triangle(ab)$.\\
With this notation the formula reads\\
\phx\qquad\qquad $\triangle(ab)\quad=\quad(\triangle a)\cdot b\,\,+\,\,a\cdot(\triangle b)\,\,+\,\,(\triangle a)\cdot(\triangle b)$.\\
Buzz phrase: The change to $ab$ \quad is equal to\\
\phx\qquad\qquad\qquad\qquad\qquad (the change to $a$) $\,\,$ times $\,\,$ $b$ \qquad\quad plus\\
\phx\qquad\qquad\qquad\qquad\qquad $a$ $\,\,$ times $\,\,$ (the change to $b$) \qquad\quad plus\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\,\,$ the product of the two changes.\\
The Precalculus Product Rule and Calculus Product Rule are similar.\\
However, in the Precalculus Product Rule, we need to remember\\
\phx\qquad\qquad that third term: \quad ``the product of the two changes''.

\subsection{Sequential limits of sums and products}

\begin{thm}
Let $s,t\in\R^\N$.\\
\phx\qquad\qquad Assume: $(\,\,s\to 2\hbox{ in }\R\,\,)\,\,\&\,\,(\,\,t\to 3\hbox{ in }\R\,\,)$.\\
\phx\qquad\qquad Then: \qquad\quad\qquad (1) $s+t\to5$ in $\R$\\
\phx\qquad\qquad\qquad\qquad and \qquad (2) $st\to6$ in $\R$.
\end{thm}

\begin{proof}
Since $s\to 2$ in $\R$, we have:
$\forall\vep>0$, $\exists L\in\N$ s.t.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge L\,\,)\,\,\Rightarrow\,\,(\,\,d_\R(s_j,2)<\vep\,\,)$.\\
\phantom{ll} Also, since $t\to 3$ in $\R$, we have:
$\forall\vep>0$, $\exists M\in\N$ s.t.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge M\,\,)\,\,\Rightarrow\,\,(\,\,d_\R(t_j,3)<\vep\,\,)$.\\
\\
{\it Proof of (1):}\\
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_\R((s+t)_j,5)<\vep\,\,)$.\\
Given $\vep>0$. \quad
Want: $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_\R((s+t)_j,5)<\vep\,\,)$.\\
Choose $L\in\N$ s.t., $\forall j\in\N$, \quad $(\,\,j\ge L\,\,)\,\,\Rightarrow\,\,(\,\,d_\R(s_j,2)<\vep/2\,\,)$.\\
Choose $M\in\N$ s.t., $\forall j\in\N$, \quad $(\,\,j\ge M\,\,)\,\,\Rightarrow\,\,(\,\,d_\R(t_j,3)<\vep/2\,\,)$.\\
Let $K:=\max\{L,M\}$. \qquad
Then $K\in\N$.\\
Want: $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_\R((s+t)_j,5)<\vep\,\,)$.\\
Assume $j\ge K$. \qquad
Want: $d_\R((s+t)_j,5)<\vep$.\\
Since $j\ge K\ge L$, by choice of $L$, we get $d_\R(s_j,2)<\vep/2$.\\
Since $j\ge K\ge M$, by choice of $M$, we get $d_\R(t_j,3)<\vep/2$.\\
Then $d_\R((s+t)_j,5)=|(s+t)_j-5|=|(s_j+t_j)-(2+3)|$\\
\phx\qquad\qquad $=|(s_j-2)+(t_j-3)|\le|s_j-2|+|t_j-3|$\\
\phx\qquad\qquad $=(d(s_j,2))+(d(t_j,3))<(\vep/2)+(\vep/2)=\vep$.\\
{\it End of proof of (1).}\\
\\
{\it Proof of (2):}\\
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_\R((st)_j,6)<\vep\,\,)$.\\
Given $\vep>0$. \quad
Want: $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_\R((st)_j,6)<\vep\,\,)$.\\
Let $\delta:=\min\{\vep/10,\sqrt{\vep/2}\}$. \qquad
Then $\delta>0$.\\
Choose $L\in\N$ s.t., $\forall j\in\N$, \quad $(\,\,j\ge L\,\,)\,\,\Rightarrow\,\,(\,\,d_\R(s_j,2)<\delta\,\,)$.\\
Choose $M\in\N$ s.t., $\forall j\in\N$, \quad $(\,\,j\ge M\,\,)\,\,\Rightarrow\,\,(\,\,d_\R(t_j,3)<\delta\,\,)$.\\
Let $K:=\max\{L,M\}$. \qquad
Then $K\in\N$.\\
Want: $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_\R((st)_j,6)<\vep\,\,)$.\\
Assume $j\ge K$. \qquad
Want: $d_\R((st)_j,6)<\vep$.\\
By definiton of $\delta$, we have: $\delta\le\vep/10$ and $\delta\le\sqrt{\vep/2}$.\\
Since $\delta<\vep/10$, we get: $5\delta\le\vep/2$.\\
Since $0<\delta\le\sqrt{\vep/2}$, we get $\delta^2\le\vep/2$.\\
Since $j\ge K\ge L$, by choice of $L$, we get $d_\R(s_j,2)<\delta$.\\
Since $j\ge K\ge M$, by choice of $M$, we get $d_\R(t_j,3)<\delta$.\\
By the Precalculus Product Rule,\\
\phx\qquad $s_jt_j-2\cdot3=(s_j-2)\cdot3+2\cdot(t_j-3)+(s_j-2)\cdot(t_j-3)$.\\
Then $d_\R((st)_j,6)=|(st)_j-6|=|s_jt_j-2\cdot3|$\\
\phx\qquad\qquad $=|(s_j-2)\cdot3+2\cdot(t_j-3)+(s_j-2)\cdot(t_j-3)|$\\
\phx\qquad\qquad $\le|s_j-2|\cdot3+2\cdot|t_j-3|+|s_j-2|\cdot|t_j-3|$\\
\phx\qquad\qquad $=(d_\R(s_j,2))\cdot3+2\cdot(d_\R(t_j,3))+(d_\R(s_j,2))\cdot(d_\R(t_j,3))$\\
\phx\qquad\qquad $<\delta\cdot3+2\cdot\delta+\delta\cdot\delta\,\,=\,\,5\delta+\delta^2\,\,\le\,\,(\vep/2)+(\vep/2)=\vep$.\\
{\it End of proof of (2).}
\end{proof}

\subsection{Continuity}

\begin{defn}
Let $X$ and $Y$ be metric spaces.\\
Let $f:X\drta Y$ and let  $p$ be an object.\\
Then by $f$ is \bfindex[continuous] at $p$ from $X$ to $Y$, we mean:\\
\phx\qquad\qquad  $[\,\,p\in\D_f\,\,]\qquad\&$\\
\phx\qquad\qquad $[\,\,\forall\vep>0,\,\,\exists\delta>0~\hbox{s.t.},\,\,\forall w\in\D_f,$\\
\phx\qquad\qquad\qquad\qquad $(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,f_p)<\vep\,\,)\,\,]$.
\end{defn}

\begin{thm}
Define $f:\R\to\R$ by: $\forall x\in\R$, $f_x=x^2$.\\
Then $f$ is continuous at $2$ from $\R$ to $\R$.
\end{thm}

\begin{proof}
Since $f_2=4\ne\frownie$, we get $2\in\D_f$.\\
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d(x,2)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_x,f_2)<\vep\,\,)$.\\
Given $\vep>0$. \quad
Want: $\exists\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d(x,2)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_x,f_2)<\vep\,\,)$.\\
Let $\delta:=\min\{1,\vep/5\}$. \quad
Then $\delta>0$.\\
Want: $\forall w\in\D_f$, $(\,\,d(x,2)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_x,f_2)<\vep\,\,)$.\\
Given $w\in\D_f$. \qquad
Want: $(\,\,d(x,2)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,f_2)<\vep\,\,)$.\\
Assume $d(w,2)<\delta$. \qquad
Want: $d(f_w,f_2)<\vep$.\\
By definition of $\delta$, we have $\delta\le1$ and $\delta\le\vep/5$.\\
Since $\delta\le\vep/5$, we get $5\delta\le\vep$.\\
We have $|w-2|=d(w,2)<\delta$.\\
Then $|w+2|=|w-2+4|\le|w-2|+4<\delta+4\le1+4=5$.\\
Since $0\le|w+2|<5$ and $0\le|w-2|<\delta$, by multiplying,\\
\phx\qquad\qquad we get $|w+2|\cdot|w-2|<5\delta$.\\
Then $d(f_w,f_2)=|f_w-f_2|=|w^2-2^2|=|(w+2)\cdot(w-2)|$\\
\phx\qquad\qquad\qquad\qquad $=|w+2|\cdot|w-2|<5\delta\le\vep$, \quad as desired.
\end{proof}

\begin{thm}
Define $f:\R\drta\R$ by: $\forall x\in\R$, $f_x=1/x$.\\
Then $f$ is continuous at $4$ from $\R$ to $\R$.
\end{thm}

\begin{proof}
Since $f_4=1/4\ne\frownie$, we get $4\in\D_f$.\\
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d(x,4)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_x,f_4)<\vep\,\,)$.\\
Given $\vep>0$. \quad
Want: $\exists\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d(x,4)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_x,f_4)<\vep\,\,)$.\\
Let $\delta:=\min\{1,12\vep\}$. \quad
Then $\delta>0$.\\
Want: $\forall w\in\D_f$, $(\,\,d(x,4)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_x,f_4)<\vep\,\,)$.\\
Given $w\in\D_f$. \qquad
Want: $(\,\,d(x,4)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,f_4)<\vep\,\,)$.\\
Assume $d(w,4)<\delta$. \qquad
Want: $d(f_w,f_4)<\vep$.\\
By definition of $\delta$, we have both $\delta\le1$ and $\delta\le12\vep$.\\
It follows both that $4-\delta\ge3$ and that $\ds{\frac{\delta}{12}\le\vep}$.\\
Since $|\bullet|$ is Lipschitz-$1$, we get $d(|w|,|4|)\le d(w,4)$.\\
Then $d(|w|,4)=d(|w|,|4|)\le d(w,4)<\delta$, \quad so $d(|w|,4)<\delta$.\\
Since $d(|w|,4)<\delta$, we get: $4-\delta<|w|<4+\delta$.\\
Then \quad $|w|>4-\delta\ge3$, \quad so \quad $|w|>3$, \quad so \quad $4\cdot|w|>12$.\\
Also, we have $|4-w|=d(w,4)<\delta$.\\
Since $0\le|4-w|<\delta$ and since $4\cdot|w|>12>0$,\\
\phx\qquad we conclude that: \quad $\ds{\,\,\,\,\frac{|4-w|}{4\cdot|w|}<\frac{\delta}{12}}$.\\
Then $\ds{d(f_w,f_4)=|f_w-f_4|=\left|\frac{1}{w}-\frac{1}{4}\right|=\left|\frac{4-w}{4w}\right|}$\\
\phx\qquad\qquad\qquad $\ds{\,=\frac{|4-w|}{|4|\cdot|w|}=\frac{|4-w|}{4\cdot|w|}<\frac{\delta}{12}\le\vep}$, \quad as desired.
\end{proof}

\begin{defn}
Let $X$ and $Y$ be metric spaces.\\
Let $f:X\drta Y$ and let $S$ be a set.\\
Then $f$ is \bfindex[continuous] on $S$ means:\\
\phx\qquad\qquad $\forall p\in S$, \quad $(\,\,f\hbox{ is continuous at }p\hbox{ from }X\hbox{ to }Y\,\,)$.
\end{defn}

\begin{defn}
Let $X$ and $Y$ be metric spaces, $f:X\drta Y$.\\
Then $f$ is \bfindex[continuous] means: \qquad $f$ is continuous on $\D_f$.
\end{defn}

\begin{thm}
Define $f:\R\drta\R$ by: $\forall x\in\R$, $f_x=1/x$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,$ Then $f$ is continuous from $\R$ to $\R$.
\end{thm}

\begin{proof}
Want: $f$ is continuous on $\D_f$ from $\R$ to $\R$.\\
We have $\D_f=\R_0^\times$. \quad
Want: $f$ is continuous on $\R_0^\times$ from $\R$ to $\R$.\\
Want: $\forall p\in\R_0^\times$, $f$ is continuous at $p$ from $\R$ to $\R$.\\
Given $p\in\R_0^\times$. \quad
Want: $f$ is continuous at $p$ from $\R$ to $\R$.\\
We have $p\in\R_0^\times=\D_f$.\\
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,f_p)<\vep\,\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,f_p)<\vep\,\,)$.\\
Since $p\in\R_0^\times$, we get $|p|>0$, and so $\ds{\frac{|p|}{2}>0}$.\\
Since $p\in\R_0^\times$, we get $\ds{\frac{p^2}{2}>0}$. \quad
So, since $\vep>0$, we get $\ds{\frac{p^2}{2}\cdot\vep>0}$.\\
Let $\ds{\delta:=\min\left\{\frac{|p|}{2},\frac{p^2}{2}\cdot\vep\right\}}$. \qquad
Then $\delta>0$.\\
Want: $\forall w\in\D_f$, $\,\,$ $(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,f_p)<\vep\,\,)$.\\
Given $w\in\D_f$. \qquad Want: $(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,f_p)<\vep\,\,)$.\\
Assume $d(w,p)<\delta$. \qquad Want: $d(f_w,f_p)<\vep$.\\
By definition of $\delta$, we have both $\ds{\delta\le\frac{|p|}{2}}$ and $\ds{\delta\le\frac{p^2}{2}\cdot\vep}$.\\
It follows both that $\ds{|p|-\delta\ge|p|-\frac{|p|}{2}}$ and that $\ds{\frac{2}{p^2}\cdot\delta\le\vep}$.\\
Since $|\bullet|$ is Lipschitz-$1$, we get $d(\,|p|\,,\,|w|\,)\,\le\,d(\,p\,,\,w\,)$.\\
Since $d(\,|p|\,,\,|w|\,)\,\le\,d(\,p\,,\,w\,)\,<\,\delta$, \quad
we get $|p|-\delta<|w|<|p|+\delta$.\\
Then $\ds{|w|>|p|-\delta>|p|-\frac{|p|}{2}=\frac{|p|}{2}}$, so $\ds{|w|>\frac{|p|}{2}}$, so $\ds{|w|\cdot|p|>\frac{|p|^2}{2}}$.\\
So, since $|p|^2=p^2$, we get $\ds{|w|\cdot|p|>\frac{p^2}{2}}$.\\
Also, we have $|p-w|=d(w,p)<\delta$.\\
Since $0\le|p-w|<\delta$ and since $\ds{|w|\cdot|p|>\frac{p^2}{2}>0}$,\\
\phx\qquad we conclude that: \quad $\ds{\,\,\,\,\frac{|p-w|}{|p|\cdot|w|}<\frac{\delta}{p^2/2}}$.\\
Then $\ds{d(f_w,f_p)=|f_w-f_p|=\left|\frac{1}{w}-\frac{1}{p}\right|=\left|\frac{p-w}{wp}\right|}$\\
\phx\qquad\qquad\qquad $\ds{\,=\frac{|p-w|}{|w|\cdot|p|}<\frac{\delta}{p^2/2}=\frac{2}{p^2}\cdot\delta\le\vep}$, \quad as desired.
\end{proof}

\subsection{Uniform continuity}

\begin{thm}
Let $X$ and $Y$ be metric spaces, $f:X\drta Y$.\\
Then: \quad $\,(\,\,f\hbox{ is continuous from }X\hbox{ to }Y\,\,)$\\
\phx\qquad $\Leftrightarrow\,\,(\,\,\forall p\in\D_f,\,f\hbox{ is continuous at }p\hbox{ from }X\hbox{ to }Y\,\,)$\\
\phx\qquad $\Leftrightarrow\,\,(\,\,\forall p\in\D_f,\,\forall\vep>0,\,\exists\delta>0~\hbox{s.t.},\,\forall q\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad $(\,d(p,q)<\delta\,)\,\Rightarrow\,(\,d(f_p,f_q)<\vep\,)\,\,)$\\
\phx\qquad $\Leftrightarrow\,\,(\,\,\forall\vep>0,\,\forall p\in\D_f,\,\exists\delta>0~\hbox{s.t.},\,\forall q\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad $(\,d(p,q)<\delta\,)\,\Rightarrow\,(\,d(f_p,f_q)<\vep\,)\,\,)$\\
\end{thm}

\begin{defn}
Let $X$ and $Y$ be metric spaces, $f:X\drta Y$.\\
By $f$ is \bfindex[uniformly continuous] from $X$ to $Y$, we mean:\\
\phx\qquad\qquad $\,\,\forall\vep>0,\,\exists\delta>0~\hbox{s.t.},\,\forall p\in\D_f,\,\forall q\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad $(\,d(p,q)<\delta\,)\,\Rightarrow\,(\,d(f_p,f_q)<\vep\,)$\\
\end{defn}

In homework, you'll show that Lipschitz implies uniformly continuous implies continuous.\\

\subsection{Sequential continuity}

\begin{defn}
Let $X$ and $Y$ be metric spaces.\\
Let $f:X\drta Y$ \qquad and \qquad let $p$ be an object.\\
By $f$ is \bfindex[sequentially continuous] at $p$ from $X$ to $Y$, we mean:\\
\phx\qquad $[\,\,p\in\D_f\,\,]\,\,\,\,\&\,\,\,\,[\,\,\forall s\in\D_f^\N,\,\,(\,s\to p\hbox{ in }X\,)\,\Rightarrow\,(\,f\circ s\to f_p\hbox{ in }Y\,)\,\,]$.
\end{defn}

\begin{thm}\wrlab{thm-cont-implies-scont}
Let $X,Y$ be metric spaces, $f:X\drta Y$, $p\in X$.\\
Assume that $f$ is continuous at $p$ from $X$ to $Y$.\\
Then $f$ is sequentially continuous at $p$ from $X$ to $Y$.
\end{thm}

\begin{proof}
Since $f$ is continuous at $p$ from $X$ to $Y$, \quad
we see that $p\in\D_f$.\\
It remains to show: $\forall s\in\D_f^\N$, $(\,s\to p\hbox{ in }X\,)\,\Rightarrow\,(\,f\circ s\to f_p\hbox{ in } Y\,)$.\\
Given $s\in\D_f^\N$. \quad 
Want: $(\,s\to p\hbox{ in }X\,)\,\Rightarrow\,(\,f\circ s\to f_p\hbox{ in } Y\,)$.\\
Assume $s\to p\hbox{ in }X$. \qquad\qquad
Want: $f\circ s\to f_p\hbox{ in } Y$.\\
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,d((f\circ s)_j,f_p)<\vep\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,d((f\circ s)_j,f_p)<\vep\,)$.\\
Since $f$ is continuous at $p$ from $X$ to $Y$,\\
\phx\quad choose $\delta>0$ s.t., $\forall w\in\D_f,\,\,(\,d(w,p)<\delta\,)\,\Rightarrow\,(\,d(f_w,f_p)<\vep\,)$.\\
Since $s\to p$ in $X$, choose $K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,d(s_j,p)<\delta\,)$.\\
Then $K\in\N$. \quad
Want: $\forall j\in\N,\,\,(\,j\ge K\,)\,\Rightarrow\,(\,d((f\circ s)_j,f_p)<\vep\,)$.\\
Given $j\in\N$. \qquad
Want: $(\,j\ge K\,)\,\Rightarrow\,(\,d((f\circ s)_j,f_p)<\vep\,)$.\\
Assume $j\ge K$. \qquad
Want: $d((f\circ s)_j,f_p)<\vep$.\\
Since $j\ge K$, by choice of $K$, we have: $d(s_j,p)<\delta$.\\
Since $s\in\D_f^\N$, we conclude that $s_j\in\D_f$.\\
Let $w:=s_j$. \qquad Then $w\in\D_f$.\\
So, since $d(w,p)=d(s_j,p)<\delta$, by choice of $\delta$, we have: $d(f_w,f_p)<\vep$.\\
We have $(f\circ s)_j=f(s_j)=f(w)$.\\
Then $d((f\circ s)_j,f_p)=d(f_w,f_p)<\vep$, as desired.\\
\end{proof}

\begin{thm}\wrlab{thm-disttozero-implies-converges}
Let $X$ be a metric space, $s\in X^\N$, $p\in X$.\\
Assume: $\forall j\in\N$, $d(s_j,p)\le1/j$. \qquad
Then $s\to p$ in $X$.
\end{thm}

\begin{proof}
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,d(s_j,p)<\vep\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,j\ge K\,)\,\Rightarrow\,(\,d(s_j,p)<\vep\,)$.\\
By the Archimedean Principle, choose $K\in\N$ s.t.~$K>1/\vep$.\\
Then $K\in\N$. \qquad
Want: $\forall j\in\N$, $(\,j\ge K\,)\,\Rightarrow\,(\,d(s_j,p)<\vep\,)$.\\
Given $j\in\N$. \qquad
Want: $(\,j\ge K\,)\,\Rightarrow\,(\,d(s_j,p)<\vep\,)$.\\
Assume $j\ge K$. \qquad
Want: $d(s_j,p)<\vep$.\\
By hypothesis, $d(s_j,p)\le1/j$.\\
Since $j\ge K>0$, we get: $1/j\le1/K$.\\
Since $K>1/\vep>0$, we get: $1/K<\vep$.\\
Then $d(s_j,p)\le1/j\le1/K<\vep$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-contin-equiv-scontin}
Let $X,Y$ be metric spaces, $f:X\drta Y$, $p\in X$.\\
\phx Then: \qquad $\,\,\,(\,\,f\hbox{ is continuous at }p\hbox{ from }X\hbox{ to }Y\,\,)$\\
\phx\qquad\qquad $\Leftrightarrow\,\,(\,\,f\hbox{ is sequentially continuous at }p\hbox{ from }X\hbox{ to }Y\,\,)$.
\end{thm}

\begin{proof}
By \tref{thm-cont-implies-scont}, we have $\Rightarrow$. \quad
Want: $\Leftarrow$.\\
Assume: $f$ is sequentially continuous at $p$ from $X$ to $y$.\\
Want: $f$ is continuous at $p$ from $X$ to $Y$.\\
Assume: $f$ is not continuous at $p$ from $X$ to $Y$. \quad
Want: Contradiction.\\
Choose $\vep>0$ s.t., $\forall\delta>0$, $\exists w\in\D_f$ s.t.~$(\,d(w,p)<\delta\,)\,\&\,(\,d(f_w,f_p)\ge\vep\,)$.\\
Then $\forall j\in\N$, $\exists w\in\D_f$ s.t.~$(\,d(w,p)<1/j\,)\,\&\,(\,d(f_w,f_p)\ge\vep\,)$.\\
By the Axiom of Choice,\\
\phx\qquad choose $s\in\D_f^\N$ s.t., $\forall j\in\N$, $(\,d(s_j,p)<1/j\,)\,\&\,(\,d(f_{s_j},f_p)\ge\vep\,)$.\\
Then $\forall j\in\N$, $d(s_j,p)<1/j$,\\
\phx\qquad and so, by \tref{thm-disttozero-implies-converges}, we see that: $s\to p$ in $X$.\\
So, since $s\in\D_f^\N$ and since $f$ is sequentially continuous at $p$,\\
\phx\qquad we conclude that: $f\circ s\to f_p$ in $Y$.\\
So choose $K\in\N$ s.t., $\forall j\in\N$, $(\,j\ge K\,)\,\Rightarrow\,(\,d((f\circ s)_j,f_p)<\vep\,)$.\\
So, since $K\ge K$, we get $d((f\circ s)_K,f_p)<\vep$, and so $\vep>d((f\circ s)_K,f_p)$.\\
By the choice of $s$, we have: $d(f_{s_K},f_p)\ge\vep$.\\
Then $\vep>d((f\circ s)_K,f_p)=d(f_{s_K},f_p)\ge\vep$,\\
\phx\qquad\qquad and so $\vep>\vep$. \qquad\qquad
Contradiction.
\end{proof}

\subsection{Arithmetic of functionals}

\begin{defn}
Let $f$ and $g$ be functionals.\\
Then $f+g\,,\,f-g\,,\,fg\,,\,f/g$ are the functionals defined by: $\forall x$,\\
\phx\qquad $(f+g)_x=f_x+g_x$ \qquad\qquad $(f-g)_x=f_x-g$,\\
\phx\qquad $(fg)_x=f_x\cdot g_x$ \qquad\qquad\quad $\,\,\,\,(f/g)_x=f_x/g$.
\end{defn}

\begin{defn}
Let $f$ be a functional and let $a\in\R$.\\
Then $af$ is the functional defined by: $\forall x$,\\
\phx\qquad\qquad\ $(af)_x=a\cdot f_x$.
\end{defn}

\begin{thm}
Let $f$ and $g$ be functionals.\\
Then:\qquad\qquad $\D_{f+g}\quad=\quad\D_{f-g}\quad=\quad\D_{fg}\quad=\quad\D_f\,\bigcap\,\D_g$.\\
Also, \qquad\qquad $\D_{f/g}\quad=\quad\D_f\,\bigcap\,\D_g\,\bigcap\,(g^*(\R_0^\times))$.
\end{thm}

\begin{thm}
Let $f$ be a functional and let $a\in\R$.\\
Then:\qquad\qquad $\D_{af}\quad=\quad\D_f$.\\
\end{thm}

\subsection{Pairing of functions}
\quad
\vskip.1in\noindent

Frownie is infective.
We make the convention that:\\
\phx\qquad\qquad $\forall t$, \quad $\left(\begin{tabular}{c}$1\mapsto t$\\$2\mapsto\frownie$\end{tabular}\right)=\frownie=\left(\begin{tabular}{c}$1\mapsto\frownie$\\$2\mapsto t$\end{tabular}\right)$.\\
Recall that: \qquad $\forall x,y$, \quad $(x,y)=\left(\begin{tabular}{c}$1\mapsto x$\\$2\mapsto y$\end{tabular}\right)$.\\
It follows that: \quad $\forall t$, \quad $(t,\frownie)=\frownie=(\frownie,t)$. \quad
Frownie is infective.\\

\begin{defn}
Let $f$ and $g$ be functions.\\
then $(f,g)^{{\rm fn}}$ is the function defined by:\\
\phx\qquad\qquad\qquad\qquad $\forall x$, \quad $(f,g)_x^{{\rm fn}}=(f_x,g_x)$
\end{defn}

We are almost always sloppy and write $(f,g)$ for $(f,g)^{{\rm fn}}$.
As a consequence, when we write $(f,g)$, the reader must discern, from context,
whether we mean \qquad $(f,g)^{{\rm fn}}$ \quad or \quad $\left(\begin{tabular}{c}$1\mapsto f$\\$2\mapsto g$\end{tabular}\right)$.\\
Hint: When $f$ and $g$ are functions, by $(f,g)$, we usually mean $(f,g)^{{\rm fn}}$.

\begin{thm}
Define $f,g:\R\drta\R$ by:\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\forall t\in\R$, \quad $(\,(\,f_t=\sqrt{t}\,)\,\&\,(\,g_t=\sqrt{3-t}\,)\,)$.\\
Let $h:=(f,g)$. \qquad
Then:\\
\phx\qquad\qquad $\!\D_f=[0;\infty)$ \quad and \quad $\D_g=(-\infty;3]$ \quad and \quad $\D_h=[0;3]$\\
\phx\qquad\qquad and \quad $h:\R\drta\R^2$ \quad and \quad $\forall t\in\R$, $h_t=(\sqrt{t},\sqrt{3-t})$.
\end{thm}

\begin{thm}
$\forall$functions $f,g$, \quad $\D_{(f,g)}=\D_f\bigcap\D_g$.
\end{thm}

\subsection{Properties of continuity}

\begin{thm}
Let $X$ be a metric space, $f,g:X\drta\R$, $p\in X$.\\
Assume $f$ and $g$ are both continuous at $p$ from $X$ to $\R$.\\
Then $f+g$ and $fg$ are both continuous at $p$ from $X$ to $\R$.
\end{thm}

\begin{proof}
By \tref{thm-contin-equiv-scontin},\\
\phx\quad $f$ and $g$ are both sequentially continuous at $p$ from $X$ to $\R$.\\
Also, by \tref{thm-contin-equiv-scontin}, it suffices to show:\\
\phx\quad  $f+g$ and $fg$ are both sequentially continuous at $p$ from $X$ to $\R$.\\
Since $f$ and $g$ are both sequentially continuous at $p$ from $X$ to $\R$,\\
\phx\quad we get: $p\in\D_f$ and $p\in\D_g$, \quad so $p\in\D_f\bigcap\D_g$.\\
We have: $\D_{f+g}=\D_f\bigcup\D_g=\D_{fg}$. \quad
Then: $p\in\D_{f+g}$ and $p\in\D_{fg}$.\\
It suffices to show: $\forall s\in(\D_f\bigcap\D_g)^\N$, \quad $(\,\,$ $s\to p$ in $X$ $\,\,)\quad\Rightarrow$\\
\phx\qquad $(\,\,$ $(f+g)\circ s\to(f+g)_p$ in $\R$ \quad and \quad $(fg)\circ s\to(fg)_p$ in $\R$ $\,\,)$.\\
Given $s\in(\D_f\bigcap\D_g)^\N$. \qquad Want: $(\,\,$ $s\to p$ in $X$ $\,\,)\quad\Rightarrow$\\
\phx\qquad $(\,\,$ $(f+g)\circ s\to(f+g)_p$ in $\R$ \quad and \quad $(fg)\circ s\to(fg)_p$ in $\R$ $\,\,)$.\\
Assume: \qquad $s\to p$ in $X$.\\
Want: \qquad $(f+g)\circ s\to(f+g)_p$ in $\R$ \quad and \quad $(fg)\circ s\to(fg)_p$ in $\R$.\\
We have: \quad $(f+g)_p=f_p+g_p$ \quad and \quad $(fg)_p=f_p\cdot g_p$.\\
Want: \qquad $(f+g)\circ s\to f_p+g_p$ in $\R$ \quad and \quad $(fg)\circ s\to f_p\cdot g_p$ in $\R$.\\
We have $s\in(\D_f\bigcap\D_g)^\N\subseteq\D_f^\N$ and $s\in(\D_f\bigcap\D_g)^\N\subseteq\D_g^\N$.\\
So, since $s\to p$ in $X$ and\\
\phx\quad since $f$ and $g$ are both sequentially continuous at $p$ from $X$ to $\R$,\\
we get: \quad $f\circ s\to f_p\hbox{ in }\R$ \quad and \quad $g\circ s\to g_p\hbox{ in }\R$.\\
Then, by HW\#4-3 and HW\#4-4, we get:\\
\phx\quad $(f\circ s)+(g\circ s)\to f_p+g_p\hbox{ in }\R$ \quad and \quad $(f\circ s)\cdot(g\circ s)\to f_p\cdot g_p\hbox{ in }\R$.\\
Want: $(f\circ s)+(g\circ s)=(f+g)\circ s$ \quad and \quad $(f\circ s)\cdot(g\circ s)=(fg)\circ s$.\\
Want: $\forall j\in\N$, \quad $((f\circ s)+(g\circ s))_j=((f+g)\circ s)_j$ \qquad and\\
\phx\qquad\qquad\qquad\quad $((f\circ s)\cdot(g\circ s))_j=((fg)\circ s)_j$.\\
Given $j\in\N$. \quad Want:\\
\phx $((f\circ s)+(g\circ s))_j=((f+g)\circ s)_j$ \quad and \quad $((f\circ s)\cdot(g\circ s))_j=((fg)\circ s)_j$.\\
We have $((f\circ s)+(g\circ s))_j=(f\circ s)_j+(g\circ s)_j=f_{s_j}+g_{s_j}$\\
\phx\quad\qquad\qquad\qquad\qquad\qquad $\,\,\,\,=(f+g)_{s_j}=((f+g)\circ s)_j$.\\
Want: $((f\circ s)\cdot(g\circ s))_j=((fg)\circ s)_j$.\\
We have $((f\circ s)\cdot(g\circ s))_j=(f\circ s)_j\cdot(g\circ s)_j=f_{s_j}\cdot g_{s_j}$\\
\phx\quad\qquad\qquad\qquad\qquad\qquad $\,=(fg)_{s_j}=((fg)\circ s)_j$, \quad as desired.
\end{proof}

\begin{thm}
Let $X$ and $Y$ be metric spaces, $a\in Y$.\\
Then $C_X^a$ is continuous from $X$ to $Y$.
\end{thm}

\begin{proof}
Since $C_X^a$ is Lipschitz-$0$ from $X$ to $Y$,\\
\phx\qquad\qquad\qquad we see that $C_X^a$ is Lipschitz from $X$ to $Y$.\\
Then, by HW\#5-1, $C_X^a$ is uniformly continuous from $X$ to $Y$.\\
Then, by HW\#5-2, $C_X^a$ is continuous from $X$ to $Y$, as desired.
\end{proof}

\begin{thm}
Let $X$ be a metric space.\\
Let \quad $f:X\drta\R$,\quad $p\in X$, \quad $a\in\R$.\\
Assume that $f$ is continuous at $p$ from $X$ to $\R$.\\
Then $af$ is continuous at $p$ from $X$ to $\R$.
\end{thm}

\begin{proof}
Since $C_X^a$ is continuous from $X$ to $\R$ \quad and \quad $\D_{C_X^a}=X$,\\
\phx\qquad we see that $C_X^a$ is continuous on $X$ from $X$ to $\R$.\\
So, since $p\in X$, we see that $C_X^a$ is continuous at $p$ from $X$ to $\R$.\\
So, since $f$ is also continuous at $p$ from $X$ to $\R$,\\
\phx\qquad we see that $C_X^a\cdot f$ is continuous at $p$ from $X$ to $\R$.\\
It therefore suffices to show: \quad $C_X^a\cdot f=af$.\\
Want: $\forall q\in X$, \quad $(C_X^a\cdot f)_q=(af)_q$.\\
Given $q\in X$. \qquad
Want: $(C_X^a\cdot f)_q=(af)_q$.\\
We have: $(C_X^a\cdot f)_q=(C_X^a)_q\cdot f_q=a\cdot f_q=(af)_q$, \quad as desired.
\end{proof}

\begin{thm}
Let $X$ be a metric space, $f,g:X\drta\R$, $p\in X$.\\
Assume that $f$ and $g$ are both continuous at $p$ from $X$ to $\R$.\\
Then $f-g$ is continuous at $p$ from $X$ to $\R$.
\end{thm}

\begin{proof}
Since $g$ is continuous at $p$ from $X$ to $\R$,\\
\phx\quad we get: $(-1)\cdot g$ is continuous at $p$ from $X$ to $\R$.\\
So, since $f$ is also continuous at $p$ from $X$ to $\R$,\\
\phx\quad we get: $f+(-1)\cdot g$ is continuous at $p$ from $X$ to $\R$.\\
It therefore suffices to show: \quad $f+(-1)\cdot g=f-g$.\\
Want: $\forall q\in X$, \quad $(f+(-1)\cdot g)_q=(f-g)_q$.\\
Given $q\in X$. \qquad
Want: $(f+(-1)\cdot g)_q=(f-g)_q$.\\
We have: $(f+(-1)\cdot g)_q=f_q+((-1)\cdot g)_q=f_q+(-1)\cdot g_q$\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,=f_q-g_q=(f-g)_q$, \quad as desired.
\end{proof}

The following theorem is called
\bfindex[Absoluteness of Continuity]:

\begin{thm}
Let $X$ and $Y$ be metric spaces.\\
Let $A\subseteq X$, $B\subseteq Y$, $f:A\drta B$.\\
Then: \quad $\,(\,\,f\hbox{ is continuous from }A\hbox{ to }B\,\,)$\\
\phx\qquad $\Leftrightarrow\,\,(\,\,f\hbox{ is continuous from }X\hbox{ to }Y\,\,)$.
\end{thm}

\begin{proof}
{\it Proof of $\Rightarrow$:}\\
Assume: $f\hbox{ is continuous from }A\hbox{ to }B$.\\
Want: $f\hbox{ is continuous from }X\hbox{ to }Y$.\\
Want: $f$ is continuous on $\D_f$ from $X$ to $Y$.\\
Want: $\forall p\in\D_f$, $f$ is continuous at $p$ from $X$ to $Y$.\\
Given $p\in\D_f$. \quad
Want: $f$ is continuous at $p$ from $X$ to $Y$.\\
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall w\in X$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,d_X(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(f_w,f_p)<\vep\,\,)$.\\
Given $\vep>0$. \quad
Want: $\exists\delta>0$ s.t., $\forall w\in X$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,d_X(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(f_w,f_p)<\vep\,\,)$.\\
Since $f$ is continuous from $A$ to $B$,\\
\phx\quad we get: $f$ is continuous on $\D_f$ from $A$ to $B$.\\
So, since $p\in\D_f$, we get: $f$ is continuous at $p$ from $A$ to $B$.\\
So choose $\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,d_A(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_B(f_w,f_p)<\vep\,\,)$.\\
Want: $\forall w\in\D_f$, $[\,\,(\,\,d_X(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(f_w,f_p)<\vep\,\,)\,\,]$.\\
Given $w\in\D_f$. \quad
Want: $(\,\,d_X(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(f_w,f_p)<\vep\,\,)$.\\
Assume: $d_X(w,p)<\delta$. \quad
Want: $d_Y(f_w,f_p)<\vep$.\\
Since $f:A\drta B$, we get $\D_f\subseteq A$ and $\I_f\subseteq B$.\\
Then $w,x\in\D_f\subseteq A$ and $f_w,f_x\in\I_f\subseteq B$.\\
So, since $d_A=d_X|(A\times A)$ and since $d_B=d_Y|(B\times B)$,\\
\phx\qquad we get: $d_A(w,p)=d_X(w,p)$ and $d_B(f_w,f_p)=d_Y(f_w,f_p)$.\\
Then $d_A(w,p)=d_X(w,p)<\delta$,\\
\phx\qquad so, by choice of $\delta$, we get: $d_B(f_w,f_p)<\vep$.\\
Then $d_Y(f_w,f_p)=d_B(f_w,f_p)<\vep$, as desired.\\
{\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:}\\
Assume: $f\hbox{ is continuous from }X\hbox{ to }Y$.\\
Want: $f\hbox{ is continuous from }A\hbox{ to }B$.\\
Want: $f$ is continuous on $\D_f$ from $A$ to $B$.\\
Want: $\forall p\in\D_f$, $f$ is continuous at $p$ from $A$ to $B$.\\
Given $p\in\D_f$. \quad
Want: $f$ is continuous at $p$ from $A$ to $B$.\\
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall w\in X$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,d_A(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_B(f_w,f_p)<\vep\,\,)$.\\
Given $\vep>0$. \quad
Want: $\exists\delta>0$ s.t., $\forall w\in X$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,d_A(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_B(f_w,f_p)<\vep\,\,)$.\\
Since $f$ is continuous from $X$ to $Y$,\\
\phx\quad we get: $f$ is continuous on $\D_f$ from $X$ to $Y$.\\
So, since $p\in\D_f$, we get: $f$ is continuous at $p$ from $X$ to $Y$.\\
So choose $\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,d_X(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(f_w,f_p)<\vep\,\,)$.\\
Want: $\forall w\in\D_f$, $[\,\,(\,\,d_A(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_B(f_w,f_p)<\vep\,\,)\,\,]$.\\
Given $w\in\D_f$. \quad
Want: $(\,\,d_A(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_B(f_w,f_p)<\vep\,\,)$.\\
Assume: $d_A(w,p)<\delta$. \quad
Want: $d_B(f_w,f_p)<\vep$.\\
Since $f:A\drta B$, we get $\D_f\subseteq A$ and $\I_f\subseteq B$.\\
Then $w,x\in\D_f\subseteq A$ and $f_w,f_x\in\I_f\subseteq B$.\\
So, since $d_A=d_X|(A\times A)$ and since $d_B=d_Y|(B\times B)$,\\
\phx\qquad we get: $d_A(w,p)=d_X(w,p)$ and $d_B(f_w,f_p)=d_Y(f_w,f_p)$.\\
Then $d_X(w,p)=d_A(w,p)<\delta$,\\
\phx\qquad so, by choice of $\delta$, we get: $d_Y(f_w,f_p)<\vep$.\\
Then $d_B(f_w,f_p)=d_Y(f_w,f_p)<\vep$, as desired.\\
{\it End of proof of $\Leftarrow$.}\\
\end{proof}

\begin{thm}
Let $f,g,h$ be functions.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,$ Then $(h\circ g)\circ f=h\circ(g\circ f)$.
\end{thm}

\begin{proof}
Want: $\forall x,\,\,((h\circ g)\circ f)_x=(h\circ(g\circ f))_x$.\\
Given $x$. \qquad Want: $((h\circ g)\circ f)_x=(h\circ(g\circ f))_x$.\\
We have: $((h\circ g)\circ f)_x=(h\circ g)(f_x)=h(g(f_x))$\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,\,\,=h((g\circ f)_x)=(h\circ(g\circ f))_x$, \quad as desired.
\end{proof}

\begin{thm}\wrlab{thm-comp-contin-at-pt}
Let $X,Y,Z$ be metric spaces, $p\in X$.\\
Let $f:X\drta Y$ and $g:Y\drta Z$.\\
Assume: $f$ is continuous at $p$ from $X$ to $Y$.\\
Assume: $g$ is continuous at $f_p$ from $Y$ to $Z$.\\
Then $g\circ f$ is continuous at $p$ from $X$ to $Z$.
\end{thm}

\begin{proof}
We have: $f$ is sequentially continuous at $p$ from $X$ to $Y$.\\
Also, $g$ is sequentially continuous at $f_p$ from $Y$ to $Z$.\\
Want: $g\circ f$ is sequentially continuous at $p$ from $X$ to $Z$.\\
Want: $\forall s\in\D_{g\circ f}^\N$, $(\,s\to p\hbox{ in }X\,)\,\Rightarrow\,(\,(g\circ f)\circ s\to(g\circ f)_p\hbox{ in }Z\,)$.\\
Given $s\in\D_{g\circ f}^\N$. $\,$
Want: $(\,s\to p\hbox{ in }X\,)\,\Rightarrow\,(\,(g\circ f)\circ s\to(g\circ f)_p\hbox{ in }Z\,)$.\\
Assume: $s\to p\hbox{ in }X$. \qquad
Want: $(g\circ f)\circ s\to(g\circ f)_p\hbox{ in }Z$.\\
We have: $\forall j\in\N$,\\
\phx\quad $s_j\in\D_{g\circ f}$, so $(g\circ f)(s_j)\ne\frownie$,\\
\phx\quad so $g(f(s_j))=(g\circ f)(s_j)\ne\frownie$,\\
\phx\quad so both $f(s_j)\ne\frownie$ and $g((f\circ s)_j)=g(f(s_j))\ne\frownie$,\\
\phx\quad so both $s_j\in\D_f$ and $(f\circ s)_j\in\D_g$.\\
Then $s\in\D_f^\N$ and $f\circ s\in\D_g^\N$.\\
Since $s\in\D_f^\N$ and since $s\to p$ in $X$\\
\phx\quad and since $f$ is sequentially continuous at $p$ from $X$ to $Y$,\\
\phx\quad we get: $f\circ s\to f_p$ in $Y$.\\
Since $f\circ s\in\D_g^\N$ and since $f\circ s\to f_p$ in $Y$\\
\phx\quad and since $g$ is sequentially continuous at $f_p$ from $Y$ to $Z$,\\
\phx\quad we get: $g\circ(f\circ s)\to g_{f_p}$ in $Z$.\\
So, since $(g\circ f)\circ s=g\circ(f\circ s)$ and since $(g\circ f)_p=g_{f_p}$,\\
\phx\quad we get: $(g\circ f)\circ s\to(g\circ f)_p$ in $Z$, as desired.
\end{proof}

\noindent
{\bf Alternate proof:}

\begin{proof}
Want: $\forall\vep>0$, $\exists\gamma>0$ s.t., $\forall v\in\D_{g\circ f}$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,d(v,p)<\gamma\,\,)\,\,\Rightarrow\,\,(\,\,d((g\circ f)_v,(g\circ f)_p)<\vep\,\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists\gamma>0$ s.t., $\forall v\in\D_{g\circ f}$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,d(v,p)<\gamma\,\,)\,\,\Rightarrow\,\,(\,\,d((g\circ f)_v,(g\circ f)_p)<\vep\,\,)$.\\
Since $g$ is continuous at $f_p$ from $Y$ to $Z$, choose $\delta>0$ s.t., $\forall w\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,d(w,f_p)<\gamma\,\,)\,\,\Rightarrow\,\,(\,\,d(g_w,g_{f_p})<\vep\,\,)$.\\
Since $f$ is continuous at $p$ from $X$ to $Y$, choose $\gamma>0$ s.t., $\forall v\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,d(v,p)<\gamma\,\,)\,\,\Rightarrow\,\,(\,\,d(f_v,f_p)<\delta\,\,)$.\\
Want: $\forall v\in\D_{g\circ f}$, $(\,\,d(v,p)<\gamma\,\,)\,\,\Rightarrow\,\,(\,\,d((g\circ f)_v,(g\circ f)_p)<\vep\,\,)$.\\
Given $v\in\D_{g\circ f}$. \quad
Want: $(\,\,d(v,p)<\gamma\,\,)\,\,\Rightarrow\,\,(\,\,d((g\circ f)_v,(g\circ f)_p)<\vep\,\,)$.\\
Assume: $d(v,p)<\gamma$. \qquad
Want: $d((g\circ f)_v,(g\circ f)_p)<\vep$.\\
Since $v\in\D_{g\circ f}$, we get: $(g\circ f)_v\ne\frownie$.\\
So, since $g(f_v)=(g\circ f)_v$, we get: $g(f_v)\ne\frownie$.\\
Then $f_v\in\D_g$. \quad
Then $f_v\ne\frownie$. \quad
Then $v\in\D_f$.\\
Since $v\in\D_f$ and $d(v,p)<\gamma$, by choice of $\gamma$, we get: $d(f_v,f_p)<\delta$.\\
Since $f_v\in\D_g$ and $d(f_v,f_p)<\delta$, by choice of $\delta$, we get $d(g_{f_v},g_{f_p})<\vep$.\\
Then $d((g\circ f)_v,(g\circ f)_p)=d(g_{f_v},g_{f_p})<\vep$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-comp-contin-full-stop}
Let $X,Y,Z$ be metric spaces.\\
Let $f:X\drta Y$ and let $g:Y\drta Z$.\\
Assume: $f$ is continuous from $X$ to $Y$.\\
Assume: $g$ is continuous from $Y$ to $Z$.\\
Then $g\circ f$ is continuous from $X$ to $Z$.
\end{thm}

\begin{proof}
Want: $g\circ f$ is continuous on $\D_{g\circ f}$ from $X$ to $Z$.\\
Want: $\forall p\in\D_{g\circ f}$, $g\circ f$ is continuous at $p$ from $X$ to $Z$.\\
Given $p\in\D_{g\circ f}$. \qquad
Want: $g\circ f$ is continuous at $p$ from $X$ to $Z$.\\
Since $p\in\D_{g\circ f}$, we get $(g\circ f)_p\ne\frownie$.\\
Then $g(f_p)=(g\circ f)_p\ne\frownie$, so $f_p\in\D_g$.\\
Since $f_p\in\D_g$, we get $f_p\ne\frownie$ and so $p\in\D_f$.\\
Since $f$ is continuous from $X$ to $Y$, we get:\\
\phx\qquad\qquad  $f$ is continuous on $\D_f$ from $X$ to $Y$.\\
So, since $p\in\D_f$, we get: $f$ is continuous at $p$ from $X$ to $Y$.\\
Since $g$ is continuous from $Y$ to $Z$, we get:\\
\phx\qquad\qquad $g$ is continuous on $\D_g$ from $Y$ to $Z$.\\
So, since $f_p\in\D_g$, we get: $g$ is continuous at $f_p$ from $Y$ to $Z$.\\
Then, by \tref{thm-comp-contin-at-pt}, $g\circ f$ is continuous at $p$ from $X$ to $Z$.
\end{proof}

\begin{thm}\wrlab{thm-quot-contin-at-pt}
Let $X$ be a metric space, $f,g:X\drta\R$, $p\in X$.\\
Assume that $f$ and $g$ are both continuous at $p$ from $X$ to $\R$.\\
Assume that $g_p\ne0$. \qquad
Then: $f/g$ is continuous at $p$ from $X$ to $\R$.
\end{thm}

\begin{proof}
Define $h:\R\drta\R$ by: $\forall x\in\R$, $h_x=1/x$.\\
Then $h$ is continuous from $\R$ to $\R$.\\
That is, $h$ is continuous on $\D_h$ from $\R$ to $\R$.\\
So, since $g_p\in\R_0^\times=\D_h$, we conclude:\\
\phx\quad $h$ is continuous at $g_p$ from $\R$ to $\R$.\\
So, since $g$ is continuous at $p$ from $X$ to $\R$, we conclude:\\
\phx\quad $h\circ g$ is continuous at $p$ from $X$ to $\R$.\\
So, since $f$ is continuous at $p$ from $X$ to $\R$, we conclude:\\
\phx\quad $f\cdot(h\circ g)$ is continuous at $p$ from $X$ to $\R$.\\
It therefore suffices to show: $f\cdot(h\circ g)=f/g$.\\
Want: $\forall w\in X$, $\,\,$ $(f\cdot(h\circ g))_w=(f/g)_w$.\\
Given $w\in X$. \qquad
Want: $(f\cdot(h\circ g))_w=(f/g)_w$.\\
We have $(f\cdot(h\circ g)_w=f_w\cdot((h\circ g)_w)=f_w\cdot h_{g_w}=f_w\cdot(1/g_w)$\\
\phx\qquad\qquad\qquad\qquad\quad $=f_w/g_w=(f/g)_w$, \quad as desired.
\end{proof}

\begin{thm}\wrlab{thm-four-functions-preserve-contin}
Let $X$ be a metric space, $f,g:X\drta\R$.\\
Assume: $f$ and $g$ are both continuous from $X$ to $\R$.
\quad Then:\\
\phx\qquad (1) $f+g$ is continuous from $X$ to $\R$ \qquad and\\
\phx\qquad (2) $f-g$ is continuous from $X$ to $\R$ \qquad and\\
\phx\qquad (3) $fg$ is continuous from $X$ to $\R$ \qquad and\\
\phx\qquad (4) $f/g$ is continuous from $X$ to $\R$.
\end{thm}

\begin{proof}
{\it Proof of (1),(2),(3):}\\
Unassigned HW.\\
{\it End of proof of (1),(2),(3).}\\
\\
{\it Proof of (4):}\\
Want: $f/g$ is continuous on $\D_{f/g}$ from $X$ to $\R$.\\
Want: $\forall p\in\D_{f/g}$, $f/g$ is continuous at $p$ from $X$ to $\R$.\\
Given $p\in\D_{f/g}$. \qquad
Want: $f/g$ is continuous at $p$ from $X$ to $\R$.\\
Since $p\in\D_{f/g}$, we get $(f/g)_p\ne\frownie$.\\
Then $f_p/g_p=(f/g)_p\ne\frownie$, so $f_p\ne\frownie$ and $g_p\ne\frownie$ and $g_p\ne0$.\\
Since $f_p\ne\frownie$, we get $p\in\D_f$. \quad
Since $g_p\ne\frownie$, we get $p\in\D_g$.\\
Since $f$ is continuous from $X$ to $\R$, we get:\\
\phx\qquad\qquad  $f$ is continuous on $\D_f$ from $X$ to $\R$.\\
So, since $p\in\D_f$, we get: $f$ is continuous at $p$ from $X$ to $\R$.\\
Since $g$ is continuous from $X$ to $\R$, we get:\\
\phx\qquad\qquad $g$ is continuous on $\D_g$ from $X$ to $\R$.\\
So, since $p\in\D_g$, we get: $g$ is continuous at $p$ from $X$ to $\R$.\\
Then, by \tref{thm-quot-contin-at-pt}, $f/g$ is continuous at $p$ from $X$ to $Z$.\\
{\it End of proof of (4).}
\end{proof}

\begin{thm}\wrlab{thm-restr-contin-at-pt}
Let $X$ and $Y$ be metric spaces.\\
Let $f:X\drta Y$, let $S\subseteq\D_f$ and let $p\in S$.\\
Assume that $f$ is continuous at $p$ from $X$ to $Y$.\\
Then $f|S$ is continuous at $p$ from $X$ to $Y$.
\end{thm}

\begin{proof}
We have $p\in S=\D_{f|S}$.\\
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall w\in\D_{f|S}$,\\
\phx\qquad\qquad\qquad\qquad $(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(\,(f|S)_w\,,\,(f|S)_p\,)\,<\,\vep\,\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists\delta>0$ s.t., $\forall w\in\D_{f|S}$,\\
\phx\qquad\qquad\qquad\qquad $(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(\,(f|S)_w\,,\,(f|S)_p\,)\,<\,\vep\,\,)$.\\
Since $f$ is continuous at $p$ from $X$ to $Y$, choose $\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad $(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,f_p\,)<\vep\,\,)$.\\
Then $\delta>0$. \quad
Want: $\forall w\in\D_{f|S}$,\\
\phx\qquad\qquad\qquad\qquad $(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(\,(f|S)_w\,,\,(f|S)_p\,)\,<\,\vep\,\,)$.\\
Given $w\in\D_{f|S}$.
Want: $(\,d(w,p)<\delta\,)\,\Rightarrow\,(\,d(\,(f|S)_w\,,\,(f|S)_p\,)\,<\,\vep\,)$.\\
Assume $d(w,p)<\delta$. \quad
Want: $d(\,(f|S)_w\,,\,(f|S)_p\,)\,<\,\vep$.\\
We have: $w\in\D_{f|S}=S$. \quad
By hypothesis, $S\subseteq\D_f$. \quad
Then $w\in\D_f$.\\
Since $(\,w\in\D_f\,)\,\&\,(\,d(w,p)<\delta\,)$, by choice of $\delta$,\\
\phx\qquad we get: $d(f_w,f_p)<\vep$.\\
We have $p\in S$. \quad
Also, by hypothesis, $w\in S$.\\
Then: \quad $(f|S)_p=f_p$ \quad and \quad $(f|S)_w=f_w$.\\
Then: $d(\,(f|S)_w\,,\,(f|S)_p\,)\,=\,d(f_w,f_p)\,<\,\vep$, as desired.
\end{proof}

\noindent
The converse of the preceding theorem is not true:\\
Let $S:=[4;\infty)$ and $f:=\chi_S^\R$.\\
Then $f|S=C_S^1$, so $f|S$ is continuous,\\
\phx\qquad so $f|S$ is continuous at $4$ from $\R$ to $\R$.\\
However, by HW\#8-2, $f$ is not continuous at $4$ from $\R$ to $\R$.

\begin{thm}\wrlab{thm-restr-contin}
Let $X$ and $Y$ be metric spaces.\\
Let $f:X\drta Y$ and let $S\subseteq\D_f$.\\
Assume that $f$ is continuous from $X$ to $Y$.\\
Then $f|S$ is continuous from $X$ to $Y$.
\end{thm}

\begin{proof}
Want: $f|S$ is continuous on $\D_{f|S}$ from $X$ to $Y$.\\
Want: $\forall p\in\D_{f|S}$, $f|S$ is continuous at $p$ from $X$ to $Y$.\\
Given $p\in\D_{f|S}$. \qquad
Want: $f|S$ is continuous at $p$ from $X$ to $Y$.\\
We have: $p\in\D_{f|S}=S$. \quad
By hypothesis, $S\subseteq\D_f$. \quad
Then $p\in\D_f$.\\
By hypothesis, $f$ is continuous on $\D_f$ from $X$ to $Y$.\\
So, since $p\in\D_f$, we get: $f$ is continuous at $p$ from $X$ to $Y$.\\
Then, by \tref{thm-restr-contin-at-pt}, $f|S$ is continuous at $p$ from $X$ to $Y$.
\end{proof}

\begin{thm}
Let $f,g:\R\to\R$, $p\in\R$.\\
Assume $f$ and $g$ are both continuous at $p$ from $\R$ to $\R$.\\
Then $(f,g)$ is continuous at $p$ from $\R$ to $\R$.
\end{thm}

\begin{proof}
Let $h:=(f,g)$. \quad
Want: $h$ is continuous at $p$ from $\R$ to $\R$.\\
Since $f$ and $g$ are both continuous at $p$ from $\R$ to $\R$,\\
\phx\qquad we get: $p\in\D_f$ and $p\in\D_g$.\\
We have $\D_h=D_f\bigcap\D_g$. \quad
Then $p\in\D_f\bigcap\D_g=\D_h$.\\
It remains to show: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall w\in\D_h$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d_\R(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_{\R^2}(h_w,h_p)<\vep\,\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists\delta>0$ s.t., $\forall w\in\D_h$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d_\R(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_{\R^2}(h_w,h_p)<\vep\,\,)$.\\
Let $\lambda:=\vep/\sqrt{2}$.\\
Since $f$ is continuous at $p$ from $\R$ to $\R$, choose $\alpha>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d_\R(w,p)<\alpha\,\,)\,\,\Rightarrow\,\,(\,\,d_\R(f_w,f_p)<\lambda\,\,)$.\\
Since $g$ is continuous at $p$ from $\R$ to $\R$, choose $\beta>0$ s.t., $\forall w\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d_\R(w,p)<\beta\,\,)\,\,\Rightarrow\,\,(\,\,d_\R(g_w,g_p)<\lambda\,\,)$.\\
Let $\delta:=\min\{\alpha,\beta\}$. \qquad
Then $\delta>0$.\\
Want: $\forall w\in\D_h$, \quad $(\,\,d_\R(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_{\R^2}(h_w,h_p)<\vep\,\,)$.\\
Given $w\in\D_h$. \qquad
Want: $(\,\,d_\R(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_{\R^2}(h_w,h_p)<\vep\,\,)$.\\
Assume: $d_\R(w,p)<\delta$. \qquad
Want: $d_{\R^2}(h_w,h_p)<\vep$.\\
Since $w\in\D_h=\D_{(f,g)}=\D_f\bigcap\D_g$, we get: \quad
$w\in\D_f$ \quad and \quad $w\in\D_g$.\\
Since $\delta:=\min\{\alpha,\beta\}$, we get: \quad
$\delta\le\alpha$ \quad and \quad $\delta\le\beta$.\\
Since $w\in\D_f$ and since $d_\R(w,p)<\delta\le\alpha$,
by choice of $\alpha$,\\
\phx\qquad\qquad\qquad\qquad we get: \qquad\qquad $d_\R(f_w,f_p)<\lambda$.\\
Since $|f_w-f_p|=d_\R(f_w,f_p)<\lambda$, we get: $|f_w-f_p|<\lambda$.\\
Since $0\le|f_w-f_p|<\lambda$, we get: $|f_w-f_p|^2<\lambda^2$.\\
So, since $|f_w-f_p|^2=(f_w-f_p)^2$, we get: $(f_w-f_p)^2<\lambda^2$.\\
Since $w\in\D_g$ and since $d_\R(w,p)<\delta\le\beta$,
by choice of $\beta$,\\
\phx\qquad\qquad\qquad\qquad we get: \qquad\qquad $d_\R(g_w,g_p)<\lambda$.\\
Since $|g_w-g_p|=d_\R(g_w,g_p)<\lambda$, we get: $|g_w-g_p|<\lambda$.\\
Since $0\le|g_w-g_p|<\lambda$, we get: $|g_w-g_p|^2<\lambda^2$.\\
So, since $|g_w-g_p|^2=(g_w-g_p)^2$, we get: $(g_w-g_p)^2<\lambda^2$.\\
We compute: $\,\,d_{\R^2}(h_w,h_p)\,\,=\,\,d_{\R^2}(\,(f,g)_w\,,\,(f,g)_p\,)$\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\,\,=\,\,d_{\R^2}(\,(f_w,g_w)\,,\,(f_p,g_p\,)$\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\,\,=\,\,|\,(f_w,g_w)\,-\,(f_p,g_p)\,|_{\R^2}$\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\,\,=\,\,|\,(\,f_w-f_p\,,\,g_w-g_p\,)\,|_{\R^2}$\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\,\,=\,\,\sqrt{(f_w-f_p)^2+(g_w-g_p)^2}$.\\
Since $\lambda=\vep/2$, we get $\sqrt{2}\cdot\lambda=\vep$.\\
Since $(f_w-f_p)^2<\lambda^2$ and $(g_w-g_p)^2<\lambda^2$,\\
\phx\qquad we get: \qquad\qquad $(f_w-f_p)^2+(g_w-g_p)^2<\lambda^2+\lambda^2$.\\
Since $0\le (f_w-f_p)^2+(g_w-g_p)^2<\lambda^2+\lambda^2$,\\
\phx\qquad we get: \qquad\qquad $\sqrt{(f_w-f_p)^2+(g_w-g_p)^2}<\sqrt{\lambda^2+\lambda^2}$.\\
So, \qquad since \quad $d_{\R^2}(h_w,h_p)=\sqrt{(f_w-f_p)^2+(g_w-g_p)^2}$\\
\phx $\,\,\,\,\,$ and since \quad $\sqrt{\lambda^2+\lambda^2}=\sqrt{2\lambda^2}=\sqrt{2}\cdot\lambda=\vep$,\\
\phx\qquad we get: \qquad\qquad $d_{\R^2}(h_w,h_p)<\vep$, \quad as desired.
\end{proof}

\subsection{Comparison of functions}
\quad
\vskip.1in\noindent

\noindent
We make the following convention: $\forall x\in\R$,\\
\phx\qquad $\neg(\frownie\le x)$ \quad and \quad $\neg(\frownie\ge x)$ \quad and\\
\phx\qquad $\neg(\frownie<x)$ \quad and \quad $\neg(\frownie>x)$ \quad and\\
\phx\qquad $\neg(x\le\frownie)$ \quad and \quad $\neg(x\ge\frownie)$ \quad and\\
\phx\qquad $\neg(x<\frownie)$ \quad and \quad $\neg(x>\frownie)$.\\
We also make the following convention:\\
\phx\qquad $\neg(\frownie<\frownie)$ \quad and \quad $\neg(\frownie>\frownie)$.\\
Also, since $\frownie=\frownie$, we conclude:\\
\phx\qquad $\frownie\le\frownie$ \quad and \quad $\neg(\frownie\ge\frownie)$.\\

\begin{defn}
Let $f$ and $g$ be functions and let $S$ be a set.\\
\phx\qquad\qquad\qquad By \quad $f=g$ on $S$, \quad we mean: \quad $\forall x\in S$, $f_x=g_x$.
\end{defn}

For any functions $f$ and $g$, for any set $S$, we have:\\
\phx\qquad $($ $f=g$ on $S$ $)\,\,\,\Rightarrow\,\,\,(\,\,\D_f\cap S=\D_g\cap S\,\,)$.

\begin{defn}
Let $f$ and $g$ be functionals and let $S$ be a set.\\
\phx\qquad\qquad\qquad By \quad $f\le g$ on $S$, \quad we mean: \quad $\forall x\in S$, $f_x\le g_x$.\\
\phx\qquad\qquad\qquad By \quad $f\ge g$ on $S$, \quad we mean: \quad $\forall x\in S$, $f_x\ge g_x$.
\end{defn}

\noindent
For any functionals $f$ and $g$, for any set $S$, we have:\\
\phx\qquad $($ $f\le g$ on $S$ $)\,\,\,\Rightarrow\,\,\,(\,\,\D_f\cap S=\D_g\cap S\,\,)$.\\
For any functionals $f$ and $g$, for any set $S$, we have:\\
\phx\qquad $($ $f\ge g$ on $S$ $)\,\,\,\Rightarrow\,\,\,(\,\,\D_f\cap S=\D_g\cap S\,\,)$.

\begin{defn}
Let $f$ and $g$ be functionals and let $S$ be a set.\\
\phx\qquad\qquad\qquad By \quad $f<g$ on $S$, \quad we mean: \quad $\forall x\in S$, $f_x<g_x$.\\
\phx\qquad\qquad\qquad By \quad $f>g$ on $S$, \quad we mean: \quad $\forall x\in S$, $f_x>g_x$.
\end{defn}

\noindent
For any functionals $f$ and $g$, for any set $S$, we have:\\
\phx\qquad $($ $f<g$ on $S$ $)\,\,\,\Rightarrow\,\,\,(\,\,(S\subseteq\D_f)\,\&\,(S\subseteq\D_g)\,\,)$.\\
For any functionals $f$ and $g$, for any set $S$, we have:\\
\phx\qquad $($ $f>g$ on $S$ $)\,\,\,\Rightarrow\,\,\,(\,\,(S\subseteq\D_f)\,\&\,(S\subseteq\D_g)\,\,)$.\\

\begin{defn}
Let $f$ be a function, $a$ an object, $S$ a set.\\
\phx\qquad\qquad\qquad By \quad $f\equiv a$ on $S$, \quad we mean: \quad $\forall x\in S$, $f_x=a$.
\end{defn}

\noindent
For any function $f$, for any object $a$, for any set $S$, we have:\\
\phx\qquad $(\,\,(\,f\equiv a\hbox{ on }S\,)\,\&\,(\,a\ne\frownie\,)\,\,)\,\,\Rightarrow\,\,(\,\,S\subseteq\D_f\,\,)$.\\
For any function $f$, for any set $S$, we have:\\
\phx\qquad $(\,\,f\equiv\frownie\hbox{ on }S)\,\,)\,\,\Rightarrow\,\,(\,\,S\bigcap\D_f=\emptyset\,\,)$.

\begin{defn}
Let $f$ be a functional, $a\in\R$, $S$ a set.\\
\phx\qquad\qquad\qquad By \quad $f\le a$ on $S$, \quad we mean: \quad $\forall x\in S$, $f_x\le a$.\\
\phx\qquad\qquad\qquad By \quad $f\ge a$ on $S$, \quad we mean: \quad $\forall x\in S$, $f_x\ge a$.\\
\phx\qquad\qquad\qquad By \quad $f<a$ on $S$, \quad we mean: \quad $\forall x\in S$, $f_x<a$.\\
\phx\qquad\qquad\qquad By \quad $f>a$ on $S$, \quad we mean: \quad $\forall x\in S$, $f_x>a$.\\
\phx\qquad\qquad\qquad By \quad $a\le f$ on $S$, \quad we mean: \quad $\forall x\in S$, $a\le f_x$.\\
\phx\qquad\qquad\qquad By \quad $a\ge f$ on $S$, \quad we mean: \quad $\forall x\in S$, $a\ge f_x$.\\
\phx\qquad\qquad\qquad By \quad $a<f$ on $S$, \quad we mean: \quad $\forall x\in S$, $a<f_x$.\\
\phx\qquad\qquad\qquad By \quad $a>f$ on $S$, \quad we mean: \quad $\forall x\in S$, $a>f_x$.
\end{defn}

\begin{defn}
Let $X$ be a metric space.\\
Let $f$ and $g$ be functions and let $p\in X$.\\
By $f=g$ near $p$ in $X$, we mean: $\exists B\in\scrb_X(p)$ s.t.~$f=g$ on $B$.
\end{defn}

\begin{defn}
Let $X$ be a metric space.\\
Let $f$ and $g$ be functionals and let $p\in X$.\\
By $f\le g$ near $p$ in $X$, we mean: $\exists B\in\scrb_X(p)$ s.t.~$f\le g$ on $B$.\\
By $f\ge g$ near $p$ in $X$, we mean: $\exists B\in\scrb_X(p)$ s.t.~$f\ge g$ on $B$.\\
By $f<g$ near $p$ in $X$, we mean: $\exists B\in\scrb_X(p)$ s.t.~$f<g$ on $B$.\\
By $f>g$ near $p$ in $X$, we mean: $\exists B\in\scrb_X(p)$ s.t.~$f>g$ on $B$.
\end{defn}

\begin{defn}
Let $X$ be a metric space.\\
Let $f$ be a functions, let $a$ be an object and let $p\in X$.\\
By $f\equiv a$ near $p$ in $X$, we mean: $\exists B\in\scrb_X(p)$ s.t.~$f\equiv a$ on $B$.
\end{defn}

\begin{defn}
Let $X$ be a metric space.\\
Let $f$ be a functional, let $a\in\R$ and let $p\in X$.\\
By $f\le a$ near $p$ in $X$, we mean: $\exists B\in\scrb_X(p)$ s.t.~$f\le a$ on $B$.\\
By $f\ge a$ near $p$ in $X$, we mean: $\exists B\in\scrb_X(p)$ s.t.~$f\ge a$ on $B$.\\
By $f<a$ near $p$ in $X$, we mean: $\exists B\in\scrb_X(p)$ s.t.~$f<a$ on $B$.\\
By $f>a$ near $p$ in $X$, we mean: $\exists B\in\scrb_X(p)$ s.t.~$f>a$ on $B$.
\end{defn}

\begin{thm}
Let $X$ and $Y$ be metric spaces.\\
Let $f,g:X\drta Y$ and let $p\in X$. \qquad
Assume $f=g$ near $p$ in $X$.\\
Assume $g$ is continuous at $p$ from $X$ to $Y$.\\
Then $f$ is continuous at $p$ from $X$ to $Y$.
\end{thm}

\begin{proof}
Since $g$ is continuous at $p$ from $X$ to $Y$,\\
\phx\qquad we conclude that $p\in\D_g$, \quad so $g_p\ne\frownie$.\\
Since $f=g$ near $p$ in $X$,\\
\phx\qquad choose $B\in\scrb_X(p)$ s.t.~$f=g$ on $B$.\\
Since $B\in\scrb_X(p)$, choose $\mu>0$ s.t.~$B=B_X(p,\mu)$.\\
Then $p\in B_X(p,\mu)=B$, \qquad so $p\in B$.\\
So, since $f=g$ on $B$, we get: $f_p=g_p$.\\
Then $f_p=g_p\ne\frownie$, and so $p\in\D_f$.\\
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,f_p)<\vep\,\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,f_p)<\vep\,\,)$.\\
Since $g$ is continuous at $p$ from $X$ to $Y$,\\
\phx\qquad choose $\lambda>0$ s.t., $\forall w\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d(w,p)<\lambda\,\,)\,\,\Rightarrow\,\,(\,\,d(g_w,g_p)<\vep\,\,)$.\\
Let $\delta:=\min\{\lambda,\mu\}$. \qquad
Then $\delta>0$.\\
\phx\quad Want: \qquad $\,\,\forall w\in\D_f$, \quad $\,(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,f_p)<\vep\,\,)$.\\
Given $w\in\D_f$. \qquad $\,\,$
Want: $(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,f_p)<\vep\,\,)$.\\
Assume $d(w,p)<\delta$. \qquad
Want: $d(f_w,f_p)<\vep$.\\
Since $d(w,p)<\delta\le\mu$, we get: \quad $w\in B_X(p,\mu)$.\\
So, since $B_X(p,\mu)=B$, we get: \quad $w\in B$.\\
So, since $f=g$ on $B$, we get: \quad $f_w=g_w$.\\
Since $w\in\D_f$, we get $f_w\ne\frownie$.\\
Then $g_w=f_w\ne\frownie$, and so $w\in\D_g$.\\
So, since $d(w,p)<\delta<\lambda$, by choice of $\lambda$, we get: \quad $d(g_w,g_p)<\vep$.\\
So, since $f_w=g_w$ and $f_p=g_p$, we get: \quad $d(f_w,f_p)<\vep$, as desired.
\end{proof}

\subsection{Limits of functions between metric spaces}

\begin{defn}
Let $f$ be a function and let $p$ and $q$ be objects.\\
Then $\adj_p^q\,f$ is the function defined by:\\
\phx\qquad\qquad $\forall x$, \quad $(\adj_p^q\,f)_x=\begin{cases}f_x,&\hbox{if }x\ne p\\q,&\hbox{if }x=p.\end{cases}$
\end{defn}

\begin{thm}
Let $f:=\adj_1^3(C_\R^2)$.\\
\phx\quad Then: \qquad $\forall x\in\R$, \quad $f_x=\begin{cases}2,&\hbox{if }x\ne1\\3,&\hbox{if }x=1.\end{cases}$
\end{thm}

\begin{thm}
Let $f:=\adj_3^2(C_\R^1)$.\\
\phx\quad Then: \qquad $\forall x\in\R$, \quad $f_x=\begin{cases}1,&\hbox{if }x\ne2\\2,&\hbox{if }x=3.\end{cases}$\\
Also, $\adj_3^1\,f=C_\R^1$.
\end{thm}

\begin{defn}
Let $X$ and $Y$ be metric spaces.\\
Let $f:X\drta Y$, let $p\in X$ and let $q\in Y$.\\
By $f\to q$ near $p$ from $X$ to $Y$, we mean:\\
\phx\qquad $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad $(\,\,0<d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,q)<\vep\,\,)$.
\end{defn}

\begin{thm}
Let $f:=\adj_1^3(C_\R^2)$.\\
Then: \qquad $f_1=3$ \quad and \quad $f\to2$ near $1$ from $\R$ to $\R$.
\end{thm}

\begin{proof}
We have: $f_1=(\adj_1^3(C_\R^2))_1=3$.\\
Want: $f\to2$ near $1$ from $\R$ to $\R$.\\
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,0<d(w,1)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,2)<\vep\,\,)$.
Given $\vep>0$. \quad
Want: $\exists\delta>0$ s.t., $\forall w\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,0<d(w,1)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,2)<\vep\,\,)$.
Let $\delta=6$.\qquad Then $\delta>0$.\\
Want: $\forall w\in\D_f$, \quad $(\,\,0<d(w,1)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,2)<\vep\,\,)$.\\
Given $w\in\D_f$. \quad 
Want: $(\,\,0<d(w,1)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,2)<\vep\,\,)$.\\
Assume: $0<d(w,1)<\delta$. \qquad
Want: $d(f_w,2)<\vep$.\\
Since $d(w,1)>0$, \quad we get $d(w,1)\ne0$,\\
\phx\qquad so $w\ne1$, \quad so $(\adj_1^3(C_\R^2))_w=(C_\R^2)_w$.\\
Since $w\in\R$, we have $(C_\R^2)_w=2$.\\
Then $f_w=(\adj_1^3(C_\R^2))_w=(C_\R^2)_w=2$.\\
Then $d(f_w,2)=d(2,2)=0<\vep$, as desired.
\end{proof}

\begin{thm}
Let $X$, $Y$ and $Z$ be metric spaces.\\
Let $f:X\drta Y$, $g:Y\drta Z$. \quad
Let $a\in X$, $b\in Y$.\\
Assume that \quad $f\to b$ near $a$ from $X$ to $Y$\\
\phx\quad $\,$ and that \quad $g$ is continuous at $b$ from $Y$ to $Z$.\\
Then: \quad $g\circ f\to g_b$ near $a$ from $X$ to $Z$.
\end{thm}

\begin{proof}
Want: $\forall\vep>0$, $\exists\gamma>0$ s.t., $\forall v\in\D_{g\circ f}$,\\
\phx\qquad\qquad\qquad\quad $(\,\,0\,<\,d\,(\,v\,,\,a\,)\,<\,\gamma\,\,)\,\,\Rightarrow\,\,(\,\,d\,(\,(g\circ f)_v\,,\,g_b\,)\,<\,\vep\,\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists\gamma>0$ s.t., $\forall v\in\D_{g\circ f}$,\\
\phx\qquad\qquad\qquad\quad $(\,\,0\,<\,d\,(\,v\,,\,a\,)\,<\,\gamma\,\,)\,\,\Rightarrow\,\,(\,\,d\,(\,(g\circ f)_v\,,\,g_b\,)\,<\,\vep\,\,)$.\\
Since $g$ is continuous at $b$ from $Y$ to $Z$,\\
\phx\quad choose $\delta>0$ s.t., $\forall w\in\D_g$,\\
\phx\qquad\qquad\qquad\quad $(\,\,\,d\,(\,w\,,\,b\,)\,<\,\delta\,\,)\,\,\Rightarrow\,\,(\,\,d\,(\,g_w\,,\,g_b\,)\,<\,\vep\,\,)$.\\
Since $f\to b$ near $a$ from $Z$ to $Y$,\\
\phx\quad choose $\gamma>0$ s.t., $\forall v\in\D_f$,\\
\phx\qquad\qquad\qquad\quad $(\,\,\,0\,<\,d\,(\,v\,,\,a\,)\,<\,\gamma\,\,)\,\,\Rightarrow\,\,(\,\,d\,(\,f_v\,,\,b\,)\,<\,\delta\,\,)$.\\
Then $\gamma>0$. \quad Want: $\forall v\in\D_{g\circ f}$,\\
\phx\qquad\qquad\qquad\quad $(\,\,0\,<\,d\,(\,v\,,\,a\,)\,<\,\gamma\,\,)\,\,\Rightarrow\,\,(\,\,d\,(\,(g\circ f)_v\,,\,g_b\,)\,<\,\vep\,\,)$.\\
Given $v\in\D_{g\circ f}$.\\
\phx\qquad\qquad Want: $(\,\,0\,<\,d\,(\,v\,,\,a\,)\,<\,\gamma\,\,)\,\,\Rightarrow\,\,(\,\,d\,(\,(g\circ f)_v\,,\,g_b\,)\,<\,\vep\,\,)$.\\
Assume $0\,<\,d\,(\,v\,,\,a\,)\,<\,\gamma$. \qquad
Want: $d\,(\,(g\circ f)_v\,,\,g_b\,)\,<\,\vep$.\\
Since $v\in D_{g\circ f}$, we get $(g\circ f)_v\ne\frownie$.\\
Then $g(f_v)=(g\circ f)_v\ne\frownie$, and so $f_v\in\D_g$.\\
Since $f_v\in\D_g$, we see that $f_v\ne\frownie$, and so $v\in\D_f$.\\
Then $v\in\D_f$ and $0\,<\,d\,(\,v\,,\,a\,)\,<\,\gamma$,\\
\phx\qquad\qquad\qquad\qquad so, by choice of $\gamma$, we get: \quad $d\,(\,f_v\,,\,b\,)\,<\,\delta$.\\
Let $w:=f_v$. \quad
Then $w=f_v\in\D_g$ and $d\,(\,w\,,\,b\,)\,=\,d\,(\,f_v\,,\,b\,)\,<\,\delta$,\\
\phx\qquad\qquad\qquad\qquad so, by choice of $\delta$, we get: \quad $d\,(\,g_w\,,\,g_b\,)\,<\,\vep$.\\
Then: \quad $d\,(\,(g\circ f)_v\,,\,g_b\,)\,=\,d\,(\,g_{f_v}\,,\,g_b\,)\,=\,d\,(\,g_w\,,\,g_b\,)\,<\,\vep$.
\end{proof}

\subsection{The Hausdorff property for metric spaces}

\begin{thm}\wrlab{thm-Hproperty}
Let $Y$ be a metric space, $q,r\in Y$.\\
Assume $q\ne r$. \qquad
Then $\exists\vep>0$ s.t.~$(B_X(q,\vep))\bigcap(B_X(r,\vep))=\emptyset$.
\end{thm}

\begin{proof}
Since $q\ne r$, we get: \quad $d(q,r)\ne0$.\\
So, since $d(q,r)\ge0$, we get: \quad $d(q,r)>0$.\\
Let $s:=d(q,r)$. \quad Then $s>0$. \quad
Let $\vep:=s/2$. \quad Then $\vep>0$.\\
Want: $(B_X(q,\vep))\bigcap(B_X(r,\vep))=\emptyset$.\\
Assume $(B_X(q,\vep))\bigcap(B_X(r,\vep))\ne\emptyset$. \qquad
Want: Contradiction.\\
Choose $w\in(B_X(q,\vep))\bigcap(B_X(r,\vep))$.\\
Since \quad $w\in B_X(q,\vep)$, \quad we get: \quad $d(q,w)<\vep$.\\
Since \quad $w\in B_X(r,\vep)$, \quad we get: \quad $d(w,r)<\vep$.\\
Then $s=d(q,r)\le(d(q,w))+(d(w,r))<\vep+\vep=2\vep$, \quad so $s<2\vep$.\\
Since \quad $\vep=s/2$, \quad we get $2\vep=s$.\\
Then \quad $s<2\vep=s$, \quad so $s<s$. \qquad\qquad
Contradiction.
\end{proof}

\subsection{Uniqueness of limits of sequences}

\begin{thm}\wrlab{thm-uniqueness-of-limits}
Let $Y$ be a metric space, $s\in Y^\N$, $q,r\in Y$.\\
Assume \quad $s\to q$ in $Y$ \quad and \quad $s\to r$ in $Y$. \qquad\qquad
Then $q=r$.
\end{thm}

\begin{proof}
Assume $q\ne r$. \quad
Want: Contradiction.\\
By \tref{thm-Hproperty}, choose $\vep>0$ s.t.~$(B_X(q,\vep))\bigcap(B_X(r,\vep))=\emptyset$.\\
Since $s\to q$ in $Y$, choose $L\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge L\,\,)\,\,\Rightarrow\,\,(\,\,d(s_j,q)<\vep\,\,)$.\\
Since $s\to r$ in $Y$, choose $M\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge M\,\,)\,\,\Rightarrow\,\,(\,\,d(s_j,r)<\vep\,\,)$.\\
Let $j:=\max\{L,M\}$. \quad
Then: $(\,j\in\N\,)\,\&\,(\,j\ge L\,)\,\&\,(\,j\ge M\,)$.\\
Since $j\in\N$ and $j\ge L$, by choice of $L$,\\
\phx\qquad we get: $d(s_j,q)<\vep$, \qquad so $s_j\in B_X(q,\vep)$.\\
Since $j\in\N$ and $j\ge M$, by choice of $M$,\\
\phx\qquad we get: $d(s_j,r)<\vep$, \qquad so $s_j\in B_X(r,\vep)$.\\
Then $s_j\in(B_X(q,\vep))\bigcap(B_X(r,\vep))=\emptyset$,\\
\phx\qquad\qquad so \quad $s_j\in\emptyset$. \qquad\qquad
Contradiction.
\end{proof}

\begin{thm}
Let $g:=C_{[1;2]\bigcup\{9\}}^3$.\\
\phx\qquad\qquad\qquad Then: \quad $\forall q\in\R$, $\,$ $g\to q$ near $9$.
\end{thm}

\begin{proof}
Given $q\in\R$. \qquad Want: $g\to q$ near $9$.\\
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,0<d(x,9)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(g_x,q)<\vep\,\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists\delta>0$ s.t., $\forall x\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,0<d(x,9)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(g_x,q)<\vep\,\,)$.\\
Let $\delta:=7$. \qquad
Then $\delta>0$.\\
Want: $\forall x\in\D_g$, $\,$ $(\,\,0<d(x,9)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(g_x,q)<\vep\,\,)$.\\
Given $x\in\D_g$. \qquad
Want: $(\,\,0<d(x,9)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(g_x,q)<\vep\,\,)$.\\
Since $x\in\D_g=[1;2]\bigcup\{9\}$, we get: \quad
$(\,x\in[1;2]\,)\,\vee\,(\,x\in\{9\}\,)$.\\
Then $(\,1\le x\le2\,)\,\vee\,(\,x=9\,)$.\\
Then $(\,9-x\ge7\,)\,\vee\,(\,9-x=0\,)$.\\
Then $(\,|9-x|\ge7\,)\,\vee\,(\,|9-x|=0\,)$.\\
Then $(\,d(x,9)\ge7\,)\,\vee\,(\,d(x,9)=0\,)$.\\
Then $\neg(0<d(x,9)<7)$. \qquad
Then $\neg(0<d(x,9)<\delta)$.\\
Then $(\,\,0<d(x,9)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(g_x,q)<\vep\,\,)$, \quad
as desired.
\end{proof}

\subsection{Limits and continuity}

\begin{thm}
Let $X$ and $Y$ be metric spaces.\\
Let $g:X\drta Y$ and let $p\in\D_g$.\\
Assume: $g\to g_p$ near $p$ from $X$ to $Y$.\\
Then: $g$ is continuous at $p$ from $X$ to $Y$.
\end{thm}

\begin{proof}
By hypothesis $p\in\D_g$.\\
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall w\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d_X(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(g_w,g_p)<\vep\,\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists\delta>0$ s.t., $\forall w\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d_X(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(g_w,g_p)<\vep\,\,)$.\\
Since $g\to g_p$ near $p$ from $X$ to $Y$,\\
\phx\quad choose $\delta>0$ s.t., $\forall w\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad $\,(\,\,0<d_X(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(g_w,g_p)<\vep\,\,)$.\\
Then $\delta>0$. \qquad
Want: $\forall w\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d_X(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(g_w,g_p)<\vep\,\,)$.\\
Given $w\in\D_g$. \qquad $\,\,\,$
Want: $(\,\,d_X(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(g_w,g_p)<\vep\,\,)$.\\
Assume: $d_X(w,p)<\delta$. \qquad
Want: $d_Y(g_w,g_p)<\vep$.\\
Exactly one of the following is true:\\
\phx\qquad (1) $d_X(w,p)=0$ \qquad or \qquad (2) $d_X(w,p)\ne0$.\\
\\
{\it Case 1:}\\
Since $p\in\D_g$, we get $g_p\in\I_g$. \quad
Also, $g:X\drta Y$, so $\I_g\subseteq Y$.\\
Then $g_p\in\I_g\subseteq Y$, so $d_Y(g_p,g_p)=0$.\\
Since $d_X(w,p)=0$, we get $w=p$, and so $g_w=g_p$.\\
We have $d_Y(g_w,g_p)=d_Y(g_p,g_p)=0<\vep$, as desired.\\
{\it End of Case 1.}\\
\\
{\it Case 2:}\\
Since $d_X(w,p)\ge0$ and $d_X(w,p)\ne0$, we get $d_X(w,p)>0$.\\
Then $w\in\D_g$ and $0<d_X(w,p)<\delta$.\\
So, by choice of $\delta$, we have: $d_Y(g_w,g_p)<\vep$, as desired.\\
{\it End of Case 2.}
\end{proof}

\subsection{The metric space $\N^*$}

\begin{defn}
Define $a:\N^*\to\R$ by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\forall j\in\N^*$, \quad $a_j=\begin{cases}1/j,&\hbox{if }j\ne\infty\\0,&\hbox{if }j=\infty.\end{cases}$\\
Define $d_*:\N^*\times\N^*\to[0;\infty)$ by: $\forall j,k\in\N^*$, \quad $d_*(j,k)=|a_k-a_j|$.
\end{defn}

\begin{thm}
We have: \quad $d_*(5,7)=|(1/5)-(1/7)|=2/35$\\
\phx\qquad\qquad\qquad\qquad\qquad\quad\quad and \quad $d_*(3,\infty)=|(1/3)-0|=1/3$\\
\phx\qquad\qquad\qquad\qquad\qquad\quad\quad and \quad $d_*(\infty,\infty)=|0-0|=0$.
\end{thm}

\begin{thm}
We have: $d_*\in\scrm(\N^*)$.
\end{thm}

\noindent
We refer to $d_*$ as the \bfindex[compatible metric] on $\N^*$.\\
We are sometimes sloppy and use $\N^*$ to denote the metric space $(\N^*,d_*)$.\\
Note that $d_{\N^*}=d_{(\N^*,d_*)}=(\N^*,d_*)_2=d_*$.

\begin{thm}
Let $X$ be a set and let $s\in X^\N$.\\
Then $s:\N\to X$ and $s:\N^*\drta X$.
\end{thm}

\begin{thm}
Let $X$ be a metric space, $s\in X^\N$, $q\in X$.\\
Assume: $s\to q$ in $X$. \qquad
Then: $s\to q$ near $\infty$ from $\N^*$ to $X$.
\end{thm}

\begin{proof}
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall j\in\D_s$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,0<d_{\N^*}(j,\infty)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_j,q)<\vep\,\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists\delta>0$ s.t., $\forall j\in\D_s$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,0<d_{\N^*}(j,\infty)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_j,q)<\vep\,\,)$.\\
Since $s\to q$ in $X$, choose $K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_j,q)<\vep\,\,)$.\\
Let $\delta:=1/K$. \quad Then $\delta>0$.\\
\phx\qquad Want: \quad $\forall j\in\D_s$, $\,\,\,$ $(\,\,0<d_{\N^*}(j,\infty)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_j,q)<\vep\,\,)$.\\
Given $j\in\D_s$. \qquad
Want: $(\,\,0<d_{\N^*}(j,\infty)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_j,q)<\vep\,\,)$.\\
Assume: $0<d_{\N^*}(j,\infty)<\delta$. \qquad
Want: $d_X(s_j,q)<\vep\,\,)$.\\
We have \quad $\,\,\,d_{\N^*}(j,\infty)=|(1/j)-0|=|1/j|=1/j$,\\
\phx\quad and so \quad $d_{\N^*}(j,\infty)=1/j$.\\
Then $1/j=d_{\N^*}(j,\infty)<\delta=1/K$, \quad and so $1/j<1/K$.\\
Since $0<1/j<1/K$, we get: $j>K$. \quad
Then $j\ge K$.\\
So, since $j\in\D_s=\N$, by choice of $K$, we get: $d(s_j,q)<\vep$.
\end{proof}

\subsection{The metric space $\R^*$}

\begin{defn}
Define $\alpha:\R^*\to[-1;1]$ by:\\
\phx\qquad\qquad  $\forall x\in\R^*$, \qquad
$\alpha_x\,\,\,=\,\,\,\begin{cases}-1,&\hbox{if }x=-\infty\\x/\sqrt{1+x^2},&\hbox{if }-\infty<x<\infty\\1,&\hbox{if }x=\infty.\end{cases}$\\
Define $d^*:\R^*\times\R^*\to[0;\infty)$ by: $\forall x,y\in\R^*$, $d^*(x,y)=|\alpha_y-\alpha_x|$.
\end{defn}

\begin{thm}
$d^*(5,7)=(7/\sqrt{50})-(5/\sqrt{26})$ \qquad and\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,d^*(3,\infty)=1-(3/\sqrt{10})$ \qquad\qquad $\,$ and\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,d^*(-\infty,6)=(6/\sqrt{37})-(-1)$\\
\phx\qquad\qquad\qquad\qquad\quad\qquad\qquad\quad $=(6/\sqrt{37})+1$ \qquad\quad $\,\,\,$ and\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,d^*(-\infty,\infty)=1-(-1)=2$.
\end{thm}

\begin{thm}
We have: $d^*\in\scrm(\R^*)$.
\end{thm}

\noindent
We refer to $d^*$ as the \bfindex[compatible metric] on $\R^*$.\\
We are sometimes sloppy and use $\R^*$ to denote the metric space $(\R^*,d^*)$.\\
Note that $d_{\R^*}=d_{(\R^*,d^*)}=(\R^*,d^*)_2=d^*$.

\subsection{The Intermediate Value Theorem}

\begin{thm}\wrlab{thm-greaterat-implies-greaternear}
Let $X$ be a metric space.\\
Let $f:X\drta\R$, let $p\in X$ and let $a\in\R$.\\
Assume: $(\,\,f\hbox{ is continuous at }p\hbox{ from }X\hbox{ to }\R\,\,)\,\,\&\,\,(\,\,f_p>a\,\,)$.\\
Then $\exists C\in\scrb_X(p)$ s.t.~$(\,\,f>a\hbox{ on }C\cap\D_f\,\,)$.
\end{thm}

\begin{proof}
Let $\vep:=f_p-a$. \qquad Then $\vep>0$.\\
Since $f$ is continuous at $p$ from $X$ to $\R$,\\
\phx\quad choose $\delta>0$ s.t., $\forall w\in\D_f$, \quad $(\,\,d(w,p)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_w,f_p)<\vep\,\,)$.\\
Let $C:=B_X(p,\delta)$. \qquad Then $C\in\scrb_X(p)$. \qquad
Want: $f>a$ on $C\cap\D_f$.\\
Want: $\forall w\in C\cap\D_f$, $f_w>a$. \quad
Given $w\in C\cap\D_f$. \quad
Want: $f_w>a$.\\
We have $w\in C\cap\D_f\subseteq C=B_X(p,\delta)$, \quad so $d(w,p)<\delta)$.\\
We have $w\in C\cap\D_f\subseteq\D_f$, \quad so $w\in\D_f$.\\
Since $w\in\D_f$ and $d(w,p)<\delta$,
by choice of $\delta$, we get: $d(f_w,f_p)<\vep$.\\
Then $f_p-\vep<f_w<f_p+\vep$, and so $f_w>f_p-\vep$.\\
Since $\vep=f_p-a$, we get $a=f_p-\vep$. \quad
Then $f_w>f_p-\vep=a$.
\end{proof}

\begin{thm}\wrlab{thm-lessat-implies-lessnear}
Let $X$ be a metric space.\\
Let $f:X\drta\R$, let $p\in X$ and let $a\in\R$.\\
Assume: $(\,\,f\hbox{ is continuous at }p\hbox{ from }X\hbox{ to }\R\,\,)\,\,\&\,\,(\,\,f_p<a\,\,)$.\\
Then $\exists C\in\scrb_X(p)$ s.t.~$(\,\,f<a\hbox{ on }C\cap\D_f\,\,)$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}\wrlab{thm-IVT1-prepre}
Let \quad $a\in\R$ and $b>a$ \quad and let \quad $I:=[a;b]$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Let \quad $f:\R\drta\R$ \quad and let \quad $y\in\R$.\\
Assume: \quad $f$ is continuous on $I$ from $\R$ to $\R$ \quad and \quad $f_a<y<f_b$.\\
Let \quad $S\,\,:=\,\,\{\,v\in I\,|\,f_v<y\,\}$ \quad and let \quad $x:=\sup S$.\\
Then: \quad $x\in I$ \quad and \quad $f_x\ge y$.
\end{thm}

\begin{proof}
Since $a\in[a;b]=I$ and $f_a<y$, by definition of $S$, we get: $a\in S$.\\
Then \quad $a\in S\le\sup S$, \quad and so \quad $a\le\sup S$.\\
Since \quad $S\subseteq I=[a;b]\le b$, \quad we get \quad $S\le b$, \quad and so \quad $\sup S\le b$.\\
Then: \quad $a\le\sup S\le b$. \qquad\qquad\qquad
Then: \quad $\,\,\,\sup S\in[a;b]$.\\
By  hypothesis, $x=\sup S$ and $I=[a;b]$. \quad
Then: \quad $x\in I$.\\
It remains to show: $f_x\ge y$. \quad
Assume $f_x<y$. \quad
Want: Contradiction.\\
Since \quad $f$ is continuous on $I$ from $\R$ to $\R$ \quad and \quad since $x\in I$,\\
\phx\qquad we conclude: \quad $f$ is continuous at $x$ from $\R$ to $\R$.\\
So, since $f_x<y$, by \tref{thm-lessat-implies-lessnear},\\
\phx\qquad choose $C\in\scrb_\R(x)$ s.t. $(f<y\hbox{ on }C\cap\D_f)$.\\
Choose $\lambda>0$ s.t.~$C=B_\R(x,\lambda)$.\\
Recall that $f_x<y$. \qquad
By hypothesis, we have $y<f_b$.\\
Then $f_x<y<f_b$, \quad so $f_x<f_b$, \quad so $f_x\ne f_b$, \quad so $x\ne b$.\\
Since $x\in I$ and $x\ne b$, \quad we get $x\in I_b^\times$.\\
Then \quad $x\,\,\in\,\,I_b^\times\,\,=\,\,[a;b]_b^\times\,\,=\,\,[a;b)$, \quad
and so \quad $a\le x<b$.\\
Let $\mu:=b-x$. \quad
Since $x<b$, we get $\mu>0$.\\
Let $\delta:=\min\{\,\lambda/2\,,\,\mu\,\}$. \quad
Then: $(\,\delta>0\,)\,\&\,(\,\delta\le\lambda/2\,)\,\&\,(\,\delta\le\mu\,)$.\\
Let $w:=x+\delta$. \quad
Then: $(\,w>x\,)\,\&\,(\,w\le x+(\lambda/2)\,)\,\&\,(\,w\le x+\mu\,)$.\\
Since \quad $w>x$, \quad we get: \quad $x<w$.\\
Since $\lambda>0$, \quad we get \quad $x-\lambda<x$ \quad and \quad $x+(\lambda/2)<x+\lambda$.\\
Then $x-\lambda<x<w\le x+(\lambda/2)<x+\lambda$,\\
\phx\qquad and so \quad $x-\lambda<w<x+\lambda$, \quad
and so \quad $w\in(x-\lambda;x+\lambda)$.\\
Then $w\in(x-\lambda;x+\lambda)=B_\R(x,\lambda)=C$, \qquad and so \quad $w\in C$.\\
Since \quad $\mu=b-x$, \quad we get: \quad $x+\mu=b$. \qquad Recall: $x<w$.\\
We have $a\le x<w\le x+\mu=b$, \quad so \quad $a<w\le b$.\\
Then $w\in(a;b]\subseteq[a;b]=I$, \quad so \quad $w\in I$.\\
By hypothesis, $f$ is continuous on $I$ from $\R$ to $\R$, \quad so $I\subseteq\D_f$.\\
Then $w\in I\subseteq\D_f$, so, as $w\in C$, we get $w\in C\cap\D_f$.\\
So, since $f<y$ on $C\cap\D_f$, we get $f_w<y$.\\
So, since $w\in I$, by definition of $S$, we get $w\in S$.  \qquad Recall: $x<w$.\\
Then $w\in S\le\sup S=x<w$, so $w<w$. \qquad
Contradiction.
\end{proof}

\begin{thm}\wrlab{thm-IVT1-pre}
Let \quad $a\in\R$ and $b>a$ \quad and let \quad $I:=[a;b]$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Let \quad $f:\R\drta\R$ \quad and let \quad $y\in\R$.\\
Assume: \quad $f$ is continuous on $I$ from $\R$ to $\R$ \quad and \quad $f_a<y<f_b$.\\
Then: \quad $\exists x\in I$ s.t.~$f_x=y$.
\end{thm}

\begin{proof}
Let \quad $S\,\,:=\,\,\{\,v\in I\,|\,f_v<y\,\}$ \quad and let \quad $x:=\sup S$.\\
By \tref{thm-IVT1-prepre}, we have $x\in I$. \qquad Want: $f_x=y$.\\
By \tref{thm-IVT1-prepre}, we have $f_x\ge y$. \qquad Want: $f_x\le y$.\\
Assume $f_x>y$. \qquad Want: Contradiction.\\
Since \quad $f$ is continuous on $I$ from $\R$ to $\R$ \quad and \quad $x\in I$,\\
\phx\quad we conclude that $f$ is continuous at $x$ from $\R$ to $\R$.\\
So, since $f_x>y$, by \tref{thm-greaterat-implies-greaternear},\\
\phx\qquad choose $C\in\scrb_\R(x)$ s.t.~$f>y$ on $C\cap\D_f$.\\
Choose $\lambda>0$ s.t.~$C=B_\R(x,\lambda)$.\\
Since $\lambda>0$, we get $x-\lambda<x<x+\lambda$.\\
Since $\sup S=x>x-\lambda$, we get $\sup S>x-\lambda$.\\
Then $\neg(\sup S\le x-\lambda)$, so $\neg(S\le x-\lambda)$,\\
\phx\qquad so choose $w\in S$ s.t.~$w>x-\lambda$. \qquad Then $x-\lambda<w$.\\
Then $x-\lambda<w\in S\le\sup S=x<x+\lambda$,\\
\phx\qquad so $x-\lambda<w<x+\lambda$, \qquad so $w\in(x-\lambda;x+\lambda)$.\\
Then \quad $w\in(x-\lambda;x+\lambda)=B_\R(x,\lambda)=C$, \quad so \quad $w\in C$.\\
By hypothesis, $f$ is continuous on $I$ from $\R$ to $\R$, \quad so $I\subseteq\D_f$.\\
Then $w\in S\subseteq I\subseteq\D_f$, so, as $w\in C$, we get $w\in C\cap\D_f$.\\
So, since $f>y$ on $C\cap\D_f$, we get $f_w>y$.\\
Since $w\in S$, by definition of $S$, we get: $f_w<y$. \qquad Contradiction.
\end{proof}

The next theorem is our first version of
the Intermediate Value Theorem, which we label \bfindex[IVT\#1].

\begin{thm}\wrlab{thm-IVT1}
Let \quad $a\in\R$ and $b>a$ \quad and let \quad $I:=[a;b]$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Let \quad $f:\R\drta\R$ \quad and let \quad $y\in\R$.\\
Assume: \quad $f$ is continuous on $I$ from $\R$ to $\R$ \quad and \quad $f_a<y<f_b$.\\
Then: \quad $\exists x\in(a;b)$ s.t.~$f_x=y$.
\end{thm}

\begin{proof}
By \tref{thm-IVT1}, choose $x\in I$ s.t.~$f_x=y$.\\
As $a<y<b$, we get $f_a<f_x<f_b$,
so $f_a\ne f_x\ne f_b$, so $a\ne x\ne b$.\\
Then $x\in I_{\{a,b\}}^\times=[a;b]_{\{a,b\}}^\times=(a;b)$. \qquad Want: $f_x=y$.\\
By choice of $x$, we have $f_x=y$, as desired.
\end{proof}

The next theorem is our second version of
the Intermediate Value Theorem, which we label \bfindex[IVT\#2].

\begin{thm}\wrlab{thm-IVT2}
Let $a\in\R$, $b\ge a$ and $f:\R\drta\R$.\\
Assume: \quad $f$ is continuous on $[a;b]$ from $\R$ to $\R$ \quad and \quad $f_a\le f_b$.\\
Then: \quad $[f_a;f_b]\,\,\subseteq\,\,f_*(\,[a;b]\,)$.
\end{thm}

\begin{proof}
Want: $\forall y\in[f_a;f_b]$, $\,$ $y\in f_*(\,[a;b]\,)$.\\
Given $y\in[f_a;f_b]$. \qquad
Want: $y\in f_*(\,[a;b]\,)$.\\
By definition of $f_*(\,[a;b]\,)$, \quad
want: $\exists x\in[a;b]\bigcap\D_f$ s.t.~$f_x=y$.\\
By hypothesis, $f$ is continuous on $[a;b]$ from $\R$ to $\R$,\\
\phx\quad so $[a;b]\subseteq\D_f$, \quad so $[a;b]\bigcap\D_f=[a;b]$.\\
Want: $\exists x\in[a;b]$ s.t.~$f_x=y$.\\
Since $y\in[f_a;f_b]$, we get $f_a\le y\le f_b$.\\
Exactly one of the following is true:\\
\phx\quad (1) $y=f_a$ \quad or \quad (2) $y=f_b$ \quad or \quad (3) $f_a<y<f_b$.\\
\\
{\it Case 1:}\\
Let $x:=a$. Then $x\in[a;b]$. Want: $f_x=y$. Know: $y=f_a$.\\
We have $f_x=f_a=y$, as desired.\\
{\it End of Case 1.}\\
\\
{\it Case 2:}\\
Let $x:=b$. Then $x\in[a;b]$. Want: $f_x=y$. Know: $y=f_b$.\\
We have $f_x=f_b=y$, as desired.\\
{\it End of Case 2.}\\
\\
{\it Case 3:}\\
By hypothesis, $b\ge a$. \quad
Let $I:=[a;b]$. \quad
Want: $\exists x\in I$ s.t.~$f_x=y$.\\
Since $f_a<y<f_b$, we get $f_a<f_b$, so $f_a\ne f_b$, so $a\ne b$.\\
Sicne $b\ge a$ and $a\ne b$, we see that $b>a$.\\
Then, by \tref{thm-IVT1-pre}, \qquad we conclude: \qquad $\exists x\in I$ s.t.~$f_x=y$.\\
{\it End of Case 3.}
\end{proof}

\begin{defn}
Let $a,b\in\R^*$, $\alpha:=\min\{a,b\}$, $\beta:=\max\{a,b\}$.\\
\phx\qquad\qquad\qquad Then \quad $[a|b]:=[\alpha;\beta]$ \quad and \quad $(a|b):=(\alpha;\beta)$.
\end{defn}

\begin{thm}
We have $[8|6]=[6|8]=[6;8]$\\
\phx\qquad\qquad\qquad\qquad\qquad and \quad $\,\,\,[9|2]=[2|9]=[2;9]$\\
\phx\qquad\qquad\qquad\qquad\qquad and \quad $\,\,\,(3|5)=(5|3)=(3;5)$.
\end{thm}

The next theorem is our third version of
the Intermediate Value Theorem, which we label \bfindex[IVT\#3].

\begin{thm}\wrlab{thm-IVT3}
Let $a\in\R$, $b\ge a$ and $f:\R\drta\R$.\\
Assume: \quad $f$ is continuous on $[a;b]$ from $\R$ to $\R$.
Then: \quad $[f_a|f_b]\,\,\subseteq\,\,f_*(\,[a;b]\,)$.
\end{thm}

\begin{proof}
At least one of the following is true:\\
\phx\quad (1) $f_a\le f_b$ \quad or \quad (2) $b_a\ge f_b$.\\
\\
{\it Case 1:}\\
By \tref{thm-IVT2}, \quad $[f_a;f_b]\,\,\subseteq\,\,f_*(\,[a;b]\,)$.\\
Since $f_a\le f_b$, we get: $[f_a|f_b]=[f_a;f_b]$.\\
Then $[f_a|f_b]\,\,=\,\,[f_a;f_b]\,\,\subseteq\,\,f_*(\,[a;b]\,)$, as desired.\\
{\it End of Case 1.}\\
\\
{\it Case 2:} \qquad
Let $g:=-f$.\\
Then \quad $g$ is continuous on $[a;b]$ from $\R$ to $\R$ \quad and \quad $g_a\le g_b$.\\
Then, by \tref{thm-IVT2}, \quad $[g_a;g_b]\,\,\subseteq\,\,g_*(\,[a;b]\,)$.\\
Multiplying by $-1$, we get $-(\,[g_a;g_b]\,)\,\,\subseteq\,\,-(\,g_*(\,[a;b]\,)\,)$.\\
Then $[-g_b;-g_a]\,\,\subseteq\,\,(-g_*)(\,[a;b]\,)$.\\
So, since $f_b=-g_b$ and $f_a=-g_a$ and $f=-g$,\\
\phx\qquad we get $[f_b;f_a]\,\,\subseteq\,\,f_*(\,[a;b]\,)$.\\
Since $f_a\ge f_b$, we get: $[f_a|f_b]=[f_b;f_a]$.\\
Then $[f_a|f_b]\,\,=\,\,[f_b;f_a]\,\,\subseteq\,\,f_*(\,[a;b]\,)$, as desired.\\
{\it End of Case 2.}
\end{proof}

\noindent
The next theorem is our final version of\\
\phx\qquad\qquad\qquad the \bfindex[Intermediate Value Theorem].

\begin{thm}\wrlab{thm-IVT-final}
Let \quad $a,b\in\R$ \quad and \quad $f:\R\drta\R$.\\
Assume: \quad $f$ is continuous on $[a|b]$ from $\R$ to $\R$.\\
Then: \quad $[f_a|f_b]\,\,\subseteq\,\,f_*(\,[a|b]\,)$.
\end{thm}

\begin{proof}
At least one of the following is true:\\
\phx\qquad\qquad\qquad (1) $a\le b$ \quad or \quad (2) $a\ge b$.\\
\\
{\it Case 1:}\\
By \tref{thm-IVT3}, \quad $[f_a|f_b]\,\,\subseteq\,\,f_*(\,[a;b]\,)$.\\
Since $a\le b$, we get: $[a|b]=[a;b]$.\\
Then $[f_a|f_b]\,\,=\,\,f_*(\,[a;b]\,)\,\,=\,\,f_*(\,[a|b]\,)$, as desired.\\
{\it End of Case 1.}\\
\\
{\it Case 2:} \qquad
Let $\alpha:=b$ and $\beta:=a$. \qquad
Then $\alpha\le\beta$.\\
By \tref{thm-IVT3}, \quad $[f_\alpha|f_\beta]\,\,\subseteq\,\,f_*(\,[\alpha;\beta]\,)$.\\
Since $a\ge b$, we see that $[a|b]=[b;a]$.\\
Then $[a|b]\,\,=\,\,[\alpha;\beta]$. \qquad
Then $f(\,[a|b]\,)\,\,=\,\,f(\,[\alpha;\beta]\,)$.\\
Also, $[f_\alpha|f_\beta]\,\,=\,\,[f_\beta|f_\alpha]\,\,=\,\,[f_a|f_b]$.\\
Then $[f_a|f_b]\,\,=\,\,[f_\alpha|f_\beta]\,\,\subseteq\,\,f_*(\,[\alpha;\beta]\,)\,\,=\,\,f_*(\,[a|b]\,)$, as desired.\\
{\it End of Case 2.}
\end{proof}

\section{Compactness and the Extreme Value Theorem}

\subsection{Increasing and decreasing}

\begin{defn}
Let $f:\R^*\drta\R^*$, $S\subseteq\D_f$.\\
By $f$ is \bfindex[strictly-increasing] on $S$, we mean:\\
\phx\qquad\qquad\qquad $\forall w,x\in S$, \quad $(w<x)\Rightarrow(f_w<f_x)$.\\
By $f$ is \bfindex[strictly-decreasing] on $S$, we mean:\\
\phx\qquad\qquad\qquad $\forall w,x\in S$, \quad $(w<x)\Rightarrow(f_w>f_x)$.\\
By $f$ is \bfindex[semi-increasing] on $S$, we mean:\\
\phx\qquad\qquad\qquad $\forall w,x\in S$, \quad $(w\le x)\Rightarrow(f_w\le f_x)$.\\
By $f$ is \bfindex[semi-decreasing] on $S$, we mean:\\
\phx\qquad\qquad\qquad $\forall w,x\in S$, \quad $(w\le x)\Rightarrow(f_w\ge f_x)$.
\end{defn}

\begin{defn}
Let $f:\R^*\drta\R^*$.\\
By $f$ is {\bf strictly-increasing}, we mean: $f$ is strictly-increasing on $\D_f$.\\
By $f$ is {\bf strictly-decreasing}, we mean: $f$ is strictly-decreasing on $\D_f$.\\
By $f$ is {\bf semi-increasing}, we mean: $f$ is semi-increasing on $\D_f$.\\
By $f$ is {\bf semi-decreasing}, we mean: $f$ is semi-decreasing on $\D_f$.
\end{defn}

We discussed increasing/decreasing and secant slopes.

\begin{thm}
Define $f:\R\to\R$ by: $\forall x\in\R$, $f_x=x^2$.\\
\phx\qquad\qquad Then $f$ is strictly-decreasing on $(-\infty;0]$ \qquad and\\
\phx\qquad\qquad\qquad $\,\,\,f$ is strictly-increasing on $[0;\infty)$.
\end{thm}

\begin{thm}
$\,(1,1/2,1/3,\ldots)$ is strictly-decreasing \qquad and\\
\phx\qquad\qquad\qquad\qquad\quad $(1,1,2,2,3,3,4,4,\ldots)$ is semi-increasing.
\end{thm}

\begin{thm}
Let $s\in\R^\N$.\\
\phx\qquad\qquad Assume: $\forall j\in\N$, \quad $s_j<s_{j+1}$.\\
\phx\qquad\qquad Then: $s$ is strictly-increasing.
\end{thm}

\begin{proof}
Since $s$ is a sequence, we get: $\D_s=\N$.\\
Want: $s$ is strictly-increasing on $\D_s$.\\
Want: $s$ is strictly-increasing on $\N$.\\
Want: $\forall j,k\in\N$, \quad $(j<k)\Rightarrow(s_j<s_k)$.\\
Given $j,k\in\N$. \qquad
Want: $(j<k)\Rightarrow(s_j<s_k)$.\\
Assume $j<k$. \qquad
Want: $s_j<s_k$.\\
Since $j,k\in\Z$, we get $k-j\in\Z$. \quad
Also, since $j<k$, we get $k-j>0$.\\
Let $\ell:=k-j$. \quad
Then $\ell\in\Z$ and $\ell>0$. \quad
Then $\ell\in\N$.\\
Also, $k=j+\ell$. \quad
Want: $s_j<s_{j+\ell}$.\\
Want: $\forall m\in\N$, \quad $s_j<s_{j+m}$.\\
Let $T:=\{m\in\N\,|\,s_j<s_{j+m}\}$. \quad
Want $T=\N$.\\
By assumption, $\forall i\in\N$, $s_i<s_{i+1}$. \quad
Then $s_j<s_{j+1}$.\\
Then $1\in T$. \quad
By the PMI, want: $\forall m\in\N$, $m+1\in\N$.\\
Given $m\in\N$. \qquad
Want: $m+1\in\N$.\\
Since $m\in\N$, we get $s_j<s_{j+m}$. \quad
Want: $s_j<s_{j+m+1}$.\\
By assumption, $\forall i\in\N$, $s_i<s_{i+1}$. \quad
Then $s_{j+m}<s_{j+m+1}$.\\
Then $s_j<s_{j+m}<s_{j+m+1}$, as desired.
\end{proof}

There are three more theorems that are similar to the last:

\begin{thm}
Let $s\in\R^\N$.\\
\phx\qquad\qquad Assume: $\forall j\in\N$, \quad $s_j>s_{j+1}$.\\
\phx\qquad\qquad Then: $s$ is strictly-decreasing.
\end{thm}

\begin{proof}
Unassigned HW
\end{proof}

\begin{thm}
Let $s\in\R^\N$.\\
\phx\qquad\qquad Assume: $\forall j\in\N$, \quad $s_j\le s_{j+1}$.\\
\phx\qquad\qquad Then: $s$ is semi-increasing.
\end{thm}

\begin{proof}
Unassigned HW
\end{proof}

\begin{thm}
Let $s\in\R^\N$.\\
\phx\qquad\qquad Assume: $\forall j\in\N$, \quad $s_j\ge s_{j+1}$.\\
\phx\qquad\qquad Then: $s$ is semi-decreasing.
\end{thm}

\begin{proof}
Unassigned HW
\end{proof}

\begin{thm}\wrlab{thm-incr-comp-incr-is-incr}
Let $f,g:\R\drta\R$.\\
\phx\qquad\qquad\qquad Assume: $f$ and $g$ are both strictly-increasing.\\
\phx\qquad\qquad\qquad Then: $g\circ f$ is strictly-increasing.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\subsection{Subsequences}

\begin{defn}
Let $s$ and $t$ be sequences.\\
\phx\qquad\qquad By $t$ is a \bfindex[subsequence] of $s$, we mean:\\
\phx\qquad\qquad\qquad\qquad $\exists$strictly-increasing $\ell\in\N^\N$ s.t.~$t=s\circ\ell$.
\end{defn}

\begin{thm}
Let $\,$ $s:=(2,4,6,8,\ldots)$, $\,$ $t:=(2^2,4^2,6^2,8^2,\ldots)$,\\
\phx\qquad\quad $u:=(1,2,3,4,\ldots)$, \quad $v:=(4,2,8,6,12,10,16,14,20,18,\ldots)$.\\
Then \quad $t$ is a subsequence of $s$ \quad and \quad $s$ is a subsequence of $u$ \quad and\\
\phx\qquad\quad  $u$ is NOT a subsequence of $s$ \quad and\\
\phx\qquad\quad  $v$ is NOT a subsequence of $u$.
\end{thm}

\begin{thm}\wrlab{thm-subsubseq-is-subseq}
Let $s,t,u$ be sequences.\\
\phx\qquad Assume: $(\,u\hbox{ is a subsequence of }t\,)\,\&\,(\,t\hbox{ is a subsequence of }s\,)$.\\
\phx\qquad Then: $u\hbox{ is a subsequence of }s$.
\end{thm}

\begin{proof}
Choose strictly-increasing $\ell,m\in\N^\N$ s.t.~$u=t\circ\ell$ and $t=s\circ m$.\\
Then $m\circ\ell\in\N^\N$. \quad
Also, by \tref{thm-incr-comp-incr-is-incr}, $m\circ\ell$ is strictly-increasing.\\
It therefore suffices to show: $u=s\circ(m\circ\ell)$.\\
We have $u=t\circ\ell=(s\circ m)\circ\ell=s\circ(m\circ\ell)$, as desired.
\end{proof}

\begin{defn}
Let $X$ be a metric space and let $s\in X^\N$.\\
By $s$ is \bfindex[convergent] in $X$, we mean: \quad
$\exists p\in X$ \quad s.t.~\quad $s\to p$ in $X$.
\end{defn}

\begin{defn}
Let $X$ be a metric space and let $s\in X^\N$.\\
By $s$ is \bfindex[subconvergent] in $X$, we mean:\\
\phx\qquad\qquad $\exists$subsequence $t$ of $s$ \quad s.t.~\quad $t$ is convergent in $X$.
\end{defn}

\begin{thm}
Let $s:=(-1,1,-1,1,-1,1,-1,1,-1,1,\ldots)$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,$ Then $s$ is subconvergent in $\R$.
\end{thm}

\begin{proof}
Want: $\exists$subsequence $t$ of $s$ s.t.~$t$ is convergent in $\R$.\\
Let $\ell:=(2,4,6,8,\ldots)$. \quad
Then $\ell\in\N^\N$ and $\ell$ is strictly-increasing.\\
Let $t:=s\circ\ell$. \quad
Then $t$ is a subsequence of $s$.\\
Want: $t$ is convergent in $\R$. \quad
Want: $\exists p\in\R$ s.t.~$t\to p$ in $\R$.\\
Let $p:=1$. \quad
Then $p\in\R$. \quad
Want: $t\to p$ in $\R$.\\
We have: $\forall j\in\N$, $t_j=(s\circ\ell)_j=s_{\ell_j}=s_{2j}=1$.\\
Then $t=C_\N^1$. \quad
Then $t\to 1$ in $\R$. \quad
Then $t\to p$ in $\R$.
\end{proof}

\begin{thm}
Let $s:=(2,4,6,8,\ldots)$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,$ Then $s$ is not subconvergent in $\R$.
\end{thm}

\begin{proof}
Assume $s$ is subconvergent in $\R$. \quad
Want: Contradiction.\\
Choose a subsequence $t$ of $s$ s.t.~$t$ is convergent in $\R$.\\
Chose $p\in\R$ s.t.~$t\to p$ in $\R$.\\
Choose a strictly-increasing $\ell\in\N^\N$ s.t.~$t=s\circ\ell$.\\
Since $t\to p$ in $\R$, choose $K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad$(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(t_j,p)<1\,\,)$.\\
By the Archimedean Principle,\\
\phx\qquad choose $j\in\N$ s.t.~$j\,\ge\,\max\{\,(p+1)/2\,,\,K\,\}$.\\
Then $j\ge(p+1)/2$ and $j\ge K$.\\
Since $j\in\N$ and $j\ge K$, by choice of $K$, we get: $d(t_j,p)<1$.\\
Then $p-1<t_j<p+1$. \qquad
Then $p+1>t_j$.\\
We have $t_j=(s\circ\ell)_j=s_{\ell_j}$, \quad so $t_j=s_{\ell_j}$.\\
By definition of $s$, we know: $\forall k\in\N$, $s_k=2k$. \qquad
Then $s_{\ell_j}=2\ell_j$.\\
By HW\#8-3, $\ell_j\ge j$. \qquad
Then $2\ell_j\ge 2j$.\\
Since $j\ge(p+1)/2$, we get $2j\ge p+1$.\\
Then $p+1>t_j=s_{\ell_j}=2\ell_j\ge2j\ge p+1$,\\
\phx\qquad so $p+1>p+1$. \quad
Contradiction.
\end{proof}

\subsection{Product metrics and product metric spaces}

\begin{thm}
Let $X$ and $Y$ be metric spaces, $Z:=X\times Y$.\\
Define $d:Z\times Z\to[0;\infty)$ by $\forall p,q\in\Z$,\\
\phx\qquad $d(p,q)\quad=\quad\sqrt{\,\,(d_X(p_1,q_1))^2\,\,+\,\,(d_Y(p_2,q_2))^2\,\,}$.\\
Then $d\in\scrm(Z)$.
\end{thm}

\begin{defn}
Let $X$ and $Y$ be metric spaces, $Z:=X\times Y$.\\
Define $d:Z\times Z\to[0;\infty)$ by $\forall p,q\in\Z$,\\
\phx\qquad $d(p,q)\quad=\quad\sqrt{\,\,(d_X(p_1,q_1))^2\,\,+\,\,(d_Y(p_2,q_2))^2\,\,}$.\\
Then $X\,\,\times_{MS}\,\,Y\quad:=\quad(X\times Y,d)$.
\end{defn}

We are often sloppy and write $X\times Y$ for $X\,\,\times_{MS}\,\,Y$.

\subsection{Stereographic projection}

\begin{thm}
Let $V:=\R^2$, $C:=S_V(0_V,1)$.\\
Let \quad $p:=(1,0)$, \quad $q\in C_p^\times$, \quad $t\in\R$.\\
Assume: $p$ and $q$ and $(0,t)$ collinear ({\it i.e.}, on a line).\\
Then: \qquad $\ds{t\,\,=\,\,\frac{q_2}{1-q_1}}$ \qquad and \qquad $\ds{q\,\,=\,\,\left(\,\frac{t^2-1}{t^2+1}\,,\,\frac{2t}{t^2+1}\,\,\right)}$.
\end{thm}

The preceding is a precalculus theorem that we proved in detail in~class.
It describes the importance of the ``stereographic projection'' maps given by:\\
\phx\qquad\qquad\qquad $\ds{q\,\,\,\mapsto\qquad\,\,\,\,\,\,\,\frac{q_2}{1-q_1}\!\!\qquad\qquad\quad:\qquad C_p^\times\to\R}$\\
\phx\qquad and \qquad $\ds{\,t\,\,\,\mapsto\,\,\,\left(\,\frac{t^2-1}{t^2+1}\,,\,\frac{2t}{t^2+1}\,\,\right)\qquad:\qquad\R\to C_p^\times}$.\\
These two functions are called $f$ and $g$ in \tref{thm-stereo-homeom}, below.

\subsection{Two basic facts}

\begin{thm}
Let $a,b\ge0$. \quad
Then $\sqrt{a+b}\le\sqrt{a}+\sqrt{b}$.
\end{thm}

\begin{proof}
Let $s:=\sqrt{a}$ and $t:=\sqrt{b}$. \quad
Want: $\sqrt{a+b}\le s+t$.\\
Since $s^2=a$ and $t^2=b$, we get $s^2+t^2=a+b$.\\
We have $s,t\ge0$, so $2st\ge0$. \quad
Then $s^2+t^2\le s^2+2st+t^2$.\\
So, since $s^2+t^2=a+b$ and $s^2+2st+t^2=(s+t)^2$,\\
\phx\qquad we see that $a+b\le(s+t)^2$.\\
Since $a,b\ge0$, we get $a+b\ge0$, so $0\le a+b$.\\
Since $0\le a+b\le(s+t)^2$, we see that $\sqrt{a+b}\le s+t$, as desired.
\end{proof}

\begin{thm}
Define $\alpha:\R^*\to[-1;1]$ by: $\forall x\in\R^*$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\alpha_x\quad=\quad\begin{cases}-1,&\hbox{if }x=-\infty\\x/\sqrt{1+x^2},&\hbox{if }-\infty<x<\infty\\1,&\hbox{if }x=\infty.\end{cases}$
Define $\beta:[-1;1]\to\R^*$ by: $\forall y\in[-1;1]$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\beta_y\quad=\quad\begin{cases}-\infty,&\hbox{if }y=-1\\y/\sqrt{1-y^2},&\hbox{if }-1<y<1\\\infty,&\hbox{if }y=1.\end{cases}$\\
Then: \qquad $\,\,\,\,\alpha$ and $\beta$ are both strictly-increasing \qquad\qquad and\\
\phx\qquad\qquad $\alpha\circ\beta=\id_{[-1;1]}$ \qquad and \qquad $\beta\circ\alpha=\id_{\R^*}$.\\
Also: \quad $\alpha:\R^*\hookrightarrow>[-1;1]$ \quad and \quad $\beta:[-1;1]\hookrightarrow>\R^*$ \qquad and\\
\phx\qquad\qquad\qquad $\,\,\,\alpha^{-1}=\beta$ \qquad and \qquad $\beta^{-1}=\alpha$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\subsection{Limits involving infinite quantities}

\begin{thm}\wrlab{thm-closeto-minusinfty-implies-very-neg}
$\forall N\in\R$, $\exists\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d^*(x,-\infty)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,x<N\,\,)$.
\end{thm}

\begin{proof}
Given $N\in\R$. \qquad
Want:$\exists\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d^*(x,-\infty)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,x<N\,\,)$.\\
Define $\alpha:\R^*\to[-1;1]$ by: $\forall x\in\R^*$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\alpha_x\quad=\quad\begin{cases}-1,&\hbox{if }x=-\infty\\x/\sqrt{1+x^2},&\hbox{if }-\infty<x<\infty\\1,&\hbox{if }x=\infty.\end{cases}$
Define $\beta:[-1;1]\to\R^*$ by: $\forall y\in[-1;1]$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\beta_y\quad=\quad\begin{cases}-\infty,&\hbox{if }y=-1\\y/\sqrt{1-y^2},&\hbox{if }-1<y<1\\\infty,&\hbox{if }y=1.\end{cases}$\\
Since $N>-\infty$ and $\alpha$ is strictly-increasing, we get: $\alpha_N>\alpha_{-\infty}$.\\
Let $\delta:=1+\alpha_N$. \quad
Then $\delta>1+\alpha_{-\infty}=1+(-1)=0$, so $\delta>0$.\\
Want: $\forall x\in\R$, $(\,\,d^*(x,-\infty)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,x<N\,\,)$.\\
Given $x\in\R$. \qquad
Want: $(\,\,d^*(x,-\infty)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,x<N\,\,)$.\\
Assume: $d^*(x,-\infty)<\delta$. \qquad
Want: $x<N$.\\
Since $x>-\infty$ and $\alpha$ is strictly-increasing, we get: $\alpha_x>\alpha_{-\infty}$.\\
Then $\alpha_x-\alpha_{-\infty}>0$, and so $|\alpha_x-\alpha_{-\infty}|=\alpha_x-\alpha_{-\infty}$.\\
Then $\alpha_x-\alpha_{-\infty}=|\alpha_x-\alpha_{-\infty}|=d^*(x,-\infty)<\delta$.\\
Then $\alpha_x<\delta+\alpha_{-\infty}=\delta+(-1)=(1+\alpha_N)+(-1)=\alpha_N$,\\
\phx\qquad so $\alpha_x<\alpha_N$.\\
So, since $\beta$ is strictly-increasing, $\beta_{\alpha_x}<\beta_{\alpha_N}$.\\
Then $x=(\beta\circ\alpha)_x=\beta_{\alpha_x}<\beta_{\alpha_N}=(\beta\circ\alpha)_N=N$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-veryneg-to-verypos-implies-nearinfty-nearminusinfty}
Let $f:\R\drta\R$.\\
Assume: $\forall M\in\R$, $\exists N\in\R$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,x<N\,\,)\,\,\Rightarrow\,\,(\,\,f_x>M\,\,)$.\\
Then $f\to\infty$ near $-\infty$ from $(\R^*,d^*)$ to $(\R^*,d^*)$.
\end{thm}

\begin{proof}
Want: $\forall\vep>0$, $\exists\delta>0$ s.t, $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,d^*(x,-\infty)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d^*(f_x,\infty)<\vep\,\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists\delta>0$ s.t, $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,d^*(x,-\infty)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d^*(f_x,\infty)<\vep\,\,)$.\\
By HW\#9-4, choose $M\in\R$ s.t., $\forall y\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,y>M\,\,)\,\,\Rightarrow\,\,(\,\,d^*(y,\infty)<\delta\,\,)$.\\
By hypothesis, choose $N\in\R$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,x<N\,\,)\,\,\Rightarrow\,\,(\,\,f_x>M\,\,)$.\\
By \tref{thm-closeto-minusinfty-implies-very-neg}, choose $\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,d^*(x,-\infty)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,x<N\,\,)$.\\
Want: $\forall x\in\D_f$, $(\,\,d^*(x,-\infty)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d^*(f_x,\infty)<\vep\,\,)$.\\
Given $x\in\D_f$. \quad
Want: $(\,\,d^*(x,-\infty)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d^*(f_x,\infty)<\vep\,\,)$.\\
Assume: $d^*(x,-\infty)<\delta$. \qquad
Want: $d^*(f_x,\infty)<\vep$.\\
By hypothesis, $f:\R\drta\R$, \qquad so \qquad $\D_f\subseteq\R$ \quad and \quad $\I_f\subseteq\R$.\\
Since $x\in\D_f\subseteq\R$ and $d^*(x,-\infty)<\delta$, by choice of $\delta$, we get: $x<N$.\\
Since $x\in\D_f$ and $x<N$, by choice of $N$, we get $f_x>M$.\\
Let $y:=f_x$. \qquad\qquad
Then $y>M$.\\
Since $x\in\D_f$, \qquad it follows that $f_x\in\I_f$.\\
Since $y=f_x\in\I_f\subseteq\R$ and $y>M$, by choice of $M$, we get $d^*(y,\infty)<\delta$.\\
Then $d^*(f_x,\infty)=d^*(y,\infty)<\vep$, as desired.
\end{proof}

The converse of \tref{thm-veryneg-to-verypos-implies-nearinfty-nearminusinfty} is also true:

\begin{thm}
Let $f:\R\drta\R$.\\
Assume: $f\to\infty$ near $-\infty$ from $(\R^*,d^*)$ to $(\R^*,d^*)$.\\
Then: $\forall M\in\R$, $\exists N\in\R$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,x<N\,\,)\,\,\Rightarrow\,\,(\,\,f_x>M\,\,)$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\noindent
The preceding two theorems concern limits of $\infty$ near $-\infty$.\\
There are also two theorems concerning limits of $-\infty$ near $-\infty$.\\
There are also two theorems concerning limits of $\infty$ near $\infty$.\\
There are also two theorems concerning limits of $-\infty$ near $\infty$.\\
We leave it to the reader to forumlate and prove all these theorems.

\subsection{Isometries and homeomorphisms}

\begin{defn}
Let $X$ and $Y$ be metric spaces, $f$ an object.\\
By $f$ is an \bfindex[isometry] from $X$ to $Y$, we mean:\\
\phx\qquad\qquad $f:X\hookrightarrow>Y$ \qquad\qquad\qquad\qquad\qquad and\\
\phx\qquad\qquad $\forall p,q\in X$, $d_Y(f_p,f_q)=d_X(p,q)\,$ \qquad and\\
\phx\qquad\qquad $\forall p,q\in Y$, $d_X(f_p^{-1},f_q^{-1})=d_X(p,q)$.
\end{defn}

\begin{thm}
Let $X$ and $Y$ be metric spaces, $f:X\to>Y$.\\
Assume: $\forall p,q\in X$, $d_Y(f_p,f_q)=d_X(p,q)$.\\
Then $f$ is an isometry from $X$ to $Y$.
\end{thm}

\begin{proof}\ \qquad
{\it Claim: $f$ is 1-1.}\\
\\
{\it Proof of Claim:} \quad
Want: $\forall p,q\in X$, $(\,f_p=f_q\,)\,\Rightarrow\,(\,p=q\,)$.\\
Given $p,q\in X$. \qquad
Want: $(\,f_p=f_q\,)\,\Rightarrow\,(\,p=q\,)$.\\
Assume $f_p=f_q$. \qquad
Want: $p=q$.\\
Since $f_p=f_q$, we get $d_Y(f_p,f_q)=0$.\\
By hypothesis, $d_Y(f_p,f_q)=d_X(p,q)$.\\
Then $d_X(p,q)=0$. \qquad
Then $p=q$, as desired.\\
{\it End of proof of Claim.}\\
\\
By the Claim $f$ is 1-1. \quad
By hypothesis, $f:X\to>Y$. \\
Then $f:X\hookrightarrow>Y$.\\
By hypothesis, $\forall p,q\in X$, $d_Y(f_p,f_q)=d_X(p,q)$.\\
It remains to show: $\forall p,q\in Y$, $d_X(f_p^{-1},f_q^{-1})=d_Y(p,q)$.\\
Given $p,q\in Y$. \qquad
Want: $d_X(f_p^{-1},f_q^{-1})=d_Y(p,q)$.\\
Let $s:=f_p^{-1}$ and $t:=f_q^{-1}$. \qquad
Then $f_s=p$ and $f_q=t$.\\
Since $s,t\in X$, by hypothesis, we get $d_Y(f_s,f_t)=d_X(s,t)$.\\
Then $d_X(f_p^{-1},f_q^{-1})=d_X(s,t)=d_Y(f_s,f_t)=d_Y(p,q)$.
\end{proof}

\begin{defn}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ By $X$ and $Y$ are \bfindex[isometric], we mean:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\exists f$ s.t.~$($ $f$ is an isometry from $X$ to $Y$ $)$.
\end{defn}

\begin{thm}
$(\R^*,d^*)$ and $[-1;1]$ are isometric.
\end{thm}

\begin{proof}
Want: $\exists\alpha$ s.t.~$($ $\alpha$ is an isometry from $(\R^*,d^*)$ to $[-1;1]$ $)$.
Define $\alpha:\R^*\to[-1;1]$ by:\\
\phx\qquad\qquad  $\forall x\in\R^*$, \qquad
$\alpha_x\,\,\,=\,\,\,\begin{cases}-1,&\hbox{if }x=-\infty\\x/\sqrt{1+x^2},&\hbox{if }-\infty<x<\infty\\1,&\hbox{if }x=\infty.\end{cases}$\\
Want: $\alpha$ is an isometry from $(\R^*,d^*)$ to $[-1;1]$.\\
Unassigned HW: Show $\alpha$ is an isometry from $(\R^*,d^*)$ to $[-1;1]$.
\end{proof}

\begin{defn}
Let $X$ and $Y$ be metric spaces, $f$ an object.\\
By $f$ is a \bfindex[homeomorphism] from $X$ to $Y$, we mean:\\
\phx\qquad\qquad $f:X\hookrightarrow>Y$ \qquad\qquad\qquad\qquad\quad and\\
\phx\qquad\qquad $f$ is continuous from $X$ to $Y$ \qquad and\\
\phx\qquad\qquad $f^{-1}$ is continuous from $Y$ to $X$.
\end{defn}

\begin{thm}\wrlab{thm-contin-inject-not-necess-homeom}
Let $X:=[1;2)\bigcup(3;4]$ and $Y:=[7;9]$.\\
Define $f:X\to Y$ by: $\forall w\in X$, $f_w=\begin{cases}w+6,&\hbox{if }w<2\\w+5,&\hbox{if }w>2.\end{cases}$\\
\phx\quad Then $\,\,\,f:X\hookrightarrow>Y$ \qquad\qquad\qquad\qquad\qquad\qquad\quad and\\
\phx\qquad\qquad $f$ is continuous from $X$ to $Y\,$ \qquad\qquad\qquad and\\
\phx\qquad\qquad $f^{-1}$ is not continuous at $8$ from $Y$ to $X$.
\end{thm}

The idea of the preceding theorem is expressed by:\\
\phx\qquad\qquad ``attaching is continuous, but tearing apart is not''.\\
The function $f$ attaches $[1;2)$ to $[3;4]$ to make $[7;9]$.\\
The function $f^{-1}$ tears apart $[7;9]$ to make $[1;2)$ and $[3;4]$.

\begin{defn}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad $\,$ By $X$ and $Y$ are \bfindex[homeomorphic], we mean:\\
\phx\qquad\qquad\qquad\qquad $\exists f$ s.t.~$($ $f$ is an homeomorphism from $X$ to $Y$ $)$.
\end{defn}

\begin{thm}
Let $S$ and $T$ be sets.\\
\phx\qquad\qquad\qquad Let $f:S\to T$ and $g:T\to S$.\\
\phx\qquad\qquad\qquad Assume: $(\,\,g\circ f=\id_S\,\,)\,\,\&\,\,(\,\,f\circ g=\id_T\,\,)$.\\
\phx\qquad\qquad\qquad Then: \qquad $(\,\,f:S\hookrightarrow>T\,\,)\,\,\&\,\,(\,\,g:T\hookrightarrow>S\,\,)$\\
\phx\qquad\qquad\quad\qquad\qquad\qquad $\,\,\,\,\&\,\,(\,\,f^{-1}=g\,\,)\,\,\&\,\,(\,\,g^{-1}=f\,\,)$.\\
\end{thm}

\begin{thm}\wrlab{thm-sqrt-uc-not-Lip}
Define $s:\R\drta\R$ by $s_x=\sqrt{x}$.\\
Then: \qquad\qquad\qquad\qquad\qquad $\D_s=[0;\infty)=\I_s$\\
\phx\qquad\qquad\qquad and \quad $s$ is uniformly continuous from $\R$ to $\R$\\
\phx\qquad\qquad\qquad and \quad $s$ is not Lipschitz from $\R$ to $\R$.
\end{thm}

\begin{proof}
Unassigned HW.\\
{\it Hint for uniformly continuity:} Let $\delta:=\vep^2$.
\end{proof}

\begin{thm}
Define $s:\R\drta\R$ by $s_x=\sqrt{x}$.\\
Then: \qquad\qquad\qquad $\,\,\,s$ is continuous from $\R$ to $\R$\\
\phx\qquad\qquad\qquad and \quad $s$ is continuous from $[0;\infty)$ to $[0;\infty)$.
\end{thm}

\begin{proof}
By \tref{thm-sqrt-uc-not-Lip}, $s$ is uniformly continuous from $\R$ to $\R$.\\
It follows that $s$ is continuous from $\R$ to $\R$.\\
Want: $s$ is continuous from $[0;\infty)$ to $[0;\infty)$.\\
By \tref{thm-sqrt-uc-not-Lip}, we have: $\D_s=[0;\infty)=\I_s$.\\
Then $s:[0;\infty)\to>[0;\infty)$, \quad and so $s:[0;\infty)\drta[0;\infty)$.\\
Then, by Absoluteness of Continuity,\\
\phx\quad we get: \qquad $s$ is continuous $[0;\infty)$ to $[0;\infty)$, \quad as desired.
\end{proof}

\begin{thm}
$(-1;1)$ and $\R$ are homeomorphic.
\end{thm}

\begin{proof}
Want: $\exists f$ s.t.~$f$ is a homeomorphism from $(-1;1)$ onto $\R$.\\
Define $f:(-1;1)\to\R$ by: \qquad $\forall x\in(-1;1)$, \quad $\ds{f_x=\frac{x}{\sqrt{1-x^2}}}$.\\
Want: $f$ is a homeomorphism from $(-1;1)$ onto $\R$.\\
Want: \quad $\,\,(\,f:(-1;1)\hookrightarrow>\R\,)\qquad\&$\\
\phx\qquad\qquad $(\,f\hbox{ is continuous from }(-1;1)\hbox{ to }\R\,)\qquad\&$\\
\phx\qquad\qquad $(\,f^{-1}\hbox{ is continuous from }\R\hbox{ to }(-1;1)\,)$.\\
Define $g:\R\to\R$ by: \qquad $\forall x\in\R$, \quad $\ds{g_x=\frac{x}{\sqrt{1+x^2}}}$.\\
Then: \quad $(\,\I_g\subseteq(-1;1)\,)\quad\&\quad(\,g\circ f=\id_{(-1;1)}\,)\quad\&\quad(\,f\circ g=\id_\R\,)$.\\
It follows that: $f:(-1;1)\hookrightarrow>\R$ and $g=f^{-1}$.\\
Want: \quad $\,\,(\,f\hbox{ is continuous from }(-1;1)\hbox{ to }\R\,)\qquad\&$\\
\phx\qquad\qquad $(\,g\hbox{ is continuous from }\R\hbox{ to }(-1;1)\,)$.\\
By properties of limits, we conclude that:\\
\phx\qquad\qquad $\,\,(\,f\hbox{ is continuous from }\R\hbox{ to }\R\,)\qquad\&$\\
\phx\qquad\qquad $(\,g\hbox{ is continuous from }\R\hbox{ to }\R\,)$.\\
Then, by Absoluteness of Continuity, we have:\\
\phx\qquad\qquad $\,\,(\,f\hbox{ is continuous from }(-1;1)\hbox{ to }\R\,)\qquad\&$\\
\phx\qquad\qquad $(\,g\hbox{ is continuous from }\R\hbox{ to }(-1;1)\,)$,\\
as desired.
\end{proof}

\begin{thm}\wrlab{thm-stereo-homeom}
Let $V:=\R^2$, $C:=S_V(0_V,1)$, $p:=(1,0)$.\\
Then $C_p^\times$ and $\R$ are homeomorphic.
\end{thm}

\begin{proof}
Want: $\exists f$ s.t.~$f$ is a homeomorphism from $C_p^\times$ onto $\R$.\\
Define $\phi:V\drta\R$ by: \qquad $\forall q\in V$, \quad $\ds{\phi_q=\frac{q_2}{1-q_1}}$.\\
By properties of continuity, $\phi$ is continuous from $V$ to $\R$.\\
Also, $C_p^\times\subseteq\{q\in V\,|\,q_1<1\}\subseteq\D_\phi$. \quad
Let $f:=\phi|C_p^\times$.\\
Want: $f$ is a homeomorphism from $C_p^\times$ onto $\R$.\\
Since $\phi$ is continuous from $V$ to $\R$,\\
\phx\qquad it follows that $f$ is continuous from $V$ to $\R$.\\
Then, by Absoluteness of Continuity, $f$ is continuous from $C_p^\times$ to $\R$.\\
Want: \quad $f:C_p^\times\hookrightarrow>\R$ \quad and \quad $f^{-1}$ is continuous from $\R$ to $C_p^\times$.\\
Define $\lambda,\mu:\R\to\R$ by: \quad $\forall t\in\R$, \quad $\ds{\lambda_t=\frac{t^2-1}{t^2+1}}$ \quad and \quad $\ds{\mu_t=\frac{2t}{t^2+1}}$.\\
By properties of continuiy, $\lambda$ and $\mu$ are both continuous from $\R$ to $\R$,\\
\phx\qquad and so $(\lambda,\mu)$ is continuous from $\R$ to $\R^2$.\\
Let $g:=(\lambda,\mu)$. $\,\,$
Recall: $V=\R^2$. $\,\,$
Then $g$ is continuous from $\R$ to $V$.\\
{\it Unassigned HW:} \qquad $\forall t\in\R$, \qquad $g_t\in C_p^\times$ \quad and \quad $f_{g_t}=t$\\
\phx\qquad\qquad\qquad and \qquad $\forall q\in C_p^\times$, \qquad $g_{f_q}=q$.\\
Then \quad $\I_g\subseteq C_p^\times$ \quad and \quad $f\circ g=\id_\R$ \quad and \quad $g\circ f=\id_{C_p^\times}$.\\
Then \quad $f:C_p^\times\hookrightarrow>\R$ \quad and \quad $f^{-1}=g$.\\
Want: $f^{-1}$ is continuous from $\R$ to $C_p^\times$.\\
Since \quad $g$ is continuous from $\R$ to $V$ \quad and \quad  $\I_g\subseteq C_p^\times$,\\
\phx\quad by Absoluteness of Continuity, \quad $g$ is continuous from $\R$ to $C_p^\times$.\\
So, since $f^{-1}=g$, we get: $f^{-1}$ is continuous from $\R$ to $C_p^\times$.
\end{proof}

\subsection{Boundedness and compactness}
\quad
\vskip.1in\noindent

NOTE FOR NEXT YEAR: Define $S$ is $a$-wide to mean:
$\forall p,q\in S$, $d(p,q)<a$.
Then bounded means finitely wide, {\it i.e.},
$\exists a\in\R$ s.t.~$S$ is $a$-wide.
Also, Cauchy means: $\forall\vep>0$,
$\exists\vep$-wide tail.
Just use bounded on metric spaces, no ``bounded in $\dots$''.

\begin{defn}
Let $X$ be a metric space and let $S\subseteq X$.\\
\phx\qquad\qquad\qquad By $S$ is \bfindex[bounded] in $X$, we mean:\\
\phx\qquad\qquad\qquad\qquad $(\,\,\exists B\in\scrb_X~\hbox{s.t.}~S\subseteq B\,\,)\,\,\vee\,\,(\,\,S=X=\emptyset\,\,)$.
\end{defn}

\begin{thm}\ \\
\phx\qquad $[1;2)$ is bounded in $\R$ \quad and \quad $(-\infty;0)$ is NOT bounded in $\R$.
\end{thm}

\begin{thm}
Let $V:=\R^2$, $C:=S_V(0_v,1)$, $p:=(0,1)$.\\
\phx\qquad\qquad\qquad\qquad\quad Then $C$ and $C_p^\times$ are both bounded in $\R^2$.
\end{thm}

\begin{thm}
$\R\times\{0\}$ is NOT bounded in $\R^2$.
\end{thm}

\begin{thm}
Let $X$ be a metric space.\\
\phx\qquad\qquad\qquad\qquad\quad Then $\emptyset$ is bounded in $X$.
\end{thm}

\begin{defn}
Let $X$ be a metric space.\\
\phx\qquad\qquad\qquad By $X$ is \bfindex[bounded], we mean:\\
\phx\qquad\qquad\qquad\qquad $X$ is bounded in $X$.
\end{defn}

\begin{thm}\ \\
\phx\qquad $[1;2)$ is bounded \quad and \quad $(-\infty;0)$ is NOT bounded.
\end{thm}

\begin{thm}
Let $V:=\R^2$, $C:=S_V(0_v,1)$, $p:=(0,1)$.\\
\phx\qquad\qquad\qquad\qquad\quad Then $C$ and $C_p^\times$ are both bounded.
\end{thm}

\begin{thm}
$\R\times\{0\}$ is NOT bounded.
\end{thm}

\begin{thm}
$\emptyset$ is bounded.
\end{thm}

\begin{thm}
Both $(\N^*,d_*)$ and $(\R^*,d^*)$ are bounded.
\end{thm}

\noindent
We have shown that\\
\phx\qquad the bounded metric space $(-1;1)$\\
\phx\qquad\qquad is homeomorphic to\\
\phx\qquad  the unbounded metric space $\R$.\\
\\
Let $V:=\R^2$, $C:=S_V(0_V,1)$, $p:=(0,1)$.\\
We have shown that\\
\phx\qquad the bounded metric space $C_p^\times$\\
\phx\qquad\qquad is homeomorphic to\\
\phx\qquad  the unbounded metric space $\R$.\\
Moreover, \qquad\qquad $\R\,\,\,\,\,\,\,$ is isometric to $\,\,\,\,\,\,\,\R\times\{0\}$,\\
\phx\qquad so \qquad\qquad\quad $\R$ is homeomorhpic to $\R\times\{0\}$.\\
We conclude that\\
\phx\qquad the bounded metric space $C_p^\times$\\
\phx\qquad\qquad is homeomorphic to\\
\phx\qquad the unbounded metric space $\R\times\{0\}$.\\
\\
Based on these observations, one might think that\\
\phx\qquad\qquad boundedness is not a useful concept to topology.\\
And yet, it is possible for a metric spaces $X$ to be SO bounded that\\
\phx\qquad\qquad any metric space homeomorphic to $X$ is bounded.\\
Let's say that a metric space $X$ is ``super-bounded'' if:\\
\phx\qquad\qquad any metric space homeomorphic to $X$ is bounded.\\
We will show: $C$ is super-bounded, although $C_p^\times$ is not.

\begin{defn}
Let $X$ be a metric space.\\
\phx\qquad By $X$ is \bfindex[compact], we mean: $\forall s\in X^\N$, $s$ is subconvergent in $X$.
\end{defn}

Let $V:=\R^2$, $C:=S_V(0_V,1)$, $p:=(0,1)$.\\
We will show: $C$ is compact, although $C_p^\times$ is not.
We will also show:\\
\phx\qquad that any metric space\\
\phx\qquad\qquad that is homeomorphic to a compact metric space\\
\phx\qquad is compact.\\
In particular, any metric space homeomorphic to $C$ is compact.\\
We will also show:\\
\phx\qquad any compact metric space is bounded.\\
Then any metric space homeomorphic to $C$ is bounded.\\
That is, $C$ is super-bounded.\\

\noindent
In fact, any compact metric space is super-bounded,\\
\phx\qquad and it's possible to show the converse:\\
\phx\qquad\qquad\qquad any super-bounded metric space is compact,\\
although that's beyond our scope.

\subsection{Basic properties of compactness}

\begin{thm}\wrlab{thm-cpt-implies-bdd}
Let $X$ be a compact metric space.\\
\phx\qquad\qquad\qquad\qquad\quad Then $X$ is bounded.
\end{thm}

\begin{proof}
Assume $X$ is unboudned. \quad Want: Contradiction.\\
Since $\emptyset$ is bounded, we get: $X\ne\emptyset$.\\
Choose $p\in X$. \qquad
$\forall j\in\N$, let $B_j:=B_X(p,j)$.\\
Then, $\forall j\in\N$, we have: $B_j\in\scrb_X$.\\
So, since $X$ is unbounded, we conclude: \quad $\forall j\in\N$,\\
\phx\qquad\qquad $X\not\subseteq B_j$ \qquad and so \qquad $\exists q\in X$ s.t.~$q\notin B_j$.\\
So, by the Axiom of Choice, choose $s\in X^\N$ s.t., $\forall i\in\N$, $s_i\notin B_i$.\\
Since $X$ is compact, we see that $s$ is subconvergent.\\
Choose a subsequence $t$ of $s$ s.t.~$t$ is convergent in $X$.\\
Since $t$ is convergent in $X$,\\
\phx\qquad choose $z\in X$ s.t.~$t\to z$ in $X$.\\
Since $t$ is a subsequence of $s$,\\
\phx\qquad choose a strictly-increasing $\ell\in\N^\N$ s.t.~$t=s\circ\ell$.\\
Since $t\to z$ in $X$,
\phx\qquad choose $K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(t_j,z)<1\,\,)$.\\
By the Archimedean Property, choose $j\in\N$ s.t.\\
\phx\qquad\qquad $j\,\,\,\,>\,\,\,\,\max\,\,\{\,\,K\,\,,\,\,1+(d(z,p))\,\,\}$.\\
Then \quad $j>K$ \quad and \quad $j>1+(d(z,p))$,\\
\phx so \quad\quad $j\ge K$ \quad and \quad $j\ge1+(d(z,p))$.\\
Since $j\ge K$, by choice of $K$, we get: $d(t_j,z)<1$.\\
Recall: $\forall i\in\N$, $s_i\notin B_i$. $\,$
Then $s_{\ell_j}\notin B_{\ell_j}$. $\,$
Then $t_j=(s\circ\ell)_j=s_{\ell_j}\notin B_{\ell_j}$.\\
Since $t_j\notin B_{\ell_j}=B_X(p,\ell_j)$, we get: $d(t_j,p)\ge\ell_j$.\\
By HW\#8-3, we get $\ell_j\ge j$. \quad Then $d(t_j,p)\ge\ell_j\ge j$, $\,$ so $j\le d(t_j,p)$.\\
Recall: $j\ge1+(d(z,p))$. \quad
Then $1+(d(z,p))\le j$.\\
Recall: $d(t_j,z)<1$. \quad
Then $(d(t_j,z))+(d(z,p))<1+(d(z,p))$.\\
By the triangle inequality, we have: \qquad $d(t_j,p)\le(d(t_j,z))+(d(z,p))$.\\
Then $\,j\le d(t_j,p)\le(d(t_j,z))+(d(z,p))<1+(d(z,p))\le j$,\\
\phx\quad so $j<j$. \qquad
Contradiction.
\end{proof}

\begin{thm}\wrlab{thm-lifting-sequences}
Let $X$ and $Y$ be sets.\\
Let $f:X\to>Y$ and let $s\in Y^\N$.\\
Then: \quad $\exists\sigma\in X^\N$ s.t.~$f\circ\sigma=s$.
\end{thm}

\begin{proof}
We have: $\forall j\in\N$, \quad $s_j\in Y=\I_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad and so \quad $\exists q\in X$ s.t.~$f_q=s_j$.\\
So, by the Axiom of Choice, choose $\sigma\in X^\N$ s.t., $\forall j\in\N$, $f_{\sigma_j}=s_j$.\\
Want: $f\circ\sigma=s$. \qquad
Want: $\forall j\in\N$, $(f\circ\sigma)_j=s_j$.\\
Given $j\in\N$. \qquad
Want: $(f\circ\sigma)_j=s_j$.\\
We have $(f\circ\sigma)_j=f_{\sigma_j}=s_j$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-subseq-pushes-forward}
Let $X$ and $Y$ be sets.\\
Let $f:X\to Y$ and let $\sigma,\tau\in X^\N$.\\
Assume: \quad $\tau$ is a subsequence of $\sigma$.\\
Then: \qquad $f\circ\tau$ is a subsequence of $f\circ\sigma$.
\end{thm}

\begin{proof}
Choose a strictly-increasing $\ell\in\N^\N$ s.t.~$\tau=\sigma\circ\ell$.\\
Then $f\circ\tau=f\circ(\sigma\circ\ell)=(f\circ\sigma)\circ\ell$,\\
\phx\qquad and so $f\circ\tau$ is a subsequence of $f\circ\sigma$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-convergence-pushes-forward}
Let $X$ and $Y$ be metric spaces.\\
Let $f:X\to Y$ and let $\tau\in X^\N$.\\
Assume: $\,$ $f$ is continuous from $X$ to $Y$ $\,$ and $\,$ $\tau$ is convergent in $X$.\\
Then: \quad $f\circ\tau$ is convergent in $Y$.
\end{thm}

\begin{proof}
Choose $p\in X$ s.t.~$\tau\to p$ in $X$.\\
Since \quad $p\in X=\D_f$ \quad and \quad $f$ is continuous on $\D_f$ from $X$ to $Y$,\\
\phx\qquad we conclude that: \quad $f$ is continuous at $p$ from $X$ to $Y$.\\
So, \quad since $\tau\to p$ in $X$, \quad we get: \quad $f\circ\tau\to f_p$ in $Y$.\\
Then $f\circ\tau$ is convergent in $Y$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-subconvergence-pushes-forward}
Let $X$ and $Y$ be metric spaces.\\
Let $f:X\to Y$ and let $\sigma\in X^\N$.\\
Assume: $\,f$ is continuous from $X$ to $Y\,$ and $\,\sigma$ is subconvergent in $X$.\\
Then: \quad $f\circ\sigma$ is subconvergent in $Y$.
\end{thm}

\begin{proof}
Choose a subsequence $\tau$ of $\sigma$ s.t.~$\tau$ is convergent in $X$.\\
By \tref{thm-subseq-pushes-forward}, $f\circ\tau$ is a subsequence of $f\circ\sigma$.\\
It therefore suffices to show: $f\circ\tau$ is convergent in $Y$.\\
By \tref{thm-convergence-pushes-forward}, $f\circ\tau$ is convergent in $Y$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-continsurj-fromcpt-image-cpt}
Let $X$ and $Y$ be metric spaces, $f:X\to>Y$.\\
Assume: \quad $X$ is compact \quad and \quad $f$ is continuous from $X$ to $Y$.\\
Then : \quad $Y$ is compact.
\end{thm}

\begin{proof}
Want: \quad $\,\,\forall s\in Y^\N$, \quad $s$ is subconvergent in $Y$.\\
Given $s\in Y^\N$. \qquad $\,\,\,\,$
Want: $s$ is subconvergent in $Y$.\\
By \tref{thm-lifting-sequences}, choose $\sigma\in X^\N$ s.t.~$f\circ\sigma=s$.\\
Since $X$ is compact and $\sigma\in X^\N$, \quad $\sigma$ is subconvergent in $X$.\\
Then, \qquad by \tref{thm-subconvergence-pushes-forward}, \quad $f\circ\sigma$ is subconvergent in $Y$.\\
So, since $f\circ\sigma=s$, we get: \qquad\quad $\,\,\,\,s$ is subconvergent in $Y$.
\end{proof}

\begin{thm}\wrlab{thm-inf-sup-extreals}
$\forall S\subseteq\R^*$, \quad $\inf S\,,\,\sup S\,\,\in\,\,\R^*$.
\end{thm}

\begin{thm}
$\inf\,\emptyset=\infty>-\infty=\sup\,\emptyset$.
\end{thm}

\begin{thm}
Let $S\subseteq\R^*$. \qquad
Assume $S\ne\emptyset$.\\
\phx\qquad\qquad\qquad\qquad\quad Then $\inf S\le\sup S$.
\end{thm}

\begin{proof}
Choose $a\in S$. \qquad
Then $a\le\sup S$ and $a\ge\inf S$.\\
Then $\inf S\le a\le\sup S$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-nonempty-bdd-implies-realsup}
Let $S\subseteq\R$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Assume: $(\,\,S\ne\emptyset\,\,)\,\,\&\,\,(\,\,S\hbox{ is bounded in }\R\,\,)$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: $\inf S\,,\,\sup S\,\,\in\,\,\R$.
\end{thm}

\begin{proof}
By \tref{thm-inf-sup-extreals}, we have: $\inf S\,,\,\sup S\,\,\in\,\,\R^*$.\\
Choose $B\in\B_\R$ s.t.~$S\subseteq B$.\\
Choose $p\in\R$ and $a>0$ s.t.~$B=B_\R(p,a)$.\\
Since $S\subseteq B=B_\R(p,a)=(p-a;p+a)$,\\
\phx\qquad we see that: $p-a<S<p+a$.\\
Then $p-a\le S\le p+a$.\\
Then $-\infty<p-a\le\inf S\le\sup S\le p+a<\infty$,\\
\phx\qquad so $-\infty<\inf S<\infty$ and $-\infty<\sup S<\infty$,\\
\phx\qquad so $\inf S\notin\{-\infty,\infty\}$ and $\sup S\notin\{-\infty,\infty\}$,\\
Then $\inf S\,,\,\sup S\,\,\in\,\,(\R^*)_{\{-\infty,\infty\}}^\times=\R$, as desired.
\end{proof}

\subsection{Basics of topology in metric spaces}

\begin{defn}
Let $X$ be a metric space and $A\subseteq X$.\\
Then $\partial_XA:=\{p\in X\,|\,\forall B\in\scrb_X(p),\,B\bigcap A\ne\emptyset\ne B\bigcap(X\bksl A)\}$.
\end{defn}

In the preceding definition, note that $B\bigcap(X\bksl A)=B\bksl A$.

\begin{thm}
Let $X:=\R^2$, $U:=B_X(0_X,1)$, $C:=S_X(0_X,1)$.\\
Let $H:=[0;\infty)\times\R$, $A:=U\bigcup(C\bigcap H)$. \qquad
Then $\partial_XA=C$.
\end{thm}

\begin{defn}
Let $X$ be a metric space and $A\subseteq X$. \quad Then:\\
\phx\qquad $\partial_X^\seq A\quad:=\quad\{\quad p\in X\qquad|\quad\,\,\,(\exists s\in A^\N\qquad\hbox{s.t.}~s\to p\hbox{ in }X)$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\,\,\&\,\,(\exists t\in(X\bksl A)^\N~\hbox{s.t.}~t\to p\hbox{ in }X)\quad\}$.\\
\end{defn}

\begin{thm}
Let $X:=\R^2$, $U:=B_X(0_X,1)$, $C:=S_X(0_X,1)$.\\
Let $H:=[0;\infty)\times\R$, $A:=U\bigcup(C\bigcap H)$. \qquad
Then $\partial_X^\seq A=C$.
\end{thm}

\begin{thm}\wrlab{thm-boundary-equals-seqvbndry}
Let $X$ be a metric space and $A\subseteq X$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then $\partial_XA=\partial_X^\seq A$.
\end{thm}

\begin{proof}
By HW\#10-3, we have: $\partial_XA\supseteq\partial_X^\seq A$. \qquad
Want: $\partial_XA\subseteq\partial_X^\seq A$.\\
Want: $\forall p\in\partial_XA$, $p\in\partial_X^\seq A$. \qquad
Given $p\in\partial_XA$. \qquad
Want: $p\in\partial_X^\seq A$.\\
Define $B:\N\to2^X$ by: \quad $\forall j\in\N$, $B_j=B_X(p,1/j)$.\\
Since $p\in\partial_XA$, we know:\\
\phx\qquad $\forall B\in\scrb_X(p)$, $B\bigcap A\ne\emptyset\ne B\bigcap(X\bksl A)$.\\
We know, $\forall j\in\N$, that $B_j\in\scrb_X(p)$.\\
Then: $\forall j\in\N$, $B_j\bigcap A\ne\emptyset\ne B_j\bigcap(X\bksl A)$.\\
We wish to show:\\
\phx\qquad (1) $\exists s\in A^\N$ \qquad s.t.~$s\to p$ in $X$ \qquad and\\
\phx\qquad (2) $\exists t\in (X\bksl A)^\N$ s.t.~$t\to p$ in $X$\\
\\
{\it Proof of (1):}\\
Define $s\in A^\N$ by: $\forall j\in\N$, $s_j=\CH(B_j\bigcap A)$.\\
Then $s\in A^\N$. \qquad Want: $s\to p$ in $X$.\\
We have: $\forall j\in\N$, $s_j\in B_j=B_X(p,1/j)$.\\
Then: $\forall j\in\N$, $d(s_j,p)<1/j$.\\
Then: $\forall j\in\N$, $d(s_j,p)\le1/j$.\\
Then, by \tref{thm-disttozero-implies-converges}, we get $s\to p$ in $X$, as desired.\\
{\it End of proof of (1).}\\
\\
{\it Proof of (2):}\\
Define $t\in(X\bksl A)^\N$ by: $\forall j\in\N$, $t_j=\CH(B_j\bigcap(X\bksl A))$.\\
Then $t\in(X\bksl A)^\N$. \qquad Want: $t\to p$ in $X$.\\
We have: $\forall j\in\N$, $t_j\in B_j=B_X(p,1/j)$.\\
Then: $\forall j\in\N$, $d(t_j,p)<1/j$.\\
Then: $\forall j\in\N$, $d(t_j,p)\le1/j$.\\
Then, by \tref{thm-disttozero-implies-converges}, we get $t\to p$ in $X$, as desired.\\
{\it End of proof of (2).}
\end{proof}

\begin{defn}
Let $X$ be a metric space and $A\subseteq X$. \quad Then:\\
\phx\qquad\qquad\qquad $\,\Cl_XA\,\,:=\,\,A\,\bigcup\,\partial_XA$\\
\phx\qquad and \qquad $\Int_XA\,\,:=\,\,A\,\,\,\,\bksl\,\,\,\partial_XA$.
\end{defn}

\begin{thm}
Let $X$ be a metric space and $A\subseteq X$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \qquad $\Int_XA\,\,\subseteq\,\,A\,\,\subseteq\,\,\Cl_XA$.
\end{thm}

\begin{thm}
Let $X:=\R^2$, $U:=B_X(0_X,1)$, $C:=S_X(0_X,1)$.\\
Let $H:=[0;\infty)\times\R$, $A:=U\bigcup(C\bigcap H)$.\\
Then: \qquad\qquad $\,\Cl_XA\,\,=\,\,U\bigcup C\,\,=\,\,\barB_X(0_X,1)$\\
\phx\qquad\quad and \quad $\,\,\,\Int_XA\,\,=\,\,U\,\,=\,\,B_X(0_X,1)$.
\end{thm}

\begin{thm}
Let $X:=\R$ and $A:=[3;5)$.\\
\phx\qquad Then: \qquad $\Cl_XA\,\,=\,\,[3;5]$ \quad and \quad $\Int_XA=(3;5)$.
\end{thm}

\begin{thm}\wrlab{thm-quantif-equiv-for-closure-and-interior}
Let $X$ be a metric space, $A\subseteq X$, $p\in X$.\\
Then: (1) $(\,p\in\Cl_XA\,)\,\Leftrightarrow\,(\,\forall B\in\scrb_X(p),\,B\bigcap A\ne\emptyset\,)$.\\
Also: (2) $(\,p\in\Int_XA\,)\,\Leftrightarrow\,(\,\exists B\in\scrb_X(p),\,B\subseteq A\,)$.
\end{thm}

\begin{proof}
{\it Proof of (1):} \quad
Unassigned HW. \quad
{\it End of proof of (1).}\\
\\
{\it Proof of $\Rightarrow$ in (2):}\\
Assume: $p\in\Int_XA$. \qquad
Want: $\exists B\in\scrb_X(p),\,B\subseteq A\,)$.\\
Since $p\in\Int_XA=A\bksl(\partial_XA)$,
we get: $p\in A$ and $p\notin\partial_XA$.\\
Since $p\notin\partial_XA$, by definition of $\partial_XA$,\\
\phx\qquad we conclude $\neg(\,\forall B\in\scrb_X(p),\,B\bigcap A\ne\emptyset\ne B\bigcap(X\bksl A)\,)$.\\
Choose $B\in\scrb_X(p)$ s.t.~$\neg(\,B\bigcap A\ne\emptyset\ne B\bigcap(X\bksl A)\,)$.\\
Then $B\in\scrb_X(p)$, and we wish to show: $B\subseteq A$.\\
We have: $(\,B\bigcap A=\emptyset\,)\,\vee\,(\,\emptyset=B\bigcap(X\bksl A)\,)$.\\
Since $B\in\scrb_X(p)$, we get: $p\in B$.\\
So, since $p\in A$, we get: $p\in B\bigcap A$. \qquad
Then $B\bigcap A\ne\emptyset$.\\
So, since $(\,B\bigcap A=\emptyset\,)\,\vee\,(\,\emptyset=B\bigcap(X\bksl A)\,)$,\\
\phx we conclude that $\emptyset=B\bigcap(X\bksl A)$.\\
Then $B=B\bigcap X=B\bigcap(A\bigcup(X\bksl A))=(B\bigcap A)\bigcup(B\bigcap(X\bksl A))$\\
\phx\qquad\quad $\,=(B\bigcap A)\bigcup\emptyset=B\bigcap A\subseteq A$, as desired.\\
{\it End of proof of $\Rightarrow$ in (2).}\\
\\
{\it Proof of $\Leftarrow$ in (2):} \quad
Unassigned HW. \quad
{\it End of proof of $\Leftarrow$ in (2).}
\end{proof}

\begin{thm}\wrlab{thm-complement-interior-is-closure-complement}
Let $X$ be a metric space and $A\subseteq X$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then $X\bksl(\Int_XA)=\Cl_X(X\bksl A)$.
\end{thm}

\begin{proof}
{\it Proof of $\subseteq$:}\\
Want: $\forall p\in X\bksl(\Int_XA)$, $p\in\Cl_X(X\bksl A)$.\\
Given $p\in X\bksl(\Int_XA)$. \qquad
Want: $p\in\Cl_X(X\bksl A)$.\\
Want: $\forall B\in\scrb_X(p)$, $B\bigcap(X\bksl A)\ne\emptyset$.\\
Given $B\in\scrb_X(p)$. \qquad Want: $B\bigcap(X\bksl A)\ne\emptyset$.\\
Assume: $B\bigcap(X\bksl A)=\emptyset$. \qquad
Want: Contradiction.\\
We have $B=B\bigcap X=B\bigcap(A\bigcup(X\bksl A))=(B\bigcap A)\bigcup(B\bigcap(X\bksl A))$\\
\phx\qquad\qquad $\,\,\,\,=(B\bigcap A)\bigcup\emptyset=B\bigcap A\subseteq A$, \qquad
so $B\subseteq A$.\\
Since $B\in\scrb_X(p)$ and $B\subseteq A$, we conclude,\\
\phx\qquad by (2) of \tref{thm-quantif-equiv-for-closure-and-interior}, that $p\in\Int_XA$.\\
Since $p\in X\bksl(\Int_XA)$, we conclude that $p\notin\Int_XA$. \qquad
Contradiction.\\
{\it End of proof of $\subseteq$.}\\
\\
{\it Proof of $\supseteq$:} \quad
Unassigned HW. \qquad
{\it End of proof of $\supseteq$.}
\end{proof}

\begin{thm}\wrlab{thm-complement-closure-is-interior-complement}
Let $X$ be a metric space and $A\subseteq X$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then $X\bksl(\Cl_XA)=\Int_X(X\bksl A)$.
\end{thm}

\begin{proof}
Let $B:=X\bksl A$. \quad
Want: $X\bksl(\Cl_XA)=\Int_XB$.\\
Let $U:=\Int_XB$. \qquad\quad $\,$
Want: $X\bksl(\Cl_XA)=U$.\\
Since $A\subseteq X$, we get: $X\bksl(X\bksl A)=A$. \qquad
Then $X\bksl B=A$.\\
By \tref{thm-complement-interior-is-closure-complement},
$X\bksl(\Int_XB)=\Cl_X(X\bksl B)$.\\
Then $X\bksl U=X\bksl(\Int_XB)=\Cl_X(X\bksl B)=\Cl_XA$,\\
\phx\qquad so $X\bksl U=\Cl_XA$, \quad so $X\bksl(X\bksl U)=X\bksl(\Cl_XA)$.\\
Since $U\subseteq X$, we get: $X\bksl(X\bksl U)=U$.\\
Then $X\bksl(\Cl_XA)=X\bksl(X\bksl U)=U$, as desired.
\end{proof}

\begin{defn}
Let $X$ be a metric space and $A\subseteq X$.\\
By $A$ is \bfindex[closed] in $X$, we mean: $\Cl_XA=A$.\\
By $A$ is \bfindex[open] in $X$, we mean: $\Int_XA=A$.
\end{defn}

The word ``clopen'' means ``closed and open''.
As set is \bfindex[clopen] in a metric space if it is both closed and open in it.

We noted that, in $\R$, there are exactly two clopen sets: $\emptyset$ and $\R$.

We noted that, in $\R^2$, there are exactly two clopen sets: $\emptyset$ and $\R^2$.

Let $S:=[1;2]$ and $T:=[3;4]$ and $X:=S\bigcup T$.
We noted that, in~the metric space $X$ (with the relative metric inherited from $\R$),
there are exactly four clopen sets:; $\emptyset$, $S$, $T$ and $X$.

\begin{thm}
Let $X:=\R^2$, $U:=B_X(0_X,1)$, $C:=S_X(0_X,1)$.\\
Let $H:=[0;\infty)\times\R$, $A:=U\bigcup(C\bigcap H)$.\\
Then: \qquad\qquad $U$ is open in $X$ and not closed in $X$\\
\phx\qquad\quad and \quad $\,\,\,U\bigcup C$ is closed in $X$ and not open in $X$\\
\phx\qquad\quad and \quad $\,\,\,A$ is neither open nor closed in $X$\\
\phx\qquad\quad and \quad $\,\,\,X$ is both open and closed in $X$.\\
\end{thm}

\begin{thm}
Let $X:=\R$.\\
Then: \qquad\qquad $(3;5)$ is open in $X$ and not closed in $X$\\
\phx\qquad\quad and \quad $\,\,\,[3;5]$ is closed in $X$ and not open in $X$\\
\phx\qquad\quad and \quad $\,\,\,[3;5)$ is neither open nor closed in $X$\\
\phx\qquad\quad and \quad $\,\,\,X$ is both open and closed in $X$.\\
\end{thm}

\begin{thm}\wrlab{thm-open-iff-complement-closed}
Let $X$ be a metric space, $T\subseteq X$, $S\subseteq T$.\\
\phx\qquad\qquad\qquad Then: $(\,\,A\hbox{ is open in }X\,\,)\,\,\Leftrightarrow\,\,(\,\,X\bksl A\hbox{ is closed in }X\,\,)$.
\end{thm}

\begin{proof}
{\it Proof of $\Rightarrow$:}\\
Assume: $A\hbox{ is open in }X$. \qquad
Want: $X\bksl A\hbox{ is closed in }X$.\\
Want: $\Cl_X(X\bksl A)=X\bksl A$. \quad
Since $A$ is open in $X$, we get: $\Int_XA=A$.\\
It follows that $X\bksl(\Int_XA)=X\bksl A$.\\
By \tref{thm-complement-interior-is-closure-complement}, we have: $X\bksl(\Int_XA)=\Cl_X(X\bksl A)$.\\
Then $\Cl_X(X\bksl A)=X\bksl(\Int_XA)=X\bksl A$, as desired.\\
{\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:}\\
Assume: $X\bksl A\hbox{ is closed in }X$.  \qquad
Want: $A\hbox{ is open in }X$.\\
Want: $\Int_XA=A$. $\,$
As $X\bksl A$ is closed in $X$, we get: $\Cl_X(X\bksl A)=X\bksl A$.\\
By \tref{thm-complement-interior-is-closure-complement}, we have: $X\bksl(\Int_XA)=\Cl_X(X\bksl A)$.\\
Then $X\bksl(\Int_XA)=\Cl_X(X\bksl A)=X\bksl A$, \quad
so $X\bksl(\Int_XA)=X\bksl A$,\\
\phx\qquad\qquad and it follows that: $X\bksl(X\bksl(\Int_XA))=X\bksl(X\bksl A)$.\\
Since $\Int_XA\subseteq X$ and $A\subseteq X$, we conclude:\\
\phx\qquad\qquad $X\bksl(X\bksl(\Int_XA))=\Int_XA$ \quad and \quad $X\bksl(X\bksl A)=A$.\\
Then $\Int_XA=X\bksl(X\bksl(\Int_XA))=X\bksl(X\bksl A)=A$, as desired.\\
{\it End of proof of $\Leftarrow$.}
\end{proof}

\begin{thm}\wrlab{thm-closed-iff-complement-open}
Let $X$ be a metric space, $T\subseteq X$, $S\subseteq T$.\\
\phx\qquad\qquad\qquad Then: $(\,\,A\hbox{ is closed in }X\,\,)\,\,\Leftrightarrow\,\,(\,\,X\bksl A\hbox{ is open in }X\,\,)$.
\end{thm}

\begin{proof}
Since $A\subseteq X$, we get $X\bksl(X\bksl A)=A$.\\
Let $B:=X\bksl A$. \qquad
Then $X\bksl B=A$.\\
By \tref{thm-open-iff-complement-closed}, we have:\\
\phx\qquad $\,\,\,\,\,$ $(\,\,B\hbox{ is open in }X\,\,)\,\,\Leftrightarrow\,\,(\,\,X\bksl B\hbox{ is closed in }X\,\,)$.\\
Then $(\,\,X\bksl A\hbox{ is open in }X\,\,)\,\Leftrightarrow\,(\,\,A\hbox{ is closed in }X\,\,)$.\\
Then $\,$ $\,(\,\,A\hbox{ is closed in }X\,\,)\,\,\Leftrightarrow\,\,(\,\,X\bksl A\hbox{ is open in }X\,\,)$, $\,$ as desired.
\end{proof}

\begin{thm}\wrlab{thm-interiors-are-open}
Let $X$ be a metric space and $A\subseteq X$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: $\Int_XA$ is open in $X$.
\end{thm}

\begin{proof}
Let $U:=\Int_XA$. \quad
Want: $U$ is open in $X$. \quad
Want $\Int_XU=U$.\\
We have $\Int_XU\subseteq U$. \quad
Want: $U\subseteq\Int_XU$.\\
Want: $\forall p\in U$, $p\in\Int_XU$.\\
Given $p\in U$. \qquad
Want: $p\in\Int_XU$.\\
Want: $\exists B\in\scrb_X(p)$ s.t.~$B\subseteq U$.\\
Since $p\in U=\Int_XA$, choose $B\in\scrb_X(p)$ s.t.~$B\subseteq A$.\\
Then $B\in\scrb_X(p)$. \qquad
Want: $B\subseteq U$.\\
Since $B\subseteq A$, we get: $\Int_XB\subseteq\Int_XA$.\\
By HW\#10-5, $B$ is open in $X$, so $\Int_XB=B$.\\
Then $B=\Int_XB\subseteq\Int_XA=U$, as desired.
\end{proof}

\begin{thm}
Let $X$ be a metric space and $A\subseteq X$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: $\Cl_XA$ is closed in $X$.
\end{thm}

\begin{proof}
Let $C:=\Cl_XA$. \qquad Want: $C$ is closed in $X$.\\
By \tref{thm-complement-closure-is-interior-complement}, we have: $X\bksl(\Cl_XA)=\Int_X(X\bksl A)$.\\
Let $B:=X\bksl A$. \qquad
Then $X\bksl C=X\bksl(\Cl_XA)=\Int_X(X\bksl A)=\Int_XB$.\\
By \tref{thm-interiors-are-open}, $\Int_XB$ is open in $X$.\\
So, since $X\bksl C=\Int_XB$, we see that: $X\bksl C$ is open in $X$.\\
Then, by \tref{thm-closed-iff-complement-open}, $C$ is closed in $X$, as desired.
\end{proof}

\subsection{Convergence of bounded semi-monotonic sequencs}

\begin{defn}
Let $X$ be a metric space, $S$ a set.\\
\phx\qquad\qquad Then: \qquad\qquad
$X_\bi^S\quad:=\quad\{f\in X^S\,|\,\I_f\hbox{ is bounded in }X\}$.
\end{defn}

\begin{thm}\wrlab{thm-semi-incr-bdd-implies-convergent}
Let $s\in\R_\bi^\N$. \quad
Assume that $s$ is semi-increasing.\\
\phx\qquad\qquad\qquad\qquad\quad Then $s$ is convergent in $\R$.
\end{thm}

\begin{proof}
Want: $\exists q\in\R$ s.t.~$s\to q$ in $\R$.\\
Since $\D_s=\N\ne\emptyset$, we get: $\I_s\ne\emptyset$.\\
So, since $\I_s$ is bounded in $\R$, by \tref{thm-nonempty-bdd-implies-realsup}, we get: $\sup\I_s\in\R$.\\
Let $q:=\sup\I_s$. \quad Then $q\in\R$. \quad Want: $s\to q$ in $\R$.\\
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$, \quad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_j,q)<\vep\,\,)$.\\
Given $\vep>0$.
Want: $\exists K\in\N$ s.t., $\forall j\in\N$, $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_j,q)<\vep\,\,)$.\\
We have $\neg(q\le q-\vep)$. \quad
So, as $q=\sup\I_s$, we get $\neg(\sup\I_s\le q-\vep)$.\\
Then $\neg(\I_s\le q-\vep)$, so choose $y\in\I_s$ s.t.~$y>q-\vep$.\\
Since $y\in\I_s$, choose $K\in\D_s$ s.t.~$y=s_K$.\\
Then $K\in\D_s=\N$. \qquad
Want: $\forall j\in\N$, $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_j,q)<\vep\,\,)$.\\
Given $j\in\N$. \qquad
Want: $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_j,q)<\vep\,\,)$.\\
Assume: $j\ge K$. \quad
Want: $d(s_j,q)<\vep$. \quad
Want: $q-\vep<s_j<q+\vep$.\\
We have $s_j\in\I_s\le\sup\I_s=q<q+\vep$. \quad $\,\,$
Want: $q-\vep<s_j$.\\
Since $s$ is semi-increasing and $K\le j$, we get: $s_K\le s_j$.\\
By choice of $y$, we have: $q-\vep<y$. \quad
By choice of $K$, we have $y=s_K$.\\
Then $q-\vep<y=s_K\le s_j$, as desired.
\end{proof}

\begin{defn}
Let $f:\R^*\drta\R^*$.\\
By $f$ is strictly-monotone, we mean:\\
\phx\qquad\qquad $f$ is strictly-increasing \qquad or \qquad $f$ is strictly-decreasing.\\
By $f$ is semi-monotone, we mean:\\
\phx\qquad\qquad $f$ is semi-increasing \qquad or \qquad $f$ is semi-decreasing.\\
\end{defn}

\begin{thm}\wrlab{thm-conv-of-bdd-semimonotone}
Let $s\in\R_\bi^\N$. \quad
Assume that $s$ is semi-monotone.\\
\phx\qquad\qquad\qquad\qquad\quad Then $s$ is convergent in $\R$.
\end{thm}

\begin{proof}
Either \quad (1) $s$ is semi-increasing \quad or \quad (2) $s$ is semi-decreasing.\\
\\
{\it Case (1):}\\
By \tref{thm-semi-incr-bdd-implies-convergent}, $s$ is convergent in $\R$, as desired.\\
{\it End of Case (1).}\\
\\
{\it Case (2):}\\
Let $t:=-s$. \quad
Then $t\in\R_\bi^\N$ and $t$ is semi-increasing.\\
By \tref{thm-semi-incr-bdd-implies-convergent}, $t$ is convergent in $\R$.\\
Then $-t$ is convergent in $\R$.\\
So, since $-t=-(-s)=s$, we see that $s$ is convergent in $\R$, as desired.
{\it End of Case (2).}
\end{proof}

\subsection{Basic dynamical systems}

\begin{defn}
Let $f$ be a function and let $j\in\N$.\\
Then $f_\circ^j$ is the function defined by: $\forall x$,\\
\phx\quad $f_\circ^j(x)\,\,=\,\,\UE\,\{\,\,y\in\I_f\,\,|\,\,\exists s\in\I_f^n~\hbox{s.t.}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,s_1=f(x)\,)\,\&$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\forall i\in[2..j],\,\,s_i=f(s_{j-1})\,)\,\&$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,y=s_j\,)\qquad\qquad\qquad\qquad\quad\}$.
\end{defn}

Let $f$ be a function.
Then\\
\phx\quad $f_\circ^1=f$, \quad $f_\circ^2=f\circ f$, \quad $f_\circ^3=f\circ f\circ f$, \quad $f_\circ^4=f\circ f\circ f\circ f$, \quad {\it etc.}

\begin{thm}\wrlab{f-to-ellplus1-is-f-circ-f-to-ell}
Let $f$ be a function and let $j\in\N$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \qquad $f_\circ^{j+1}\,\,\,=\,\,\,f\,\,\circ\,\,f_\circ^j$.
\end{thm}

\begin{thm}\wrlab{thm-f-advances-orbit}
Let $P$ be a set, $f:P\to P$, $m\in P$.\\
Define $\ell\in P^\N$ by: $\forall j\in\N$, $\ell_j=f_\circ^j(m)$.\\
Then: $\forall j\in\N$, $\ell_{j+1}=f(\ell_j)$.
\end{thm}

\begin{proof}
Given $j\in\N$. \quad
Want: $\ell_{j+1}=f(\ell_j)$.\\
By \tref{f-to-ellplus1-is-f-circ-f-to-ell}, we have: quad $f_\circ^{j+1}\,\,\,=\,\,\,f\,\,\circ\,\,f_\circ^j$.\\
We have $\ell_{j+1}\,\,=\,\,f_\circ^{j+1}(m)\,\,=\,\,(\,f\,\,\circ\,\,f_\circ^j\,)(m)$\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\,\,\,=\,\,f(\,\,f_\circ^j(m)\,\,)\,\,=\,\,f(\ell_j)$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-infinite-sets-posints-admit-str-incr-seqs}
Let $P\in\N$. \qquad
Assume $P$ is infinite.\\
\phx\qquad\qquad Then $\exists\ell\in P^\N$ s.t.~$\ell$ is strictly-increasing.
\end{thm}

\begin{proof}
By hypothesis, the set $P$ is infinite.\\
For all $k\in P$, since $[1..k]=\{1,\ldots,k\}$, we see that $[1..k]$ is finite,\\
\phx\qquad\qquad so, since $P$ is infinite, we see that $P\bksl[1..k]$ is infinite,\\
\phx\qquad\qquad so, as $\emptyset$ is finite, we get $P\bksl[1..k]\ne\emptyset$,\\
\phx\qquad\qquad so, by the Well-Ordering Axiom, $\min(P\bksl[1..k])\ne\frownie$,\\
\phx\qquad\qquad and it follows that $\min(P\bksl[1..k])\in P\bksl[1..k]$.\\
Define $f:P\drta P$ by: $\forall k\in P$, $f(k)=\min(P\bksl[1..k])$.\\
Then: $\forall k\in P$, $f(k)\ne\frownie$ and $f(k)\in P\bksl[1..k]$.\\
Since $f:P\drta P$ we get $\D_f\subseteq P$.\\
For all $k\in P$, since $f(k)\ne\frownie$, we see that $k\in\D_f$. \quad
Then $P\subseteq\D_f$.\\
So, since $\D_f\subseteq P$, we get: $\D_f=P$. \quad
Then $f:P\to P$.\\
Since $P$ is infinite and as $\emptyset$ is finite, we get $P\ne\emptyset$,\\
\phx\qquad\qquad so, by the Well-Ordering Axiom, $\min P\ne\frownie$,\\
\phx\qquad\qquad and it follows that $\min P\in P$.\\
Let $m:=\min P$. \quad Then $m\in P$.\\
Define $\ell\in P^\N$ by: $\forall j\in\N$, $\ell_j=f_\circ^j(m)$.\\
Then $\ell\in P^\N$, and we wish to show: $\ell$ is strictly-increasing.\\
Want: $\forall j\in\N$, $\ell_{j+1}>\ell_j$.\\
Given $j\in\N$. \qquad
Want: $\ell_{j+1}>\ell_j$.\\
By \tref{thm-f-advances-orbit}, we have: $\ell_{j+1}=f(\ell_j)$.\\
Since $\ell\in P^\N$, it follows that $\ell_j\in P$.\\
Recall: $\forall k\in P$, $f(k)\in P\bksl[1..k]$.\\
Let $k:=\ell_j$. \quad
Then $k\in P$, so $f(k)\in P\bksl[1..k]$.\\
Then $\ell_{j+1}=f(\ell_j)=f(k)\in P\bksl[1..k]\subseteq\N\bksl[1..k]>k=\ell_j$.
\end{proof}

\begin{thm}\wrlab{thm-past-the-last}
Let $P\subseteq\N$. \qquad
Assume $P$ is finite.\\
\phx\qquad\qquad Then $\exists m\in\N_0$ s.t.~$(m..\infty)\subseteq\N\bksl P$.
\end{thm}

\begin{proof}
Since $P\subseteq\N$, we get: $P_0^+\subseteq\N_0^+=\N_0$.\\
Since $P_0^+$ is a nonempty finite subset of $\R$, we get: $\max\,P_0^+\ne\frownie$.\\
Let $m:=\max\,P_0^+$. \quad
Then $m\ne\frownie$, so $m\in P_0^+\le m$.\\
Want: $(m..\infty)\subseteq\N\bksl P$. \qquad
Want: $\,$ $\forall j\in(m..\infty)$, $\,$ $j\in\N\bksl P$.\\
Given $j\in(m..\infty)$. \qquad
Want: $j\in\N\bksl P$.\\
Since $m\in\N_0\ge0$, we get $(m..\infty)\subseteq(0..\infty)$.\\
Then $j\in(m..\infty)\subseteq(0..\infty)=\N$, \quad and it remains to show: $j\notin P$.\\
Since $j\in(m..\infty)$, we see that $j>m$.\\
Then $P\subseteq P_0^+\le m<j$, so $P<j$. \quad
Then $j\notin P$.
\end{proof}

\subsection{Properness of the reals}

\begin{thm}\wrlab{thm-semimonotone-subseq}
Let $s\in\R^\N$.\\
\phx\qquad\qquad\quad Then $\exists$subsequence $t$ of $s$ s.t.~$t$ is semi-monotone.
\end{thm}

\begin{proof}
Let $P:=\{j\in\N\,|\,\forall k\in(j..\infty),\,s_j\ge s_k\}$.\\
We know: either \quad (1) $P$ is infinite \quad or \quad (2) $P$ is finite.\\
\\
{\it Case 1:}\\
By \tref{thm-infinite-sets-posints-admit-str-incr-seqs},
choose $\ell\in P^\N$ s.t.~$\ell$ is strictly-increasing.\\
Let $t:=s\circ\ell$. \qquad
Then $t$ is a subsequence of $s$.\\
Want: $t$ is semi-monontone. \qquad
Want: $t$ is semi-decreasing.\\
Want: $\,$ $\forall i\in\N$, $\,$ $t_i\ge t_{i+1}$.\\
Given $i\in\N$. \qquad
Want: $t_i\ge t_{i+1}$.\\
Since $\ell\in P^\N$, we see that $\ell_i\in P$.\\
Let $j:=\ell_i$. \quad Then $j\in P$.\\
Then, by definition of $P$, we have: $\forall k\in(j..\infty),\,s_j\ge s_k$.\\
Since $\ell$ is strictly-increasing, we have $\ell_{i+1}>\ell_i$.\\
Also, since $\ell\in P^\N$, we have $\ell_{i+1}\in P$. \qquad
Let $k:=\ell_{i+1}$.\\
Since $k=\ell_{i+1}>\ell_i=j$ and since $k=\ell_{i+1}\in P\subseteq\N$,\\
\phx\qquad we see that $k\in(j..\infty)$. \qquad
Then $s_j\ge s_k$.\\
Then $t_i=(s\circ\ell)_i=s_{\ell_i}=s_j\ge s_k=s_{\ell_{i+1}}=(s\circ\ell)_{i+1}=t_{i+1}$.\\
{\it End of Case 1.}\\
\\
{\it Case 2:}\\
By \tref{thm-past-the-last},
choose $m\in\N_0$ s.t.~$(m..\infty)\subseteq\N\bksl P$.\\
Then, $\forall j\in(m..\infty)$, because $j\notin P$, we see, by definition of $P$, that:\\
\phx\qquad\qquad\qquad\qquad\quad $\neg(\,\forall k\in(j..\infty),\,s_j\ge s_k\,)$.\\
Thus, \quad $\forall j\in(m..\infty)$, \quad $\,\,\exists k\in(j..\infty)$ s.t.~$s_j<s_k$.\\
Then, \quad $\forall j\in(m..\infty)$, \quad $\,\{\,k\in(j..\infty)\,|\,s_j<s_k\,\}\,\ne\,\emptyset$.\\
Then, by the Well-Ordering Axiom, we have:\\
\phx\qquad $\,\,\,\forall j\in(m..\infty)$, \quad $\min\{\,k\in(j..\infty)\,|\,s_j<s_k\,\}\,\ne\,\frownie$.\\
Define $f:(m..\infty)\to\Z$ by:\\
\phx\qquad $\,\,\,\forall j\in(m..\infty)$, \quad $f(j)=\min\{\,k\in(j..\infty)\,|\,s_j<s_k\,\}$.\\
Then, $\,\forall j\in(m..\infty)$, \quad $f(j)\in\{\,k\in(j..\infty)\,|\,s_j<s_k\,\}$,\\
\phx\qquad\qquad and so \quad $f(j)\in(j..\infty)$ \quad and \quad $s_j<s_{f(j)}$.\\
Also, $\,\forall j\in(m..\infty)$, \quad $j>m$, \quad and so $(j..\infty)\subseteq(m..\infty)$.\\
Then, $\,\forall j\in(m..\infty)$, \quad $f(j)\in(j..\infty)\subseteq(m..\infty)$.\\
So, since $\D_f=(m..\infty)$, we get: $\forall j\in\D_f$, $f_j\in(m..\infty)$.\\
Then $\I_f\subseteq(m..\infty)$. \quad
Then $f:(m..\infty)\to(m..\infty)$.\\
Define $\ell\in(m..\infty)^\N$ by: $\forall i\in\N$, $\ell_i=f_\circ^i(m+1)$.\\
Then: \qquad $\forall i\in\N$, $\,$ $f(\ell_i)\,=\,\ell_{i+1}$.\\
We have: $\forall j\in(m..\infty)$, $f(j)\in(j..\infty)>j$, so $f(j)>j$.\\
Then, $\forall i\in\N$, $\ell_{i+1}=f(\ell_i)>\ell_i$. \qquad
Then $\ell$ is strictly-increasing.\\
Let $t:=s\circ\ell$. \qquad
Then $t$ is a subsequence of $s$.\\
Want: $t$ is semi-monontone. \qquad
Want: $t$ is strictly-increasing.\\
Want: $\,$ $\forall i\in\N$, $\,$ $t_i<t_{i+1}$. \qquad
Given $i\in\N$. \qquad
Want: $t_i<t_{i+1}$.\\
Recall: $\forall j\in(m..\infty)$, $s_j<s_{f(j)}$. \qquad
Let $j:=\ell_i$.\\
Then $j\in\I_\ell=(m..\infty)$, \quad so $s_j<s_{f(j)}$. \qquad
Also, $f(j)=f(\ell_i)=\ell_{i+1}$.\\
Then $t_i=(s\circ\ell)_i=s_{\ell_i}=s_j<s_{f(j)}=s_{\ell_{j+1}}=(s\circ\ell)_{j+1}=t_{j+1}$.\\
{\it End of Case 2.}
\end{proof}

\begin{thm}\wrlab{thm-image-composite}
Let $f$ and $g$ be functions.
Then $\I_{g\circ f}\subseteq\I_g$.
\end{thm}

\begin{proof}
Want: $\forall p\in\I_{g\circ f}$, $p\in\I_f$. \qquad
Given $p\in\I_{g\circ f}$. \qquad
Want: $p\in\I_g$.\\
Since $p\in\I_{g\circ f}$, choose $x\in\D_{g\circ f}$ s.t.~$(g\circ f)_x=p$.\\
Let $y:=f_x$. \quad
Then $g_y=g_{f_x}=(g\circ f)_x=p$.\\
Since $p\in\I_{g\circ f}$, we get $p\ne\frownie$. \qquad
Then $g_y=p\ne\frownie$, so $g_y\in\I_g$.\\
Then $p=g_y\in\I_g$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-subseq-bi-is-bi}
Let $X$ be a metric space and let $s\in X_\bi^\N$.\\
\phx\qquad\qquad\qquad $\,\,$ Let $t$ be a subsequence of $s$. \qquad
Then $t\in X_\bi^\N$.
\end{thm}

\begin{proof}
Choose a strictly-increasing $\ell\in\N^\N$ s.t.~$t=s\circ\ell$.\\
By \tref{thm-image-composite}, $\I_{s\circ\ell}\subseteq\I_s$. \qquad
Then $\I_t=\I_{s\circ\ell}\subseteq\I_s$, \quad so $\I_t\subseteq\I_s$.\\
Since $s\in\R_\bi^\N$, we get: $\I_s$ is bounded in $\R$.\\
So, since $\I_t\subseteq\I_s$, we get: $\I_t$ is bounded in $X$. \qquad
Then $t\in X_\bi^\N$.
\end{proof}

\begin{defn}
Let $X$ be a metric space.\\
\phx\quad By $X$ is \bfindex[proper], we mean: \quad $\forall s\in X_\bi^\N$, $\,$ $s$ is subconvergent in $X$.
\end{defn}

\begin{thm}\wrlab{thm-R-proper}
$\R$ is proper.
\end{thm}

\begin{proof}
Want: $\forall s\in\R_\bi^\N$, $s$ is subconvergent in $\R$.\\
Given $s\in\R_\bi^\N$. \qquad
Want: $s$ is subconvergent in $\R$.\\
Want: $\exists$subsequence $t$ of $s$ s.t.~$t$ is convergent in $\R$.\\
By \tref{thm-semimonotone-subseq}, choose a subsequence $t$ of $s$ s.t.~$t$ is semi-monotone.\\
Then $t$ is a subsequence of $s$. \qquad
Want: $t$ is convergent in $\R$.\\
By \tref{thm-subseq-bi-is-bi}, we get: $t\in\R_\bi^\N$.\\
Then, by \tref{thm-conv-of-bdd-semimonotone}, we conclude: $\,\,\,\,t$ is convergent in $\R$.
\end{proof}

\subsection{Properness of products}

\begin{thm}\wrlab{thm-Lip-sends-bdd-to-bdd}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad\qquad\qquad Let $f:X\drta Y$ and let $T\subseteq X$.\\
Assume: $(\,\,T\hbox{ is bounded in }X\,\,)\,\,\,\,\&(\,\,f\hbox{ is Lipschitz from }X\hbox{ to }Y\,\,)$.\\
Then $f_*(T)$ is bounded in $Y$.
\end{thm}

\begin{proof}
Exactly one of the following must be true:\\
\phx\qquad\qquad (1) $\D_f=\emptyset$ \qquad or \qquad $\D_f\ne\emptyset$.\\
\\
{\it Case (1):}\\
We have $f_*(T)=\{f_p\,|\,p\in T\cap\D_f\}=\{f_p\,|\,p\in\emptyset\}=\emptyset$.\\
So, since $\emptyset$ is bounded in $Y$, we get: $f_*(T)$ is bounded in $Y$.\\
{\it End of Case (1).}\\
\\
{\it Case (2):}\\
Want: $\exists C\in\scrb_Y$ s.t.~$f_*(T)\subseteq C$.\\
Since $\D_f\ne\emptyset$, choose $z\in\D_f$. \qquad
Since $z\in\D_f\subseteq X$, we get $X\ne\emptyset$.\\
So, since $T$ is bounded in $X$, choose $A\in\scrb_X$ s.t.~$T\subseteq A$.\\
By the Superset Recentering Lemma, choose $B\in\scrb_X(z)$ s.t.~$A\subseteq B$.\\
Since $B\in\scrb_X(z)$, choose $r>0$ s.t.~$B=B_X(z,r)$.\\
Since $f$ is Lipschitz, choose $K\ge0$ s.t.~$f$ is Lipschitz-$K$.\\
Let $C:=B_Y(f_z,Kr+1)$. \qquad
Then $C\in\scrb_Y$. \qquad
Want: $f_*(T)\subseteq C$.\\
Want: $\forall q\in f_*(T)$, $q\in C$. \qquad
Given $q\in f_*(T)$. \qquad
Want: $q\in C$.\\
Since $q\in f_*(T)$, choose $p\in T\cap\D_f$ s.t.~$q=f_p$.\\
We have $p\in T\subseteq A\subseteq B=B_X(z,r)$, \quad
so $d_X(p,z)<r$.\\
Since $f$ is Lipschitz-$K$, we have $d_Y(f_p,f_z)\le K\cdot(d_X(p,z))$.\\
Since $K\ge0$ and $d_X(p,z)<r$, we get $K\cdot(d_X(p,z))\le Kr$.\\
Then \quad $\,d_Y(q,f_z)=d_Y(f_p,f_z)\le K\cdot(d_X(p,z))\le Kr<Kr+1$,\\
\phx\qquad so $d_Y(q,f_z)<Kr+1$, \qquad so $q\in B_Y(f_z,Kr+1)$.\\
Then $q\in B_Y(f_z,Kr+1)=C$, as desired.\\
{\it End of Case (2).}
\end{proof}

\begin{thm}\wrlab{thm-Lip-sends-bddim-to-bddim}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad\qquad\qquad Let $f:X\to Y$ and let $s\in X_\bi^\N$.\\
Assume: $f$ is Lipschitz from $X$ to $Y$. \qquad
Then $f\circ s\in Y_\bi^\N$.
\end{thm}

\begin{proof}
Since $s\in X_\bi^\N$, we get: $s:\N\to X$ and $\I_s$ is bounded in $X$.\\
Since $s:\N\to X$ and $f:X\to Y$, we get $f\circ s:\N\to Y$.\\
Then $f\circ s\in Y^\N$. \qquad
Want: $\I_{f\circ s}$ is bounded in $Y$.\\
Since $\I_s$ is bounded in $X$ and since $f$ is Lipschitz from $X$ to $Y$,\\
\phx\qquad by \tref{thm-Lip-sends-bdd-to-bdd}, we get: $f_*(\I_s)$ is bounded in $Y$.\\
So, since $\I_{f\circ s}=f_*(\I_s)$, we get: $\I_{f\circ s}$ is bounded in $Y$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-conv-to-q-implies-subseq-conv-to-q}
Let $X$ be a metric space, $s\in X^\N$ and $q\in X$.\\
\phx\qquad\qquad\qquad\qquad\qquad Let $t$ be a subsequence of $s$.\\
\phx\qquad\qquad\quad Assume: $s\to q$ in $X$. \qquad Then: $t\to q$ in $X$.
\end{thm}

\begin{proof}
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(t_j,q)<\vep\,\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(t_j,q)<\vep\,\,)$.\\
Since $s\to q$ in $X$, choose $K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_j,q)<\vep\,\,)$.\\
Then $K\in\N$. \qquad Want: $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(t_j,q)<\vep\,\,)$.\\
Given $j\in\N$. \qquad Want: $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(t_j,q)<\vep\,\,)$.\\
\phx\qquad Assume $j\ge K$. \qquad Want: $d(t_j,q)<\vep$.\\
Since $t$ is a subsequence of $s$,\\
\phx\qquad choose a strictly-increasing $\ell\in\N^\N$ s.t.~$t=s\circ\ell$.\\
Then $\ell_j\ge j\ge K$, so, by choice of $K$, we get: $d(s_{\ell_j},q)<\vep$.\\
Then $d(t_j,q)=d((s\circ\ell)_j,q)=d(s_{\ell_j},q)<\vep$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-conv-implies-subseq-conv}
Let $X$ be a metric space, $s\in X^\N$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Let $t$ be a subsequence of $s$.\\
\phx\quad Assume: $s$ is convergent in $X$. \quad Then: $t$ is convergent in $X$.
\end{thm}

\begin{proof}
Since $s$ is convergent in $X$, choose $q\in X$ s.t.~$s\to q$ in $X$.\\
By \tref{thm-conv-to-q-implies-subseq-conv-to-q}, $t\to q$ in $X$. \qquad
Then $t$ is convergent in $X$.
\end{proof}

\begin{thm}\wrlab{thm-X-proper-gives-subseq-in-totalsp}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Let $Z:=X\times Y$ and let $s\in Z_\bi^\N$.\\
Assume $X$ is proper. \qquad
Define $\alpha:Z\to X$ by: $\forall p\in Z$, $\alpha_p=p_1$.\\
Then: $\exists$subsequence $t$ of $s$ s.t.~$\alpha\circ t$ is convergent in $X$.
\end{thm}

\begin{proof}
Since $\alpha$ is Lipschitz-1 from $Z$ to $X$
and since $s\in Z_\bi^\N$,\\
\phx\qquad\qquad\qquad by \tref{thm-Lip-sends-bddim-to-bddim}, we get: \quad $\alpha\circ s\in X_\bi^\N$.\\
So, since $X$ is proper, we get: $\alpha\circ s$ is subconvergent in $X$.\\
Choose a subsequence $v$ of $\alpha\circ s$ s.t.~$v$ is convergent in $X$.\\
Choose a strictly-increasing $\ell\in\N^\N$ s.t.~$v=(\alpha\circ s)\circ\ell$.\\
Let $t:=s\circ\ell$. \qquad\quad
Then $t$ is a subsequence of $s$.\\
Want: $\alpha\circ t$ is convergent in $X$.\\
We have $\alpha\circ t=\alpha\circ(s\circ\ell)=(\alpha\circ s)\circ\ell=v$.\\
So, since $v$ is convergent in $X$, we get:\\
\phx\qquad $\,\alpha\circ t$ is convergent in $X$, \qquad as desired.
\end{proof}

\begin{thm}\wrlab{thm-Y-proper-gives-subseq-in-totalsp}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Let $Z:=X\times Y$ and let $t\in Z_\bi^\N$.\\
Assume $Y$ is proper. \qquad
Define $\beta:Z\to Y$ by: $\forall p\in Z$, $\beta_p=p_2$.\\
Then: $\exists$subsequence $u$ of $t$ s.t.~$\beta\circ u$ is convergent in $X$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}\wrlab{thm-image-subseq-is-subseq}
Let $f$ be a function and let $t\in\D_f^\N$.\\
\phx\quad Let $u$ be a subsequence of $t$. \quad
Then $f\circ u$ is a subsequence of $f\circ t$.
\end{thm}

\begin{proof}
Want: $\exists$strictly-increasing $\ell\in\N^\N$ s.t.~$f\circ u=(f\circ t)\circ\ell$.\\
Since $u$ is a subsequence of $t$,\\
\phx\qquad\qquad choose a strictly-increasing $\ell\in\N^\N$ s.t.~$u=t\circ\ell$.\\
Then $\ell$ is strictly-increasing and $\ell\in\N^\N$. \quad
Want: $f\circ u=(f\circ t)\circ\ell$.\\
We have $f\circ u=f\circ(t\circ\ell)=(f\circ t)\circ\ell$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-projs-convergent-implies-total-convergent}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Let $Z:=X\times Y$ and let $s\in Z^\N$.\\
Define $\alpha:Z\to X$ by: $\forall p\in Z$, $\alpha_p=p_1$.\\
Define $\beta:Z\to Y$ by: $\forall p\in Z$, $\beta_p=p_2$.\\
Assume $\alpha\circ u$ is convergent in $X$
and $\beta\circ u$ is convergent in $Y$.\\
Then $u$ is convergent in $Z$.
\end{thm}

\begin{proof}
Let $s:=\alpha\circ u$ and let $t:=\beta\circ u$. \qquad
Then $(s,t)=u$.\\
Also, $s$ is convergent in $X$, so choose $a\in X$ s.t. $s\to a$ in $X$.\\
Also, $t$ is convergent in $Y$, so choose $b\in Y$ s.t. $t\to b$ in $Y$.\\
Then, by HW\#9-3, we get: \quad $(s,t)\to(a,b)$ in $X\times Y$.\\
So, \quad since $u=(s,t)$ \quad and \quad since $Z=X\times Y$,\\
\phx\qquad we conclude: \quad $u\to(a,b)$ in $Z$. \quad
Then $u$ is convergent in $Z$.
\end{proof}

\begin{thm}\wrlab{thm-prod-proper-is-proper}
Let $X$ and $Y$ be proper metric spaces.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then $X\times Y$ is proper.
\end{thm}

\begin{proof}
Let $Z:=X\times Y$. \qquad
Want: $Z$ is proper.\\
Want: $\forall s\in Z_\bi^\N$, $s$ is subconvergent in $Z$.\\
Given $s\in Z_\bi^\N$. \qquad
Want: $s$ is subconvergent in $Z$.\\
Want: $\exists$subsequence $u$ of $s$ s.t.~$u$ is convergent in $Z$.\\
Define $\alpha:Z\to X$ by: $\forall p\in Z$, $\alpha_p=p_1$.\\
By \tref{thm-X-proper-gives-subseq-in-totalsp},
choose a subsequence $t$ of $s$ s.t.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\alpha\circ t$ is convergent in $X$.\\
Define $\beta:Z\to Y$ by: $\forall p\in Z$, $\beta_p=p_2$.\\
Since $t$ is a subsequence of $s$ and since $s\in Z_\bi^\N$,\\
\phx\qquad by \tref{thm-subseq-bi-is-bi}, \quad we conclude: $t\in Z_\bi^\N$.\\
By \tref{thm-Y-proper-gives-subseq-in-totalsp},
choose a subsequence $u$ of $t$ s.t.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\beta\circ u$ is convergent in $Y$.\\
Since $u$ is a subsequence of $t$ and since $t$ is a subsequence of $s$,\\
\phx\quad by \tref{thm-subsubseq-is-subseq}, we see that $u$ is a subsequence of $s$.\\
Want: $u$ is convergent in $Z$.\\
Since $u$ is a subsequence of $t$, by \tref{thm-image-subseq-is-subseq},\\
\phx\qquad\qquad  we get: $\alpha\circ u$ is a subsequence of $\alpha\circ t$.\\
So, since $\alpha\circ t$ is convergent in $X$, by \tref{thm-conv-implies-subseq-conv}, we get:\\
\phx\qquad\quad $\,\,\,\,\alpha\circ u$ is convergent in $X$.\\
So, since $\beta\circ u$ is convergent in $Y$, we get: $u$ is convergent in $Z$.
\end{proof}

\begin{thm}\wrlab{thm-R2-proper}
$\R^2$ is proper.
\end{thm}

\begin{proof}
By \tref{thm-R-proper}, $\R$ is proper.\\
So, by \tref{thm-prod-proper-is-proper}, $\R\times\R$ is proper.\\
So, since $\R^2=\R\times\R$, we see that $\R^2$ is proper.
\end{proof}

\begin{thm}\wrlab{thm-R3-proper}
$\R^3$ is proper.
\end{thm}

\begin{proof}
By \tref{thm-R-proper}, $\R$ is proper.\\
By \tref{thm-R2-proper}, $\R^2$ is proper.\\
Then, by \tref{thm-prod-proper-is-proper}, $\R^2\times\R$ is proper.\\
So, since $\R^3=\R^2\times\R$, we see that $\R^3$ is proper.
\end{proof}

Using induction, we may show: $\forall k\in\N$, $\R^k$ is proper.

\begin{thm}\wrlab{thm-cpt-iff-proper-bdd}
Let $X$ be a metric space.\\
Then: \quad $($ $X$ is compact $)$ $\Leftrightarrow$ $($ $X$ is proper and bounded $)$.
\end{thm}

\begin{proof}
{\it Proof of $\Rightarrow$:}\\
Assume: $X$ is compact. \quad Want: $X$ is proper and bounded.\\
By \tref{thm-cpt-implies-bdd}, $X$ is bounded. \quad Want: $X$ is proper.\\
Want: $\forall s\in X_\bi^\N$, $s$ is subconvergent in $X$.\\
Given $s\in X_\bi^\N$. \qquad Want: $s$ is subconvergent in $X$.\\
Since $s\in X^\N$ and $X$ is compact, $s$ is subconvergent in $X$, as desired.\\
{\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:}\\
Assume: $X$ is proper and bounded. \quad Want: $X$ is compact.\\
Want: $\forall s\in X^\N$, $s$ is subconvergent in $X$.\\
Given $s\in X^\N$. \qquad Want: $s$ is subconvergent in $X$.\\
Since $X$ is bounded and $\I_s\subseteq X$, we get: $\I_s$ is bounded. \quad
Then $s\in X_\bi^\N$.\\
So, since $X$ is proper, we see that $s$ is subconvergent in $X$, as desired.
{\it End of proof of $\Leftarrow$.}
\end{proof}

\begin{defn}
Let $X$ be a metric space and let $a\in\R$.\\
By $X$ is $a$-\bfindex[bounded], we mean: \qquad $\forall p,q\in X$, $d(p,q)<a$.
\end{defn}

Note that a sequence in a metric space is Cauchy \quad iff\\
\phx\qquad $\forall\vep>0$, \qquad the sequence has an $\vep$-bounded tail,\\
by which we mean\\
\phx\qquad $\forall\vep>0$, \quad some tail of the sequence has $\vep$-bounded image.

\begin{thm}
Let $X$ be a metric space.\\
Then: \quad $($ $X$ is bounded $)$ $\Leftrightarrow$ $($ $\exists a\ge0$ s.t.~$X$ is $a$-bounded $)$.
\end{thm}

\begin{thm}\wrlab{thm-prod-bdd-is-bdd}
Let $X$ and $Y$ be bounded metric spaces.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,\,$ Then $X\times Y$ is bounded.
\end{thm}

\begin{proof}
Choose $a,b\ge0$ s.t.~$X$ is $a$-bounded and $Y$ is $b$-bounded.\\
Let $c:=\sqrt{a^2+b^2}$. \qquad Then $X\times Y$ is $c$-bounded.\\
Then $X\times Y$ is bounded.
\end{proof}

\begin{thm}\wrlab{thm-prod-cpt-is-cpt}
Let $X$ and $Y$ be compact metric spaces.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,\,$ Then $X\times Y$ is compact.
\end{thm}

\begin{proof}
By $\Rightarrow$ of \tref{thm-cpt-iff-proper-bdd}, $X$ and $Y$ are proper and bounded.\\
So, by \tref{thm-prod-proper-is-proper} and \tref{thm-prod-bdd-is-bdd},\\
\phx\qquad\qquad we see that:  $X\times Y$ is proper and bounded.\\
Then, by $\Leftarrow$ of \tref{thm-cpt-iff-proper-bdd}, we get: $X\times Y$ is compact.
\end{proof}

\subsection{Density of $\Q$ in $\R$}

\begin{defn}
Let $X$ be a metric space, $T\subseteq X$, $\vep>0$.\\
By $T$ is $\vep$-\bfindex[net] in $X$, we mean: $\forall w\in X$, $\exists s\in T$ s.t.~$d(w,s)<\vep$.
\end{defn}

\begin{thm}
$\Z$ is $1$-net in $\R$.
\end{thm}

\begin{proof}
Want: $\forall w\in\R$, $\exists s\in\Z$ s.t.~$d(w,s)<\vep$.\\
Given $w\in\R$. \qquad
Want: $\exists s\in\Z$ s.t.~$d(w,s)<\vep$.\\
By the Archimedean Principle, choose $j\in\N$ s.t.~$j>1-w$.\\
Then $w+j>1$. \quad
Let $x:=w+j$. \quad
Then $x>1$. \quad
Also, $x-j=w$.\\
Let $A:=\{i\in\N\,|\,i>x\}$. \qquad
By the Archimedan Principle, $A\ne\emptyset$.\\
So, since $A\subseteq\N$, by the Well-Ordering Axiom,
we have: $\min\,A\ne\frownie$. \quad
Then $\min\,A\in A$. \qquad
Let $k:=\min\,A$. \qquad
Then $k\in A$.\\
So, by definition of $A$, we have: $k\in\N$ and $k>x$.\\
Since $k>x>1$ and $k\in\N$, we get $k\in(1..\infty)$, so $k-1\in(0..\infty)$.\\
Since $k-1<k=\min A\le A$, we get: $k-1\notin A$.\\
So, since $k\in(0..\infty)=\N$, we get: $\neg(k-1>x)$. \quad
Then $k-1\le x$.\\
Let $s:=k-j-1$. \quad
As $j,k\in\N\subseteq\Z$, $s\in\Z$. \quad
Want: $d(w,s)<1$.\\
Since $k-1\le x<k$, we get: $k-j-1\le x-j<k-j$.\\
So, since \quad $k-j-1=s$ \quad and \quad $x-j=w$ \quad and \quad $k-j=s+1$,\\
\phx\qquad\qquad\qquad\qquad $\,\,$ we get \qquad\qquad $\,\,\,s\le w<s+1$,\\
\phx\qquad\qquad\qquad\qquad $\,\,$ and so \qquad\qquad $\,\,\,0\le w-s<1$.\\
Since $w-s\ge0$, we get: $|w-s|=w-s$.\\
Then $d(w,s)=|w-s|=w-s<1$, as desired.
\end{proof}

\begin{thm}
Let $X$ be a metric space, $U\subseteq X$, $T\subseteq U$, $\vep>0$.\\
Assume $T$ is $\vep$-net in $X$. \qquad
Then $U$ is $\vep$-net in $X$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{defn}
Let $X$ be a metric space, $T\subseteq X$.\\
By $T$ is \bfindex[dense] in $X$, we mean: $\Cl_XT=X$.
\end{defn}

\begin{thm}
Let $X$ be a metric space, $U\subseteq X$, $T\subseteq U$.\\
Assume $T$ is dense in $X$. \qquad
Then $U$ is dense in $X$.
\end{thm}

\begin{proof}
Want: $\Cl_XU=X$. \quad
Since $\Cl_XU\subseteq X$, we want: $X\subseteq\Cl_XU$.\\
Since $T\subseteq U$, $\Cl_XT\subseteq\Cl_XU$. \quad
Since $T$ is dense in $X$, $\Cl_XT=X$.\\
Then $X=\Cl_XT\subseteq\Cl_XU$, as desired.
\end{proof}

\begin{thm}
Let $X$ be a metric space and let $A\subseteq X$.\\
\phx\qquad Then \qquad $\Cl_X^\seq\,A\,\,\,:=\,\,\,\{\,\,p\in X\,\,|\,\,\exists s\in A^\N~\hbox{s.t.}~s\to p\hbox{ in }X\,\,\,\}$.
\end{thm}

\begin{thm}\wrlab{thm-set-in-seq-closure}
Let $X$ be a metric space and let $A\subseteq X$.\\
\phx\qquad Then \qquad $A\,\,\,\subseteq\,\,\,\Cl_X^\seq A$.
\end{thm}

\begin{proof}
Want: $\forall p\in A$, $p\in\Cl_X^\seq A$. \quad
Given $p\in A$. $\,$
Want: $p\in\Cl_X^\seq A$.\\
We have: $C_\N^p\to p$ in $X$. \qquad
Want: $\exists s\in A^\N$ s.t.~$s\to p$ in $X$.\\
Let $s:=C_\N^p$. \quad
Since $p\in A$, we get $s\in A^\N$. \quad
Want $s\to p$ in $X$.\\
Since $C_\N^p\to p$ in $X$ and since $s=C_\N^p$, we get:
$s\to p$ in $X$ as desired.
\end{proof}

\begin{thm}\wrlab{thm-seq-bdry-in-seq-closure}
Let $X$ be a metric space and let $A\subseteq X$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $\partial_X^\seq X\,\,\,\subseteq\,\,\,\Cl_X^\seq A$.
\end{thm}

\begin{proof}
Want: $\forall p\in\partial_X^\seq X$, $p\in\Cl_X^\seq A$.\\
Given $p\in\partial_X^\seq X$. \qquad Want: $p\in\Cl_X^\seq A$.\\
Want: $\exists s\in A^\N$ s.t.~$s\to p$ in $X$.\\
Since $p\in\partial_X^\seq X$, we know:\\
\phx\qquad $\,\,\exists s\in A^\N$ s.t.~$s\to p$ in $X$ \quad and \quad $\exists t\in(X\bksl A)^\N$ s.t.~$t\to p$ in $X$.\\
Then: $\exists s\in A^\N$ s.t.~$s\to p$ in $X$, \quad as desired.
\end{proof}

\begin{thm}\wrlab{thm-seq-cl-is-cl}
Let $X$ be a metric space and let $A\subseteq X$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $\Cl_X^\seq X\,\,\,=\,\,\,\Cl_XA$.
\end{thm}

\begin{proof}
By HW\#11-5, $\Cl_X^\seq X\,\subseteq\,\Cl_XA$. $\,\,\,$
Want: $\Cl_XA\,\subseteq\,\Cl_X^\seq A$.\\
By \tref{thm-set-in-seq-closure}, we have: \qquad\qquad\qquad $\,\,\,\,A\subseteq\Cl_X^\seq A$.\\
By \tref{thm-boundary-equals-seqvbndry}, we have: \quad $\,\,\,\,\,\partial_XA=\partial_X^\seq A$.\\
By \tref{thm-seq-bdry-in-seq-closure}, we have: \quad\qquad\qquad $\,\partial_X^\seq A\subseteq\Cl_X^\seq A$.\\
Since $A\subseteq\Cl_X^\seq A$ and $\partial_XA=\partial_X^\seq A\subseteq\Cl_XA$, we get:\\
\phx\qquad\qquad\qquad $\!A\bigcup(\partial_XA)\subseteq\Cl_X^\seq A$.\\
Then $\Cl_XA=A\bigcup(\partial_XA)\subseteq\Cl_X^\seq A$, as desired.
\end{proof}

\subsection{Compact vs closed and bounded}

\begin{thm}\wrlab{thm-relative-ball}
Let $Y$ be a metric space, $X\subseteq Y$, $p\in X$, $r>0$.\\
Then $B_X(p,r)=(B_Y(p,r))\cap X$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

The next result is \bfindex[Absoluteness of Bounded]:

\begin{thm}
Let $Y$ be a metric space, $X\subseteq Y$, $K\subseteq X$.\\
Then: $($ $K$ is bounded in $X$ $)$ $\Leftrightarrow$ $($ $K$ is bounded in $Y$ $)$.
\end{thm}

\begin{proof}
{\it Proof of $\Rightarrow$:} Unassigned HW. {\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:}\\
Assume $K$ is bounded in $Y$. \quad
Want: $K$ is bounded in $X$.\\
Assume $K$ is not bounded in $X$. \quad
Want: Contradiction.\\
Since $K$ is not bounded in $X$, while $\emptyset$ is bounded in $X$,\\
\phx\qquad we conclude that $K\ne\emptyset$.\\
So, since $K$ is bounded in $Y$, choose $B\in\scrb_Y$ s.t.~$K\subseteq B$.\\
Since $X\supseteq K\ne\emptyset$, we get: $X\ne\emptyset$, \qquad so choose $p\in X$.\\
By the Superset Recentering Theorem, choose $C\in\scrb_Y(p)$ s.t.~$B\subseteq C$.\\
Since $C\in\scrb_Y(p)$, choose $r>0$ s.t.~$C=B_Y(p,r)$.\\
Let $A:=B_X(p,r)$. \qquad
By \tref{thm-relative-ball}, we have: $A=C\cap X$.\\
By hypothesis, we have: $\,K\subseteq X$.\\
Since $\,\,K\subseteq B\subseteq C\,\,$ and $\,\,K\subseteq X\,\,$, we get: \quad
$K\subseteq C\cap X$.\\
Then $K\subseteq C\cap X=A$.\\
So, since $A\in\scrb_X(p)\subseteq\scrb_X$, we see that $K$ is bounded in $X$.\\
Recall that $K$ is not bounded in $X$. \qquad
Contradiction.\\
{\it End of proof of $\Rightarrow$.}
\end{proof}

The next result is called \bfindex[Absoluteness of Limit]:

\begin{thm}\wrlab{thm-abs-limit}
Let $Y$ be a metric space and let $X\subseteq Y$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Let $s\in X^\N$ and let $q\in X$.\\
\phx\quad Then: \quad $($ $s\to q$ in $X$ $)$ $\quad\Leftrightarrow\quad$ $($ $s\to q$ in $Y$ $)$.
\end{thm}

\begin{proof}
{\it Proof of $\Rightarrow$:} Unassigned HW. {\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:}\\
Assume $s\to q$ in $Y$. \quad Want: $s\to q$ in $X$.\\
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_j,q)<\vep\,\,)$.\\
Given $\vep>0$. \qquad Want: $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_j,q)<\vep\,\,)$.\\
Since $s$ is Cauchy in $Y$, choose $K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(s_j,q)<\vep\,\,)$.\\
Then $K\in\N$. \qquad
Want: $\forall j\in\N$,  $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_j,q)<\vep\,\,)$.\\
Given $j\in\N$. \qquad Want: $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_j,q)<\vep\,\,)$.\\
Assume $j\ge K$. \qquad Want: $d_X(s_j,q)<\vep$.\\
Since $s\in X^\N$, we get $s_j\in X$. \quad By hypothesis, we have $q\in X$.\\
Since $s_j,q\in X$, we conclude that $d_X(s_j,q)=d_Y(s_j,q)$.\\
Then $d_X(s_j,q)=d_Y(s_j,q)<\vep$, as desired.\\
{\it End of proof of $\Leftarrow$.}
\end{proof}

\begin{thm}\wrlab{thm-conv-subspace-implies-conv-ambient}
Let $Y$ be a metric space, $X\subseteq Y$, $s\in X^\N$.\\
Assume $s$ is convergent in $X$. \quad
Then $s$ is convergent in $Y$.
\end{thm}

\begin{proof}
Sicne $s$ is convergent in $X$, choose $q\in X$ s.t.~$s\to q$ in $X$.\\
Then, by Absoluteness of Limit, we have: $s\to q$ in $Y$.\\
So, since $q\in X\subseteq Y$, we get: $s$ is convergent in $Y$, as desired.
\end{proof}

\begin{thm}
Let \quad $Y:=\R$ \quad and let \quad $X:=(0;\infty)$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Define $s\in Y^\N$ by: \qquad $\forall j\in\N$, $\,$ $s_j=1/j$.\\
Then \quad $s$ is convergent in $Y$ \quad and \quad $s$ is not convergent in $X$.
\end{thm}

\begin{defn}
Let $X$ be a metric space.\\
Then $\scrt_X:=\{U\subseteq X\,|\,U\hbox{ is open in }X\}$ \qquad and\\
\phx\qquad $\scrt'_X:=\{C\subseteq X\,|\,C\hbox{ is closed in }X\}$.
\end{defn}

\begin{thm}\wrlab{thm-conv-closed-subspace-iff-conv-ambient}
Let $Y$ be a metric space, $X\in\scrt'_Y$, $s\in X^\N$.\\
Then: \quad $(\,\,s\hbox{ is convergent in }X\,\,)\,\,\Leftrightarrow\,\,(\,\,s\hbox{ is convergent in }Y\,\,)$.
\end{thm}

\begin{proof}
By \tref{thm-conv-subspace-implies-conv-ambient}, we have:\\
\phx\qquad\quad $\,\,\,\,(\,\,s\hbox{ is convergent in }X\,\,)\,\,\Rightarrow\,\,(\,\,s\hbox{ is convergent in }Y\,\,)$.\\
Want: $\,\,\,\,\,\,(\,\,s\hbox{ is convergent in }X\,\,)\,\,\Leftarrow\,\,(\,\,s\hbox{ is convergent in }Y\,\,)$.\\
Assume: $s$ is convergent in $Y$. \quad
Want: $s$ is convergent in $X$.\\
Since $s$ is convergent in $Y$, choose $q\in Y$ s.t.~$s\to q$ in $Y$.\\
Since \quad $s\in X^\N$ \quad and \quad $s\to q$ in $Y$, \quad we get: \quad $q\in\Cl_Y^\seq X$.\\
Since $X\in\scrt'_Y$, we see that $X$ is closed in $Y$, so $\Cl_YX=X$.\\
By \tref{thm-seq-cl-is-cl}, $\Cl_Y^\seq X=\Cl_YX$. \quad
Then $q\in\Cl_Y^\seq X=\Cl_YX=X$.\\
Since \qquad $s\to q$ in $Y$ \qquad and \qquad $s\in X^\N$ \qquad and \qquad $q\in X$,\\
\phx\quad it follows, by Absoluteness of Limit, that: $s\to q$ in $X$.
\end{proof}

\begin{thm}\wrlab{thm-subconv-closed-subspace-iff-subconv-ambient}
Let $Y$ be a metric space, $X\in\scrt'_Y$, $s\in X^\N$.\\
Then: \quad $(\,\,s\hbox{ is subconvergent in }X\,\,)\,\,\Leftrightarrow\,\,(\,\,s\hbox{ is subconvergent in }Y\,\,)$.
\end{thm}

\begin{proof}
{\it Proof of $\Rightarrow$:} Unassigned HW. {\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:}\\
Want: \quad $(\,\,s\hbox{ is subconvergent in }X\,\,)\,\,\Leftarrow\,\,(\,\,s\hbox{ is subconvergent in }Y\,\,)$.\\
Assume: $s$ is subconvergent in $Y$. \qquad
Want: $s$ is subconvergent in $X$.\\
Since $s$ is subconvergent in $Y$,\\
\phx\qquad choose a subsequence $t$ of $s$ \quad s.t. \quad $t$ is convergent in $Y$.\\
Since \quad $\D_t=\N$ \quad and \quad $\I_t\subseteq\I_s\subseteq X$, \quad we get: $t\in X^\N$.\\
Then, by \tref{thm-conv-closed-subspace-iff-conv-ambient}, we conclude: \quad $t$ is convergent in $X$.\\
So, since $t$ is a subsequence of $s$, we see that\\
\phx\qquad $s$ is subconvergent in $X$, \qquad as desired.\\
{\it End of proof of $\Rightarrow$.}
\end{proof}

\begin{thm}\wrlab{thm-cpt-implies-closed-bounded}
Let $X$ be a metric space and let $K\subseteq X$.\\
\phx\quad Assume $K$ is compact. \qquad
Then $K$ is closed and bounded in $X$.
\end{thm}

\begin{proof}
Since $K$ is compact, by \tref{thm-cpt-implies-bdd}, $K$ is bounded.\\
Then $K$ is bounded in $K$, so, by Absoluteness of Bounded,\\
\phx\qquad $K$ is bounded in $X$. \quad
Want: $K$ is closed in $X$.\\
Want: $\Cl_XK=K$. \qquad
Since $K\subseteq\Cl_XK$, we want: $\Cl_XK\subseteq K$.\\
Want: $\forall p\in\Cl_XK$, $p\in K$. \qquad
Given $p\in\Cl_XK$. \qquad
Want: $p\in K$.\\
Since $p\in\Cl_XK=\Cl_X^\seq K$, choose $s\in K^\N$ s.t.~$s\to p$ in $X$.\\
Since $s\in K^\N$ and since $K$ is compact, \quad
$s$ is subconvergent in $K$,\\
\phx\qquad so choose a subsequence $t$ of $s$ s.t.~$t$ is convergent in $K$.\\
Since $t$ is convergent in $K$, choose $q\in K$ s.t.~$t\to q$ in $K$.\\
Since $q\in K$ it suffices to show: $p=q$.\\
Since \quad $s\to p$ in $X$ \quad and \quad since $t$ is a subsequence of $s$,\\
\phx\qquad by \tref{thm-conv-to-q-implies-subseq-conv-to-q}, we get: \quad $t\to p$ in $X$.\\
Since \quad $t\to p$ in $X$ \quad and \quad $t\to q$ in $X$,\\
\phx\qquad by \tref{thm-uniqueness-of-limits}, we get: \quad $p=q$, \qquad as desired.
\end{proof}

\begin{thm}\wrlab{thm-seq-in-bdd-is-bddim}
Let $X$ be a metric space and let $K\subseteq X$.\\
\phx\qquad Assume $K$ is bounded in $X$. \qquad
Then $K^\N\subseteq X_\bi^\N$.
\end{thm}

\begin{proof}
Want: $\forall s\in K^\N$, $s\in X_\bi^\N$. \quad
Given $s\in K^\N$. \quad
Want: $s\in X_\bi^\N$.\\
Since $s\in K^\N\subseteq X^\N$, \qquad we want: \quad $\I_s$ is bounded in $X$.\\
Since \quad $\I_s\subseteq K$ \quad and since \quad $K$ is bounded in $X$,\\
\phx\qquad we conclude that $\I_s$ is bounded in $X$, \qquad as desired.
\end{proof}

\begin{thm}\wrlab{thm-inproper-cpt-iff-closed-bounded}
Let $X$ be a proper metric space and let $K\subseteq X$.\\
\phx\qquad Then: \quad $(\,\,K\hbox{ is compact}\,\,)\,\,\Leftrightarrow\,\,(\,\,K\hbox{ is closed and bounded in }X\,\,)$.\\
\end{thm}

\begin{proof}
By \tref{thm-cpt-implies-closed-bounded}, we have:\\
\phx\qquad\quad $\,\,\,\,(\,\,K\hbox{ is compact}\,\,)\,\,\Rightarrow\,\,(\,\,K\hbox{ is closed and bounded in }X\,\,)$.\\
Want: $\,\,\,\,\,\,(\,\,K\hbox{ is compact}\,\,)\,\,\Leftarrow\,\,(\,\,K\hbox{ is closed and bounded in }X\,\,)$.\\
Assume: $K$ is closed and bounded in $X$. \quad
Want: $K$ is compact.\\
Want: \quad $\forall s\in K^\N$, $\,$ $s$ is subconvergent in $K$.\\
Given $s\in K^\N$. \qquad
Want: $s$ is subconvergent in $K$.\\
Since $K$ is closed in $X$, we get $K\in\scrt'_X$.\\
So, by \tref{thm-subconv-closed-subspace-iff-subconv-ambient},
it suffices to show: \quad $s$ is subconvergent in $X$.\\
Since $K$ is bounded in $X$, by \tref{thm-seq-in-bdd-is-bddim}, we get: \quad $K^\N\subseteq X_\bi^\N$.\\
Since \quad $s\in K^\N\subseteq X_\bi^\N$ \quad and since \quad $X$ is proper,\\
\phx\qquad it follows that: \quad $s$ is subconvergent in $X$, \qquad as desired.
\end{proof}

\subsection{The unit circle is compact}

\begin{defn}
Let $X$ be a metric space and let $z\in X$.\\
Then $d_X(z,\bullet):X\to[0;\infty)$ is defined by:\\
\phx\qquad\qquad\qquad $\forall p\in X$, \quad $(d_X(z,\bullet))_p=d_X(z,p)$.
\end{defn}

By sloppiness, we sometimes write $d(z,\bullet)$ for $d_X(z,\bullet)$.

\begin{thm}\wrlab{thm-dist-to-pt-lip1}
Let $X$ be a metric space and let $z\in X$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then $d_X(z,\bullet)$ is Lipschitz-1 from $X$ to $\R$.
\end{thm}

\begin{proof}
Let $f:=d_X(z,\bullet)$. \quad
Want: $f$ is Lipschitz-1 from $X$ to $\R$.\\
Want: \quad $\forall p,q\in\D_f$, $\,$ $d_\R(f_p,f_q)\le d_X(p,q)$.\\
Given $p,q\in\D_f$. \qquad
Want: $d_\R(f_p,f_q)\le d_X(p,q)$.\\
Since $f=d_X(z,\bullet)$, we get $\D_f=X$. \quad
Then $p,q\in\D_f=X$.\\
Let $\vep:=d_X(p,q)$. \qquad
Want: $d_\R(f_p,f_q)\le\vep$.\\
Want: $(\,\,f_p\le f_q+\vep\,\,)\,\,\&\,\,(\,\,f_q\le f_p+\vep\,\,)$.\\
We have $f_p=(d_X(z,\bullet)_p=d_X(z,p)\le(d_X(z,q))+(d_X(q,p))$\\
\phx\qquad\qquad $\,\,\,\,=(d_X(z,\bullet))_q+(d_X(p,q))=f_q+\vep$. \quad
Want: $f_q\le f_p+\vep$.\\
We have $f_q=(d_X(z,\bullet)_q=d_X(z,q)\le(d_X(z,p))+(d_X(p,q))$\\
\phx\qquad\qquad $\,\,\,\,=(d_X(z,\bullet))_p+(d_X(p,q))=f_p+\vep$, \quad as desired.
\end{proof}

\begin{thm}\wrlab{thm-contin-implies-preim-open-is-open}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad Let $f:X\to Y$ and let $U\in\scrt_Y$.\\
Assume that $f$ is continuous from $X$ to $Y$. \qquad
Then $f^*U\in\scrt_X$.
\end{thm}

\begin{proof}
Want: $f^*U$ is open in $X$. \qquad
Want: $\Int_X(f^*U)=f^*U$.\\
We have: $\Int_X(f^*U)\subseteq f^*U$. \qquad
Want: $f^*U\subseteq \Int_X(f^*U)$.\\
Want: \quad $\forall p\in f^*U$, $\,$ $p\in\Int_X(f^*U)$.\\
Given $p\in f^*U$. \qquad
Want: $p\in\Int_X(f^*U)$.\\
Want: $\exists A\in\scrb_X(p)$ s.t.~$A\subseteq f^*U$.\\
Since $U\in\scrt_Y$, we see that $U$ is open in $Y$, so $\Int_YU=U$.\\
Since $p\in f^*U$, we get: $f_p\in U$. \quad
Then $f_p\in U=\Int_YU$.\\
Since $f_p\in\Int_YU$, choose $B\in\scrb_Y(f_p)$ s.t.~$B\subseteq U$.\\
Since $B\in\scrb_Y(f_p)$, choose $\vep>0$ s.t.~$B=B(f_p,\vep)$.\\
By hypothesis, $f$ is continuous from $X$ to $Y$,\\
\phx\qquad so $f$ is continuous on $\D_f$ from $X$ to $Y$.\\
As $p\in f^*U$, we get: $p\in\D_f$. \quad
Then $f$ is continuous at $p$ from $X$ to $Y$,\\
\phx\qquad so choose $\delta>0$ s.t.~$\forall q\in\D_f$, $\,$ $(\,\,d(p,q)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(f_p,f_q)<\vep\,\,)$.\\
Let $A:=B_X(p,\delta)$. \qquad
Then $A\in\scrb_X(p)$. \qquad
Want: $A\subseteq f^*U$.\\
Want: \quad $\forall q\in A$, \, $q\in f^*U$. \qquad\qquad
Given $q\in A$. \quad
Want: $q\in f^*U$.\\
Since $q\in A=B_X(p,\delta)$, we see that $q\in X$ and that $d(p,q)<\delta$.\\
Since $f:X\to Y$, we get $\D_f=X$. \qquad
Then $q\in X=\D_f$.\\
Since $q\in\D_f$ and since $d(p,q)<\delta$, \quad by choice of $\delta$,\\
\phx\qquad we see that $d(f_p,f_q)<\vep$, \quad and so $f_q\in B_Y(f_p,\vep)$.\\
Then $f_q\in B_Y(f_p,\vep)=B\subseteq U$, and so $q\in f^*U$, \quad as desired.
\end{proof}

\begin{thm}
Let $f$ be a function.
Let $A$ and $B$ be sets.\\
Then \qquad\quad $f^*(A\bigcup B)=(f^*A)\bigcup(f^*B)$ \qquad\qquad and\\
\phx\qquad\qquad\quad $\,f^*(A\bigcap B)=(f^*A)\bigcap(f^*B)$ \qquad\qquad and\\
\phx\qquad\qquad\quad $\,f^*(A\bksl B)=(f^*A)\bksl(f^*B)$.\\
Also, \qquad\quad $f_*(A\bigcup B)=(f_*A)\bigcup(f_*B)$.
\end{thm}

\begin{thm}\wrlab{thm-contin-implies-preim-closed-is-closed}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad Let $f:X\to Y$ and let $C\in\scrt'_Y$.\\
Assume that $f$ is continuous from $X$ to $Y$. \qquad
Then $f^*C\in\scrt'_X$.
\end{thm}

\begin{proof}
Let $U:=Y\bksl C$. \quad
Since $C\in\scrt'_Y$, we get: $U\in\scrt_Y$.\\
Then, by \tref{thm-contin-implies-preim-open-is-open}, we get: $f^*U\in\scrt_X$.\\
It follows that $X\bksl(f^*U)\in\scrt'_X$. \qquad
Want: $f^*C=X\bksl(f^*U)$.\\
As $C\subseteq Y$, we get: $Y\bksl(Y\bksl C)=C$. $\,$
As $f:X\to Y$, we get: $f^*Y=X$.\\
We have $Y\bksl U=Y\bksl(Y\bksl C)=C$, \quad and so $C=Y\bksl U$.\\
Then $f^*C=f^*(Y\bksl U)=(f^*Y)\bksl(f^*U)=X\bksl(f^*U)$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-preim-open-is-open-iff-contin}
Let $X$ and $Y$ be metric spaces, $f:X\to Y$.\\
Then: $($ $\forall U\in\scrt_Y$, $f^*U\in\scrt_X$ $)$ $\Leftrightarrow$ $($ $f$ is continuous from $X$ to $Y$ $)$.
\end{thm}

\begin{proof}
By \tref{thm-contin-implies-preim-open-is-open}, we have $\Leftarrow$. \quad
Want: $\Rightarrow$.\\
By HW\#12-2, we have $\Rightarrow$.
\end{proof}

\begin{thm}\wrlab{thm-preim-closed-is-closed-implies-contin}
Let $X$ and $Y$ be metric spaces, $f:X\to Y$.\\
Then: $($ $\forall C\in\scrt'_Y$, $f^*C\in\scrt'_X$ $)$ $\Leftrightarrow$ $($ $f$ is continuous from $X$ to $Y$ $)$.
\end{thm}

\begin{proof}
By \tref{thm-contin-implies-preim-closed-is-closed}, we have $\Leftarrow$. \quad
Want: $\Rightarrow$.\\
Assume: $\forall C\in\scrt'_Y$, $f^*C\in\scrt'_X$. \quad
Want: $f$ is continuous from $X$ to $Y$.\\
Then, by HW\#12-2, it suffices to show: $\forall U\in\scrt_Y$, $f^*U\in\scrt_X$.\\
Given $U\in\scrt_Y$. \qquad Want: $f^*U\in\scrt_X$.\\
Since $U\in\scrt_Y$, we get: $Y\backslash U\in\scrt'_Y$. \quad
Let $C:=Y\backslash U$. \quad
Then $C\in\scrt'_Y$.\\
Then $f^*C\in\scrt'_X$, so $X\backslash(f^*C)\in\scrt_X$. \quad
Want: $X\backslash(f^*C)=f^*U$.\\
Since $C=Y\backslash U$, we get $f^*C=(f^*Y)\backslash(f^*U)$.\\
So, since $f^*Y=X$, we get $f^*C=X\backslash(f^*U)$.\\
Since $f^*U\subseteq X$, we conclude that: $X\backslash(X\backslash(f^*U))=f^*U$.\\
Then $X\backslash(f^*C)=X\backslash(X\backslash(f^*U))=f^*U$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-singletons-closed}
Let $X$ be a metric space, $p\in X$.
Then $\{p\}\in\scrt'_X$.
\end{thm}

\begin{proof}
Let $A:=\{p\}$. \qquad
Want: $A\in\scrt'_X$.\\
Want: $A$ is closed in $X$. \qquad
Want: $\Cl_XA=A$.\\
We have: $A\subseteq\Cl_XA$. \qquad
Want: $\Cl_XA\subseteq A$.\\
Want: $\,$ $\forall q\in\Cl_XA$, $\,q\in A$. \qquad
Given $q\in\Cl_XA$. \qquad Want: $q\in A$.\\
We have $q\in\Cl_XA=\Cl_X^\seq A$, \quad
so choose $s\in A^\N$ s.t.~$s\to q$ in $X$.\\
We have: $\forall j\in\N$, $s_j\in A=\{p\}$, so $s_j=p$. \qquad
Then $s=C_\N^p$.\\
So, since $C_\N^p\to p$ in $X$, we see that $s\to p$ in $X$.\\
So, since $s\to q$ in $X$, we get: $p=q$. \qquad
Then $q=p\in\{p\}=A$.
\end{proof}

\begin{thm}
Let $X:=\R^2$ and let $C:=S_X(0_X,1)$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then $C$ is compact.
\end{thm}

\begin{proof}
Since $X$ is proper, by \tref{thm-inproper-cpt-iff-closed-bounded},\\
\phx\qquad it suffices to show: \quad $C$ is closed and bounded in $X$.\\
Since $C=S_X(0_X,1)\subseteq B_X(0_x,2)$, we see that $C$ is bounded in $X$.\\
Want: $C$ is closed in $X$. \qquad
Want: $C\in\scrt'_X$. \qquad
Let $f:=d_X(0_X,\bullet)$.\\
By \tref{thm-dist-to-pt-lip1}, \quad $\,\,\,\,\,$ $f$ is Lipschitz-$1$ from $X$ to $\R$,\\
\phx\quad and it follows that: \quad $f$ is continuous from $X$ to $\R$.\\
By \tref{thm-singletons-closed}, we get: $\{1\}\in\scrt'_\R$.\\
Then, by \tref{thm-contin-implies-preim-closed-is-closed}, we get: $f^*(\{1\})\in\scrt'_X$. \quad
Want: $f^*(\{1\})=C$.\\
We have \quad $f^*(\{1\})\,\,\,=\,\,\,\{\,\,p\in X\,\,|\,\,f_p\in\{1\}\,\,\}$\\
\phx\qquad\qquad\qquad\qquad $\,\,\,=\,\,\,\{\,\,p\in X\,\,|\,\,f_p=1\,\,\}$\\
\phx\qquad\qquad\qquad\qquad $\,\,\,=\,\,\,\{\,\,p\in X\,\,|\,\,(d(0_X,\bullet))_p=1\,\,\}$\\
\phx\qquad\qquad\qquad\qquad $\,\,\,=\,\,\,\{\,\,p\in X\,\,|\,\,d(0_X,p)=1\,\,\}$\\
\phx\qquad\qquad\qquad\qquad $\,\,\,=\,\,\,S_X(0_X,1)\,\,\,=\,\,\,C$, as desired.
\end{proof}

\subsection{The Extreme Value Theorem}

\begin{thm}\wrlab{thm-x-biggest-in-A-implies-x-max-A}
Let $A\subseteq\R$ and let $x\in\R$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Assume: $A\le x\in A$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: $x=\max\,A$.
\end{thm}

\begin{proof}
Want: $x=\UE(A\bigcap\UB_A)$. \qquad
Want: $A\bigcap\UB_A=\{x\}$.\\
As $A\le x$, we get: $X\in\UB_A$. \quad
So, since $x\in A$, we get: $x\in A\bigcap\UB_A$.\\
Then $\{x\}\subseteq A\bigcap\UB_A$. \qquad
Want: $A\bigcap\UB_A\subseteq\{x\}$.\\
Given $y\in A\bigcap\UB_A$. \qquad Want: $y\in\{x\}$. \qquad Want: $\,y=x$.\\
Since $y\in A\le x$, we get: $y\le x$. \qquad\qquad\qquad
Want: $\,x\le y$.\\
Since $y\in\UB_A$, we get: $A\le y$. \quad
Then $x\in A\le y$, so $x\le y$.
\end{proof}

\begin{thm}\wrlab{thm-max-cpt-ne-frownie}
Let $K\subseteq\R$. Say $K$ is compact and nonempty.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: $\max\,K\ne\frownie$.
\end{thm}

\begin{proof}
Since $K$ is compact, by \tref{thm-cpt-implies-closed-bounded}, we conclude:\\
\phx\qquad\qquad\qquad\quad $K$ is closed and bounded in $\R$.\\
By hypothesis, we have: $K\ne\emptyset$.\\
So, since $K$ is bounded in $\R$, by \tref{thm-nonempty-bdd-implies-realsup}, we get: $\sup\,K\in\R$.\\
Let $x:=\sup\,K$. \quad Then $x\in\R$, so $x\ne\frownie$. \qquad Want: $x=\max\,K$.\\
By \tref{thm-x-biggest-in-A-implies-x-max-A}, it suffices to show: $K\le x\in K$.\\
We have $K\le\sup\,K=x$, and so $K\le x$. \qquad\qquad\quad
Want: $x\in K$.\\
Since $K$ is closed in $\R$, we get: $\Cl_\R K=K$.\\
Want: $x\in\Cl_\R K$. \qquad
Want: $\forall B\in\scrb_\R(x)$, $B\cap K\ne\emptyset$.\\
Given $B\in\scrb_\R(x)$. \quad Want: $B\cap K\ne\emptyset$. \quad
Want: $\exists y\in K$ s.t.~$y\in B$.\\
Since $B\in\scrb_\R(x)$, choose $\vep>0$ s.t.~$B=B_\R(x,\vep)$.\\
We have $x>x-\vep$, so $\neg(x\le x-\vep)$.\\
So, as $x=\sup\,K$, we get $\neg(\sup\,K\le x-\vep)$.\\
Then $\neg(K\le x-\vep)$, so choose $y\in K$ s.t.~$y>x-\vep$.\\
Then $y\in K$. \qquad Want: $y\in B$.\\
We have $y\in K\le\sup\,K=x<x+\vep$, so $y<x+\vep$.\\
Then $x-\vep<y<x+\vep$, so $y\in(x-\vep;x+\vep)$.\\
Then $y\in(x-\vep;x+\vep)=B_\R(x,\vep)=B$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-im-cpt-under-contin-is-cpt}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Let $f:X\drta Y$ and let $K\subseteq\D_f$.\\
\phx\qquad Assume: $\,(\,\,K\hbox{ is compact}\,\,)\,\,\&\,\,(\,\,f\hbox{ is continuous from }X\hbox{ to }Y\,\,)$.\\
\phx\qquad\quad Then: $f_*K$ is compact.
\end{thm}

\begin{proof}
Let $g:=f|K$. \quad
By \tref{thm-restr-contin},\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $g$ is continuous from $X$ to $Y$.\\
So, since $g:K\to\I_g$, by Absoluteness of Continuity,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $g$ is continuous from $K$ to $\I_g$.\\
So, since $g:K\to>\I_g$, by \tref{thm-continsurj-fromcpt-image-cpt},
we have: $\I_g$ is compact.\\
It suffices to show: $\I_g=f_*K$. \quad
Since $K\subseteq\D_f$, we get $K\cap\D_f=K$.\\
Since $g=f|K$, we conclude: $(\,\,\D_g=K\,\,)\,\,\&\,\,(\,\,\forall p\in K$, $f_p=g_p\,\,)$.\\
Then $\I_g\,\,=\,\,\{\,\,g_p\,\,|\,\,p\in\D_g\,\,\}\,\,=\,\,\{\,\,g_p\,\,|\,\,p\in K\,\,\}\,\,=\,\,\{\,\,f_p\,\,|\,\,p\in K\,\,\}$\\
\phx\qquad\quad $\,\,\,=\,\,\{\,\,f_p\,\,|\,\,p\in K\cap\D_f\,\,\}\,\,=\,\,f_*K$, \qquad as desired.
\end{proof}

\begin{thm}\wrlab{thm-min-from-max}
Let $K\subseteq\R$. \qquad Then $\min\,K=-(\max(-K))$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}\wrlab{thm-maxmin-cpt-ne-frownie}
Let $K\subseteq\R$. Say $K$ is compact and nonempty.
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: $\min\,K\ne\frownie\ne\max\,K$.
\end{thm}

\begin{proof}
By \tref{thm-max-cpt-ne-frownie}, $\frownie\ne\max\,K$. \qquad
Want: $\min\,K\ne\frownie$.\\
Define $f:\R\to\R$ by $\forall x\in\R$, $f_x=-x$.\\
Then \quad $f$ is continuous from $\R$ to $\R$ \quad and \quad $K\subseteq\R=\D_f$.\\
Then, by \tref{thm-im-cpt-under-contin-is-cpt}, $f_*K$ is compact.\\
So, since $f_*K=-K$, we see that $-K$ is compact.\\
Then, by \tref{thm-max-cpt-ne-frownie}, we see that $\max(-K)\ne\frownie$,\\
\phx\qquad\qquad\qquad\qquad and it follows that $\max(-K)\in-K$.\\
By \tref{thm-min-from-max}, we have: $\min\,K=-(\max(-K))$.\\
Then $\min\,K=-(\max(-K))\in-(-K)$, so $\min\,K\ne\frownie$.
\end{proof}

\begin{thm}\wrlab{thm-seq-ge-implies-lim-ge}
Let $s\in\R^\N$ and let $a,p\in\R$.\\
Assume that \quad $s\to p$ in $\R$ \quad and that \quad $\forall j\in\N$, $s_j\ge a$.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\,$ Then $p\ge a$.\\
\end{thm}

\begin{proof}
Assume $p<a$. \qquad Want: Contradiction.\\
Let $\vep:=a-p$. \quad Then $\vep>0$.\\
Since $s\to p$ in $\R$, choose $K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_j,p)<\vep\,\,)$.\\
Then $d(s_K,p)<\vep$, so $p-\vep<s_k<p+\vep$.\\
By hypothesis, we have: $\forall j\in\N$, $s_j\ge a$. \quad Then $s_K\ge a$.\\
Then $a\le s_K<p+\vep=p+(a-p)=a$, so $a<a$. \quad
Contradiction.
\end{proof}

\begin{thm}\wrlab{thm-seq-le-implies-lim-le}
Let $s\in\R^\N$ and let $b,p\in\R$.\\
Assume that \quad $s\to p$ in $\R$ \quad and that \quad $\forall j\in\N$, $s_j\le b$.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\,$ Then $p\le b$.\\
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}\wrlab{thm-ab-is-cpt}
Let $b\in\R$, $a\le b$. \quad Then $[a;b]$ is compact.
\end{thm}

\begin{proof}
Let $K:=[a;b]$. \qquad Want: $K$ is compact.\\
By \tref{thm-R-proper}, $\R$ is proper.\\
So, by \tref{thm-inproper-cpt-iff-closed-bounded},
we want: $K$ is closed and bounded in $\R$.\\
Let $\rho:=b-a+1$. \qquad\qquad
Then $a+\rho=b+a$.\\
Since $a\le b$, we get $\rho\ge1$. \qquad\qquad
Then $\rho>0$, so $a-\rho<a$.\\
Since $a-\rho<a$ and $b<b+1$, we get $[a;b]\subseteq(a-\rho;b+1)$.\\
Then $K=[a;b]\subseteq(a-\rho;b+1)=(a-\rho;a+\rho)=B_\R(a,\rho)$.\\
Then $K$ is bounded in $\R$. $\,$
Want: $K$ is closed in $\R$. $\,$
Want: $\Cl_\R K=K$.\\
Since $K\subseteq\Cl_\R K$, it suffices to show: $\Cl_\R K\subseteq K$.\\
Want: \quad $\forall p\in\Cl_\R K$, $\,$ $p\in K$. \qquad
Given $p\in\Cl_\R K$. \qquad Want: $p\in K$.\\
Since $p\in\Cl_\R K=\Cl_\R^\seq K$, choose $s\in K^\N$ s.t.~$s\to p$ in $\R$.\\
We have: $\forall j\in\N$, $s_j\in K=[a;b]\ge a$, $\,$ so $s_j\ge a$.\\
\phx\quad Then, by \tref{thm-seq-ge-implies-lim-ge}, \qquad we get: $p\ge a$.\\
We have: $\forall j\in\N$, $s_j\in K=[a;b]\le b$, $\,$ so $s_j\le b$.\\
\phx\quad Then, by \tref{thm-seq-ge-implies-lim-ge}, \qquad we get: $p\le b$.\\
Since $a\le p\le b$, we get: $p\in[a;b]$. \qquad
Then $p\in[a;b]=K$.
\end{proof}

The next theorem is the \bfindex[Extreme Value Theorem]:

\begin{thm}\wrlab{thm-EVT}
Let $X$ be a metric space.\\
\phx\qquad Let $f:X\drta\R$, \quad let $K\subseteq\D_f$ \quad and \quad let $L:=f_*K$.\\
\phx\qquad Assume: \quad $K$ is compact and nonempty.\\
\phx\qquad Assume: \quad $f$ is continuous from $X$ to $\R$.\\
\phx\qquad\quad Then: \qquad $\,$ $\min\,L\ne\frownie\ne\max\,L$.
\end{thm}

\begin{proof}
By \tref{thm-im-cpt-under-contin-is-cpt}, we get: \quad $L$ is compact.\\
Then, by \tref{thm-maxmin-cpt-ne-frownie}, we get: \qquad $\min\,L\ne\frownie\ne\max\,L$.
\end{proof}

\begin{thm}\wrlab{thm-EVT-in-R}
Let $f:\R\drta\R$, $b\in\R$, $a\le b$, $L:=f_*([a;b])$.\\
\phx\qquad Assume: \quad $[a;b]\subseteq\D_f $ \quad and \quad $f$ is continuous from $\R$ to $\R$.\\
\phx\qquad\quad Then: \qquad\quad $\,$ $\min\,L\ne\frownie\ne\max\,L$.
\end{thm}

\begin{proof}
Let $K:=[a;b]$. \quad
By \tref{thm-ab-is-cpt}, we get: $K$ is compact.\\
Then, by \tref{thm-EVT}, we get: $\min\,L\ne\frownie\ne\max\,L$.
\end{proof}

\subsection{Cauchy sequences and complete metric spaces}

\begin{defn}
Let $X$ be a metric space and let $s\in X^\N$.\\
Then $s$ is \bfindex[Cauchy] in $X$ means:
$\forall\vep>0$, $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<\vep\,\,)$.
\end{defn}

\begin{thm}\wrlab{thm-epsilon-le1-enough}
let $X$ be a metric space, $s\in X^\N$.\\
Assume: $\forall\vep\in(0;1]$, $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<\vep\,\,)$.
Then $s$ is Cauchy in $X$.
\end{thm}

\begin{proof}
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<\vep\,\,)$.\\
Given $\vep>0$. \qquad Want: $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<\vep\,\,)$.\\
Let $\delta:=\min\{\vep,1\}$. \quad
Then $\delta\le\vep$ and $\delta\le1$ and $\delta>0$.\\
Since $\delta\in(0;1]$, by hypothesis, choose $K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<\vep\,\,)$.\\
Then $K\in\N$. \qquad
Want: $\forall i,j\in\N$, $\,(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<\vep\,\,)$.\\
Given $i,j\in\N$. \qquad\qquad\quad $\,\,\,$ Want: $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<\vep\,\,)$.\\
Assume $i,j\ge K$. \qquad Want: $d(s_i,s_j)<\vep$.\\
Since $i,j\ge K$, by choice of $K$, we have $d(s_i,s_j)<\delta$.\\
Then $d(s_i,s_j)<\delta\le\vep$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-conv-implies-C}
Let $X$ be a metric space and $s\in X^\N$.\\
Assume $s$ is convergent in $X$. \qquad
Then $s$ is Cauchy in $X$.
\end{thm}

\begin{proof}
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<\vep\,\,)$.\\
Given $\vep>0$. Want: $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<\vep\,\,)$.\\
Choose $q\in X$ s.t.~$s\to q$ in $X$. \quad
Choose $K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_j,q)<\vep/2\,\,)$.\\
Then $K\in\N$. \qquad
Want: $\forall i,j\in\N$,  $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<\vep\,\,)$.\\
\phx\qquad\quad Given $i,j\in\N$. \qquad Want: $\,(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<\vep\,\,)$.\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\,$ Assume $i,j\ge K$. \quad Want: $d(s_i,s_j)<\vep$.\\
Since $i\ge K$, by choice of $K$, we have $d(s_i,q)<\vep/2$.\\
Since $j\ge K$, by choice of $K$, we have $d(s_j,q)<\vep/2$.\\
Then $d(q,s_j)=d(s_j,q)<\vep/2$, \qquad so $d(q,s_j)<\vep/2$.\\
Then $d(s_i,s_j)\le(d(s_i,q))+(d(q,s_j))<(\vep/2)+(\vep/2)=\vep$.
\end{proof}

The next result is called \bfindex[Absoluteness of Cauchy]:

\begin{thm}
Let $Y$ be a metric space, $X\subseteq Y$, $s\in X^\N$.\\
\phx\quad Then: \quad $($ $s$ is Cauchy in $X$ $)$ $\quad\Leftrightarrow\quad$ $($ $s$ is Cauchy in $Y$ $)$.
\end{thm}

\begin{proof}
{\it Proof of $\Rightarrow$:} Unassigned HW. {\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:}\\
Assume $s$ is Cauchy in $Y$. \quad Want: $s$ is Cauchy in $X$.\\
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_i,s_j)<\vep\,\,)$.\\
Given $\vep>0$. \qquad Want: $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_i,s_j)<\vep\,\,)$.\\
Since $s$ is Cauchy in $Y$, choose $K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(s_i,s_j)<\vep\,\,)$.\\
Then $K\in\N$. \qquad
Want: $\forall i,j\in\N$,  $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_i,s_j)<\vep\,\,)$.\\
Given $i,j\in\N$. \qquad Want: $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_X(s_i,s_j)<\vep\,\,)$.\\
Assume $i,j\ge K$. \qquad Want: $d_X(s_i,s_j)<\vep$.\\
Since $s\in X^\N$, we get $s_i,s_j\in X$. \quad
Then $d_X(s_i,s_j)=d_Y(s_i,s_j)$.\\
Then $d_X(s_i,s_j)=d_Y(s_i,s_j)<\vep$, as desired.\\
{\it End of proof of $\Leftarrow$.}
\end{proof}

\begin{defn}
Let $X$ be a metric space.\\
By $X$ is \bfindex[complete], we mean:\\
\phx\quad $\forall s\in X^\N$, \quad $($ $s$ is Cauchy in $X$ $)$ $\,\Rightarrow\,$ $($ $s$ is convergent in $X$ $)$.
\end{defn}

\begin{thm}
$\R_0^\times$ is not complete.
\end{thm}

\begin{proof}
Let $X:=\R_0^\times$. \qquad Want: $X$ is not complete.\\
Define $s\in X^\N$ by: \quad $\forall j\in\N$, $s_j=1/j$. \qquad Then $s\to0$ in $\R$.\\
Want: \quad $s$ is Cauchy in $X$ \quad and \quad $s$ is not convergent in $X$.\\
Since $s\to0$ in $\R$, we see that $s$ is convergent in $\R$.\\
Then, by \tref{thm-conv-implies-C}, $s$ is Cauchy in $\R$.\\
Then, by Absoluteness of Cauchy, $s$ is Cauchy in $X$.\\
Want: $s$ is not convergent in $X$.\\
Assume: $s$ is convergent in $X$. \qquad Want: Contradiction.\\
Choose $q\in\R$ s.t.~$s\to q$ in $X$.\\
Then, by Absoluteness of Limits, $s\to q$ in $\R$.\\
So, since $s\to 0$ in $\R$, by \tref{thm-uniqueness-of-limits}, we get: $q=0$.\\
Then $0=q\in X\in\R_0^\times$, so $0\ne0$. \qquad Contradiction.
\end{proof}

\begin{thm}\wrlab{thm-Cauchy-implies-bi}
Let $X$ be a metric space, $s\in X^\N$.\\
\phx\qquad\qquad Assume $s$ is Cauchy in $X$. \qquad
Then $s\in X_\bi^\N$.
\end{thm}

\begin{proof}
Want: $\I_s$ is bounded in $X$. \qquad
Want: $\exists B\in\scrb_X$ s.t.~$\I_s\subseteq B$.\\
Since $s$ is Cauchy in $X$, choose $K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<1\,\,)$.\\
\phx\qquad\qquad\qquad\qquad\quad Then: $\forall i\in\N$, $\,\,(\,\,i\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_K)<1\,\,)$.\\
Let $q:=s_K$. \quad\qquad $\,\,\,$
Then: $\forall i\in\N$, $\,\,(\,\,i\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,q)<1\,\,)$.\\
Let $M\,\,:=\,\,\max\,\{\,\,d(s_1,q)\,,\,\ldots\,,\,d(s_K,q)\,\,\}$.\\
Then $\forall i\in\N$, $\,$ $(\,\,i\le K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,q)\le M\,\,)$.\\
Also, since $q=s_K$, we get $d(s_K,q)=0$, so $M\ge0$, so $0\le M$.\\
Let $B:=B_X(q,M+1)$. \qquad Then $B\in\scrb_X$. \qquad Want: $\I_s\subseteq B$.\\
Want: $\forall p\in\I_s$, $p\in B$. \qquad
Given $p\in\I_s$. \quad Want: $p\in B$.\\
Want: $p\in B_X(q,M+1)$. \qquad Want: $d(p,q)<M+1$.\\
Since $p\in\I_s$, choose $j\in\D_s$ s.t.~$p=s_j$. \qquad Then $j\in\D_s=\N$.\\
At least one of the following must be true:\\
\phx\qquad (1) $j\le K$ \qquad or \qquad (2) $j\ge K$.\\
\\
{\it Case 1:}\\
Recall: $\forall i\in\N$, $\,$ $(\,\,i\le K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,q)\le M\,\,)$.\\
So, \qquad\quad since $j\le K$, \quad we get $d(s_j,q)\le M$.\\
Then $d(p,q)=d(s_j,q)\le M<M+1$, as desired.\\
{\it End of Case 1.}\\
\\
{\it Case 2:}\\
Recall: $\forall i\in\N$, $\,\,(\,\,i\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,q)<1\,\,)$.\\
So, \qquad\quad since $j\ge K$, \quad we get $d(s_j,q)<1$.\\
Also, since $0\le M$, we get $1\le M+1$.\\
Then $d(p,q)=d(s_j,q)<1\le M+1$, as desired.\\
{\it End of Case 2.}
\end{proof}

\begin{thm}\wrlab{thm-C-subconv-implies-conv}
Let $X$ be a metric space and let $s\in X^\N$.\\
\phx\qquad\qquad Assume: $s$ is Cauchy and subconvergent in $X$.\\
\phx\qquad\qquad Then: $s$ is convergent in $X$.
\end{thm}

\begin{proof}
Choose a subsequence $t$ of $s$ s.t.~$t$ is convergent in $X$.\\
Choose $q\in X$ s.t.~$t\to q$ in $X$. \quad
Want: $s\to q$ in $X$.\\
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_j,q)<\vep\,\,)$.\\
Given $\vep>0$. \qquad
Want: $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_j,q)<\vep\,\,)$.\\
Since $t\to q$, choose $L\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge L\,\,)\,\,\Rightarrow\,\,(\,\,d(t_j,q)<\vep/2\,\,)$.\\
Since $s$ is Cauchy, choose $M\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,i,j\ge M\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<\vep/2\,\,)$.\\
Let $K:=\max\{L,M\}$. \qquad Then $K\in\N$.\\
Want: \quad $\forall j\in\N$, $\,$ $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_j,q)<\vep\,\,)$.\\
Given $j\in\N$. \qquad Want: $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_j,q)<\vep\,\,)$.\\
Assume $j\ge K$. \qquad Want: $d(s_j,q)<\vep$.\\
Since $t$ is a subsequence of $s$,\\
\phx\qquad\qquad choose a strictly-increasing $\ell\in\N^\N$ s.t.~$t=s\circ\ell$.\\
By HW\#8-3, we have $\ell_j\ge j$. $\,$ Then $j\ge K\ge M$ and $\ell_j\ge j\ge K\ge M$.\\
Since $j,\ell_j\ge M$, by choice of $M$, we have: \quad $d(s_j,s_{\ell_j})<\vep/2$.\\
So, since $t_j=(s\circ\ell)_j=s_{\ell_j}$, we get: \quad $d(s_j,t_j)<\vep/2$.\\
Since $j\ge K\ge L$, by choice of $L$, we have: \quad $d(t_j,q)<\vep/2$.\\
Then: \quad $d(s_j,q)\,\,\le\,\,(d(s_j,t_j))+(d(t_j,q))\,\,<\,\,(\vep/2)+(\vep/2)\,\,=\,\,\vep$.
\end{proof}

\begin{thm}\wrlab{thm-proper-implies-complete}
Let $X$ be a proper metric space.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then $X$ is complete.
\end{thm}

\begin{proof}
Want: $\forall s\in X^\N$, $(s\hbox{ is Cauchy in }X)\Rightarrow(s\hbox{ is convergent in }X)$.\\
Given $s\in X^\N$. $\,$
Want: $(s\hbox{ is Cauchy in }X)\Rightarrow(s\hbox{ is convergent in }X)$.\\
Assume $s$ is Cauchy in $X$. \qquad Want: $s$ is convergent in $X$.\\
By \tref{thm-Cauchy-implies-bi}, we have: $s\in X_\bi^\N$.\\
So, since $X$ is proper, we get: $s$ is subconvergent in $X$.\\
Then, by \tref{thm-C-subconv-implies-conv}, $s$ is convergent in $X$, as desired.
\end{proof}

\begin{thm}
$\R$ is complete.
\end{thm}

\begin{proof}
By \tref{thm-R-proper}, $\R$ is proper.\\
Then, by \tref{thm-proper-implies-complete}, $\R$ is complete.
\end{proof}

\begin{thm}
$\R^2$ is complete.
\end{thm}

\begin{proof}
By \tref{thm-R2-proper}, $\R^2$ is proper.\\
Then, by \tref{thm-proper-implies-complete}, $\R^2$ is complete.
\end{proof}

Using \tref{thm-prod-proper-is-proper} and induction,\\
\phx\qquad we may show: $\forall k\in\N$, $\R^k$ is proper.\\
Then, by \tref{thm-proper-implies-complete},
it follows that: $\forall k\in\N$, $\R^k$ is complete.

Unassigned HW: Show that\\
\phx\qquad a product of two complete metric spaces \quad is \quad complete.

\begin{defn}
Let $X$ be a metric space and let $A:=X_\set$.\\
Define $\hatd:A\times A\to[0;\infty)$ by: $\forall p,q\in A$, $\hatd(p,q)=\min\{\,d_X(p,q)\,,\,1\}$.\\
\phx\qquad\qquad Then we define: \quad $\hatX\,\,:=\,\,(\,A\,,\,\hatd\,)$.
\end{defn}

\begin{thm}
Let $X$ be a metric space and let $A:=X_\set$.\\
\phx\qquad Then $\hatX_\set=A$. \quad
Also, $\forall p\in A$, $B_\hatX(p,2)=A$.
\end{thm}

\begin{thm}\wrlab{thm-hatted-spaces-are-bounded}
$\forall$metric space $X$, we have: $\hatX$ is bounded.
\end{thm}

\begin{thm}\wrlab{thm-basic-comparison-d-dhat}
Let $X$ be a metric space. \quad Let $p,q\in X_\set$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,\,$ Let $d:=d_X$ and $\hatd:=d_\hatX$.\\
Then: \quad (1) \quad $(\,\,d(p,q)<1\,\,)\,\,\Rightarrow\,\,(\,\,\hatd(p,q)=d(p,q)\,\,)$.\\
Also: \quad $\,\,$(2) \quad $(\,\,d(p,q)\ge1\,\,)\,\,\Rightarrow\,\,(\,\,\hatd(p,q)=1\,\,)$.\\
Also: \quad $\,\,$(3) \quad $(\,\,\hatd(p,q)\ne1\,\,)\,\,\Rightarrow\,\,(\,\,d(p,q)<1\,\,)$.\\
\end{thm}

\begin{thm}\wrlab{thm-dordhat-less-than-1-implies-d-equals-dhat}
Let $X$ be a metric space. \quad Let $p,q\in X_\set$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,\,$ Let $d:=d_X$ and $\hatd:=d_\hatX$.\\
Assume: \quad either \quad (1) $d(p,q)<1$ \quad or \quad (2) $\hatd(p,q)<1$.\\
\phx\qquad\qquad\qquad\qquad\quad Then: \quad $\hatd(p,q)\,=\,d(p,q)$
\end{thm}

\begin{proof}
{\it Case 1:}\\
By (1) of \tref{thm-basic-comparison-d-dhat}, we have $\hatd(p,q)=d(p,q)$, as desired.\\
{\it End of Case 1.}\\
\\
{\it Case 2:}\\
Since $\hatd(p,q)<1$, we get: $\hatd(p,q)\ne1$.\\
Then, by (3) of \tref{thm-basic-comparison-d-dhat}, we have $d(p,q)<1$, as desired.\\
Then, by (1) of \tref{thm-basic-comparison-d-dhat}, we have $\hatd(p,q)=d(p,q)$, as desired.\\
{\it End of Case 2.}
\end{proof}

\begin{thm}\wrlab{thm-CinX-iff-CinhatX}
Let $X$ be a metric space, $s\in X_\set^\N$.\\
\phx\qquad\quad Then: \qquad $(\,\,s\hbox{ is Cauchy in }X\,\,)\,\,\Leftrightarrow\,\,(\,\,s\hbox{ is Cauchy in }\hatX\,\,)$.
\end{thm}

\begin{proof}
Let $d:=d_X$ and $\hatd:=d_\hatX$.\\
{\it Proof of $\Leftarrow$:} \quad Unassigned HW. \quad {\it End of proof of $\Leftarrow$.}\\
\\
{\it Proof of $\Rightarrow$:}\\
Assume: $s$ is Cauchy in $X$. \quad Want: $s$ is Cauchy in $\hatX$.\\
By \tref{thm-epsilon-le1-enough}, want: $\forall\vep\in(0;1]$, $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,\hatd(s_i,s_j)<\vep\,\,)$.\\
Given $\vep\in(0;1]$. \qquad Want: $\exists K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,\hatd(s_i,s_j)<\vep\,\,)$.\\
Since $s$ is Cauchy in $X$, choose $K\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(s_i,s_j)<\vep\,\,)$.\\
Then $K\in\N$. \quad Want: $\forall i,j\in\N$, \quad $\,(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,\hatd(s_i,s_j)<\vep\,\,)$.\\
Given $i,j\in\N$. \qquad\qquad\quad $\,\,$ Want: $\,(\,\,i,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,\hatd(s_i,s_j)<\vep\,\,)$.\\
Assume: $i,j\ge K$. \qquad Want: $\hatd(s_i,s_j)<\vep$.\\
Since $i,j\ge K$, by choice of $K$, we have: $d(s_i,s_j)<\vep$.\\
Then \quad $d(s_i,s_j)\,<\,\vep\,\in\,[0;1)\,<\,1$, \quad so $d(s_i,s_j)\,<\,1$,\\
\phx\qquad so, by \tref{thm-dordhat-less-than-1-implies-d-equals-dhat}, we have: \quad $\hatd(s_i,s_j)=d(s_i,s_j)$.\\
Then \qquad $\hatd(s_i,s_j)=d(s_i,s_j)<\vep$, \quad as desired.
{\it End of proof of $\Rightarrow$.}
\end{proof}

\begin{thm}\wrlab{thm-conv-in-X-iff-conv-in-hatX}
Let $X$ be a metric space, $A:=X_\set$, $s\in A^\N$.\\
\phx\qquad\quad Then: \qquad $(\,\,s\hbox{ is convergent in }X\,\,)\,\,\Leftrightarrow\,\,(\,\,s\hbox{ is convergent in }\hatX\,\,)$.
\end{thm}

\begin{proof}
We have $\hatX_\set=X_\set=A$.\\
\\
{\it Proof of $\Leftarrow$:} \quad Unassigned HW. \quad {\it End of proof of $\Leftarrow$.}\\
\\
{\it Proof of $\Rightarrow$:}\\
Assume: $s$ is convergent in $X$. \qquad
Want: $s$ is convergent in $\hat X$.\\
Since $s$ is convergent in $X$, choose $q\in X$ s.t.~$s\to q$ in $X$.\\
Then, by $\Rightarrow$ of HW\#12-5, we get: $s\to q$ in $\hatX$.\\
Then $s$ is convergent in $\hatX$, as desired.\\
{\it End of proof of $\Rightarrow$.}
\end{proof}

\begin{thm}\wrlab{thm-subconv-in-X-iff-subconv-in-hatX}
Let $X$ be a metric space, $A:=X_\set$, $s\in A^\N$.\\
\phx\qquad\quad Then: \qquad $(\,\,s\hbox{ is subconvergent in }X\,\,)\,\,\Leftrightarrow\,\,(\,\,s\hbox{ is subconvergent in }\hatX\,\,)$.
\end{thm}

\begin{proof}
We have $\hatX_\set=X_\set=A$.\\
\\
{\it Proof of $\Leftarrow$:} \quad Unassigned HW. \quad {\it End of proof of $\Leftarrow$.}\\
\\
{\it Proof of $\Rightarrow$:}\\
Assume: $s$ is subconvergent in $X$. \qquad
Want: $s$ is subconvergent in $\hat X$.\\
Since $s$ is subconvergent in $X$, choose $t\in A^\N$ s.t.~$t$ is convergent in $X$.\\
Then, by $\Rightarrow$ of \tref{thm-conv-in-X-iff-conv-in-hatX}, we get: $t$ is convergent in $X$.\\
Then $s$ is subconvergent in $\hatX$, as desired.\\
{\it End of proof of $\Rightarrow$.}
\end{proof}

\begin{thm}\wrlab{thm-X-cpt-iff-hatX-cpt}
Let $X$ be a metric space.\\
\phx\qquad\quad Then: \qquad $(\,\,X\hbox{ is compact}\,\,)\,\,\Leftrightarrow\,\,(\,\,\hatX\hbox{ is compact}\,\,)$.
\end{thm}

\begin{proof}
Let $A:=X_\set$. \quad Then $A=\hatX_\set$.\\
\\
{\it Proof of $\Leftarrow$:} \quad Unassigned HW. \quad {\it End of proof of $\Leftarrow$.}\\
\\
{\it Proof of $\Rightarrow$:}\\
Assume: $X$ is compact. \quad Want: $\hatX$ is compact.\\
Want: $\forall s\in A^\N$, $s$ is subconvergent in $\hatX$.\\
Given $s\in A^\N$. \qquad Want: $s$ is subconvergent in $\hatX$.\\
Since $X$ is compact, $s$ is subconvergent in $X$.\\
Then, by $\Rightarrow$ of \tref{thm-subconv-in-X-iff-subconv-in-hatX}, $s$ is subconvergent in $\hatX$, as desired.
{\it End of proof of $\Rightarrow$.}
\end{proof}

\begin{thm}\wrlab{thm-hatbbr-complete}
$\hatbbR$ is complete.
\end{thm}

\begin{proof}
Want: \quad $\forall s\in\hatbbR^\N$, $\,\,(s\hbox{ is Cauchy in }\hatbbR)\Rightarrow(s\hbox{ is convergent in }\hatbbR)$.\\
Given $s\in\hatbbR^\N$. \qquad Want: $(s\hbox{ is Cauchy in }\hatbbR)\Rightarrow(s\hbox{ is convergent in }\hatbbR)$.\\
Assume: $s$ is Cauchy in $\hatbbR$. \qquad
Want: $s$ is convergent in $\hatbbR$.\\
Since $s$ is Cauchy in $\hatbbR$, by $\Leftarrow$ of \tref{thm-CinX-iff-CinhatX},\\
\phx\qquad we conclude that $s$ is Cauchy in $\R$.\\
So, since $\R$ is complete, we get: $s$ is convergent in $\R$.\\
Then, by $\Rightarrow$ of \tref{thm-conv-in-X-iff-conv-in-hatX}, $s$ is convergent in $\hatbbR$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-hatbbr-nonproper}
$\hatbbR$ is nonproper.
\end{thm}

\begin{proof}
Assume $\hatbbR$ is proper. \qquad Want: Contradiction.\\
By \tref{thm-hatted-spaces-are-bounded}, we see that $\hatbbR$ is bounded.\\
Since $\hatbbR$ is proper and bounded, by $\Leftarrow$ of \tref{thm-cpt-iff-proper-bdd},\\
\phx\qquad we see that: \quad $\hatbbR$ is compact.\\
Then, by $\Leftarrow$ of \tref{thm-X-cpt-iff-hatX-cpt}, we get: \quad $\R$ is compact.\\
Then, by $\Rightarrow$ of \tref{thm-cpt-iff-proper-bdd}, we get: \quad$\R$ proper and bounded.\\
Then $\R$ is bounded. \quad Also, $\R$ is unbounded. \quad Contradiction.
\end{proof}

\subsection{Continuous injections over compacta are homeomorphisms}

\begin{thm}\wrlab{thm-closed-in-cpt-is-cpt}
Let $K$ be a compact metric space, $C\in\scrt'_K$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then $C$ is compact.
\end{thm}

\begin{proof}
Want: $\forall s\in C^\N$, $s$ is subconvergent in $C$.\\
Given $s\in C^\N$. \qquad Want: $s$ is subconvergent in $C$.\\
Since $s\in C^\N\subseteq K^\N$ and since $K$ is compact,\\
\phx\qquad we conclude: \quad $s$ is subconvergent in $K$.\\
So, since $C\in\scrt'_K$, by $\Leftarrow$ of \tref{thm-subconv-closed-subspace-iff-subconv-ambient},\\
\phx\qquad we see that: \quad $s$ is subconvergent in $C$.
\end{proof}

Recall (\tref{thm-contin-inject-not-necess-homeom}) that\\
\phx\qquad a continuous bijection may not be a homeomorphism.\\
The next result says that that doesn't happen when the domain is compact.

\begin{thm}
Let $K$, $Y$ be metric spaces. Let $f:K\hookrightarrow>Y$.\\
Assume: \quad $($ $K$ is compact $)\,\,\&\,\,($ $f$ is continuous from $K$ to $Y$ $)$.\\
\phx\qquad\qquad Then: \quad $f$ is a homeomorphism from $K$ onto $Y$.
\end{thm}

\begin{proof}
Since $f:K\hookrightarrow>Y$ and since $f$ is continuous from $K$ to $Y$,\\
\phx\qquad\qquad\quad it only remains to show: $f^{-1}$ is continuous from $Y$ to $K$.\\
Let $g:=f^{-1}$. Then $g:Y\hookrightarrow>K$. Want: $g$ is continuous from $Y$ to $K$.\\
By $\Leftarrow$ of \tref{thm-preim-closed-is-closed-implies-contin},
want: $\forall C\in\scrt'_K$, $g^*C\in\scrt'_Y$.\\
Given $C\in\scrt'_K$. \qquad Want: $g^*C\in\scrt'_Y$.\\
Since $K$ is compact and $C\in\scrt'_K$,\\
\phx\qquad by \tref{thm-closed-in-cpt-is-cpt}, we get: \quad $C$ is compact.\\
So, since $f$ is continuous from $K$ to $Y$,\\
\phx\qquad by \tref{thm-im-cpt-under-contin-is-cpt}, we get: \quad $f_*C$ is compact.\\
Then, by \tref{thm-cpt-implies-closed-bounded}, we get $f_*C$ is closed in $Y$, and so $f_*C\in\scrt'_Y$.\\
Since $g=f^{-1}$, we get: $g^*C=f_*C$. \quad
Then $g^*C=f_*C\in\scrt'_Y$.
\end{proof}

\subsection{Continuous on compact implies uniformly continuous}

\begin{thm}\wrlab{thm-sum-le-sum-squares-times-sqrt2}
Let $a,b\in\R$. \quad
Then $a+b\le\sqrt{2}\cdot\sqrt{a^2+b^2}$.
\end{thm}

\begin{proof}
We have: $\forall x\in\R$, $x\le|x|$. \qquad
Then $a+b\le|a+b|$.\\
Since $a+b\le|a+b|=\sqrt{(a+b)^2}$,\\
\phx\qquad it suffices to show: $\sqrt{(a+b)^2}\le\sqrt{2}\cdot\sqrt{a^2+b^2}$.\\
It therefore suffices to show: $0\le(a+b)^2\le2\cdot(a^2+b^2)$.\\
We have: $\forall x\in\R$, $0\le x^2$. \qquad
Then $0\le(a+b)^2$.\\
It remains to show: $(a+b)^2\le2\cdot(a^2+b^2)$.\\
We have: $\forall x\in\R$, $0\le x^2$. \qquad
Then $0\le(a-b)^2$.\\
Then $0\le a^2-2ab+b^2$, so $2ab\le a^2+b^2$.\\
Adding $a^2$ and $b^2$ to both sides gives: $a^2+2ab+b^2\le a^2+a^2+b^2+b^2$.\\
Then $(a+b)^2=a^2+2ab+b^2\le a^2+a^2+b^2+b^2=2\cdot(a^2+b^2)$.
\end{proof}

\begin{thm}\wrlab{thm-dist-is-root2-Lip}
Let $X$ be a metric space.\\
\phx\qquad\qquad\qquad Then $d_X$ is Lipschits-$\sqrt{2}$ from $X\times X$ to $\R$.
\end{thm}

\begin{proof}
Let $f:=d_X$, let $V:=X\times X$ and let $K:=\sqrt{2}$.\\
Want: $f$ is Lipschitz-$K$ from $V$ to $\R$.\\
Want: $\forall s,t\in V$, $d_\R(f_s,f_t)\le K\cdot(d_V(s,t))$.\\
Given $s,t\in V$. \qquad Want: $d_\R(f_s,f_t)\le K\cdot(d_V(s,t))$.\\
Let $\vep:=K\cdot(d_V(s,t))$. \qquad
Want: $d_\R(f_s,f_t)\le\vep$.\\
Want: $(\,\,f_s\le f_t+\vep\,\,)\,\,\&\,\,(\,\,f_t\le f_s+\vep\,\,)$.\\
Let $\,\,\,a:=d_X(s_1,t_1)$ and $b:=d_X(s_2,t_2)$.\\
Then $a=d_X(t_1,s_1)$ and $\,\,b=d_X(t_2,s_2)$.\\
Then \quad $d_V(s,t)\,\,=\,\,d_{X\times X}(\,(s_1,s_2)\,,\,(t_1,t_2)\,)$\\
\phx\qquad\qquad\qquad\quad $=\sqrt{(\,d_X(s_1,t_1)\,)^2\,+\,(\,d_X(s_2,t_2)\,)^2}\,\,=\,\,\sqrt{a^2+b^2}$.\\
By \tref{thm-sum-le-sum-squares-times-sqrt2}, we have: $a+b\le\sqrt{2}\sqrt{a^2+b^2}$.\\
So, since $\sqrt{2}=K$ and $\sqrt{a^2+b^2}=d_V(s,t)$,
we get: $a+b\le K\cdot(d_V(s,t))$.\\
So, since $K\cdot(d_V(s,t))=\vep$, we get: $a+b\le\vep$.\\
We have $f_s=f(s)=f(s_1,s_2)=d_X(s_1,s_2)$\\
\phx\quad and $\,\,\,\,f_t=f(t)=\,f(t_1,t_2)\,=\,d_X(t_1,t_2)$.\\
We compute $f_s=d_X(s_1,s_2)\le(d_X(s_1,t_1))+(d_X(t_1,t_2))+(d_X(t_2,s_2))$\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\,\,\,\,\,\,=a+f_t+b=f_t+(a+b)\le f_t+\vep$.\\
It remains to show: $f_t\le f_s+\vep$.\\
We compute $f_t\,=\,d_X(t_1,t_2)\le(d_X(t_1,s_1))+(d_X(s_1,s_2))+(d_X(s_2,t_2))$\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\,\,\,\,\,\,=a+f_s+b=f_t+(a+b)\le f_s+\vep$.
\end{proof}

\begin{thm}\wrlab{thm-both-contin-p-implies-pairing-contin-p}
Let $A$, $X$ and $Y$ be metric spaces.\\
Let $f:A\drta X$, let $g:A\drta Y$ and let $p\in A$.\\
Assume that $f$ is continuous at $p$ from $A$ to $X$.\\
Assume that $g$ is continuous at $p$ from $A$ to $Y$.\\
Then $(f,g)$ is continuous at $p$ from $A$ to $X\times Y$.\\
\end{thm}

\begin{proof}
Let $h:=(f,g)$ and let $Z:=X\times Y$.\\
We wish to show: $h$ is continuous at $p$ from $A$ to $Z$.\\
Since $f$ is continuous at $p$ from $A$ to $X$, we get: $p\in\D_f$.\\
Since $g$ is continuous at $p$ from $A$ to $Y$, we get: $p\in\D_g$.\\
Then $p\in\D_f\cap\D_g$. \quad
Since $h=(f,g)$, we get: $\D_h=\D_f\cap\D_g$.\\
Then $p\in\D_f\cap\D_g=\D_h$.\\
It remains to show: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall q\in\D_h$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\,$ $(\,\,d(p,q)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(h_p,h_q)<\vep\,\,)$.\\
Given $\vep>0$. \qquad\quad$\,$
Want: $\exists\delta>0$ s.t., $\forall q\in\D_h$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\,$ $(\,\,d(p,q)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(h_p,h_q)<\vep\,\,)$.\\
Since $f$ is continuous at $p$ from $A$ to $X$,\\
\phx\qquad\qquad\qquad\qquad\quad$\,$choose $\lambda>0$ s.t., $\forall q\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\,$ $(\,\,d(p,q)<\lambda\,\,)\,\,\Rightarrow\,\,(\,\,d(f_p,f_q)<\vep/\sqrt{2}\,\,)$.\\
Since $g$ is continuous at $p$ from $A$ to $Y$,\\
\phx\qquad\qquad\qquad\qquad\quad$\,$choose $\mu>0$ s.t., $\forall q\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\,$ $(\,\,d(p,q)<\mu\,\,)\,\,\Rightarrow\,\,(\,\,d(g_p,g_q)<\vep/\sqrt{2}\,\,)$.\\
Let $\delta:=\min\{\lambda,\mu\}$. \qquad\qquad\qquad
Then $\delta>0$.\\
Want: \qquad $\,\,\forall q\in\D_h$, \quad $(\,\,d(p,q)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(h_p,h_q)<\vep\,\,)$.\\
Given $q\in\D_h$. \quad
Want: $(\,\,d(p,q)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d(h_p,h_q)<\vep\,\,)$.\\
Assume: $d(p,q)<\delta$. \qquad Want: $d(h_p,h_q)<\vep$.\\
Since $d(p,q)<\delta\le\lambda$, by choice of $\lambda$, we get: $d(f_p,f_q)<\vep/\sqrt{2}$.\\
Since $d(p,q)<\delta\le\mu$, by choice of $\mu$, we get: $d(g_p,g_q)<\vep/\sqrt{2}$.\\
Since $0\le d(f_p,f_q)<\vep/\sqrt{2}$, we get: $(d(f_p,f_q))^2<(\vep/\sqrt{2})^2$.\\
Since $0\le d(g_p,g_q)<\vep/\sqrt{2}$, we get: $(d(g_p,g_q))^2<(\vep/\sqrt{2})^2$.\\
Then \qquad\qquad $(d(f_p,f_q))^2+(d(g_p,g_q))^2<(\vep/\sqrt{2})^2+(\vep/\sqrt{2})^2$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $=(\vep^2/2)+\vep^2/2)=\vep^2$.\\
Since \qquad $\,\,0\le(d(f_p,f_q))^2+(d(g_p,g_q))^2<\vep^2$,\\
we conclude that: \quad $\,\,\sqrt{(d(f_p,f_q))^2+(d(g_p,g_q))^2}<\sqrt{\vep^2}$.\\
Then: $\,$ $d(\,h_p\,,\,h_q\,)\,\,=\,\,d(\,(f,g)_p\,,\,(f,g)_p\,)$\\
\phx\qquad\qquad\qquad\qquad $=\,\,d(\,(f_p,g_p)\,,\,(f_q,g_q)\,)$\\
\phx\qquad\qquad\qquad\qquad $=\,\,\sqrt{(d(f_p,f_q))^2+(d(g_p,g_q))^2}\,\,<\sqrt{\vep^2}\,\,=\,\,\vep$.
\end{proof}

\begin{thm}\wrlab{thm-both-contin-implies-pairing-contin}
Let $A$, $X$ and $Y$ be metric spaces.\\
Let $f\in C^A_X$ and $g\in C^A_Y$. \qquad
Then $(f,g)\in C^A_{X\times Y}$.
\end{thm}

\begin{proof}
Let $h:=(f,g)$ and let $Z:=X\times Y$. \qquad
Want: $h\in C^A_Z$.\\
Since $f:A\to X$ and $g:A\to Y$, we get: $h:A\to Z$.\\
Want: $h$ is continuous from $A$ to $Z$.\\
Want: $h$ is continuous on $\D_h$ from $A$ to $Z$.\\
Want: $h$ is continuous on $A$ from $A$ to $Z$.\\
Want: $\forall p\in A$, $h$ is continuous at $p$ from $A$ to $Z$.\\
Given $p\in A$. \qquad Want: $h$ is continuous at $p$ from $A$ to $Z$.\\
We have: $f$ is continuous from $A$ to $X$.\\
Then: $f$ is continuous on $\D_f$ from $A$ to $X$.\\
Then: $f$ is continuous on $A$ from $A$ to $X$.\\
So, since $p\in A$, we get: $f$ is continuous at $p$ from $A$ to $X$.\\
We have: $g$ is continuous from $A$ to $Y$.\\
Then: $g$ is continuous on $\D_g$ from $A$ to $Y$.\\
Then: $g$ is continuous on $A$ from $A$ to $Y$.\\
So, since $p\in A$, we get: $g$ is continuous at $p$ from $A$ to $Y$.\\
Since \qquad $f$ is continuous at $p$ from $A$ to $X$\\
\phx\quad and \quad $g$ is continuous at $p$ from $A$ to $Y$,\\
it follows, from \tref{thm-both-contin-p-implies-pairing-contin-p}, that\\
\phx\quad\quad\quad $(f,g)$ is continuous at $p$ from $A$ to $X\times Y$.\\
So, since $h=(f,g)$ and $Z=X\times Y$, we get:\\
\phx\qquad\quad\quad $h$ is continuous at $p$ from $A$ to $Z$, \qquad
as desired.
\end{proof}

\begin{defn}
Let $f$ and $g$ be functions.\\
\phx\qquad\qquad Then \quad $f\,\,\times_\fn\,\,g$ \quad is the function defind by:\\
\phx\qquad\qquad\qquad $\forall x$, \qquad $(f\,\,\times_\fn\,\,g)_x\quad=\quad\begin{cases}(f_{x_1},g_{x_2}),&\hbox{if }x\in\D_f\times\D_g\\\quad\,\,\frownie,&\hbox{otherwise.}\end{cases}$
\end{defn}

By sloppiness, we typically denote \quad $f\,\,\times_\fn\,\,g$ \quad by: \quad $f\times g$.

\begin{thm}
Let $f$ and $g$ be functions. \quad Then:\\
\phx\qquad\qquad\qquad\quad $(\,\,\D_{(f,g)}=\D_f\bigcap\D_g\,\,)\quad\&\quad(\,\,\D_{f\times g}=\D_f\times\D_g\,\,)$.
\end{thm}

\begin{thm}
Let $S$, $T$, $X$ and $Y$ be sets, $f\in X^S$, $g\in Y^T$.\\
\phx\qquad Then: \quad $(f,g):S\bigcap T\to X\times Y$ \quad and \quad $f\times g:S\times T\to X\times Y$.
\end{thm}

\begin{thm}
Let $X$, $Y$ and $Z$ be metric spaces.\\
\phx\qquad Let $f\in C^X_Y$ and let $g\in C^Y_Z$. \qquad Then $g\circ f\in C^X_Z$.
\end{thm}

The next theorem follows from \tref{thm-comp-contin-full-stop}.

\begin{thm}\wrlab{thm-prod-contin-is-contin}
Let $S$, $T$, $X$ and $Y$ be metric spaces.\\
\phx\qquad Let $f\in C^S_X$ and let $g\in C^T_Y$. \qquad Then $f\times g\in C^{S\times T}_{X\times Y}$.
\end{thm}

\begin{proof}
Let $A:=S\times T$, $Z:=X\times Y$ and $h:=f\times g$. \quad
Want: $h\in C^A_Z$.\\
Define $\sigma:A\to S$ and $\tau:A\to T$ by: \quad $\forall p\in A$, $\sigma_p=p_1$ and $\tau_p=p_2$.\\
Then $\sigma$ is Lipschitz-$1$ from $A$ to $S$ and $\tau$ is Lipschitz-$1$ from $A$ to $T$.\\
Then $\sigma$ is continuous from $A$ to $S$ and $\tau$ is continuous from $A$ to $T$.\\
Then \qquad $\sigma\in C^A_S$ \qquad and \qquad $\tau\in C^A_T$.\\
Since $\sigma\in C^A_S$ and $f\in C^S_X$, we get: $f\circ\sigma\in C^A_X$.\\
Since $\tau\in C^A_T$ and $g\in C^T_Y$, we get: $g\circ\tau\in C^A_Y$.\\
Then, by \tref{thm-both-contin-implies-pairing-contin}, we have $(f\circ\sigma,g\circ\tau)\in C^A_{X\times Y}$.\\
So, since $Z=X\times Y$, we get: $(f\circ\sigma,g\circ\tau)\in C^A_Z$.\\
It therefore suffices to show that $(f\circ\sigma,g\circ\tau)=h$.\\
Want: $\forall p\in A$, $(f\circ\sigma,g\circ\tau)_p=h_p$.\\
Given $p\in A$. \qquad Want: $(f\circ\sigma,g\circ\tau)_p=h_p$.\\
Since $p\in A=S\times T=\D_f\times\D_g$, we get: $(f\times g)_p=(f_{p_1},g_{p_2})$.\\
Then $(\,f\circ\sigma\,,\,g\circ\tau\,)_p\,\,=\,\,(\,(f\circ\sigma)_p\,,\,(g\circ\tau)_p\,)\,\,=\,\,(\,f_{\sigma_p}\,,\,g_{\tau_p}\,)$\\
\phx\qquad\qquad\qquad\qquad\quad $\,=\,\,(\,f_{p_1}\,,\,g_{p_2}\,)\,\,=\,\,(f\times g)_p\,\,=\,\,h_p$, as desired.
\end{proof}

The next result follows from Absoluteness of Continuity.

\begin{thm}
Let $A$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad\qquad\qquad Let $X\subseteq Y$. \qquad
Then $C^A_X\subseteq C^A_Y$.
\end{thm}

The following is a Squeeze Theorem.

\begin{thm}\wrlab{thm-squeeze-to-zero}
Let $s\in\R^\N$.\\
\phx\qquad\qquad Assume: $\forall j\in\N$, $0\le s_j<1/j$. \qquad
Then $s\to0$ in $\R$.
\end{thm}

\begin{thm}
Let $V:=[0;\infty)$ and $I:=[0;\infty)$.\\
Define $\alpha,\beta\in C^V_I$ by: $\forall u\in V$, $\ds{\alpha_u=\frac{u}{1+u^2}}$ and $\beta_u=u$.\\
Then: \qquad $\forall u\in V$, \quad $\,\,\,\,[\,\,\,(\,\,\alpha_u=0\,\,)\,\,\Rightarrow\,\,(\,\,\beta_u=0\,\,)\,\,\,]$ \qquad and\\
\phx\qquad\qquad\qquad\qquad\quad $\neg[\,\,\,\forall\vep>0,\exists\delta>0~\hbox{s.t.},\forall u\in V,$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,\alpha_u<\delta\,\,)\,\,\Rightarrow\,\,(\,\,\beta_u<\vep\,\,)\,\,\,]$.
\end{thm}

\begin{thm}\wrlab{thm-alpha-small-implies-beta-small}
Let $V$ be a compact metric space.\\
\phx\qquad\qquad\qquad\qquad\qquad Let $I:=[0;\infty)$ and let $\alpha,\beta\in C^V_I$.\\
\phx\qquad Assume: \qquad $\forall u\in V$, \quad $(\,\,\alpha_u=0\,\,)\,\,\Rightarrow\,\,(\,\,\beta_u=0\,\,)$.\\
\phx\qquad Then: \qquad\quad $\,\forall\vep>0,\exists\delta>0~\hbox{s.t.},\forall u\in V,$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,\alpha_u<\delta\,\,)\,\,\Rightarrow\,\,(\,\,\beta_u<\vep\,\,)$.
\end{thm}

\begin{proof}
\qquad\qquad\qquad\qquad Assume $\neg[\,\,\,\forall\vep>0,\exists\delta>0~\hbox{s.t.},\forall u\in V,$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,\alpha_u<\delta\,\,)\,\,\Rightarrow\,\,(\,\,\beta_u<\vep\,\,)\,\,\,]$.\\
Want: Contradiction. \quad Choose $\vep>0$ s.t., $\forall\delta>0$, $\exists u\in V$ s.t.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,\alpha_u<\delta\,\,)\,\,\&\,\,(\,\,\beta_u\ge\vep\,\,)\,\,\,]$.\\
Then: $\forall j\in\N$, $\exists u\in V$ s.t.~$\,(\,\,\alpha_u<1/j\,\,)\,\,\&\,\,(\,\,\beta_u\ge\vep\,\,)\,\,\,]$.\\
By the Axiom of Choice, choose $s\in V^\N$ s.t.~$\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,\alpha_{s_j}<1/j\,\,)\,\,\&\,\,(\,\,\beta_{s_j}\ge\vep\,\,)\,\,\,]$.\\
Since $V$ is compact, $s$ is subconvergent in $V$.\\
Choose a subsequence $t$ of $s$ s.t.~$t$ is convergent in $V$.\\
Choose a strictly-increasing $\ell\in\N^\N$ s.t.~$t=s\circ\ell$.\\
By HW\#8-3, we have: $\forall j\in\N$, $\ell_j\ge j$.\\
Since \quad $\forall j\in\N$, we have $\ell_j\ge j>0$, \quad we get: \quad $\forall j\in\N$, $1/\ell_j\le1/j$.\\
Also, we have: $\forall j\in\N$, $t_j=(s\circ\ell)_j=s_{\ell_j}$.\\
By choice of $s$, we know: $\forall j\in\N$, $(\,\,\alpha(s_j)<1/j\,\,)\,\,\&\,\,(\,\,\beta(s_j)\ge\vep\,\,)$.\\
It follows that: $\forall j\in\N$, $(\,\,\alpha(s_{\ell_j})<1/\ell_j\,\,)\,\,\&\,\,(\,\,\beta(s_{\ell_j})\ge\vep\,\,)$.\\
Then $\forall j\in\N$, $(\,\,(\alpha\circ t)_j=\alpha(t_j)=\alpha(s_{\ell_j})<1/\ell_j\le1/j\,\,)\quad\&$\\
\phx\qquad\qquad\quad $\,\,\,(\,\,(\beta\circ t)_j=\beta(t_j)=\beta(s_{\ell_j})\ge\vep\,\,)$.\\
Then $\forall j\in\N$, $(\,\,(\alpha\circ t)_j<1/j\,\,)\,\,\&\,\,(\,\,(\beta\circ t)_j\ge\vep\,\,)$.\\
Since $t$ is convergent in $V$, choose $q\in V$ s.t.~$t\to q$ in $V$.\\
Since $q\in V$ and $\alpha,\beta\in C^V_I\subseteq C^V_\R$, we conclude:\\
\phx\qquad\qquad $\alpha$ and $\beta$ are both continuous at $q$ from $V$ to $\R$.\\
So, since $t\to q$ in $V$, we get:\\
\phx\qquad\qquad $\alpha\circ t\to\alpha_q$ in $\R$ \quad and \quad $\beta\circ t\to\beta_q$ in $\R$.\\
Since $\alpha\in C^V_I$, we get $\I_\alpha\subseteq I$. \quad
Then $\I_\alpha\subseteq I=[0;\infty)\ge0$.\\
Then: \quad $\forall j\in\N$, $(\alpha\circ t)_j=\alpha(t_j)\in\I_\alpha\ge0$.\\
Since \quad $\forall j\in\N$, $0\le(\alpha\circ t)_j<1/j$,\\
\phx\qquad by \tref{thm-squeeze-to-zero}, we get: $\alpha\circ t\to0$ in $\R$.\\
So, since $\alpha\circ t\to\alpha_q$ in $\R$, by Uniqueness of Limits, we have $\alpha_q=0$.\\
By hypothesis, we have: \qquad $\forall u\in V$, \quad $(\,\,\alpha_u=0\,\,)\,\,\Rightarrow\,\,(\,\,\beta_u=0\,\,)$.\\
So, since $q\in V$ and $\alpha_q=0$, we see that $\beta_q=0$.\\
Since \quad $\forall j\in\N$, $(\beta\circ t)_j\ge\vep$ \quad and \quad $\beta\circ t\to\beta_q$ in $\R$,\\
\phx\qquad by \tref{thm-seq-ge-implies-lim-ge}, we get: \quad $\beta_q\ge\vep$.\\
Then $0<\vep\le\beta_q=0$, so $0<0$. \qquad Contradiction.
\end{proof}

\begin{thm}
Let $X$ be a compact metric space.\\
\phx\qquad\qquad\qquad\qquad\qquad Let $Y$ be a metric space. \quad Let $f\in C^Y_X$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then $f$ is uniformaly continuous from $X$ to $Y$.\\
\end{thm}

\begin{proof}
Let $V:=X\times X$, $W:=Y\times Y$, $g:=f\times f$ and $I:=[0;\infty)$.\\
Since $X$ is compact, by \tref{thm-prod-cpt-is-cpt}, we get: $V$ is compact.\\
Since $f\in C^Y_X$, by \tref{thm-prod-contin-is-contin}, we get: $g\in C^V_W$.\\
By \tref{thm-dist-is-root2-Lip}, \qquad\qquad $d_X$ is Lipschitz-$\sqrt{2}$ from $V$ to $\R$\\
\phx\qquad\qquad\qquad\qquad and \qquad $\,\,\,d_Y$ is Lipschitz-$\sqrt{2}$ from $W$ to $\R$.\\
It follows that: \qquad\qquad\qquad $d_X$ is continuous from $V$ to $\R$\\
\phx\qquad\qquad\qquad\qquad and \qquad $\,\,\,d_Y$ is continuous from $W$ to $\R$.\\
So, since $\I_{d_X}\subseteq I$ and $\I_{d_Y}\subseteq I$, by Absoluteness of Continuity,\\
\phx\qquad we see that:\qquad\qquad\quad $\,d_X$ is continuous from $V$ to $I$\\
\phx\qquad\qquad\qquad\qquad and \qquad $\,\,\,d_Y$ is continuous from $W$ to $I$.\\
We conclude that: \quad $d_X\in C^V_I$ and $d_Y\in C^W_I$.\\
Since $g\in C^V_W$ and $d_Y\in C^W_I$, we see that: $d_Y\circ g\in C^V_I$.\\
Let $\alpha:=d_X$ and let $\beta:=d_Y\circ g$. \qquad
Then $\alpha,\beta\in C_I^V$.\\
\\
{\it Claim:} $\forall u\in V$, $(\,\,\alpha_u=0\,\,)\,\,\Rightarrow\,\,(\,\,\beta_u=0\,\,)$.\\
{\it Proof of Claim:}\\
Given $u\in V$. \quad Want: $(\,\,\alpha_u=0\,\,)\,\,\Rightarrow\,\,(\,\,\beta_u=0\,\,)$.\\
Assume $\alpha_u=0$. \qquad Want: $\beta_u=0$.\\
We have $d_X(u_1,u_2)=\alpha(u_1,u_2)=\alpha(u)=\alpha_u=0$, so $d_X(u_1,u_2)=0$.\\
Then $u_1=u_2$. \qquad
Then $f_{u_1}=f_{u_2}$. \qquad
Then $d_Y(f_{u_1},f_{u_2})=0$.\\
We have $g_u=(f\times f)_u=(f_{u_1},f_{u_2})$, \quad so $g_u=(f_{u_1},f_{u_2})$.\\
Then $\beta_u=(d_Y\circ g)_u=d_Y(g_u)=d_Y(f_{u_1},f_{u_2})=0$, as desired.\\
{\it End of proof of Claim.}\\
\\
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall p,q\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d_X(p,q)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(f_p,f_q)<\vep\,\,)$.\\
Given $\vep>0$. \qquad Want: $\exists\delta>0$ s.t., $\forall p,q\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,d_X(p,q)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(f_p,f_q)<\vep\,\,)$.\\
Since $V$ is compact, by the Claim and \tref{thm-alpha-small-implies-beta-small},\\
\phx\qquad choose $\delta>0$ s.t. \qquad $\forall u\in V$, \quad $(\,\,\alpha_u<\delta\,\,)\,\,\Rightarrow\,\,(\,\,\beta_u<\vep\,\,)$.\\
Then $\delta>0$. Want: $\forall p,q\in\D_f$, $(\,\,d_X(p,q)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(f_p,f_q)<\vep\,\,)$.\\
Given $p,q\in\D_f$. \qquad Want:  $(\,\,d_X(p,q)<\delta\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(f_p,f_q)<\vep\,\,)$.\\
Assume: $d_X(p,q)<\delta$. \qquad Want: $d_Y(f_p,f_q)<\vep$.\\
We have $p,q\in\D_f=X$, \qquad so $(p,q)\in X\times X$.\\
Let $u:=(p,q)$. \qquad Then $u\in X\times X=V$.\\
We have $\alpha_u=\alpha(u)=\alpha(p,q)=d_X(p,q)<\delta$.\\
Then, by choice of $\delta$, we see that: $\beta_u<\vep$.\\
Since $u=(p,q)$, we get: \qquad $u_1=p$ \quad and \quad $u_2=q$.\\
It follows that: \qquad $g_u=(f\times f)_u=(f_{u_1},f_{u_2})=(f_p,f_q)$.\\
Then \quad $d_Y(f_p,f_q)=d_Y(g_u)=(d_Y\circ g)_u=\beta_u<\vep$, \quad as desired.
\end{proof}

\subsection{Uniform limits and continuity}

\begin{defn}
Let $X$ and $Y$ be sets, $f\in(Y^X)^\N$, $p\in X$.\\
\phx\qquad\qquad Then $f_\bullet(p)\in Y^\N$ is defined by: \quad $\forall j\in\N$, $(f_\bullet(p))_j=f_j(p)$.
\end{defn}

\begin{defn}
Let $X$ be a set and let $Y$ be a metric space.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ Let $f\in(Y^X)^\N$ and let $g\in Y^X$.\\
By $f\to g$ pointwise\index[terms]{pointwise limit} from $X$ to $Y$, we mean:\\
\phx\qquad\qquad $\forall p\in X$, \quad $f_\bullet(p)\to g(p)$ in $Y$.\\
By $f\to g$ uniformly\index[terms]{uniform limit} from $X$ to $Y$, we mean:\\
\phx\qquad\qquad $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$, $\forall p\in X$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(\,f_j(p)\,,\,g(p)\,)\,<\,\vep\,\,)$.
\end{defn}

Up to reordering quantifiers, the definitions of uniform and pointwise limits are the same:

\begin{thm}
Let $X$ be a set and let $Y$ be a metric space.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ Let $f\in(Y^X)^\N$ and let $g\in Y^X$.\\
Then: \quad $[$ $f\to g$ pointwise\index[terms]{pointwise limit} from $X$ to $Y$ $]$ \qquad $\Leftrightarrow$\\
\phx\qquad\quad $\,\,\,[$ $\forall p\in X$, $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(\,f_j(p)\,,\,g(p)\,)\,<\,\vep\,\,)$ $]$.
\end{thm}

\begin{thm}
Let $X:=[0;1]$ and let $Y:=\R$.\\
Define $f\in(Y^X)^\N$ by: $\forall j\in\N$, $\forall w\in X$, $f_j(w)=w^j$. \quad
Let $g:=\chi_{\{1\}}^X$.
Then: \qquad\quad $\,\,\,\,[$ \quad $\,\,\,f\to g$ pointwise from $X$ to $Y$ $\,\,\,\,\,]$\\
\phx\qquad\quad $\&$ \qquad $[$ $\neg($ $f\to g$ uniformly from $X$ to $Y$ $)$ $]$\\
\phx\qquad\quad $\&$ \qquad $[$ $\forall j\in\N$, \quad $f_j$ is continuous at $1$ from $X$ to $Y$ $\,\,\,\,]$\\
\phx\qquad\quad $\&$ \qquad $[$ \qquad\quad $\neg\,($ $\,g$ is continuous at $1$ from $X$ to $Y$ $)$ $]$.
\end{thm}

\begin{thm}\wrlab{thm-unif-lim-contin-at-implies-contin-at}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad Let $f\in(Y^X)^\N$, \quad let $g\in Y^X$ \quad and \quad let $q\in X$.\\
Assume: \qquad\qquad $[$ $f\to g$ uniformly from $X$ to $Y$ $]$\\
\phx\qquad\qquad\quad $\&$ \quad $\,\,\,[$ $\forall j\in\N$, $f_j$ is continuous at $q$ from $X$ to $Y$ $]$\\
Then: \qquad\qquad\quad $\,[$ \qquad\quad $\,\,\,g\,$ is continuous at $q$ from $X$ to $Y$ $]$.\\
\end{thm}

\begin{proof}
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall p\in X$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,d(p,q)<\delta\,\,]\,\,\Rightarrow\,\,[\,\,d(\,g(p)\,,\,g(q)\,)\,<\,\vep\,\,]$.\\
Given $\vep>0$. \qquad Want: $\exists\delta>0$ s.t., $\forall p\in X$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,d(p,q)<\delta\,\,]\,\,\Rightarrow\,\,[\,\,d(\,g(p)\,,\,g(q)\,)\,<\,\vep\,\,]$.\\
Since $f\to g$ uniformly from $X$ to $Y$,\\
\phx\qquad\qquad choose $K\in\N$ s.t., $\forall j\in X$, $\forall z\in X$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,j\ge K\,\,]\,\,\Rightarrow\,\,[\,\,d(\,f_j(z)\,,\,g(z)\,)\,<\,\vep/3\,\,]$.\\
By hypothesis, $\forall j\in\N$,\\
\phx\qquad $\,\,f_j$ is continuous at $q$ from $X$ to $Y$.\\
Then $f_K$ is continuous at $q$ from $X$ to $Y$. \qquad Let $h:=f_K$.\\
Then $\,\,h\,\,$ is continuous at $q$ from $X$ to $Y$, so\\
\phx\qquad\qquad choose $\delta>0$ s.t., $\forall p\in X$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,d(p,q)<\delta\,\,]\,\,\Rightarrow\,\,[\,\,d(\,h(p)\,,\,h(q)\,)\,<\,\vep/3\,\,]$.\\
Then $\delta>0$. $\,$ Want: $\forall p\in X$, $[\,\,d(p,q)<\delta\,\,]\,\,\Rightarrow\,\,[\,\,d(\,g(p)\,,\,g(q)\,)\,<\,\vep\,\,]$.\\
Given $p\in X$. \qquad\quad $\,$ Want: $[\,\,d(p,q)<\delta\,\,]\,\,\Rightarrow\,\,[\,\,d(\,g(p)\,,\,g(q)\,)\,<\,\vep\,\,]$.\\
\phx\qquad\qquad\qquad Assume: $d(p,q)<\delta$. \qquad $\,$ Want: $d(\,g(p)\,,\,g(q)\,)\,<\,\vep$.\\
Since $K\ge K$, it follows, from the choice of $K$, that\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\forall z\in X$, \qquad\quad $\,\,\,d(\,f_K(z)\,,\,g(z)\,)\,<\,\vep/3$.\\
Then: \quad\quad $\,\,\,d(\,f_K(p)\,,\,g(p)\,)\,<\,\vep/3\,$ \quad and \quad $\,d(\,f_K(q)\,,\,g(q)\,)\,<\,\vep/3$.\\
So, since $f_K=h$, we get:\\
\phx\qquad\qquad\quad $\,\,d(\,h(p)\,,\,g(p)\,)\,<\,\vep/3\,\,\,$ \quad and \quad $\,\,\,d(\,h(q)\,,\,g(q)\,)\,<\,\vep/3$.\\
As $d(\,g(p)\,,\,h(p)\,)\,=\,d(\,h(p)\,,\,g(p)\,)\,<\,\vep/3$,  $\quad\,\,d(\,g(p)\,,\,h(p)\,)\,<\,\vep/3$.\\
Since $d(p,q)<\delta$, by choice of $\delta$, we have: \quad $\,\,\,\,\,d(\,h(p)\,,\,h(q)\,)\,<\,\vep/3$.\\
Then $d(\,g(p)\,,\,g(q)\,)$\\
\phx\qquad\quad $\,\,\,\le\,(\,d(\,g(p)\,,\,h(p)\,)\,)\,+\,(\,d(\,h(p)\,,\,h(q)\,)\,)\,+\,(\,d(\,h(q)\,,\,g(p)\,)\,)$\\
\phx\qquad\quad $\,\,\,<\,(\,\vep/3\,)\,+\,(\,\vep/3\,)\,+\,(\,\vep/3\,)\,=\,\vep$, \qquad\qquad as desired.
\end{proof}

\begin{thm}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Let $f\in(C^X_Y)^\N$ \quad and \quad let $\,g\in Y^X$.\\
Assume: $f\to g$ uniformly from $X$ to $Y$. \quad Then: $g\in C^X_Y$.
\end{thm}

\begin{proof}
\qquad\qquad\quad $\,$ Want: $g$ is continuous from $X$ to $Y$.\\
\phx\qquad\qquad\qquad\qquad Want: $g$ is continuous on $\D_g$ from $X$ to $Y$.\\
\phx\qquad\qquad\qquad\qquad Want: $g$ is continuous on $X$ from $X$ to $Y$.\\
\phx\qquad\qquad Want: $\forall q\in X$, $\,\,g$ is continuous at $q$ from $X$ to $Y$.\\
Given $q\in X$. \qquad Want: $\,\,\,g$ is continuous at $q$ from $X$ to $Y$.\\
\phx\quad We have: \quad $\forall j\in\N$, \qquad $f_j\in C^X_Y$.\\
\phx\quad $\,$ Then: \qquad $\forall j\in\N$, \qquad $f_j$ is continuous from $X$ to $Y$.\\
\phx\quad $\,$ Then: \qquad $\forall j\in\N$, \qquad $f_j$ is continuous on $\D_{f_j}$ from $X$ to $Y$.\\
\phx\quad $\,$ Then: \qquad $\forall j\in\N$, \qquad $f_j$ is continuous on $X$ from $X$ to $Y$.\\
\phx\quad $\,$ Then: \qquad $\forall j\in\N$, \qquad $f_j$ is continuous at $q$ from $X$ to $Y$.\\
Then, by \tref{thm-unif-lim-contin-at-implies-contin-at}, $\,\,\,\,$
$g\,$ is continuous at $q$ from $X$ to $Y$.
\end{proof}

\begin{thm}\wrlab{thm-unifconv-implies-ptwiseconv}
Let $X$ and $Y$ be metric spaces.\\
Let $f\in(Y^X)^\N$ and let $g\in Y^X$.\\
Assume $f\to g$ uniformly from $X$ to $Y$.\\
Then $f\to g$ pointwise from $X$ to $Y$.
\end{thm}

\begin{proof}
Want: $\forall p\in X$, $f_\bullet(p)\to g(p)$ in $Y$.\\
Given $p\in X$. \qquad Want: $f_\bullet(p)\to g(p)$ in $Y$.\\
Want: $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(\,(f_\bullet(p))_j\,,\,g(p)\,)\,<\,\vep\,\,)$.\\
Given $\vep>0$. \qquad Want: $\exists K\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(\,(f_\bullet(p))_j\,,\,g(p)\,)\,<\,\vep\,\,)$.\\
Since $f\to g$ uniformly from $X$ to $Y$, choose $K\in\N$ s.t., $\forall j\in\N$, $\forall q\in X$,\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d_Y(\,f_j(q)\,,\,g(q)\,)\,<\,\vep\,\,)$.\\
Then $K\in\N$. Want: $\forall j\in\N$, $(\,j\ge K\,)\,\Rightarrow\,(\,d_Y(\,(f_\bullet(p))_j\,,\,g(p)\,)\,<\,\vep\,\,)$.\\
Given $j\in\N$. \qquad\quad $\,\,$ Want: $(\,j\ge K\,)\,\Rightarrow\,(\,d_Y(\,(f_\bullet(p))_j\,,\,g(p)\,)\,<\,\vep\,\,)$.\\
\phx\qquad\qquad\quad Assume: $j\ge K$. $\,\,\,$ \qquad Want: $d_Y(\,(f_\bullet(p))_j\,,\,g(p)\,)\,<\,\vep$.\\
Since $j\in\N$, since $p\in X$ and since $j\ge K$, by choice of $K$,\\
\phx\qquad\qquad\quad $\,$ we see that: \quad $d_Y(\,f_j(p)\,,\,g(p)\,)\,<\,\vep$.\\
Then $\,d_Y(\,(f_\bullet(p))_j\,,\,g(p)\,)\,=\,d_Y(\,f_j(p)\,,\,g(p)\,)\,<\,\vep$, as desired.
\end{proof}

In HW\#13-2, we see that Lipschitz maps carry Cauchy sequences to Cauchy sequences.
Mere continuity is not enough:

\begin{thm}
Let $X:=(-1;1)$ and let $Y:=\R$.\\
Define $f:X\to Y$ by $\forall w\in X$, $f_w=w/\sqrt{1-w^2}$.\\
Define $s\in X^\N$ by $\forall j\in\N$, $s_j=j/(j+1)$.\\
Then: $f$ is continuous from $X$ to $Y$ \qquad\qquad and\\
\phx\qquad $\,\,s$ is Cauchy in $X$ \qquad\qquad\qquad\qquad and\\
\phx\qquad $\,\,f\circ s$ is not Cauchy in $Y$.
\end{thm}

\begin{defn}
Let $K$ and $Y$ be metric spaces.\\
Let $Z:=C^K_Y$ and $d:=d_Y$. \quad Assume: $K$ is compact and nonempty.\\
Then $d^K_Y:Z\times Z\to[0;\infty)$ is defined by:\\
\phx\qquad\quad $\forall f,g\in Z$, \qquad $d^K_Y(f,g)\quad=\quad\max\,\I_{d\circ(f,g)}$.
\end{defn}

Let $K$, $Y$ be metric spaces. Assume $K$ is empty. Then $C^K_Y=\{\emptyset\}$.\\
That is, the only function in $C^K_Y$ is the empty function.\\
\\
We have little use for the empty metric space, but,\\
for the sake of completeness, we define $d^K_Y$ when $K$ is empty:

\begin{defn}
Let $K$ and $Y$ be metric spaces.\\
Let $Z:=C^K_Y$. \qquad\qquad Assume: $K$ is empty.\\
Then $d^K_Y:Z\times Z\to[0;\infty)$ is defined by:\\
\phx\qquad\quad $\forall f,g\in Z$, \qquad $d^K_Y(f,g)\quad=\quad0$.
\end{defn}

\begin{thm}
Let $K$ and $Y$ be metric spaces.\\
Let $Z:=C^K_Y$. \qquad Assume: $K$ is compact. \qquad
Then $d^K_Y\in\scrm(C^K_Y)$.
\end{thm}

The metric $d^K_Y$ is sometimes called the \bfindex[uniform metric] on $C^K_Y$.\\
It is the standard metric on $C^K_Y$.

By sloppiness, we use $C^K_Y$ to denote the metric space $(C^K_Y,d^K_Y)$.\\
Note that $d_{C^K_Y}=d^K_Y$.
By slopiness, we sometimes use $d$ to denote $d^K_Y$.

\begin{thm}\wrlab{thm-conv-in-unif-metric-iff-conv-unif}
Let $K$ and $Y$ be metric spaces.\\
Let $Z:=C^K_Y$, $f\in Z^\N$, $g\in Z$. \quad
Assume $K$ is compact and nonempty.\\
Then: \quad $(\,\,f\to g\hbox{ in }Z\,\,)\,\,\Leftrightarrow\,\,(\,\,f\to g\hbox{ uniformly from }K\hbox{ to }Y\,\,)$.
\end{thm}

\begin{proof}
{\it Proof of $\Rightarrow$:}
Unassigned Homework.
{\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:} \quad Let $d:=d^K_Y$.\\
Assume: $f\to g$ uniformly from $K$ to $Y$. \quad
Want: $f\to g$ in $Z$.\\
Want: $\forall\vep>0$, $\exists L\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,j\ge L\,\,)\,\,\Rightarrow\,\,(\,\,d(\,f_j\,,\,g\,)<\vep\,\,)$.\\
Given $\vep>0$. \qquad Want: $\exists L\in\N$ s.t., $\forall j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,j\ge L\,\,)\,\,\Rightarrow\,\,(\,\,d(\,f_j\,,\,g\,)<\vep\,\,)$.\\
Since $f\to g$ uniformly from $K$ to $Y$, choose $L\in\N$ s.t., $\forall j\in\N$, $\forall p\in K$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $(\,\,j\ge L\,\,)\,\,\Rightarrow\,\,(\,\,d(\,f_j(p)\,,\,g(p)\,)<\vep\,\,)$.\\
Then $L\in\N$. \quad Want: $\forall j\in\N$, $\,(\,\,j\ge L\,\,)\,\,\Rightarrow\,\,(\,\,d(\,f_j\,,\,g\,)<\vep\,\,)$.\\
\phx\qquad Given $j\in\N$. \quad $\,\,\,$ Want: $\,(\,\,j\ge L\,\,)\,\,\Rightarrow\,\,(\,\,d(\,f_j\,,\,g\,)<\vep\,\,)$.\\
\phx\qquad\qquad\qquad\qquad $\,$ Assume: $j\ge L$. \quad Want: $d(\,f_j\,,\,g\,)<\vep$.\\
As $d:Z\times Z\to[0;\infty)$, we get $d(\,f_j\,,\,g\,)\in[0;\infty)$, and so $d(\,f_j\,,\,g\,)\ne\frownie$.\\
Let $\alpha:=d_Y\circ(f,g)$. \qquad Then $d(\,f_j\,,\,g\,)=\max\,\I_\alpha$.\\
Since $\max\,\I_\alpha=d(\,f_j\,,\,g\,)\ne\frownie$, we get: \quad $\max\,\I_\alpha\in\I_\alpha$.\\
Since $d(\,f_j\,,\,g\,)=\max\,\I_\alpha\in\I_\alpha$,\\
\phx\qquad choose $p\in\D_\alpha$ s.t.~$d(\,f_j\,,\,g\,)=\alpha_p$. \qquad\qquad Want: $\alpha_p<\vep$.\\
Since $\alpha=d_Y\circ(f_j,g)$, it follows that $\D_\alpha\subseteq\D_{(f_j,g)}$.\\
Since $f_j,g\in C^K_Y$, we get: \qquad $\D_{f_j}=K=\D_g$.\\
Then $p\in\D_\alpha\subseteq \D_{(f_j,g)}=\D_{f_j}\bigcap\D_g=K\bigcap K=K$.\\
So, since $j\ge L$, by the choice of $L$, we get: $d_Y(\,f_j(p)\,,\,g(p)\,)<\vep$.\\
Then $\,\,\,\alpha_p=(d_Y\circ(f_j,g))_p=d_Y(\,(f_j,g)_p\,)=d_Y(\,f_j(p)\,,\,g(p)\,)<\vep$.\\
{\it End of proof of $\Leftarrow$.}
\end{proof}

\begin{defn}
Let $X$ and $Y$ be sets, $S\subseteq Y^X$ and $p\in X$.\\
\phx\qquad\qquad Then $\ev^S_p:S\to Y$ is defined by: \quad $\forall f\in S$, $\,$ $(\ev^S_p)_f=f_p$.
\end{defn}

\begin{thm}\wrlab{thm-Cauchy-and-ptwiseconv-implies-unifconv}
Let $K$ and $Y$ be metric spaces.\\
Let \quad $f\in(C^K_Y)^\N$ \quad and let \quad $g\in Y^K$.\\
Assume: \qquad $K$ is compact \qquad\qquad\qquad\qquad and\\
\phx\qquad\qquad\quad $\,$ $f$ is Cauchy in $C^K_Y$ \qquad\qquad\qquad and\\
\phx\qquad\qquad\quad $\,$ $f\to g$ pointwise from $K$ to $Y$.\\
Then: \qquad $\,\,\,$ $f\to g$ uniformly from $K$ to $Y$.
\end{thm}

\begin{proof}
Want: $\forall\vep>0$, $\exists L\in\N$ s.t., $\forall i\in\N$, $\forall p\in K$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,i\ge L\,\,]\,\,\Rightarrow\,\,[\,\,d_Y(\,f_i(p)\,,\,g(p)\,)<\vep\,\,]$.\\
Given $\vep>0$. \qquad Want: $\exists L\in\N$ s.t., $\forall i\in\N$, $\forall p\in K$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,i\ge L\,\,]\,\,\Rightarrow\,\,[\,\,d_Y(\,f_i(p)\,,\,g(p)\,]<\vep\,\,)$.\\
Let $d:=d^K_Y$. \quad Since $f$ is Cauchy in $C^K_Y$, choose $L\in\N$ s.t., $\forall i,j\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,i,j\ge L\,\,]\,\,\Rightarrow\,\,[\,\,d(\,f_i\,,\,f_j\,)<\vep/2\,\,]$.\\
Then $L\in\N$. \quad Want: $\forall i\in\N$, $\forall p\in K$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,i\ge L\,\,]\,\,\Rightarrow\,\,[\,\,d_Y(\,f_i(p)\,,\,g(p)\,]<\vep\,\,)$.\\
Given $i\in\N$, $p\in K$. $\,$ Want: $[\,\,i\ge L\,\,]\,\,\Rightarrow\,\,[\,\,d_Y(\,f_i(p)\,,\,g(p)\,)<\vep\,\,]$.\\
\phx\qquad\qquad\qquad\quad $\,$ Assume: $i\ge L$. \quad Want: $d_Y(\,f_i(p)\,,\,g(p)\,)<\vep$.\\
By assumption, $f\to g$ pointwise from $K$ to $Y$, so $f_\bullet(p)\to g(p)$ in $Y$,\\
so choose $M\in\N$ s.t., $\forall j\in\N$, $[\,j\ge M\,]\Rightarrow[\,d_Y(\,(f_\bullet(p))_j\,,\,g(p)\,)<\vep/2\,]$.\\
Let $j:=\max\{L,M\}$. \qquad
Then $j\in\N$ and $j\ge L$ and $j\ge M$.\\
Since $j\ge L$, by choice of $L$, we get: \quad $d(\,f_i\,,\,f_j\,)<\vep/2$.\\
Since $j\ge M$, by choice of $M$, we get: \quad $d(\,(f_\bullet(p))_j\,,\,g(p)\,)<\vep/2$.\\
\phx\qquad\qquad\quad $\,\,$ Then: \quad $d(\,f_j(p)\,,\,g(p)\,)=d(\,(f_\bullet(p))_j\,,\,g(p)\,)<\vep/2$.\\
Let $E:=\ev^Z_p$. \qquad By HW\#13-4, $E$ is Lipschitz-$1$ from $Z$ to $Y$.\\
Then $d_Y(\,E_{f_i}\,,\,E_{f_j}\,)\le1\cdot(d(\,f_i\,,\,f_j\,))$.\\
We have: \qquad $\,E_{f_i}=E(f_i)=(\ev^K_p)(f_i)=f_i(p)$\\
\phx\qquad and \qquad $E_{f_j}=E(f_j)=(\ev^K_p)(f_j)=f_j(p)$.\\
Then $\,\,d_Y(\,f_i(p)\,,\,f_j(p)\,)=d_Y(\,E_{f_i}\,,\,E_{f_j}\,)$\\
\phx\qquad\qquad\qquad\qquad\qquad $\le1\cdot(d(\,f_i\,,\,f_j\,))=d(\,f_i\,,\,f_j\,)<\vep/2$.\\
So, since $d_Y(\,f_j(p)\,,\,g(p)\,)<\vep/2$, we conclude that:\\
\phx\qquad\qquad $[\,d_Y(\,f_i(p)\,,\,f_j(p)\,)\,]\,+\,[\,d_Y(\,f_j(p)\,,\,g(p)\,)\,]\,\,<\,\,[\,\vep/2\,]\,+\,[\,\vep/2\,]$.\\
By the Triangle Inequality,\\
\phx\qquad\quad $d_Y(\,f_i(p)\,,\,g(p)\,)\,\,\le\,\,[\,d_Y(\,f_i(p)\,,\,f_j(p)\,)\,]\,+\,[\,d_Y(\,f_j(p)\,,\,g(p)\,)\,]$.\\
Then \quad $d_Y(\,f_i(p)\,,\,g(p)\,)\,\,\le\,\,[\,d_Y(\,f_i(p)\,,\,f_j(p)\,)\,]\,+\,[\,d_Y(\,f_j(p)\,,\,g(p)\,)\,]$\\
\phx\quad\qquad\qquad\qquad\qquad\quad $\,\,\,<\,\,[\,\vep/2\,]\,+\,[\,\vep/2\,]\,\,=\,\,\vep$, $\,$ as desired.
\end{proof}

\begin{thm}\wrlab{thm-cpt-implies-complete}
Let $K$ be a compact metric space.\\
\phx\qquad\qquad\qquad\qquad\quad Then $K$ is complete.
\end{thm}

\begin{proof}
Since $K$ is compact, by $\Rightarrow$ of \tref{thm-cpt-iff-proper-bdd},\\
\phx\qquad\qquad we conclude that: \quad $K$ proper and bounded.\\
Then $K$ is proper, so, by \tref{thm-proper-implies-complete}, $K$ is complete.
\end{proof}

\begin{defn}
Let $Y$ be a metric space and $s\in Y^\N$.\\
\phx\qquad\qquad\qquad Then: \qquad $Y$\!-\,$\lim s\,\,:=\,\,\UE\{q\in Y\,|\,s\to q\hbox{ in }Y\}$.
\end{defn}

Writers often omit the ``$Y$-'' and simply write ``$\lim s$''.
Also, common is to pick an unbound ``dummy variable'', like $j$, and write ``$\ds{\lim_{j\to\infty}\,s_j}$''.

\begin{thm}
Let $Y$ be a metric space, $s\in Y^\N$, $p\in Y$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,$ Then: \quad $(\,\,s\to p\hbox{ in }Y\,\,)\,\,\Leftrightarrow\,\,(\,\,Y\!\hbox{-}\lim s=p\,\,)$.
\end{thm}

\noindent
In the preceding theorem, $\Rightarrow$ follows from uniqueness of limits.\\
In the preceding theorem, $\Leftarrow$ follows from the definition of $Y\!\hbox{-}\lim$.\\

\begin{thm}
Let $Y$ be a metric space and let $s\in Y^\N$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,$ Assume $s$ is convergent in $Y$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,$ Then $\frownie\,\,\ne\,\,Y\!\hbox{-}\lim s\,\,\in\,\,Y$.
\end{thm}

\begin{thm}
Let $X$ be a set and let $Y$ be a metric space.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,$ Let $f\in(Y^X)^\N$ and let $g\in Y^X$.\\
\phx\qquad\qquad $\,\,\,$ Then: \qquad $(\,\,f\to g\hbox{ pointwise from }X\hbox{ to }Y\,\,)$\\
\phx\qquad\qquad\qquad\quad $\,\,\,$ $\Leftrightarrow\,\,\,\,(\,\,\forall p\in X,\,\,\,f_\bullet(p)\to g(p)\hbox{ in }Y\,\,)$\\
\phx\qquad\qquad\qquad\quad $\,\,\,$ $\Leftrightarrow\,\,\,\,(\,\,\forall p\in X,\,\,\,Y\!\hbox{-}\lim(f_\bullet(p))\,\,=\,\,g(p)\,\,)$.
\end{thm}

Note that, in the following theorem, by \tref{thm-cpt-implies-complete}, $K$ is complete.
Thus both the domain $K$ and the target $Y$ are complete metric spaces,
and the conclusion is that $C^K_Y$ is a complete metric space.

We do not attempt to generalize by weakening the assumption that $K$ is compact;
in fact, in this writeup, we do not specify any standard metric $C^K_Y$, except in the case where $K$ is compact.

In a later theorem (\tref{cky-may-be-nonproper}), we show that $C^K_Y$ may be nonproper, and therefore noncompact.

\begin{thm}\wrlab{thm-completeness-of-cpt-to-complete-contin-fnspace}
Let $K$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad\qquad Assume: \quad $K$ is compact \quad and \quad $Y$ is complete.\\
\phx\qquad\qquad\qquad\qquad Then: \qquad\qquad\qquad $\,$ $C^K_Y$ is complete.
\end{thm}

\begin{proof}
Let $Z:=C^K_Y$. \quad
Want: $Z$ is complete.\\
Want: $\forall f\in Z^\N$, $\,$ $(\,f\hbox{ is Cauchy in }Z\,)\,\Rightarrow(\,f\hbox{ is convergent in }Z\,)$.\\
Given $f\in Z^\N$. $\,$ Want: $(\,f\hbox{ is Cauchy in }Z\,)\,\Rightarrow(\,f\hbox{ is convergent in }Z\,)$.\\
Assume: $f\hbox{ is Cauchy in }Z$. \qquad Want: $f\hbox{ is convergent in }Z$.\\
Want: $\exists g\in Z$ s.t.~$f\to g$ in $Z$.\\
By HW\#13-5, $\forall p\in K$, $f_\bullet(p)$ is Cauchy in $Y$.\\
So, since $Y$ is complete, we conclude:\\
\phx\qquad $\forall p\in K$, $f_\bullet(p)$ is convergent in $Y$.\\
Define $g\in Y^K$ by: \quad $\forall p\in K$, $\,$ $g(p)=Y\!\hbox{-}\lim(f_\bullet(p))$.\\
Then $g\in Y^K=Z$. \qquad Want: $f\to g$ in $Z$.\\
Then: $\forall p\in K$, $\,$ $f_\bullet(p)\to g(p)$ in $Y$.\\
Then: $f\to g$ pointwise from $K$ to $Y$.\\
So, since $f$ is Cauchy in $Z$ and since $Z=C^K_Y$, it follows,\\
\phx\quad by \tref{thm-Cauchy-and-ptwiseconv-implies-unifconv}, that $f\to g$ uniformly from $K$ to $Y$.\\
Then, by \tref{thm-conv-in-unif-metric-iff-conv-unif}, $f\to g$ in $Z$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-ptwise-limit-discontin-implies-not-unif-subconvergent}
Let $K$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad\qquad\qquad Assume: $K$ is compact.\\
\phx\qquad\qquad\qquad\qquad\qquad Let $Z:=C^K_Y$, let $f\in Z^\N$ and let $h\in Y^K\backslash Z$.\\
\phx\qquad\qquad\qquad\qquad\qquad Assume: $f\to h$ pointwise from $K$ to $Y$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: $f$ is not subconvergent in $Z$.
\end{thm}

\begin{proof}
Assume $f$ is subconvergent in $Z$. \quad Want: Contradiction.\\
Choose a subsequence $\phi$ of $f$ s.t.~$\phi$ is convergent in $Z$.\\
Choose $g\in Z$ s.t.~$\phi\to g$ in $Z$.\\
Since $g\in Z$ and $h\notin Z$, we see that: $g\ne h$.\\
Choose $p\in K$ s.t. $g(p)\ne h(p)$.\\
Since $\phi\to g$ in $Z$ and since $Z=C^K_Y$, it follows,\\
\phx\quad by \tref{thm-Cauchy-and-ptwiseconv-implies-unifconv}, that $\phi\to g$ uniformly from $K$ to $Y$.\\
Then, by \tref{thm-unifconv-implies-ptwiseconv},we have: \quad $\phi\to g$ pointwise from $K$ to $Y$.\\
\phx\qquad\qquad\qquad By hypothesis, \quad\qquad $\,f\to h$ pointwise from $K$ to $Y$.\\
Then: \quad $\phi_\bullet(p)\to g(p)$ in $Y$ \quad and \quad $f_\bullet(p)\to h(p)$ in $Y$.\\
Let $E:=\ev^Z_p$. \qquad
Recall: $\phi$ is a subsequence of $f$.\\
\phx\qquad\qquad $\,\,$ Then: \qquad $E\circ\phi$ is a subsequence of $E\circ f$.\\
So, since $E\circ\phi=\phi_\bullet(p)$ and since $E\circ f=f_\bullet(p)$,\\
\phx\qquad we conclude that $\phi_\bullet(p)$ is a subsequence of $f_\bullet(p)$.\\
So, since $f_\bullet(p)\to h(p)$ in $Y$, we see that $\phi_\bullet(p)\to h(p)$ in $Y$.\\
Since $\phi_\bullet(p)\to g(p)$ in $Y$ and since $\phi_\bullet(p)\to h(p)$ in $Y$, we conclude,\\
\phx\qquad by uniqueness of limits, that $g(p)=h(p)$.\\
By choice of $p$, we have: \qquad\qquad $\,\,g(p)\ne h(p)$. \quad Contradiction.
\end{proof}

\begin{thm}\wrlab{cky-may-be-nonproper}
Let $K:=[0;1]$ and let $Y:=\R$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: \quad $C^K_Y$ is nonproper.
\end{thm}

\begin{proof}
Let $Z:=C^K_Y$. \qquad Want: $Z$ is nonproper.\\
Want: $\exists f\in Z_\bi^\N$ s.t.~$f$ is not subconvergent in $Z$.\\
Define $f\in Z^\N$ by: \qquad $\forall j\in\N$, $\forall x\in K$, \quad $f_j(x)=x^j$.\\
Let $g:=C_K^0$. \quad Then $g\in C^K_Y=Z$.\\
Also: $\forall j\in\N$, we have $d_Z(f_j,g)=d^K_Y(f_j,g)=1$, so $f_j\in B_Z(g,1)$.\\
Then $\I_f\subseteq B_Z(g,1)$. \quad
Then $\I_f$ is bounded in $Z$.\\
So, as $f\in Z^\N$, we get $f\in Z^\N_\bi$. \quad
Want: $f$ is not subconvergent in $Z$.\\
Let $h:=\chi_{\{1\}}^K$. \qquad Then $h\in Y^K$.\\
As $h$ is not continuous at $1$ from $K$ to $Y$, we get: $h\notin Z$.\\
Then \quad $f\to h$ pointwise from $K$ to $Y$ \quad and \quad $h\in Y^X\backslash Z$.\\
Then, by \tref{thm-ptwise-limit-discontin-implies-not-unif-subconvergent}, we get: $f$ is not subconvergent in $Z$.
\end{proof}

From \tref{thm-completeness-of-cpt-to-complete-contin-fnspace} and \tref{cky-may-be-nonproper},
we see that the metric space $C^{[0;1]}_\R$ is both complete and nonproper.
So, while proper implies complete, the converse is NOT true; complete does not imply proper.

We already observed this; see \tref{thm-hatbbr-complete} and \tref{thm-hatbbr-nonproper}.

\section{Derivatives}

\subsection{$\DNZ$ and $\BNZ$ and $\CVZ$}

\begin{defn}
Let $X$ be a metric space, $f$ a function, $p\in X$.\\
\phx\quad Then \quad $f$ is \bfindex[defined near] $p$ in $X$ \quad means:\\
\phx\qquad\qquad\qquad $\exists B\in\scrb_X(p)$ s.t.~$B\subseteq\D_f$.
\end{defn}

\begin{defn}
Let $X$ and $Y$ be metric spaces.\\
\phx\qquad\qquad\qquad\qquad\qquad Let $f:X\drta Y$ and let $p\in X$.\\
\phx\quad Then \quad $f$ is \bfindex[bounded near] $p$ from $X$ to $Y$ \quad means:\\
\phx\qquad\qquad\qquad $\exists B\in\scrb_X(p)$ s.t.~$(\,\,B\subseteq\D_f\,\,)\,\,\&\,\,(\,\,f_*B\hbox{ is bounded in }Y\,\,)$.
\end{defn}

\begin{defn}
We define:\\
\phx\qquad $\DNZ\,\,:=\,\,\{\,\,f:\R\drta\R\,\,|\,\,f\hbox{ is defined near }0\hbox{ in }\R\,\,\}$\\
\phx and $\,\BNZ\,\,:=\,\,\{\,\,f:\R\drta\R\,\,|\,\,f\hbox{ is bounded near }0\hbox{ from }\R\hbox{ to }\R\,\,\}$\\
\phx and $\,\CVZ\,\,:=\,\,\{\,\,f\in\DNZ\,\,|\,\,(\,\,f\hbox{ is continuous }0\hbox{ from }\R\hbox{ to }\R\,\,)\,\,\&$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,h_0=0\,\,)\qquad\qquad\qquad\qquad\}$.
\end{defn}

\begin{thm}\wrlab{thm-bnz-quantified-equiv}
Let $f:\R\to\R$. Then:\\
\phx\quad $[\,f\in\BNZ\,]\,\Leftrightarrow\,[\,\exists\delta,K>0~\hbox{s.t.},\forall x\in\R,\,(\,|x|<\delta\,)\,\Rightarrow(\,|f_x|\le K\,)\,]$.
\end{thm}

\begin{proof}
{\it Proof of $\Leftarrow$:} Unassigned HW. {\it End of proof of $\Leftarrow$.}\\
\\
{\it Proof of $\Rightarrow$:}
Assume: $f\in\BNZ$.\\
Want: $\exists\delta,K>0~\hbox{s.t.},\forall x\in\R, (\,|x|<\delta\,)\,\Rightarrow(\,|f_x|<K\,)$.\\
Since $f\in\BNZ$, we see that $f$ is bounded near $0$ from $\R$ to $\R$,\\
\phx\qquad so choose $B\in\scrb_\R(0)$ s.t.~$(\,\B\subseteq\D_f\,\,)\,\,\&\,\,(\,\,f_*B\hbox{ is bounded in }\R\,\,)$.\\
Since $f_*B$ is bounded in $\R$, choose $C\in\scrb_\R$ s.t.~$f_*B\subseteq C$.\\
By the Recentering Theorem, choose $D\in\scrb_\R(0)$ s.t.~$C\subseteq D$.\\
Since $B,D\in\scrb_\R(0)$, choose $\delta,K>0$ s.t.\\
\phx\qquad\qquad\qquad $(\,\,B=B_\R(0,\delta)\,\,)\,\,\&\,\,(\,\,D=B_\R(0,K)\,\,)$.\\
Then $\delta,K>0$. \quad\quad $\,\,\,$ Want: $\forall x\in\R, (\,|x|<\delta\,)\,\Rightarrow(\,|f_x|<K\,)$.\\
Given $x\in\R$. \qquad\quad $\,\,\,$ Want: $(\,|x|<\delta\,)\,\Rightarrow(\,|f_x|<K\,)$.\\
Assume $|x|<\delta$. \qquad $\,$ Want: $|f_x|<K$.\\
Since $B\subseteq\D_f$, we get: $B\bigcap\D_f=B$.\\
We have $d_\R(x,0)=|x-0|=|x|<\delta$, and so $x\in B_\R(0,\delta)$.\\
Then $x\in B_\R(0,\delta)=B=B\bigcap\D_f$, so $f_x\in f_*B$.\\
Then $f_x\in f_*B\subseteq C\subseteq D=B_\R(0,K)$, so $d_\R(f_x,0)<K$.\\
So, since $d_\R(f_x,0)=|f_x-0|=|f_x|$,\\
\phx\qquad\qquad we conclude that $|f_x|<K$, as desired.\\
{\it End of proof of $\Leftarrow$.}
\end{proof}

\begin{thm}\wrlab{thm-cvz-quantified-equiv}
Let $f:\R\to\R$. Then:\\
\phx $[\,f\in\CVZ\,]\,\Leftrightarrow\,[\,\forall\vep>0,\exists\delta>0~\hbox{s.t.},\forall x\in\R, (\,|x|<\delta\,)\,\Rightarrow(\,|f_x|\le\vep\,)\,]$.
\end{thm}

\begin{proof}
This is HW\#14-1.
\end{proof}

\begin{thm}\wrlab{thm-cvz-and-output-ball-gives-input-ball}
Let $f\in\CVZ$ and let $B\in\scrb_\R(0)$.\\
Then: $\exists A\in\scrb_\R(0)$ s.t.~$(\,\,A\subseteq\D_f\,\,)\,\,\&\,\,(\,\,f_*A\subseteq B\,\,)$.
\end{thm}

\begin{proof}
Since $B\in\scrb_\R(0)$, choose $\vep>0$ s.t.~$B=B_\R(0,\vep)$.\\
By \tref{thm-cvz-quantified-equiv}, choose $\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|f_x|<\vep\,\,)$.\\
Let $A:=B_\R(0,\delta)$. $\,$ Then $A\in\scrb_\R(0)$. $\,$ Want: $(\,A\subseteq\D_f\,)\,\&\,(\,f_*A\subseteq B\,)$.\\
\\
{\it Claim:} $A\subseteq\D_f$.\\
{\it Proof of Claim:}\\
Want: $\forall x\in A$, $x\in\D_f$. \qquad Given $x\in A$. \qquad Want: $x\in\D_f$.\\
Since $x\in A=B_\R(0,\delta)$, we get $d_\R(x,0)<\delta$.\\
Then $|x|=|x-0|=d_\R(x,0)<\delta$, so, by choice of $\delta$, we get: \quad $|f_x|<\vep$.\\
Then $f_x\ne\frownie$, and so $x\in\D_f$, as desired.\\
{\it End of proof of Claim.}\\
\\
By the claim, it suffices to show: $f_*A\subseteq B$.\\
Want: $\forall y\in f_*A$, $y\in B$. \qquad
Given $y\in f_*A$. \qquad Want: $y\in B$.\\
Since $y\in f_*A$, choose $x\in A\bigcap\D_f$ s.t.~$y=f_x$.\\
Since $x\in A\bigcap\D_f\subseteq A=B_\R(0,\delta)$, we get: $d_\R(x,0)<\delta$.\\
Then $|x|=|x-0|=d_\R(x,0)<\delta$, so, by choice of $\delta$, we get: \quad $|f_x|<\vep$.\\
Then $d_\R(f_x,0)=|f_x-0|=|f_x|<\vep$, and so $f_x\in B_\R(0,\vep)$.\\
Then $y=f_x\in B_\R(0,\vep)=B$, as desired.
\end{proof}

\begin{thm}
$\DNZ\supseteq\BNZ\supseteq\CVZ$.
\end{thm}

\begin{proof}
By HW\#14-2, we have $\CVZ\subseteq\BNZ$. \quad Want: $\BNZ\subseteq\DNZ$.\\
Want: $\forall f\in\BNZ,\,f\in\DNZ$. \qquad
Given $f\in\BNZ$. \quad Want: $f\in\DNZ$.\\
Since $f\in\BNZ$, we see that $f$ is bounded near $0$ from $\R$ to $\R$,\\
\phx\qquad so choose $B\in\scrb_\R(0)$ s.t.~$(\,\B\subseteq\D_f\,\,)\,\,\&\,\,(\,\,f_*B\hbox{ is bounded in }\R\,\,)$.\\
Since $B\in\scrb_\R(0)$ and $B\subseteq\D_f$, we get: $f$ is defined near $0$ in $\R$.\\
Then $f\in\DNZ$, as desired.
\end{proof}

We showed graphs to show that\\
\phx\qquad not every element of $\BNZ$ is vanishing at $0$ \qquad\qquad and\\
\phx\qquad not every element of $\BNZ$ that vanishes at $0$\\
\phx\qquad\qquad\qquad\qquad is continuous at $0$ \qquad\qquad and\\
\phx\qquad not every element of $\DNZ$ is bounded near $0$.\\
Consequently, $\DNZ\supsetneq\BNZ\supsetneq\CVZ$.

\begin{defn}
Let $X$ be a metric space.\\
Let $f:X\drta\R$, let $g:X\to[0;\infty)$ and let $p\in X$.\\
By $g$ \bfindex[envelopes] $f$ near $p$ in $X$, we mean:\\
\phx\qquad $\exists B\in\scrb_X(p)$ \qquad s.t. \qquad $-g\le f\le g$ on $B$.
\end{defn}

\begin{thm}
Let $f:\R\drta\R$. \quad Then:\\
\phx\qquad\quad $\,[\,(\,f\in\BNZ\,)\,\Leftrightarrow\,(\,\exists K\ge0~\hbox{s.t.}~C^K_\R\hbox{ envelopes }f\hbox{ near }0\,)\,]$\\
\phx\qquad $\,\&\,[\,(\,f\in\CVZ\,)\,\Leftrightarrow\,(\,\forall\vep\ge0,\,C^\vep_\R\hbox{ envelopes }f\hbox{ near }0\,)\,]$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{defn}
Let $S\subseteq\R$ and let $T$ be a set of functionals.\\
\phx\qquad\qquad\qquad\qquad\qquad Then $\,\,S\,\,\cdot_\ss\,\,T\,\,:=\,\,\{\,a\cdot f\,|\,(\,a\in S\,)\,\&\,(\,f\in T\,)\,\}$.
\end{defn}

We typically omit the ``$\ss$'', and write ``$S\cdot T$''.

We often omit the ``$\cdot$'', and write $``ST$''.

\begin{defn}
Let $S$ and $T$ be a sets of functionals.\\
\phx\qquad\qquad\qquad\qquad Then $\,\,S\,\,+_\ss\,\,T\,\,:=\,\,\{\,f+g\,|\,(\,f\in S\,)\,\&\,(\,g\in T\,)\,\}$.\\
\phx\qquad\qquad\qquad\qquad Also, $\,\,S\,\,\cdot_\ss\,\,T\,\,:=\,\,\{\,f\cdot g\,|\,(\,f\in S\,)\,\&\,(\,g\in T\,)\,\}$.
\end{defn}

We typically omit the ``$\ss$'', and write ``$S+T$'' and ``$S\cdot T$''.

We often omit the ``$\cdot$'', and write $``ST$''.

For any two sets $S$ and $T$ of functionals, we have:\\
\phx\qquad\qquad\qquad $S+T=T+S$ \qquad and \qquad $ST=TS$.

\begin{thm}
Let \quad $f:\R\drta\R$, \qquad $g\in\DNZ$.\\
Assume: $f=g$ near $0$ in $\R$. \qquad\qquad Then: $\,\,\,f\in\DNZ$.
\end{thm}

\begin{thm}
Let \quad $f:\R\drta\R$, \qquad $g\in\BNZ$.\\
Assume: $f=g$ near $0$ in $\R$. \qquad\qquad Then: $\,\,\,f\in\BNZ$.
\end{thm}

\begin{thm}
Let \quad $f:\R\drta\R$, \qquad $g\in\CVZ$.\\
Assume: $f=g$ near $0$ in $\R$. \qquad\qquad Then: $\,\,\,f\in\CVZ$.
\end{thm}

The preceding three theorems express that $\DNZ$, $\BNZ$ and $\CVZ$ are all ``zero-local'' conditions.
That is, given the restriction $f$ to a ball around zero in $\R$,
we can determine whether or not $f$ satisfies the condition.
We do not need to know how $f$ behaves far away from zero.

\begin{defn}
Let $S$ be a set of functionals.
By $S$ is v.op.-closed, we mean: \quad $(\,\,\R\cdot S\subseteq S\,\,)\,\,\&\,\,(\,\,S+S\subseteq S\,\,)$.
\end{defn}

In the preceding definition, ``v.op.-closed'' is an abbreviation for ``vector-operation-closed''.
Unassigned HW: Show $\DNZ$ is v.op.-closed.
From HW\#13-3, $\BNZ$ and $\CVZ$ are both v.op.-closed, as well.

\begin{thm}\wrlab{thm-bnz-times-cvz-subset-cvz}
$\BNZ\cdot\CVZ\subseteq\CVZ$.
\end{thm}

\begin{proof}
Want: $\forall h\in\BNZ\cdot\CVZ$, $h\in\CVZ$.\\
Given $h\in\BNZ\cdot\CVZ$. \qquad Want: $h\in\CVZ$.\\
By HW\#14-1, it suffices to show: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|h_x|<\vep\,\,)$.\\
Given $\vep>0$. \qquad Want: $\exists\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|h_x|<\vep\,\,)$.\\
Since $h\in\BNZ\cdot\CVZ$, choose $f\in\BNZ$ and $g\in\CVZ$ s.t.~$h=fg$.\\
Since $f\in\BNZ$, by \tref{thm-bnz-quantified-equiv}, choose $\lambda,K>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,|x|<\lambda\,\,)\,\,\Rightarrow\,\,(\,\,|f_x|<K\,\,)$.\\
Since $g\in\CVZ$, by \tref{thm-bnz-quantified-equiv}, choose $\mu>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,|x|<\mu\,\,)\,\,\Rightarrow\,\,(\,\,|g_x|<\vep/K\,\,)$.\\
Let $\delta:=\min\{\lambda,\mu\}$. \quad
Then: $\delta>0$ and $\delta\le\lambda$ and $\delta\le\mu$.\\
\phx\qquad Want: \quad $\forall x\in\R$, $\,$ $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|h_x|<\vep\,\,)$.\\
Given $x\in\R$. \quad $\,\,$ Want: $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|h_x|<\vep\,\,)$.\\
Since $|x|<\delta\le\lambda$, by choice of $\lambda$, we have: $|f_x|<K$.\\
Since $|x|<\delta\le\mu$, by choice of $\mu$, we have: $|g_x|<\vep/K$.\\
Since $0\le|f_x|<K$ and $0\le|g_x|<\vep/K$,
we get: $|f_x|\cdot|g_x|<K\cdot(\vep/K)$.\\
Then $|h_x|=|(fg)_x|=|f_x\cdot g_x|=|f_x|\cdot|g_x|<K\cdot(\vep/K)=\vep$.
\end{proof}

Unassigned HW: Show $\DNZ\cdot\DNZ\subseteq\DNZ$.

From HW\#13-4, \qquad $\,\,\,\,\BNZ\cdot\BNZ\subseteq\BNZ$.

From \tref{thm-bnz-times-cvz-subset-cvz}, $\,\,\BNZ\cdot\CVZ\subseteq\CVZ$.
Also $\CVZ\subseteq\BNZ$.

Then \quad $\,\,\,\CVZ\cdot\CVZ\subseteq\BNZ\cdot\CVZ\subseteq\CVZ$,\\
\phx\qquad\qquad\qquad\qquad\quad so $\,\,\,\,\CVZ\cdot\CVZ\subseteq\CVZ$.

\subsection{Little-o and big-O functions, and homogeneous polynomials}
\quad
\vskip.1in\noindent

Recall that $|\bullet|:\R\to\R$. \qquad Also: $\forall x\in\R$, $|\bullet|_x=|x|$.

\begin{defn}
$(\bullet):=\id_\R$.
\end{defn}

\begin{thm}
$\forall x\in\R$, \quad $(\bullet)_x=x$.
\end{thm}

It is our convention, in this course, that $0^0=1$.

Then, for any functional $f$, we have $f^0=C_{\D_f}^1$.

Note, also, that, for any functional $f$, we have $f^1=f$.

\begin{thm}
$(\,\,\,|\bullet|^0=C_\R^1=(\bullet)^0\,\,\,)$\\
\phx\qquad\qquad $\,\,\&\,\,\,(\,\,\,|\bullet|^1=|\bullet|\,\,\,)\,\,\,\&\,\,\,(\,\,\,(\bullet)^1=(\bullet)\,\,\,)$\\
\phx\qquad\qquad $\,\,\&\,\,\,(\,\,\,|\bullet|^2=(\bullet)^2\,\,\,)\,\,\,\&\,\,\,(\,\,\,|\bullet|^4=(\bullet)^4\,\,\,)$\\
\phx\qquad\qquad $\,\,\&\,\,\,(\,\,\,|\bullet|^6=(\bullet)^6\,\,\,)\,\,\,\&\,\,\,(\,\,\,|\bullet|^8=(\bullet)^8\,\,\,)$.
\end{thm}

In the preceding theorem, the pattern established, for even powers continues.

We graphed $(\bullet)^3$ and $-(\bullet)^3$ and noted that\\
\phx\qquad the union of the graphs of $|\bullet|^3$ and $-|\bullet|^3$\\
is the same as\\
\phx\qquad the union of the graphs of $(\bullet)^3$ and $-(\bullet)^3$.

\begin{defn}
Let $S$ be a set of functionals, $f$ a functional.\\
\phx\qquad\qquad\qquad Then $\,\,S\,\,\cdot_\sf\,\,f\,\,:=\,\,S\cdot\{f\}$ \quad and \quad $f\,\,\cdot_\fs\,\,S\,\,:=\,\,\{f\}\cdot S$.
\end{defn}

We typically omit ``$\sf$'' and ``$\fs$'', and write ``$S\cdot f$'' and ``$f\cdot S$''.

We often omit the ``$\cdot$'', and write ``$Sf$'' and ``$fS$''.

For any set $S$ of functionals, for any functional $g$, we have:\\
\phx\qquad\qquad $Sg=\{fg\,|\,f\in S\}$ \qquad and \qquad $gS=\{gf\,|\,f\in S\}$.

For any set $S$ of functionals, for any functional $f$, we have: $Sf=fS$.

\begin{defn}
Let $j\in\N_0$. \qquad Then:\\
\phx\qquad $\scrlito_j\,\,:=\,\,\CVZ\,\cdot\,|\bullet|^j$ \qquad and \qquad $\scrbigo_j\,\,:=\,\,\BNZ\,\cdot\,|\bullet|^j$.
\end{defn}

Note that $\scrlito_0=\CVZ$ and that $\scrbigo_0=\BNZ$.

Recall that $\CVZ\subsetneq\BNZ\subsetneq\DNZ$.

Then $\scrlito_0=\CVZ\subseteq\BNZ\subseteq\DNZ$.

Also, $\scrbigo_0=\BNZ\subseteq\DNZ$, and $\scrbigo_0=\BNZ\not\subseteq\CVZ$.

From \tref{thm-bnz-times-cvz-subset-cvz}, $\BNZ\cdot\CVZ\subseteq\CVZ$.

Also, for all $j\in\N_0$, we have: $|\bullet|^j\in\BNZ$.

Then, for all $j\in\N_0$, we have: $\scrlito_j\subseteq\CVZ\cdot\BNZ=\BNZ\cdot\CVZ\subseteq\CVZ$.

Then, for all $j\in\N_0$, we have: $\scrlito_j\subseteq\CVZ\subseteq\BNZ\subseteq\DNZ$.

Also, for all $j\in\N$, we have $|\bullet|^j\in\CVZ$.

Then, for all $j\in\N$, we have: $\scrbigo_j\subseteq\BNZ\cdot\CVZ\subseteq\CVZ$.

Then, for all $j\in\N$, we have $\scrbigo_j\subseteq\CVZ\subseteq\BNZ\subseteq\DNZ$.

However, recall that $\scrbigo_0=\BNZ\not\subseteq\CVZ$.

Since $\CVZ\subseteq\BNZ$, we conclude: \quad $\forall j\in\N_0$, $\,$ $\scrlito_j\subseteq\scrbigo_j$.

\begin{thm}
Let $f:\R\drta\R$ and let $j\in\N_0$. \quad Then:\\
\phx\qquad\quad $\,[\,(\,f\in\scrbigo_j\,)\,\Leftrightarrow\,(\,\exists K\ge0~\hbox{s.t.}~K\cdot|\bullet|^j\hbox{ envelopes }f\hbox{ near }0\,)\,]$\\
\phx\qquad $\,\&\,[\,(\,f\in\scrlito_j\,)\,\Leftrightarrow\,(\,\forall\vep\ge0,\,\vep\cdot|\bullet|^j\hbox{ envelopes }f\hbox{ near }0\,)\,]$.
\end{thm}

For all $f:\R\drta\R$, for all $j\in\N$, we have $\ds{\frac{f}{|\bullet|^j}\notin\DNZ\supseteq\BNZ\supseteq\CVZ}$.

We make an adjustment to correct for this difficulty:

\begin{defn}
Let $f:\R\drta\R$ and let $j\in\Z$. \quad Then:\\
\phx\qquad $f_j^+:=\adj_0^0\left(f\cdot|\bullet|^j\right)$ \qquad and \qquad $f_j^-:=f_{-j}^+$.
\end{defn}

For any $f:\R\drta\R$, for any $j\in\Z$, we have: $\ds{f_j^-:=\adj_0^0\left(\frac{f}{|\bullet|^j}\right)}$

For any $f:\R\drta\R$, we have: $f_0^+=f=f_0^-$.

Let $f:\R\drta\R$ and let $j,k\in\Z$. Then:\\
\phx\quad\qquad\qquad $\,[\,\,(f_j^+)_k^+=f_{j+k}^+\,\,]\qquad\&\qquad[\,\,(f_j^-)_k^-=f_{j+k}^-\,\,]\qquad\&$\\
\phx\quad\quad $[\,\,(f_j^+)_k^-=f_{j-k}^+=f_{k-j}^-\,\,]\qquad\&\qquad[\,\,(f_j^-)_k^+=f_{k-j}^+=f_{j-k}^-\,\,]$.

For any $f:\R\drta\R$, we have: $(f_j^+)_j^-=\adj_0^0f=(f_j^-)_j^+$.

\begin{thm}
Let $f:\R\drta\R$, let $j\in\Z$ and let $g:=f_j^-$.\\
\phx\qquad Then $\D_g=(\D_f)^+_0$. \qquad
Also, $(\,\,0\in\D_f\,\,)\,\,\Rightarrow\,\,(\,\,\D_g=\D_f\,\,)$.
\end{thm}

For any $f\in\DNZ$, for any $j\in\Z$, we have: $\D_{f_j^+}=\D_f=\D_{f_j^-}$.

For any $f\in\DNZ$, for any $j\in\Z$, we have: $f_j^+,f_j^-\in\DNZ$.

For any $f\in\DNZ$, we have: $\adj_0^0f=f_0^+\in\DNZ$.

For any $f:\R\drta\R$, we have: $(\,\,f_0=0\,\,)\,\,\Rightarrow\,\,(\,\,\adj_0^0f=f\,\,)$.

\begin{thm}\wrlab{thm-adj-in-bnz-cvz}
We have:\\
\phx\quad $($ $\forall f\in\CVZ$, $\,\adj_0^0f=f\in\CVZ$ $)$ $\,\&\,$ $($ $\forall f\in\BNZ$, $\,\adj_0^0f\in\BNZ$ $)$.
\end{thm}

We graphed $(\bullet)/|\bullet|$ and $(\bullet)^2/|\bullet|$.

We noted that $(\bullet)^2/|\bullet|=|\bullet|^2/|\bullet|\ne|\bullet|$.

We graphed $(\bullet)_1^-$ and $((\bullet)^2)_1^-$.

We noted that $(\bullet)_1^-\notin\CVZ$ and $(\bullet)_1^-\in\BNZ\subseteq\DNZ$.

We noted that $((\bullet)^2)_1^-=(|\bullet|^2)_1^-=|\bullet|\in\CVZ\subseteq\BNZ\subseteq\DNZ$.

\begin{thm}\wrlab{thm-inoj-iff-demotion-inCVZ}
Let $f:\R\drta\R$, $j\in\N_0$. \quad Assume $f_0=0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \qquad $(\,\,f\in\scrlito_j\,\,)\,\,\Leftrightarrow\,\,(\,\,f_j^-\in\CVZ\,\,)$.
\end{thm}

\begin{proof}
Since $f_0=0$, it follows that: \quad $\adj_0^0f=f$.\\
\\
{\it Proof of $\Rightarrow$:}\\
Assume $f\in\scrlito_j$. \quad Want: $f_j^-\in\CVZ$.\\
Since $f\in\scrlito_j=\CVZ\cdot|\bullet|^j$, choose $\phi\in\CVZ$ s.t.~$f=\phi\cdot|\bullet|^j$.\\
Since $\phi\in\CVZ$, by \tref{thm-adj-in-bnz-cvz}, we get: $\adj_0^0\phi\in\CVZ$.\\
Then $f=\adj_0^0f=\adj_0^0(\phi\cdot|\bullet|^j)=\phi_j^+$.\\
Then $f_j^-=(\phi_j^+)_j^-=\adj_0^0\phi\in\CVZ$, as desired.\\
{\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:}\\
Assume $f_j^-\in\CVZ$. \quad Want: $f\in\scrlito_j$.\\
Let $g:=f_j^-\cdot|\bullet|^j$. \quad
Then $g\in\CVZ\cdot|\bullet|^j=\scrlito_j$. \quad
Want: $g=f$.\\
We have $g_0=(f_j^-\cdot|\bullet|^j)_0=(f_j^-)_0\cdot|0|^j=0$. \quad
Then $\adj_0^0g=g$.\\
Then $g=\adj_0^0g=\adj_0^0(f_j^-\cdot|\bullet|^j)=(f_j^-)_j^+=\adj_0^0f=f$, as desired.\\
{\it End of proof of $\Leftarrow$.}
\end{proof}

\begin{thm}\wrlab{thm-inOj-iff-demotion-inBNZ}
Let $f:\R\drta\R$, $j\in\N_0$. \quad Assume $f_0=0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \qquad $(\,\,f\in\scrbigo_j\,\,)\,\,\Leftrightarrow\,\,(\,\,f_j^-\in\BNZ\,\,)$.
\end{thm}

\begin{proof}
This is HW\#14-5.
\end{proof}

\begin{thm}
Let $f,g:\R\drta\R$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Assume: $f=g$ near $0$ in $\R$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: $(\,\,f_0=0\,\,)\,\,\Leftrightarrow\,\,(\,\,g_0=0\,\,)$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Also: $\forall j\in\N_0$, $f_j^-=g_j^-$ near $0$ in $\R$.
\end{thm}

\begin{thm}
Let $f,g:\R\drta\R$, $j\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Assume: $f=g$ near $0$ in $\R$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Assume $g\in\scrlito_j$. \qquad Then $f\in\scrlito_j$.
\end{thm}

\begin{thm}
Let $f,g:\R\drta\R$, $j\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Assume: $f=g$ near $0$ in $\R$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Assume $g\in\scrbigo_j$. \qquad Then $f\in\scrbigo_j$.
\end{thm}

The preceding two theorems express that:\\
\phx\qquad\quad $\forall j\in\N_0$, \quad $\scrlito_j$ and $\scrbigo_j$ are both ``zero-local'' conditions.

\begin{defn}
Define $\sqrt{\bullet}\!:\!\R\!\drta\!\R$ by: $\forall x\in\R$, $(\sqrt{\bullet})_x\!=\!\sqrt{x}$.
\end{defn}

\begin{defn}
Let $f$ be a functional.\\
\phx\qquad\qquad Then we define: \quad $(\,\,|f|:=|\bullet|\circ f\,\,)\,\,\&\,\,(\,\,\sqrt{f}:=\sqrt{\bullet}\circ f\,\,)$.
\end{defn}

\noindent
Let $f$ be a functional. \qquad
Recall: $f^0=C_\R^1$.\\
Also, $\forall j\in\N$, $f^j=f\cdots f$ is the $j$-fold product of $f$ with itself.


\begin{defn}
Let $f$ be a functional, $j\in\N$.\\
\phx\qquad\qquad Then we define: \quad $\ds{f^{-j}\,\,:=\,\,\frac{1}{f^j}}$.
\end{defn}

\begin{defn}
Let $f$ be a functional, $j\in\Z$.\\
\phx\qquad\qquad Then we define: \quad $f^{j+(1/2)}\,\,:=\,\,f^j\cdot\sqrt{f}$.
\end{defn}

\begin{thm}
Let $f$ be a functional and let $x$ be an object.\\
Then: \qquad\quad $\,\,[\,\,|f|_x=|f_x|\,\,]\quad\&\quad[\,\,(\sqrt{f})_x=\sqrt{f_x}\,\,]$\\
\phx\qquad\qquad $\&\quad[\,\,\forall j\in\Z,\,\,\,(f^j)_x=(f_x)^j\,\,]$\\
\phx\qquad\qquad $\&\quad[\,\,\forall j\in\Z,\,\,\,(f^{j+(1/2)})_x=(f_x)^{j+(1/2)}\,\,]$.
\end{thm}

\begin{thm}\wrlab{thm-jthpower-in-bigoj-not-litoj}
Let $j\in\N_0$. \quad Then $(\bullet)^j\in\scrbigo_j\backslash\scrlito_j$.
\end{thm}

\begin{proof}
Let $f:=(\bullet)^j$. \qquad Want: $f\in\scrbigo_j\backslash\scrlito_j$.\\
We have: \quad either \quad (1) $j=0$ \quad or \quad (2) $j\ne0$.\\
\\
{\it Case (1):}\\
We have $(\bullet)^j=(\bullet)^0=C_\R^1\in\BNZ\backslash\CVZ=\scrbigo_0\backslash\scrlito_0=\scrbigo_j\backslash\scrlito_j$.\\
{\it End of Case (1).}\\
\\
{\it Case (2):}\\
We have $j\in\N_0\backslash\{0\}=\N$. \qquad
Then $f(0)=0^j=0$.\\
So, by \tref{thm-inOj-iff-demotion-inBNZ} and \tref{thm-inoj-iff-demotion-inCVZ},\\
\phx\qquad it suffices to show $f_j^-\in\BNZ\backslash\CVZ$.\\
Since $\ds{f_j^-=((\bullet)^j)_j^-=\adj_0^0\left(\frac{f}{|\bullet|^j}\right)=\adj_0^0\left(\frac{(\bullet)^j}{|\bullet|^j}\right)}$,
we see that:\\
\phx\qquad $\bigg($ $\forall x>0$, \quad $\ds{f_j^-(x)=\frac{x^j}{x^j}=1}$ $\bigg)$ \quad $\&$ \quad $\bigg($ $f_0=0$ $\bigg)$\\
\phx\qquad\qquad $\&$ \qquad $\bigg($ $\forall x<0$, \quad $\ds{f_j^-(x)=\frac{x^j}{-x^j}=-1}$ $\bigg)$.\\
Then $f_j^-\in\BNZ\backslash\CVZ$, as desired.\\
{\it End of Case (2).}
\end{proof}

\begin{thm}
Let $j\in\N$. \qquad Then $\scrlito_j\subsetneq\scrbigo_j$.
\end{thm}

\begin{proof}
We have $\scrlito_j=\CVZ\cdot|\bullet|^j\subseteq\BNZ\cdot|\bullet|^j=\scrbigo_j$.\\
Want: $\exists f$ s.t.~$(\,\,f\notin\scrlito_j\,\,)\,\,\&\,\,(\,\,f\in\scrbigo_j\,\,)$.\\
Let $f:=|\bullet|^j$. \qquad Want: $(\,\,f\notin\scrlito_j\,\,)\,\,\&\,\,(\,\,f\in\scrbigo_j\,\,)$.\\
Since $f_0=0$, by \tref{thm-inOj-iff-demotion-inBNZ} and \tref{thm-inoj-iff-demotion-inCVZ},\\
\phx\qquad it suffices to show $f_j^-\in\BNZ\backslash\CVZ$.\\
Since \qquad\qquad\qquad\qquad\qquad\qquad $(f_j^-)_0=0=(\adj_0^0(C_\R^1))_0$\\
\phx and since \qquad\qquad $\forall x\in\R_0^\times$, \quad $\,\,\,(f_j^-)_x=1=(\adj_0^0(C_R^1))_x$,\\
\phx we conclude: \qquad\qquad\qquad\qquad\qquad $f_j^-=\adj_0^0(C_\R^1)$.\\
Then: \quad $f_j^-\,\,=\,\,\adj_0^0(C_\R^1)\,\,\in\,\,\BNZ\backslash\CVZ$, as desired.
\end{proof}

\begin{thm}
Let $j\in\N_0$. \qquad Then $\scrbigo_{j+1}\subsetneq\scrlito_j$.
\end{thm}

\begin{proof}
We have $|\bullet|\in\CVZ$. \quad
Recall: $\BNZ\cdot\CVZ\subseteq\CVZ$.\\
Then: \qquad $\scrbigo_{j+1}=\BNZ\cdot|\bullet|^{j+1}=\BNZ\cdot|\bullet|\cdot|\bullet|^j$\\
\phx\qquad\qquad\qquad\qquad\quad $\in\BNZ\cdot\CVZ\cdot|\bullet|^j\subseteq\CVZ\cdot|\bullet|^j=\scrlito_j$.\\
Want: $\exists f$ s.t.~$(\,\,f\notin\scrbigo_{j+1}\,\,)\,\,\&\,\,(\,\,f\in\scrlito_j\,\,)$.\\
Let $f:=|\bullet|^{j+(1/2)}$. \qquad Want: $(\,\,f\notin\scrbigo_{j+1}\,\,)\,\,\&\,\,(\,\,f\in\scrlito_j\,\,)$.\\
Since $f_0=0$, by \tref{thm-inOj-iff-demotion-inBNZ} and \tref{thm-inoj-iff-demotion-inCVZ},\\
\phx\qquad it suffices to show: \quad $(\,\,f_{j+1}^-\notin\BNZ\,\,)\,\,\&\,\,(\,\,f_j^-\in\CVZ\,\,)$.\\
Since \qquad\qquad\qquad\qquad\qquad\qquad $(f_{j+1}^-)_0=0=(\adj_0^0(|\bullet|^{-1/2}))_0$\\
\phx and since \qquad\qquad $\forall x\in\R_0^\times$, \quad $\,\,\,(f_{j+1}^-)_x=|x|^{-1/2}=(\adj_0^0(|\bullet|^{-1/2}))_x$,\\
\phx we conclude: \qquad\qquad\qquad\qquad\qquad $f_{j+1}^-=\adj_0^0(|\bullet|^{-1/2})$.\\
Then: \quad $f_{j+1}^-\,\,=\,\,\adj_0^0(|\bullet|^{-1/2})\,\,\notin\,\,\BNZ$. \qquad
Want: $f_j^-\in\CVZ$.\\
Since \qquad\qquad\qquad\qquad\qquad\qquad $(f_j^-)_0=0=(\adj_0^0(|\bullet|^{1/2}))_0$\\
\phx and since \qquad\qquad $\forall x\in\R_0^\times$, \quad $\,\,\,(f_j^-)_x=|x|^{1/2}=(\adj_0^0(|\bullet|^{1/2}))_x$,\\
\phx we conclude: \qquad\qquad\qquad\qquad\qquad $f_j^-=\adj_0^0(|\bullet|^{1/2})$.\\
Since $(|\bullet|^{1/2})_0=|0|^{1/2}=0$, \qquad we get: \quad $\,\,\adj_0^0(|\bullet|^{1/2})\,\,=\,\,|\bullet|^{1/2}$.\\
Then: \quad $f_j^-\,\,=\,\,\adj_0^0(|\bullet|^{1/2})\,\,=\,\,|\bullet|^{1/2}\,\,\in\,\,\CVZ$, as desired.
\end{proof}

\begin{thm}
Let $j\in\N_0$. \quad Then $\scrlito_j+\scrlito_j\subseteq\scrlito_j$.
\end{thm}

\begin{proof}
Recall: $\CVZ+\CVZ\subseteq\CVZ$.\\
Then: $\scrlito_j+\scrlito_j=(\CVZ\cdot|\bullet|^j)+(\CVZ\cdot|\bullet|^j)$\\
\phx\qquad\qquad\qquad $=(\CVZ+\CVZ)\cdot|\bullet|^j\subseteq\CVZ\cdot|\bullet|^j=\scrlito_j$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-bigo-lito-vopclosed}
Let $j\in\N_0$. Then $\scrlito_j$ and $\scrbigo_j$ are v.op.-closed.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}
Let $j,k\in\N_0$. \quad Then $\scrlito_j\cdot\scrbigo_k\subseteq\scrlito_{j+k}$.
\end{thm}

\begin{proof}
Recall: $\CVZ\cdot\BNZ\subseteq\CVZ$.\\
Then: $\,\scrlito_j\cdot\scrbigo_k=(\CVZ\cdot|\bullet|^j)\cdot(\BNZ\cdot|\bullet|^k)$\\
\phx\qquad\qquad\qquad $=(\CVZ\cdot\BNZ)\cdot|\bullet|^{j+k}\subseteq\CVZ\cdot|\bullet|^{j+k}=\scrlito_{j+k}$.
\end{proof}

\begin{thm}
Let $j,k\in\N_0$. Then:\\
\phx\qquad\qquad\quad $(\,\,\scrbigo_j\cdot\scrbigo_k\subseteq\scrbigo_{j+k}\,\,)\,\,\&\,\,(\,\,\scrlito_j\cdot\scrbigo_k\subseteq\scrlito_{j+k}\,\,)$\\
\phx\qquad\qquad $\&\,\,(\,\,\,\scrbigo_j\cdot\scrlito_k\subseteq\scrlito_{j+k}\,\,\,)\,\,\&\,\,(\,\,\scrlito_j\cdot\scrlito_k\,\subseteq\scrlito_{j+k}\,\,)$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{defn}
For all $j\in\N_0$, let $\scrh_j:=\R\cdot(\bullet)^j$.
\end{defn}

We have: $\forall j\in\N_0$,\\
\phx\qquad\qquad $\scrh_j\quad=\quad\{\,\,f:\R\to\R\,\,|\,\,\exists a\in\R\,\,\,\,\hbox{s.t.},\,\,\forall x\in\R,\,\,f_x=ax^j\,\,\}$.

\begin{defn}
We define:\\
\phx\qquad $\scrc:=\scrh_0$ \quad and \quad $\scrl:=\scrh_1$ \quad and \quad $\scrq:=\scrh_2$ \quad and \quad $\scrk:=\scrh_3$.
\end{defn}

\begin{thm}
$\forall j\in\N_0$, $\scrh_j$ is v.op.-closed.
\end{thm}

\begin{thm}
$\forall j,k\in\N_0$, $\scrh_j\cdot\scrh_k\subseteq\scrh_{j+k}$.
\end{thm}

\begin{defn}
We define: \quad $\bfzero:=C_\R^0$.
\end{defn}

\begin{thm}
Let $j\in\N_0$. $\,$ Then $\scrh_j\subseteq\scrbigo_j$ and $\scrh_j\bigcap\scrlito_j=\{\bfzero\}$.
\end{thm}

\begin{proof}
By \tref{thm-jthpower-in-bigoj-not-litoj}, we have: \quad $(\bullet)^j\in\scrbigo_j$.\\
By \tref{thm-bigo-lito-vopclosed}, $\scrbigo_j$ and $\scrlito_j$ are v.op.-closed,\\
\phx\qquad\qquad and so \quad $\R\cdot\scrbigo_j\subseteq\scrbigo_j$ \quad and \quad $\R\cdot\scrlito_j\subseteq\scrlito_j$.\\
Then: $\scrh_j=\R\cdot(\bullet)^j\subseteq\R\cdot\scrbigo_j\subseteq\scrbigo_j$. \qquad\qquad
Want: $\scrh_j\bigcap\scrlito_j=\{\bfzero\}$.\\
We have: $\bfzero=0\cdot(\bullet)^j\in\scrh_j$. \quad
Also: $\bfzero=C_\R^0=\C_\R^0\cdot|\bullet|^j\in\CVZ\cdot|\bullet|^j=\scrlito_j$.\\
Since $\bfzero\in\scrh_j$ and $\bfzero\in\scrlito_j$,
we get $\bfzero\in\scrh_j\bigcap\scrlito_j$,
so $\{\bfzero\}\subseteq\scrh_j\bigcap\scrlito_j$.\\
Want: $\scrh_j\bigcap\scrlito_j\subseteq\{\bfzero\}$. \qquad
Want: $\forall f\in\scrh_j\bigcap\scrlito_j$, $f\in\{\bfzero\}$.\\
Given $f\in\scrh_j\bigcap\scrlito_j$. \qquad Want: $f\in\{\bfzero\}$. \qquad Want: $f=\bfzero$.\\
Since $f=\scrh_j=\R\cdot(\bullet)^j$, choose $a\in\R$ s.t.~$f=a\cdot(\bullet)^j$.\\
Want: $a=0$. \qquad
Assume $a\ne0$. \qquad Want: Contradiction.\\
Since $1/a\in\R$ and $f\in\scrlito_j$, we get: \quad $(1/a)\cdot f\in\R\cdot\scrlito_j$.\\
Since $f=a\cdot(\bullet)^j$, we get: \quad $(1/a)\cdot f=(\bullet)^j$.\\
Then \quad $(\bullet)^j=(1/a)\cdot f\in\R\cdot\scrlito_j\subseteq\scrlito_j$, \quad and so \quad $(\bullet)^j\in\scrlito_j$.\\
By \tref{thm-jthpower-in-bigoj-not-litoj}, we have: \quad $(\bullet)^j\notin\scrlito_j$. \qquad Contradiction.
\end{proof}

\begin{thm}
We have:\\
\phx\qquad $\scrc\subseteq\scrbigo_0$ \quad and \quad $\scrc\cap\scrlito_0=\{\bfzero\}$ \quad and\\
\phx\qquad $\scrl\subseteq\scrbigo_1$ \quad and \quad $\scrl\cap\scrlito_1=\{\bfzero\}$ \quad and\\
\phx\qquad $\scrq\subseteq\scrbigo_2$ \quad and \quad $\scrq\cap\scrlito_2=\{\bfzero\}$ \quad and\\
\phx\qquad $\scrk\subseteq\scrbigo_3$ \quad and \quad $\scrk\cap\scrlito_3=\{\bfzero\}$.\\
\end{thm}

\begin{defn}
Let $S$ and $T$ be sets of functions.\\
Then: \quad $T\,\,\,\circ_\ss\,\,\,S:=\{\,\,g\circ f\,\,|\,\,(\,\,g\in T\,\,)\,\,\&\,\,(\,\,f\in S\,\,)\}$.
\end{defn}

We typically omit ``$\ss$'' and simply write ``$T\circ S$''.

\begin{defn}
Let $f$ be a function, $S$ a set of functions.\\
\phx\qquad\qquad Then: \quad\qquad\qquad $f\,\,\circ_\fs\,\,S\,\,\,\,:=\,\,\,\,\{f\}\,\,\circ\,\,\,\,S$\\
\phx\qquad\qquad\qquad\quad\quad  and \qquad $S\,\,\circ_\sf\,\,f\,\,\,\,:=\,\,\,\,\,\,S\,\,\,\,\,\circ\,\,\{f\}$.
\end{defn}

We typically omit ``$\fs$'' and ``$\sf$'', and write ``$f\circ S$'' and ``$S\circ f$''.

\begin{thm}\wrlab{thm-each-circ-cvz-subseteq-itself}
We have:\\
\phx\qquad (1) $\DNZ\circ\CVZ\subseteq\DNZ$ \qquad\qquad and\\
\phx\qquad (2) $\BNZ\circ\CVZ\subseteq\BNZ$ \qquad\qquad and\\
\phx\qquad (3) $\CVZ\circ\CVZ\subseteq\CVZ$.
\end{thm}

\begin{proof}
{\it Proof of (1):}\\
Want: $\forall h\in\DNZ\circ\CVZ$, $h\in\DNZ$.\\
Given $h\in\DNZ\circ\CVZ$. \qquad Want: $h\in\DNZ$.\\
Want: $\exists A\in\scrb_\R(0)$ s.t.~$A\subseteq\D_h$.\\
Since $h\in\DNZ\circ\CVZ$, choose $g\in\DNZ$ and $f\in\CVZ$ s.t.~$h=g\circ f$.\\
Since $g\in\DNZ$, choose $B\in\scrb_\R(0)$ s.t.~$B\subseteq\D_g$.\\
By \tref{thm-cvz-and-output-ball-gives-input-ball}, choose $A\in\scrb_\R(0)$ s.t.~$(\,\,A\subseteq\D_f\,\,)\,\,\&\,\,(\,\,f_*A\subseteq B\,\,)$.\\
Then $A\in\scrb_\R(0)$. \qquad Want: $A\subseteq\D_h$.\\
Want: $\forall x\in A$, $x\in\D_h$. \qquad Given $x\in A$. \qquad Want: $x\in\D_h$.\\
Since $x\in A\subseteq\D_f$, we get $x\in A\cap\D_f$. \qquad
Then $f_x\in f_*A$.\\
Since $f_x\in f_*A\subseteq B\subseteq\D_g$, it follows that $g_{f_x}\ne\frownie$.\\
Since $h_x=(g\circ f)_x=g_{f_x}\ne\frownie$, it follows that $x\in\D_h$, as desired.\\
{\it End of proof of (1).}\\
\\
{\it Proof of (2):} Unassigned HW. {\it End of proof of (2).}\\
\\
{\it Proof of (3).} Unassigned HW. {\it End of proof of (3).}
\end{proof}

\begin{thm}\wrlab{thm-cvz-circ-bnz-not-subseteq-dnz}
Let $g:=C_{(-1;1)}^0$.\\
\phx\qquad\qquad Define $f:\R\to\R$ by: $\forall x\in\R$, $f_x=x+2$.\\
\phx\qquad\qquad Then $f\in\BNZ\subseteq\DNZ$ and $g\in\CVZ\subseteq\BNZ\subseteq\DNZ$.\\
\phx\qquad\qquad Also, $g\circ f\notin\DNZ\supseteq\BNZ\supseteq\CVZ$.
\end{thm}

\begin{proof}
Unassigned HW: Show that $f\in\BNZ$ and that $g\in\CVZ$.\\
Then, since $\CVZ\subseteq\BNZ\subseteq\DNZ$, it remains to show: $g\circ f\notin\DNZ$.\\
We have $(g\circ f)_0=g_{f_0}=g_2=\frownie$, so $0\ne\D_{g\circ f}$.\\
Then $g\circ f$ is not defined near $0$ in $\R$, so $g\circ f\notin\DNZ$.
\end{proof}

\begin{thm}\wrlab{thm-cl-bdd-is-bdd}
Let $X$ be a metric space and let $A\subseteq X$.\\
\phx\qquad\qquad Assume that $A$ is bounded in $X$.\\
\phx\qquad\qquad Then $\Cl_XA$ is bounded in $X$.
\end{thm}

\begin{proof}
Let $C:=\Cl_XA$. \quad Want: $C$ is bounded in $X$.\\
Want: $\exists t>0$ s.t.~$C$ is $t$-bounded in $X$.\\
Since $A$ is bounded in $X$, choose $s>0$ s.t.~$A$ is $s$-bounded in $X$.\\
Let $t:=s+2$. \quad Then $t>0$. \quad Want: $C$ is $t$-bounded in $X$.\\
Want: $\forall p,q\in C$, $d_X(p,q)<t$. \quad Given $p,q\in C$. $\,$ Want: $d_X(p,q)<t$.\\
Since $p,q\in C=\Cl_XA$, we get $(B_X(p,1))\cap A\ne\emptyset\ne(B_X(q,1))\cap A$.\\
Choose $y\in(B_X(p,1))\cap A$ and $z\in(B_X(q,1))\cap A$.\\
Then $y,z\in A$. \qquad
So, since $A$ is $s$-bounded, we get: $d_X(y,z)<s$.\\
Since $y\in B_X(p,1)$, it follows that $d_X(p,y)<1$.\\
Since $z\in B_X(q,1)$, it follows that $d_X(z,q)<1$.\\
By the triangle inequality, $d_X(p,q)\le(d_X(p,y))+(d_X(y,z))+(d_X(z,q))$.\\
Then $d_X(p,q)<1+s+1=s+2=t$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-cl-bdd-is-cpt}
Let $X$ be a proper metric space and let $A\subseteq X$.\\
Assume: $A$ is bounded in $X$. \qquad Then $\Cl_XA$ is compact.
\end{thm}

\begin{proof}
By \tref{thm-cl-bdd-is-bdd}, $\Cl_XA$ is bounded in $X$.\\
So, since $\Cl_XA$ is closed in $X$ and since $X$ is proper,\\
\phx\qquad it follows, from \tref{thm-inproper-cpt-iff-closed-bounded}, that $\Cl_XA$ is compact.
\end{proof}

Recall: \quad $\forall x,y\in\R$, \qquad\qquad\qquad $\,\,d_\hatbbR(x,y)=\min\{\,|x-y|\,,\,1\,\}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad and \qquad $d_\R(x,y)\quad=\quad\,|x-y|$.

\begin{thm}
Let $f:=\id_\R$ and let $A:=\R$.\\
Then: \quad $f$ is a homeomorphism from $\hatbbR$ to $\R$ \quad and \quad $f_*A=A$.\\
Also: \quad $A$ is bounded in $\hatbbR$ \quad and \quad $f_*A$ is unbounded in $\R$.
\end{thm}

\begin{proof}
Unassigned HW. {\it Hint:} Use \tref{thm-conv-in-X-iff-conv-in-hatX}.
\end{proof}

\begin{thm}\wrlab{thm-contin-im-bddinproper-is-bdd}
Let $X$ and $Y$ be metric spaces, $A\subseteq X$, $f\in C^X_Y$.\\
Assume: \quad $X$ is proper \quad and \quad $A$ is bounded in $X$.\\
\phx\qquad\qquad\qquad Then: \qquad\qquad $f_*A$ is bounded in $Y$.
\end{thm}

\begin{proof}
By \tref{thm-cl-bdd-is-cpt}, $\Cl_XA$ is compact.\\
Let $B:=\Cl_XA$. \qquad Then $B$ is compact.\\
Since $f\in C^X_Y$, we have: $f$ is continuous from $X$ to $Y$.\\
Then, by \tref{thm-im-cpt-under-contin-is-cpt}, $f_*B$ is compact.\\
Then, by \tref{thm-cpt-implies-bdd}, $f_*B$ is bounded in $Y$.\\
Since $A\subseteq\Cl_XA=B$, we get: $f_*A\subseteq f_*B$.\\
So, since $f_*B$ is bounded in $Y$, it follows that $f_*A$ is bounded in $Y$.
\end{proof}

\begin{thm}
$C^\R_\R\,\,\circ\,\,\BNZ\,\,\subseteq\BNZ$.
\end{thm}

\begin{proof}
Want: \quad $\forall h\in C^\R_\R\,\,\circ\,\,\BNZ$, $\,$ $h\in\BNZ$.\\
Given $h\in C^\R_\R\,\,\circ\,\,\BNZ$. \qquad Want: $h\in\BNZ$.\\
Want: $\exists B\in\scrb_\R(0)$ s.t.~$(\,\,B\subseteq\D_h\,\,)\,\,\&\,\,(\,\,h_*B\hbox{ is bounded in }\R\,\,)$.\\
Since $h\in C^\R_\R\,\,\circ\,\,\BNZ$, choose $g\in\C^\R_\R$ and $f\in\BNZ$ s.t.~$h=g\circ f$.\\
Since $f\in\BNZ$, we know: $f$ is bounded near $0$ in $\R$,\\
\phx\qquad so choose $B\in\scrb_\R(0)$ s.t.~$(\,\,B\subseteq\D_f\,\,)\,\,\&\,\,(\,\,f_*B\hbox{ is bounded in }\R\,\,)$.\\
Then $B\in\scrb_\R(0)$. \qquad Want: $(\,\,B\subseteq\D_h\,\,)\,\,\&\,\,(\,\,h_*B\hbox{ is bounded in }\R\,\,)$.\\
Since $f\in\BNZ$, we get $f:\R\drta\R$, and so $\I_f\subseteq\R$.\\
Since $g\in C^\R_\R$, we get $g:\R\to\R$, and so $\D_g=\R$.\\
Since $B\subseteq\D_f$ and since $f_*B\subseteq\I_f\subseteq\R=\D_g$, we conclude: $B\subseteq\D_{g\circ f}$.\\
Recall: $h=g\circ f$. \qquad\qquad Then: \quad $B\subseteq\D_{g\circ f}=\D_h$.\\
It remains to show: \quad $h_*B$ is bounded in $\R$.\\
Since \quad $g\in C^\R_\R$ \quad and \quad $\R$ is proper \quad and \quad $f_*B$ is bounded in $\R$,\\
\phx\qquad it follows, from \tref{thm-contin-im-bddinproper-is-bdd}, that $g_*(f_*B)$ is bounded in $\R$.\\
So, since $h_*B=(g\circ f)_*B=g_*(f_*B)$, we get: $h_*B$ is bounded in $\R$.
\end{proof}

\noindent
Warning: Define $f:\R\to\R$ by: $\forall x\in\R$, $f_x=x+2$. Let $g:=C_{(-1;1)}^0$.\\
By \tref{thm-cvz-circ-bnz-not-subseteq-dnz}, we have: \quad $f\in\BNZ$ \quad and \quad $g\circ f\notin\BNZ$.\\
\\
Unassigned HW: Show, $\forall j\in\N_0$, that $|\bullet|^j\in C^\R_\R$.

\begin{thm}\wrlab{thm-abs-power-bnz-is-bnz}
Let $\phi\in\BNZ$ and let $j\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad Then $|\phi|^j\in\BNZ$.
\end{thm}

\begin{proof}
We have: \quad $|\phi|^j\,\,=\,\,|\bullet|^j\,\,\circ\,\,\phi\,\,\in\,\,C^\R_\R\,\,\circ\,\,\BNZ\,\,\subseteq\,\,\BNZ$.
\end{proof}

\noindent
By \tref{thm-cvz-circ-bnz-not-subseteq-dnz}, we have: \quad $\CVZ\circ\BNZ\not\subseteq\DNZ$.\\
Then $\scrbigo_0\circ\scrbigo_0=\BNZ\circ\BNZ\subseteq\CVZ\circ\BNZ\not\subseteq\DNZ\supseteq\BNZ=\scrbigo_0$,\\
\phx\qquad and it follows that: \quad $\scrbigo_0\circ\scrbigo_0\not\subseteq\scrbigo_0$.\\
The next theorem asserts that, for {\it positive} integers $j,k$, $\scrbigo_k\circ\scrbigo_j\subseteq\scrbigo_{j+k}$.

\begin{thm}
Let $j,k\in\N_0$. \quad
Then $\scrbigo_k\circ\scrbigo_j\subseteq\scrbigo_{j+k}$.
\end{thm}

\begin{proof}
Want: \qquad $\forall h\in\scrbigo_k\circ\scrbigo_j$, \quad $h\in\scrbigo_{j+k}$.\\
Given $h\in\scrbigo_k\circ\scrbigo_j$. \qquad\qquad Want: $h\in\scrbigo_{j+k}$.\\
Since $h\in\scrbigo_k\circ\scrbigo_j$, choose $f\in\scrbigo_j$ and $g\in\scrbigo_k$ s.t.~$h=g\circ f$.\\
Since $j\in\N$, we get: $\scrbigo_j\subseteq\scrbigo_1$.\\
\phx\qquad Then \qquad $\,f\in\scrbigo_j\subseteq\scrbigo_1\subseteq\scrlito_0=\CVZ$, \qquad\quad so $f\in\CVZ$.\\
Since $f\in\scrbigo_j=\BNZ\circ|\bullet|^j$, choose $\phi\in\BNZ$ s.t.~$f=\phi\circ|\bullet|^j$.\\
Since $g\in\scrbigo_k=\BNZ\circ|\bullet|^k$, choose $\psi\in\BNZ$ s.t.~$g=\psi\circ|\bullet|^k$.\\
Then: $\forall x\in\R$, \qquad $\,h_x\,\,=\,\,(g\circ f)_x\,\,=\,\,g(f_x)$\\
\phx\qquad\qquad\qquad\qquad\qquad $=\,\,(\psi\cdot|\bullet|^k)(f_x)$\\
\phx\qquad\qquad\qquad\qquad\qquad $=\,\,[\psi(f_x)]\,\cdot\,[\,\,|\qquad\,\,f_x\,\,\qquad\,|^k\,\,]$\\
\phx\qquad\qquad\qquad\qquad\qquad $=\,\,[\psi(f_x)]\,\cdot\,[\,\,|\,\,\,\,(\phi\cdot|\bullet|^j)_x\,\,\,\,|^k\,\,]$\\
\phx\qquad\qquad\qquad\qquad\qquad $=\,\,[\psi(f_x)]\,\cdot\,[\,\,|\,\,\,\,\,\,\phi_x\,\cdot\,|x|^j\,\,\,\,\,\,\,|^k\,\,]$\\
\phx\qquad\qquad\qquad\qquad\qquad $=\,\,[\psi(f_x)]\,\cdot\,[\,\,|\,\phi_x\,|^k\,\,]\,\cdot\,[\,\,|\,\,\,|x|^j\,\,\,|^k\,\,]$\\
\phx\qquad\qquad\qquad\qquad\qquad $=\,\,[\psi(f_x)]\,\cdot\,[\,\,|\,\phi_x\,|^k\,\,]\,\cdot\,[\quad\,|x|^{jk}\,\quad]$\\
\phx\qquad\qquad\qquad\qquad\qquad $=\,\,(\psi\circ f)_x\,\,\cdot\,\,(|\phi|^k)_x\,\,\,\cdot\,\,\,\,\,\,(|\bullet|^{jk})_x$\\
\phx\qquad\qquad\qquad\qquad\qquad $=\,\,(\,\,(\psi\circ f)\,\,\cdot\,\,|\phi|^k\,\,\cdot\,\,|\bullet|^{jk}\,\,)_x$\\
\phx\qquad\quad Then: \qquad $\,h\,\,\,\,\,\,=\,\,\,\,\,\,(\psi\circ f)\,\,\cdot\,\,|\phi|^k\,\,\cdot\,\,|\bullet|^{jk}$.\\
Since $\phi\in\BNZ$, by \tref{thm-abs-power-bnz-is-bnz}, we have: $|\phi|^k\in\BNZ$.\\
By (2) of \tref{thm-each-circ-cvz-subseteq-itself}, we have: $\BNZ\circ\CVZ\subseteq\BNZ$.\\
By HW\#13-4, we conclude that: \quad $\,\BNZ\,\cdot\,\BNZ\subseteq\BNZ$.\\
Then: \quad $h\,\,=\,\,(\,\,\,\,\psi\,\,\,\,\circ\,\,\,\,f\,\,\,\,\,)\,\cdot\,\,\,|\phi|^k\cdot|\bullet|^{jk}$\\
\phx\qquad\qquad $\,\,\,\,\,\in\,\,\,(\BNZ\circ\CVZ)\cdot\,\BNZ\cdot|\bullet|^{jk}$\\
\phx\qquad\qquad $\,\,\,\,\,\subseteq\,\,\quad\,\,\,\,\,\BNZ\,\,\,\,\,\,\quad\cdot\BNZ\cdot|\bullet|^{jk}$\\
\phx\qquad\qquad $\,\,\,\,\,\subseteq\qquad\qquad\qquad\quad\BNZ\cdot|\bullet|^{jk}$.\\
\phx\qquad\qquad\qquad Then: \qquad $\,\,h\in\BNZ\cdot|\bullet|^{jk}=\scrbigo_{jk}$, \qquad as desired.
\end{proof}

\subsection{Double-translates of functions $\R\drta\R$}

\begin{defn}
Let $f:\R\drta\R$ and let $x\in\R$.\\
\phx\qquad\qquad\qquad\qquad $\,$ Then $f_x^\T:\R\drta\R$ is defined by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\forall h\in\R$, \quad $f_x^\T(h)=[f(x+h)]-[f(x)]$.
\end{defn}

\noindent
Then: $\forall f:\R\drta\R$, $\forall x,h\in\R$, we have: \quad $(f_x^\T)_h=f_{x+h}-f_x$.

\noindent
Then: $\forall f:\R\drta\R$, $\forall x\in\R\backslash\D_f$, $\forall h\in\R$, we have: \quad $(f_x^\T)_h=\frownie$.

\noindent
Then: $\forall f:\R\drta\R$, $\forall x\in\R\backslash\D_f$, we have: \quad $f_x^\T=\emptyset$.

\noindent
Consequently, it is only useful to study $f_x^\T$ when $x\in\D_f$.

\noindent
We will call $f_x^\T$ the ``double-translate of $f$ based at $x$.

\noindent
It is not a scalar; it is a function $\R\drta\R$.

\noindent
The next theorem, in part, asserts that\\
\phx\qquad\qquad the graph of the double-translate\\
passes through the origin in $\R^2$.

\noindent
It also describes how\\
\phx\qquad\qquad the domain of the double-translate\\
is related to\\
\phx\qquad\qquad the domain of the original function.\\
It also gives two examples of how\\
\phx\qquad\qquad properties of the double translate $f_x^\T$ \qquad near $0$\\
are related to\\
\phx\qquad\qquad properties of the original function $f$ \qquad near $x$.

\begin{thm}\wrlab{thm-some-basic-dbl-transl-properties}
Let $f:\R\drta\R$, let $x\in\D_f$ and let $\phi:=f_x^\T$.\\
Then: \quad\qquad\qquad\qquad $[\,\,\phi_0=0\,\,]\qquad\&\qquad[\,\,\D_\phi=\D_f-x\,\,]$\\
\phx\qquad\qquad $\&\qquad[\,\,(\,\,\phi\hbox{ is defined near }0\hbox{ in }\R\,\,)$\\
\phx\qquad\qquad\qquad\qquad\qquad $\Leftrightarrow\,\,(\,\,f\hbox{ is defined near }x\hbox{ in }\R\,\,)\,\,]$\\
\phx\qquad\qquad $\&\qquad[\,\,(\,\,\phi\hbox{ is continuous at }0\hbox{ from }\R\hbox{ to }\R\,\,)$\\
\phx\qquad\qquad\qquad\qquad\qquad $\Leftrightarrow\,\,(\,\,f\hbox{ is continuous at }x\hbox{ from }\R\hbox{ to }\R\,\,)\,\,]$.
\end{thm}

\begin{thm}\wrlab{thm-precalc-additivity-rule}
Let $f,g:\R\drta\R$ and let $x\in\R$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,$ Then: $(f+g)_x^\T\,\,=\,\,f_x^\T\,+\,g_x^\T$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}\wrlab{thm-precalc-scalar-mult-rule}
Let $a\in\R$, $f:\R\drta\R$, $x\in\R$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,$ Then $(af)_x^\T\,\,=\,\,a\,\cdot\,f_x^\T$
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}\wrlab{thm-precalc-prod-rule}
Let $f,g:\R\drta\R$ and let $x\in\D_f\bigcap\D_g$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,$ Then $(fg)_x^\T\,\,=\,\,f_x\,\cdot\,g_x^\T+g_x\,\cdot\,f_x^\T+f_x^\T\,\cdot\,g_x^\T$.\\
\end{thm}

\begin{proof}
Let $U:=(fg)_x^\T$ and let $V:=f_x\,\cdot\,g_x^\T+g_x\,\cdot\,f_x^\T+f_x^\T\,\cdot\,g_x^\T$.\\
Want: $U=V$. \qquad Want: $\forall h\in\R$, $U_h=V_h$.\\
Given $h\in\R$. \qquad Want: $U_h=V_h$.\\
Let \quad $a:=f_x$, \quad $A:=f_{x+h}$, \quad $b:=g_x$, \quad $B:=g_{x+h}$.\\
Then: $\,$ $(f_x^\T)_h=f_{x+h}-f_x=A-a$ \quad and \quad $(g_x^\T)_h=g_{x+h}-g_x=B-b$.\\
Also, \quad $(f_x\,\cdot\,g_x^\T)_h=f_x\,\cdot\,[(g_x^\T)_h]=a(B-b)$.\\
Also, \quad $(g_x\,\cdot\,f_x^\T)_h=g_x\,\cdot\,[(f_x^\T)_h]=b(A-a)$.\\
Also, \quad $(f_x^\T\,\cdot\,g_x^\T)_h=[(f_x^\T)_h]\,\cdot\,[(g_x^\T)_h]=(A-a)(B-b)$.\\
Also, \quad $(fg)_x=f_x\cdot g_x=ab$ \quad and \quad $(fg)_{x+h}=f_{x+h}\cdot g_{x+h}=AB$.\\
We have $\,V_h=(f_x\,\cdot\,g_x^\T)_h\,+\,(g_x\,\cdot\,f_x^\T)_h\,+\,(f_x^\T\,\cdot\,g_x^\T)_h$\\
\phx\qquad\qquad\quad $=a(B-b)+b(A-a)+(A-a)(B-b)$\\
\phx\qquad\qquad\quad $=aB-ab+bA-ba+AB-A b-aB+ab$\\
\phx\qquad\qquad\quad $=\qquad\qquad\quad\,\,\,\,\,\,\,-\,ba+AB$.\\
Then $U_h=((fg)_x^\T)_h=(fg)_{x+h}-(fg)_x=AB-ab$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\,\,\,\,\,=-\,ba+AB=V_h$.
\end{proof}

\begin{thm}\wrlab{thm-precalc-chain-rule}
Let $f,g:\R\drta\R$, let $x\in\D_f$ and let $y:=f_x$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,$ Then $(g\circ f)_x^\T\,\,=\,\,g_y^\T\,\,\circ\,\,f_x^\T$.
\end{thm}

\begin{proof}
Let $U:=(g\circ f)_x^\T$ and let $V:=g_y^\T\,\,\circ\,\,f_x^\T$.\\
Want: $U=V$. \qquad Want: $\forall h\in\R$, $U_h=V_h$.\\
Given $h\in\R$. \qquad Want: $U_h=V_h$.\\
Let $k:=(f_x^\T)_h$. \qquad\qquad $\,$ Then: $\,$ $k=f_{x+h}-f_x$.\\
So, since $y=f_x$, we get: $y+k=f_x+f_{x+h}-f_x=f_{x+h}$.\\
Since $(f_x^\T)_h=k$, we get: $(g_y^\T\,\,\circ\,\,f_x^\T)_h=(g_y^\T)_k$.\\
Then $V_h=(g_y^\T\,\,\circ\,\,f_x^\T)_h=(g_y^\T)_k=g_{y+k}\,-\,g_y$, \quad so $V_h=g_{y+k}\,-\,g_y$.\\
Then $U_h=((g\circ f)_x^\T)_h=(g\circ f)_{x+h}\,-\,(g\circ f)_x=g_{f_{x+h}}\,-\,g_{f_x}$.\\
So, since $f_{x+h}=y+k$ and $f_x=y$, we get: $\,U_h=g_{y+k}\,-\,g_y$.\\
Then $U_h=g_{y+k}\,-\,g_y=V_h$, as desired.
\end{proof}

\subsection{Linearizations and the $D$-derivative}

\begin{defn}
Let $f:\R\drta\R$ and let $x\in\R$.\\
\phx\qquad\qquad Then: \quad $\,\LINS_xf\,\,:=\,\,\{\,\,L\in\scrl\,\,|\,\,f_x^\T\,-\,L\,\,\in\,\,\scrlito_1\,\,\}$.\\
\phx\qquad\qquad Also, \qquad\quad $D_xf\,\,:=\,\,\UE\,(\,\LINS_xf\,)$.
\end{defn}

\noindent
The set $\LINS_xf$ is the set of ``linearizations of $f$ at $x$''.\\
Eventually, we will show that $\LINS_xf$ has at most one element.\\
The linear function $D_xf$ is the ``$D$-\bfindex[derivative] of $f$ at $x$''.\\
It is not a scalar; it is a linear function from $\R$ to $\R$.\\
Intuition: Its graph is the tangent line to the graph of $f_x^\T$ at the origin.\\
The $D$-derivative of $f$ at $x$ is the unique linearization of $f$ at $x$,\\
\phx\qquad\qquad provided such a linearization exists.

\begin{thm}\wrlab{thm-linearization-implies-dbltrans-in-bigo1}
Let $f:\R\drta\R$ and let $x\in\R$.\\
\phx\qquad\qquad\qquad\qquad\quad Assume: $\LINS_xf\ne\emptyset$. \quad Then: $f_x^\T\,\,\in\,\,\scrbigo_1$.\\
\end{thm}

\begin{proof}
Choose $L\in\LINS_xf$. \qquad Then $(\,\,L\in\scrl\,\,)\,\,\&\,\,(\,\,f_x^\T-L\in\scrlito_1\,\,)$.\\
We have: \quad $f_x^\T-L\in\scrlito_1\subseteq\scrbigo_1$ \quad and \quad $L\in\scrl=\scrh_1\subseteq\scrbigo_1$.\\
Then $f_x^\T=(f_x^\T-L)+L\in\scrbigo_1+\scrbigo_1=\scrbigo_1$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-L-linearization-imlies-L-D-deriv}
Let $f:\R\drta\R$, let $x\in\R$ and let $L\in\scrl$.\\
\phx\qquad\qquad\qquad\qquad\quad Assume: $L\in\LINS_xf$. \quad Then: $D_xf=L$.\\
\end{thm}

\begin{proof}
Since $D_xf=\UE(\LINS_xf)$, we wish to show: $\LINS_xf=\{L\}$.\\
Since $L\in\LINS_xf$, we get $\{L\}\subseteq\LINS_xf$. \qquad Want: $\LINS_xf\subseteq\{L\}$.\\
Want: \qquad $\forall M\in\LINS_xf$, \quad $M\in\{L\}$.\\
Given $M\in\LINS_xf$. \qquad Want: $M\in\{L\}$.\\
Since $L\in\LINS_xf$, we get $\LINS_xf\ne\emptyset$, so,\\
\phx\qquad by \tref{thm-linearization-implies-dbltrans-in-bigo1}, we see that $f_x^\T\in\scrbigo_1$.\\
Then $f_x^\T\in\scrbigo_1\subseteq\scrlito_0=\CVZ\subseteq\BNZ\subseteq\DNZ$, so $f_x^\T\in\DNZ$.\\
Since $f_x^\T\in\DNZ$, it follows that: $f_x^\T\,-\,f_x^\T=0$ near $0$ in $\R$.\\
Then $(\,f_x^\T\,-\,L\,)\,-\,(\,f_x^\T\,-\,M\,)=M-L$ near $0$ in $\R$.\\
Since $L,M\in\LINS_xf$, we get $f_x^\T\,-\,L\,\in\,\scrlito_1$ and $f_x^\T\,-\,M\,\in\,\scrlito_1$.\\
Then $(\,f_x^\T\,-\,L\,)\,-\,(\,f_x^\T\,-\,M\,)\,\in\,\scrlito_1-\scrlito_1\,\subseteq\,\scrlito_1$.\\
So, since $(\,f_x^\T\,-\,L\,)\,-\,(\,f_x^\T\,-\,M\,)=M-L$ near $0$ in $\R$,\\
\phx\qquad and since $\scrlito_1$ is a zero-local condition, \quad we get: $M\,-\,L\,\in\,\scrlito_1$.\\
Also, $M-L\in\scrl-\scrl\subseteq\scrl=\scrh_1$. \qquad\qquad
Recall: $\scrh_1\bigcap\scrlito_1=\{\bfzero\}$.\\
Then $M-L\in\scrh_1\bigcap\scrlito_1=\{\bfzero\}$, so $M-L=\bfzero$, so $M=L\in\{L\}$.
\end{proof}

\subsection{The prime-derivative}

\begin{thm}
Let $m\in\R$.\\
\phx\qquad Define $L:\R\to\R$ by: \quad $\forall x\in\R$, $L_x=mx$.\\
\phx\qquad Then: \quad $L=m\cdot(\bullet)\in\R\cdot(\bullet)=\scrl$ \quad and \quad $L_1=m\cdot1=m$.
\end{thm}

\noindent
The point of the preceding theorem is that,\\
\phx\qquad to get the slope of a linear function $L$,\\
\phx\qquad\qquad\qquad you can simply compute $L_1$.

\begin{defn}
Let $f:\R\drta\R$.\\
\phx\qquad Then $f':\R\to\R$ is defined by: \quad $\forall x\in\R$, $\,$ $f'_x=(D_xf)_1$.
\end{defn}

\noindent
The function $f'$ is called the \bfindex[derivative] or \bfindex[prime-derivative] of $f$.\\
The prime-derivative of $f$ at $x$, denoted $f'_x$, is a scalar.\\
Recall that the $D$-derivative of $f$ at $x$ is not a scalar;\\
\phx\qquad  it is a linear function from $\R$ to $\R$.\\
To compute the prime-derivative of at $x$,\\
\phx\qquad you can simply take the slope of the $D$-derivative of $f$ at $x$.\\
Recall the intuition:\\
\phx\qquad The graph of the $D$-derivative of $f$ at $x$ \quad is\\
\phx\qquad the tangent line to the graph of $f_x^\T$ at the origin.\\
So, since that tangent line is parallel to\\
\phx\qquad the tangent line to the graph of $f$ at $(x,f_x)$,\\
we conclude:\\
\phx\qquad the prime-derivative of $f$ at $x$ \quad is the slope of\\
\phx\qquad the tangent line to the graph of $f$ at $(x,f_x)$.\\

\begin{thm}
Define $f:\R\to\R$ by: $\forall x\in\R$, $f_x=x^2$.\\
\phx\qquad\qquad\qquad\qquad\quad Then: $f'_3=6$.\\
\end{thm}

\begin{proof}
Define $L\in\scrl$ and $Q\in\scrq$ by:\\
\phx\qquad\qquad\qquad\qquad $\forall h\in\R$, \qquad $L_h=6h$ \quad and \quad $Q_h=h^2$.\\
We have: \qquad $\forall h\in\R$, \qquad $(f_3^\T)_h\,\,=\,\,f_{3+h}\,-\,f_3\,\,=\,\,(3+h)^2\,-\,3^2$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\,\,\,=\,\,(9+6h+h^2)\,-\,9\,\,=\,\,6h+h^2$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\,\,\,=\,\,L_h+Q_h\,\,=\,\,(L+Q)_h$.\\
Then $f_3^\T\,\,=\,\,L\,+\,Q$, so $f_3^\T\,-\,L\,\,=\,\,Q$.\\
So, since $Q\in\scrq=\scrh_2\subseteq\scrbigo_2\subseteq\scrlito_1$,\\
\phx\qquad we get: $f_3^\T\,-\,L\,\,\in\,\,\scrlito_1$, \qquad and so $L\in\LINS_3f$.\\
Then, by \tref{thm-L-linearization-imlies-L-D-deriv}, $D_3f=L$.\\
Then $f'_3=(D_3f)_1=L_1=6\cdot1=6$.
\end{proof}

\begin{thm}
Define $f:\R\to\R$ by: $\forall x\in\R$, $f_x=x^2$.\\
\phx\qquad\qquad\qquad\qquad\quad Let $x\in\R$. \qquad Then: $f'_x=x^2$.\\
\end{thm}

\begin{proof}
Unassigned HW. $\,$ {\it Hint:} Take the preceding proof and replace:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $\,\,\,\,\,\,$ $3$ by $x$, $\,$ $6$ by $2x$ $\,$ and $\,$ $9$ by $x^2$.
\end{proof}

\begin{thm}
Let $f:\R\drta\R$ and let $x\in\R$.\\
\phx\qquad\qquad\qquad\qquad\quad Assume: $\LINS_xf\ne\emptyset$. \qquad Then $x\in\D_{f'}$.
\end{thm}

\begin{proof}
Choose $L\in\LINS_xf$. \quad
By \tref{thm-L-linearization-imlies-L-D-deriv}, we have: $D_xf=L$.\\
We have \quad $L\in\LINS_xf\subseteq\scrl$, \quad so $L:\R\to\R$, \quad so $L_1\ne\emptyset$.\\
Then \quad $f'_x=(D_xf)_1=L_1\ne\frownie$, \quad so $x\in\D_{f'}$, \quad as desired.
\end{proof}

\begin{thm}\wrlab{thm-prderiv-to-dderiv}
Let $f:\R\drta\R$ and let $x\in\D_{f'}$.\\
\phx\qquad\qquad\qquad\qquad\quad Then $D_xf\in\LINS_xf\subseteq\scrl$.
\end{thm}

\begin{proof}
By definition of $\LINS_xf$, we have $\LINS_xf\subseteq\scrl$.\\
It remains to show: $D_xf\in\LINS_xf$. \qquad
Since $x\in\D_{f'}$, we get $f'_x\ne\frownie$.\\
So, since $f'_x=(D_xf)_1$, we get $(D_xf)_1\ne\frownie$, so $D_xf\ne\frownie$.\\
So, since $D_xf=\UE(\LINS_xf)$, we get $\UE(\LINS_xf)\ne\frownie$,\\
\phx\qquad and it follows that $\UE(\LINS_xf)\in\LINS_xf$.\\
Then $D_xf=\UE(\LINS_xf)\in\LINS_xf$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-prderiv-defdnear-continat}
$\,$Let $\,f:\R\drta\R$ and let $x\in\D_{f'}$.\\
\phx\qquad\qquad Then: \qquad\qquad $f$ is defined near $x$ in $\R$\\
\phx\qquad\qquad\qquad\qquad and \quad $f$ is continuous at $x$ from $\R$ to $\R$.
\end{thm}

\begin{proof}
By \tref{thm-linearization-implies-dbltrans-in-bigo1}, we have $f_x^\T\in\scrbigo_1$.\\
Then $f_x^\T\in\scrbigo_1\subseteq\scrlito_0=\CVZ$, \qquad so $f_x^\T\in\CVZ$,\\
\phx\qquad\qquad so $f_x^\T$ is continuous at $0$ from $\R$ to $\R$.\\
Then, by \tref{thm-some-basic-dbl-transl-properties}, $f$ is continuous at $x$ from $\R$ to $\R$.\\
Want: $f$ is defined near $x$ in $\R$.\\
We have $f_x^\T\in\CVZ\subseteq\BNZ\subseteq\DNZ$, \qquad so $f_x^\T\in\DNZ$,\\
\phx\qquad\qquad so $f_x^\T$ is defined near $0$ in $\R$.\\
Then, by \tref{thm-some-basic-dbl-transl-properties}, $f$ is defined near $x$ in $\R$, as desired.
\end{proof}

\begin{thm}
Let $f:=C_{[2;3]}^1$. \qquad Then $f'_2=\frownie=f'_3$.
\end{thm}

\begin{proof}
Since $\neg\,\,($ $f$ is defined near $2$ $\,)$,\\
\phx\qquad\qquad by \tref{thm-prderiv-defdnear-continat}, \qquad we conclude: $2\notin\D_{f'}$.\\
\phx Also, $\,$since $\neg\,\,($ $f$ is defined near $3$ $\,)$,\\
\phx\qquad\qquad by \tref{thm-prderiv-defdnear-continat}, \qquad we conclude: $3\notin\D_{f'}$.\\
Since $2\notin\D_{f'}$ and $3\notin\D_{f'}$, we conclude: $f'_2=\frownie=f'_3$.
\end{proof}

\begin{thm}\wrlab{thm-dom-deriv-subset-dom-fn}
Let $f:\R\drta\R$. \qquad Then $\D_{f'}\subseteq\D_f$.
\end{thm}

\begin{proof}
Want: $\forall x\in\D_{f'}$, $x\in\D_f$. \qquad
Given $x\in\D_{f'}$. $\,$ Want: $x\in\D_f$.\\
By \tref{thm-prderiv-defdnear-continat}, \quad we see that $f$ is defined near $x$,\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,$ so choose $B\in\scrb_\R(x)$ s.t.~$B\subseteq\D_f$.\\
Since $B\in\scrb_\R(x)$, we get: $x\in B$. \qquad $\,$
Then $x\in B\subseteq\D_f$.
\end{proof}

\subsection{Basic properties of $D$-derivatives}

\begin{thm}\wrlab{thm-D-add-prime-prod-rule}
Let $f,g:\R\drta\R$ and let $x\in\D_{f'}\bigcap\D_{g'}$.\\
\phx\qquad Then: \qquad\qquad (1) $D_x(f+g)\,\,=\,\,D_xf\,+\,D_xg$\\
\phx\qquad\qquad\qquad and $\,\,\,\,$ (2) $D_x(fg)\,\,=\,\,f_x\,\cdot\,D_xg\,\,\,+\,\,g_x\,\cdot\,D_xf$.\\
\end{thm}

\begin{proof}
{\it Proof of (1):} Unassigned HW. {\it End of proof of (1).}\\
\\
{\it Proof of (2):}\\
By \tref{thm-dom-deriv-subset-dom-fn}, $x\in\D_f\bigcap\D_g$.\\
Then $f_x\in\I_f\subseteq\R$ and $g_x\in\I_g\subseteq\R$.\\
By \tref{thm-prderiv-to-dderiv}, we have $D_xf\in\LINS_xf\subseteq\scrl$ and $D_xg\in\LINS_xg\subseteq\scrl$.\\
Then $\LINS_xf\ne\emptyset\ne\LINS_xg$, so, by \tref{thm-linearization-implies-dbltrans-in-bigo1},\\
\phx\qquad we get: \quad $f_x^\T\in\scrbigo_1$ \quad and \quad $g_x^\T\in\scrbigo_1$.\\
Let \quad $s:=f_x$, \quad $t:=g_x$, \quad $L:=D_xf$, \quad $M:=D_xg$.\\
Then \quad $s,t\in\R$ \quad and \quad $L,M\in\scrl$.\\
So, since $\scrl$ is v.op.-closed, we see that: \quad $\,\,\,\,sM+tL\in\scrl$.\\
\phx\quad Since \qquad\quad $\,f_x\,\cdot\,D_xg\,\,\,+\,\,g_x\,\cdot\,D_xf\,\,=\,\,sM+tL$,\\
\phx\qquad\qquad\qquad we wish to prove: $D_x(fg)=sM+tL$.\\
Then, by \tref{thm-L-linearization-imlies-L-D-deriv}, we wish to show: $sM+tL\in\LINS_x(fg)$.\\
Since $sM+tL\in\scrl$, we need only show: $(fg)_x^\T-(sM+tL)\in\scrlito_1$.\\
By \tref{thm-precalc-prod-rule}, $(fg)_x^\T\,\,=\,\,f_x\,\cdot\,g_x^\T+g_x\,\cdot\,f_x^\T+f_x^\T\,\cdot\,g_x^\T$.\\
Let $\phi:=f_x^\T$ and $\psi:=g_x^\T$. \qquad
Then $(fg)_x^\T=s\cdot\psi+t\cdot\phi+\phi\cdot\psi$.\\
Then $(fg)_x^\T-(sM+tL)=s\cdot(\psi-M)+t\cdot(\phi-M)+\phi\cdot\psi$.\\
Since $L=D_xf\in\LINS_xf$, we get $\,\,\,f_x^\T\,-\,L\,\in\,\scrlito_1$.\\
\phx\qquad\qquad\qquad Then \quad $\,\,\,\,\,\,\,\phi-L=f_x^\T\,-\,L\,\in\,\scrlito_1$.\\
Since $M=D_xg\in\LINS_xg$, we get $g_x^\T\,-\,M\,\in\,\scrlito_1$.\\
\phx\qquad\qquad\qquad Then \quad $\,\,\,\psi-M=g_x^\T\,-\,M\,\in\,\scrlito_1$.\\
Also, we have: \quad $\phi=f_x^\T\in\scrbigo_1$ \quad and \quad $\psi=g_x^\T\in\scrbigo_1$.\\
Then: \quad $\phi\,\cdot\,\psi\,\,\in\,\,\scrbigo_1\,\cdot\,\scrbigo_1\,\,\subseteq\,\,\scrbigo_2$.\\
Then $(fg)_x^\T-(sM+tL)=s\cdot(\phi-L)+t\cdot(\psi-M)+\phi\cdot\psi$\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,\,\in\,\,\,\R\,\cdot\,\scrlito_1\quad\,\,+\,\,\,\,\R\,\cdot\,\scrlito_1\quad\,\,+\,\,\scrbigo_2$\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,\subseteq\qquad\,\,\scrlito_1\quad\,\,+\qquad\,\,\scrlito_1\quad\,\,+\,\,\,\scrlito_1$\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,\subseteq\qquad\qquad\qquad\qquad\,\scrlito_1$, \quad as desired.\\
{\it End of proof of (2).}
\end{proof}

\begin{thm}\wrlab{thm-D-scalar-mult}
Let $a\in\R$, let $f:\R\drta\R$ and let $x\in\D_{f'}$.\\
\phx\qquad\qquad\qquad\qquad\quad Then: \quad $D_x(a\cdot f)\,\,=\,\,a\,\cdot\,\,D_xf$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}
Let $f$ and $g$ be functionals. Let $h$ be a function.\\
\phx\qquad\qquad\qquad\qquad\quad Then $(f+g)\circ h=(f\circ h)+(g\circ h)$.
\end{thm}

\begin{proof}
Want: $\forall x,\,((f+g)\circ h)_x=((f\circ h)+(g\circ h))_x$.\\
Given $x$. \qquad Want: $((f+g)\circ h)_x=((f\circ h)+(g\circ h))_x$.\\
We have $((f\circ h)+(g\circ h))_x=(f\circ h)_x+(g\circ h)_x=f_{h_x}+g_{h_x}$.\\
Then $((f+g)\circ h)_x=(f+g)_{h_x}=f_{h_x}+g_{h_x}=((f\circ h)+(g\circ h))_x$.
\end{proof}

\begin{thm}
Let $f:=C_\R^3$, $g:=C_\R^4$, $h:=(\bullet)^2$.\\
\phx\qquad\qquad\qquad\qquad\quad Then $h\circ(f+g)\ne(h\circ f)+(h\circ g)$.
\end{thm}

\begin{proof}
We have $(f+g)_1=f_1+g_1=3+4=7$.\\
We have $(h\circ f)_1=h_{f_1}=h_3=9$ and $(h\circ g_1=h_4=16$.\\
Then $((h\circ f)+(h\circ g))_1=(h\circ f)_1+(h\circ g)_1=9+16=25$.\\
Then $(h\circ(f+g))_1=h_{(f+g)_1}=h_7=49\ne25=((h\circ f)+(h\circ g))_1$, so $(h\circ(f+g))_1\ne((h\circ f)+(h\circ g))_1$.\\
Then $h\circ(f+g)\ne(h\circ f)+(h\circ g)$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-linear-implies-additive}
Let $L\in\scrl$ and $s,t\in\R$. \quad Then $L_{s+t}=L_s+L_t$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}\wrlab{thm-lin-circ-sum}
Let $f$ and $g$ be functionals. Let $L\in\scrl$.\\
\phx\qquad\qquad\qquad\qquad\quad Then $L\circ(f+g)=(L\circ f)+(L\circ g)$.
\end{thm}

\begin{proof}
Want: $\forall x$, $(L\circ(f+g))_x=((L\circ f)+(L\circ g))_x$.\\
Given $x$. \qquad Want: $(L\circ(f+g))_x=((L\circ f)+(L\circ g))_x$.\\
Let \quad $s:=f_x$ \quad and \quad $t:=g_x$.\\
Then $((L\circ f)+(L\circ g))_x=(L\circ f)_x+(L\circ g)_x=L_{f_x}+L_{g_x}=L_s+L_t$.\\
Also, $(f+g)_x=f_x+g_x=s+t$. \quad
By \tref{thm-linear-implies-additive}, $L_{s+t}=L_s+L_t$.\\
Then $(L\circ(f+g))_x=L_{(f+g)_x}=L_{s+t}=L_s+L_t=((L\circ f)+(L\circ g))_x$.
\end{proof}

\begin{thm}\wrlab{thm-D-chain-rule}
Let $f,g:\R\drta\R$, let $x\in\D_{f'}$ and let $y:=f_x$.\\
\phx\qquad\quad Assume: $y\in\D_{g'}$. \qquad\quad Then: $D_x(g\circ f)=(D_yg)\circ(D_xf)$.
\end{thm}

\begin{proof}
By \tref{thm-dom-deriv-subset-dom-fn}, $x\in\D_f$ and $y\in\D_g$.\\
Then $f_x\in\I_f\subseteq\R$ and $g_y\in\I_g\subseteq\R$.\\
By \tref{thm-prderiv-to-dderiv}, we have $D_xf\in\LINS_xf\subseteq\scrl$ and $D_yg\in\LINS_yg\subseteq\scrl$.\\
Then $\LINS_xf\ne\emptyset$, so, by \tref{thm-linearization-implies-dbltrans-in-bigo1}, we get: $f_x^\T\in\scrbigo_1$.\\
Let \quad $L:=D_xf$, \quad $M:=D_yg$. \qquad Then \quad $L,M\in\scrl$.\\
Then $M\circ L\in\scrl\circ\scrl=\scrh_1\circ\scrh_1\subseteq\scrh_1=\scrl$.\\
Since $(D_yg)\circ(D_xf)=M\circ L$, we wish to prove: $D_x(g\circ f)=M\circ L$.\\
Then, by \tref{thm-L-linearization-imlies-L-D-deriv}, we wish to show: $M\circ L\in\LINS_x(g\circ f)$.\\
Since $M\circ L\in\scrl$, we need only show: $(g\circ f)_x^\T-(M\circ L)\in\scrlito_1$.\\
By \tref{thm-precalc-chain-rule}, $(g\circ f)_x^\T\,\,=\,\,g_y^\T\,\,\circ\,\,f_x^\T$.\\
Let $\phi:=f_x^\T$ and $\psi:=g_y^\T$. \qquad
Then $(g\circ f)_x^\T=\psi\circ\phi$.\\
Since $L=D_xf\in\LINS_xf$, we get $\,\,\,f_x^\T\,-\,L\,\in\,\scrlito_1$.\\
\phx\qquad\qquad\qquad Then \quad $\,\,\,\,\,\,\,\phi-L=f_x^\T\,-\,L\,\in\,\scrlito_1$.\\
Since $M=D_yg\in\LINS_yg$, we get $g_y^\T\,-\,M\,\in\,\scrlito_1$.\\
\phx\qquad\qquad\qquad Then \quad $\,\,\,\psi-M=g_y^\T\,-\,M\,\in\,\scrlito_1$.\\
Let $R:=\phi-L$ and $S:=\psi-M$. \qquad Then $R,S\in\scrlito_1$.\\
Also, \quad $L+R=\phi$ \quad and \quad $M+S=\psi$.\\
Since $M\in\scrl$, by \tref{thm-lin-circ-sum},\\
\phx\qquad we get: \quad $\,\,M\circ(L+R)=(M\circ L)+(M\circ R)$.\\
Then $\,\,\,\,\psi\circ\phi=(M+S)\circ\phi$\\
\phx\qquad\qquad\quad $\,\,=(M\circ\phi)+(S\circ\phi)$\\
\phx\qquad\qquad\quad $\,\,=(M\circ(L+R))+(S\circ\phi)$\\
\phx\qquad\qquad\quad $\,\,=(M\circ L)+(M\circ R)+(S\circ\phi)$.\\
We have $M\circ R\,\,\in\,\,\scrl\circ\scrlito_1\,\,=\,\,\scrh_1\circ\scrlito_1\,\,\subseteq\,\,\scrbigo_1\circ\scrlito_1\,\,\subseteq\,\,\scrlito_1$.\\
Recall: $f_x^\T\in\scrbigo_1$. \quad
Then $S\circ\phi\,\,=\,\,S\circ f_x^\T\,\,\in\,\,\scrlito_1\circ\scrbigo_1\,\,\subseteq\,\,\scrlito_1$.\\
Then:  $(\psi\circ\phi)\,-\,(M\circ L)\,\,=\,\,(M\circ R)\,+\,(S\circ\phi)\,\,\in\,\,\scrlito_1\,+\,\scrlito_1\,\,\subseteq\,\,\scrlito_1$.
\end{proof}

\subsection{Basic properties of prime-derivatives}

\begin{thm}\wrlab{thm-prime-add-prime-prod-rule}
Let $f,g:\R\drta\R$ and let $x\in\D_{f'}\bigcap\D_{g'}$.\\
\phx\qquad Then: \qquad\qquad (1) $(f+g)'_x\,\,=\,\,f'_x\,+\,g'_x$\\
\phx\qquad\qquad\qquad and $\,\,\,\,$ (2) $(fg)'_x\,\,=\,\,f_x\,\cdot\,g'_x\,\,\,+\,\,g_x\,\cdot\,f'_x$.\\
\end{thm}

\begin{proof}
{\it Proof of (1):} Unassigned HW. {\it End of proof of (1).}\\
\\
{\it Proof of (2):}\\
By \tref{thm-D-add-prime-prod-rule}, we have $D_x(fg)\,\,=\,\,f_x\,\cdot\,D_xg\,\,\,+\,\,g_x\,\cdot\,D_xf$.\\
Let \quad $s:=f_x$, \quad $t:=g_x$, \quad $L:=D_xf$, \quad $M:=D_xg$.\\
Then $D_x(fg)\,\,=\,\,f_x\,\cdot\,D_xg\,\,\,+\,\,g_x\,\cdot\,D_xf\,\,=\,\,s\,\cdot\,M\,\,+\,\,t\,\cdot\,L$.\\
Also, we have: \quad  $f'_x=(D_xf)_1=L_1$ \quad and \quad $g'_x=(D_xg)_1=M_1$.\\
Then $(fg)'_x=(D_x(fg))_1=\,(s\,\cdot\,M\,\,+\,\,t\,\cdot\,L)_1$\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,=\,\,s\,\cdot\,M_1\,\,+\,\,t\,\cdot\,L_1$\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,=\,\,f_x\,\cdot\,g'_x\,\,\,+\,\,g_x\,\cdot\,f'_x$.\\
{\it End of proof of (2).}
\end{proof}

\begin{thm}\wrlab{thm-prime-scalar-mult}
Let $a\in\R$, let $f:\R\drta\R$ and let $x\in\D_{f'}$.\\
\phx\qquad\qquad\qquad\qquad\quad Then: \quad $(a\cdot f)'_x\,\,=\,\,a\,\cdot\,\,f'_x$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}\wrlab{thm-compose-linears}
Let $L,M\in\scrl$. \quad Then $(M\circ L)_1=M_1\cdot L_1$.
\end{thm}

\begin{proof}
Since $M\in\scrl=\R\cdot(\bullet)$, choose $t\in\R$ s.t.~$M=t\cdot(\bullet)$.\\
Then: \quad $M_1=t\cdot1=t$ \quad and \quad $M_{L_1}=t\cdot L_1$.\\
Then $(M\circ L)_1=M_{L_1}=t\cdot L_1=M_1\cdot L_1$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-prime-chain-rule}
Let $f,g:\R\drta\R$, let $x\in\D_{f'}$ and let $y:=f_x$.\\
\phx\qquad\quad Assume: $y\in\D_{g'}$. \qquad\quad Then: $(g\circ f)'_x=g'_y\cdot f'_x$.
\end{thm}

\begin{proof}
By \tref{thm-D-chain-rule}, we have $D_x(g\circ f)\,\,=\,\,(D_yg)\circ(D_xf)$.\\
Let \qquad $L:=D_xf$ \qquad and \qquad $M:=D_yg$.\\
Then \qquad $D_x(g\circ f)\,\,=\,\,(D_yg)\circ(D_xf)\,\,=\,\,M\circ L$.\\
By \tref{thm-compose-linears}, we have: \quad $(M\circ L)_1=M_1\cdot L_1$.\\
Also, we have: \quad  $f'_x=(D_xf)_1=L_1$ \quad and \quad $g'_y=(D_yg)_1=M_1$.\\
Then: \quad $(g\circ f)'_x=(D_x(g\circ f))_1=(M\circ L)_1=M_1\cdot L_1=g'_y\cdot f'_x$.
\end{proof}

\subsection{Fermat's Theorem}

\begin{defn}
Let $f$ be a functional and let $p\in\D_f$.\\
\phx\quad Then $f$ has a \bfindex[global strict-maximum] means: \quad
$f<f_p$ on $(\D_f)_p^\times$.\\
\phx\quad Also, $f$ has a \bfindex[global strict-minimum] means: \quad
$f>f_p$ on $(\D_f)_p^\times$.\\
\phx\quad Also, $f$ has a \bfindex[global semi-maximum] means: \quad
$f\le f_p$ on $(\D_f)_p^\times$.\\
\phx\quad Also, $f$ has a \bfindex[global semi-minimum] means: \quad
$f\ge f_p$ on $(\D_f)_p^\times$.\\
\phx\quad Also, $f$ has a \bfindex[global strict-extremum] means:\\
\phx\qquad\quad $f$ has a global strict-maximum or global strict-minimum at $p$.\\
\phx\quad Also, $f$ has a \bfindex[global semi-extremum] means:\\
\phx\qquad\quad $f$ has a global semi-maximum or global semi-minimum at $p$.\\
\end{defn}

\begin{defn}
Let $X$ be a metric space, $fX\drta\R$, $p\in\D_f$.\\
\phx\quad Then $f$ has a \bfindex[local strict-maximum] means:\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\exists B\in\scrb_X(p)$ s.t.~$f<f_p$ on $B_p^\times$.\\
\phx\quad Also, $f$ has a \bfindex[local strict-minimum] means:\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\exists B\in\scrb_X(p)$ s.t.~$f>f_p$ on $B_p^\times$.\\
\phx\quad Also, $f$ has a \bfindex[local semi-maximum] means:\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\exists B\in\scrb_X(p)$ s.t.~$f\le f_p$ on $B_p$.\\
\phx\quad Also, $f$ has a \bfindex[local semi-minimum] means:\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\exists B\in\scrb_X(p)$ s.t.~$f\ge f_p$ on $B_p$.\\
\phx\quad Also, $f$ has a \bfindex[local strict-extremum] means:\\
\phx\qquad\quad $f$ has a local strict-maximum or local strict-minimum at $p$.\\
\phx\quad Also, $f$ has a \bfindex[local semi-extremum] means:\\
\phx\qquad\quad $f$ has a local semi-maximum or local semi-minimum at $p$.\\
\end{defn}

\begin{thm}
Let $j\in\N_0$, $\phi\in\scrlito_j$, $\vep>0$.\\
\phx\qquad\qquad Then $|\phi|\le\vep\cdot|\bullet|^j$ near $0$ in $\R$.
\end{thm}

\begin{proof}
Want: $\exists B\in\scrb_\R(0)$ s.t.~$|\phi|\le\vep\cdot|\bullet|^j$ on $B$.\\
Since $\phi\in\scrlito_j=\CVZ\cdot|\bullet|^j$,
choose $\psi\in\CVZ$ s.t.~$\phi=\psi\cdot|\bullet|^j$.\\
Since $\psi\in\CVZ$, by \tref{thm-cvz-quantified-equiv}, choose $\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,|\psi_x|<\vep\,\,\,)$.\\
Let $B:=B_\R(0,\delta)$. \quad Then $B\in\scrb_\R(0)$. \quad Want: $|\phi|\le\vep\cdot|\bullet|^j$ on $B$.\\
\phx\quad Want: \quad $\forall x\in B$, \qquad $|\phi|_x\le(\vep\cdot|\bullet|^j)_x$.\\
Given $x\in B$. \qquad Want: $|\phi|_x\le(\vep\cdot|\bullet|^j)_x$.\\
Since $x\in B=B_\R(0,\delta)$, we get: \quad $d_\R(x,0)<\delta$.\\
Since $|x|=|x-0|=d_\R(x,0)<\delta$, by choice of $\delta$, we get: \quad $|\psi_x|<\vep$.\\
Then $|\phi|_x=|\phi_x|=|(\psi\cdot|\bullet|^j)_x|=|\,\,\psi_x\cdot|x|^j\,\,|=|\psi_x|\cdot|x|^j$.\\
Since $|x|\ge0$ and since $|\psi_x|<\vep$, we get $|\psi_x|\cdot|x|^j\le\vep\cdot|x|^j$.\\
Then $|\phi|_x=|\psi_x|\cdot|x|^j\le\vep\cdot|x|^j=(\vep\cdot|\bullet|^j)_x$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-pre-fermat}
Let $f:\R\drta\R$ and let $c\in\D_{f'}$.\\
\phx\qquad\qquad Assume: \quad $f$ has a local semi-maximum at $c$ in $\R$.\\
\phx\qquad\qquad Then: \qquad $f'_c=0$.
\end{thm}

\begin{proof}
Since $c\in\D_{f'}$, we get: $f'_c\ne\frownie$.\\
Let $L:=D_cf$. \quad Then, since $L_1=f'_c$, we get $L_1\ne\frownie$. \quad Then $L\ne\frownie$.\\
Since $\UE(\LINS_cf)=D_cf=L\ne\frownie$, we get $\UE(\LINS_cf)\in\LINS_cf$. \quad
Then \qquad $L=D_cf=\UE(\LINS_cf)\in\LINS_cf$, \qquad so \qquad $L\in\LINS_cf$.\\
Since $L\in\LINS_cf$ and since $f$ has a local semi-maximum at $c$ in $\R$,\\
\phx\qquad it follows, from, HW\#1-3, that: \quad $L=\bfzero$.\\
Then $f'_c=L_1=\bfzero_1=0$, as desired.
\end{proof}

The next result is called \bfindex[Fermat's Theorem].

\begin{thm}\wrlab{thm-fermat}
Let $f:\R\drta\R$ and let $c\in\D_{f'}$.\\
\phx\qquad\qquad Assume: \quad $f$ has a local semi-extremum at $c$ in $\R$.\\
\phx\qquad\qquad Then: \qquad $f'_c=0$.
\end{thm}

\begin{proof}
Since $f$ has a local semi-extremum at $c$ in $\R$,\\
\phx\qquad\qquad\quad at least one of the following must be true:\\
\phx\qquad\qquad\qquad\qquad $\,\,$ (1) $f$ has a local semi-maximum at $c$ in $\R$\\
\phx\qquad\qquad\qquad or \quad $\,$ (2) $f$ has a local semi-minimum at $c$ in $\R$.\\
\\
{\it Case 1:}
By \tref{thm-pre-fermat}, $f'_c=0$, as desired.
{\it End of Case 1.}\\
\\
{\it Case 2:}
Let $g:=-f$.\\
Since $f$ has a local semi-minimum at $c$ in $\R$,\\
\phx\qquad\qquad\qquad it follows that $g$ has a local semi-maximum at $c$ in $\R$.\\
Since $c\in\D_{f'}$ and $g=-f$, we get $g'_c=-f'_c$.
\qquad Want: $g'_c=0$.\\
Since $c\in\D_{f'}$, we get $f'_c\in\I_{f'}$. \quad
So, since $\I_{f'}\subseteq\R$, we get $f'_c\in\R$.\\
Since $f'_c\in\R$, we see that $-f'_c\in\R$, and so $-f'_c\ne\frownie$.\\
Since $g'_c=-f'_c\ne\frownie$, we get $c\in\D_{g'}$. \quad
Then, by \tref{thm-pre-fermat}, $g'_c=0$.\\
{\it End of Case 2.}
\end{proof}

\subsection{The Second Derivative Tests}
\quad
\vskip.1in\noindent

The next result is the \bfindex[Quadratic Taylor Theorem]:

\begin{thm}\wrlab{thm-quadratic-taylor}
Let $f:\R\drta\R$, $x\in\D_{f''}$, $m:=f'_x$ and $a:=f''_x/2$.\\
Let $L:=m\cdot(\bullet)$ and let $Q:=a\cdot(\bullet)^2$. \qquad
Then: $f_p^\T-L-Q\,\,\in\scrlito_2$.
\end{thm}

\begin{proof}
THIS WILL BE PROVED IN A LATER CLASS.
\end{proof}

The next result is the \bfindex[Positive Second Derivative Test]:

\begin{thm}
Let $f:\R\drta\R$ and let $x\in\D_{f''}$.\\
\phx\qquad\qquad\qquad Assume: \quad $(\,\,f'_x=0\,\,)\,\,\&\,\,(\,\,f''_x>0\,\,)$.\\
\phx\qquad\qquad\qquad Then: \qquad $f$ has a local strict-minimum at $x$ in $\R$.\\
\end{thm}

\begin{proof}
Let $\phi:=f_p^\T$. \quad
Want: $\phi$ has a local strict-minimum at $0$ in $\R$.\\
Let $m:=f'_x$ and let $a:=f''_x/2$. \quad
Then: $(\,m=0\,)\,\&\,(\,a>0\,)$.\\
Let $L:=m\cdot(\bullet)$ and $Q:=a\cdot(\bullet)^2$. \quad
By \tref{thm-quadratic-taylor}, $f_p^\T-L-Q\in\scrlito_2$.\\
Since $m=0$, we get: \qquad $L=m\cdot(\bullet)=0\cdot(\bullet)=\bfzero$.\\
Then $\phi-Q=\phi-\bfzero-Q=f_p^\T-L-Q\in\scrlito_2$, \quad so $\phi-Q\in\scrlito_2$.\\
Then, by HW\#1-4, $\phi$ has a local strict-minimum at $0$ in $\R$.
\end{proof}

The next result is the \bfindex[Negative Second Derivative Test]:

\begin{thm}
Let $f:\R\drta\R$ and let $x\in\D_{f''}$.\\
\phx\qquad\qquad\qquad Assume: \quad $(\,\,f'_x=0\,\,)\,\,\&\,\,(\,\,f''_x<0\,\,)$.\\
\phx\qquad\qquad\qquad Then: \qquad $f$ has a local strict-maximum at $x$ in $\R$.\\
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\subsection{Basic properties of derivatives}

\begin{thm}\wrlab{thm-deriv-const-is-zero}
Let $s\in\R$. \quad Then $(C_\R^s)'=\bfzero$.
\end{thm}

\begin{proof}
Let $f:=C_\R^s$. \quad Want: $f'=\bfzero$.\\
Want: $\forall x\in\R$, $f'_x=\bfzero_x$. \qquad\quad
Given $x\in\R$. \quad Want: $f'_x=\bfzero_x$.\\
We have: $\forall h\in\R$, $(f_x^\T)_h=f_{x+h}-f_x=s-s=0=\bfzero_h$.\\
Then $f_x^\T=\bfzero$. \qquad Then $f_x^\T-\bfzero=\bfzero-\bfzero=\bfzero\in\scrlito_1$, so $f_x^\T-\bfzero\in\scrlito_1$.\\
So, since $\bfzero=0\cdot(\bullet)\in\scrl$, we conclude that $\bfzero\in\LINS_xf$.\\
Then $D_xf=\bfzero$. \quad Then $f'_x=(D_xf)_1=\bfzero_1=0=\bfzero_x$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-equal-near-implies-derivs-equal-at}
Let $f,g:\R\drta\R$ and let $c\in\R$.\\
\phx Assume: $f=g$ near $c$ in $\R$. \qquad Then: $(\,\,D_cf=D_cg\,\,)\,\,\&\,\,(\,\,f'_c=g'_c\,\,)$.
\end{thm}

\begin{proof}
By HW\#1-5, we have: $\LINS_cf\subseteq\LINS_cg$.\\
Also, by HW\#1-5 (interchanging $f$ and $g$), we have: $\LINS_cg\subseteq\LINS_cf$.\\
Then $\LINS_cf=\LINS_cg$. \quad Then $\UE(\LINS_cf)=\UE(\LINS_cg)$.\\
Then $D_cf=\UE(\LINS_cf)=\UE(\LINS_cg)=D_cg$.\\
It remains only to show: $f'_c=g'_c$.\\
We have $f'_c=(D_cf)_1=(D_cg)_1=g'_c$, as desired.
\end{proof}

\begin{thm}
Let $f:\R\drta\R$ and let $c,s\in\R$.\\
\phx\quad Assume: $\,$ $f=s$ near $c$ in $\R$. \qquad\qquad Then: $\,$ $f'_c=0$.
\end{thm}

\begin{proof}
Since $f=s$ near $c$ in $\R$, we conclude: $f=C_\R^s$ near $c$ in $\R$.\\
Then, by \tref{thm-equal-near-implies-derivs-equal-at}, it follows that: \quad $f'_c=(C_\R^s)'_c$.\\
By \tref{thm-deriv-const-is-zero}, we see that: \quad $(C_\R^s)'=\bfzero$.\\
Then: \quad $f'_c=(C_\R^s)'_c=\bfzero'_c=0$, \quad as desired.\\
\end{proof}

\subsection{Rolle's Theorem}

\begin{thm}\wrlab{thm-contin-on-cpt-implies-attains-maxmin}
Let $K$ be a nonempty compact metric space.\\
Let $g:K\to\R$. \quad
Assume $g$ is continuous from $K$ to $\R$.\\
Then: \quad $\exists p,q\in K$ \quad s.t. \quad $g_p\le g\le g_q$ on $K$.
\end{thm}

\begin{proof}
Let $L:=\I_g$. \qquad Then $g:K\to>L$.\\
So, since $g$ is cointuous from $K$ to $L$,\\
\phx\qquad by \tref{thm-continsurj-fromcpt-image-cpt}, we see: \qquad $L$ is compact,\\
\phx\qquad and it follows that: \qquad $L$ is closed and bounded in $\R$.\\
Also, since $K\ne\emptyset$, it follows that: \quad $L\ne\emptyset$.\\
Then, by \tref{thm-maxmin-cpt-ne-frownie}, we get: \quad $\min\,L\ne\frownie\ne\max\,L$.\\
Let $s:=\min\,L$ and $t:=\max\,L$. \qquad Then $s\ne\frownie\ne t$.\\
Then $s,t\in L$ and $s\le L\le t$. \qquad Then $L\le t$ and $L\ge s$.\\
Since $s,t\in L=\I_g$, choose $p,q\in\D_g$ s.t.~$g_p=s$ and $g_q=t$.\\
Then $p,q\in\D_g=K$. \qquad Want: $g_p\le g\le g_q$ on $K$.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,$ Want: $s\le g\le t$ on $K$.\\
\phx\qquad\qquad\qquad $\,$ Want: $\forall x\in K$, $\,$ $s\le g_x\le t$.\\
\phx\qquad Given $x\in K$. \qquad $\,$ Want: $s\le g_x\le t$.\\
Since $x\in K=\D_g$, it follows that $g_x\in\I_g$.\\
Then $g_x\in\I_g=L\le t$. \qquad Want $g_x\ge s$.\\
We have $g_x\in\I_g=L\ge s$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-endpt-equal-gives-global-extr-in-interior}
Let $f:\R\drta\R$, $b\in\R$, $a<b$, $g:[a;b]\to\R$.\\
Assume: \quad $g$ is continuous from $\R$ to $\R$ \quad and \quad $g_a=g_b$.\\
Then: \quad $\exists c\in(a;b)$ s.t.~$g$ has a global semi-extremum at $c$.
\end{thm}

\begin{proof}
Let $K:=[a;b]$. \quad
Since $a<b$, we get: $K\ne\emptyset$.\\
Since $K$ is closed and bounded in $\R$ and since $\R$ is proper,\\
\phx\qquad we conclude that $K$ is compact.\\
By hypothesis, we have: \qquad $g:K\to\R$.\\
By absoluteness of continuity, $g$ is continuous from $K$ to $\R$.\\
By \tref{thm-contin-on-cpt-implies-attains-maxmin}, choose $p,q\in K$ s.t.~$g_p\le g\le g_q$ on $K$.\\
Exactly one of the following is true:\\
\phx\qquad (1) $\{p,q\}\subseteq\{a,b\}$ \qquad or \qquad (2) $\{p,q\}\not\subseteq\{a,b\}$.\\
\\
\\
{\it Case 1:}\\
Let $c:=(a+b)/2$. \quad Then $a<c<b$, so $c\in(a;b)$.\\
Want: $g$ has a global semi-extremum at $c$.\\
Want: $g$ has a global semi-maximum at $c$.\\
Want: $g\le g_c$ on $\D_g$. \qquad
Want: $g=g_c$ on $\D_g$.\\
Want: $\forall x\in\D_g$, $g_x=g_c$. \qquad
Given $x\in\D_g$. \qquad Want: $g_x=g_c$.\\
We have: $\,p\in\{p,q\}\subseteq\{a,b\}$, so $g_p\in\{g_a,g_b\}$.\\
Also: \qquad $q\in\{p,q\}\subseteq\{a,b\}$, so $g_q\in\{g_a,g_b\}$.\\
Let $y:=g_a$. \qquad Then $y=g_a=g_b$. \qquad Then $\{g_a,g_b\}=\{y\}$.\\
Then \quad $g_p,g_q\in\{g_a,g_b\}=\{y\}$, \quad so \quad $g_p=y=g_q$.\\
So, \quad since $g_p\le g\le g_q$ on $K$, \quad we get: $y\le g\le y$ on $K$.\\
Then $g=y$ on $K$. \qquad\qquad That is, \quad $\forall s\in K$, $\,$ $g_s=y$.\\
So, \quad since $x\in\D_g=K$ \quad and \quad since $c\in(a;b)\subseteq[a;b]=K$,\\
\phx\qquad we conclude: \quad $g_x=y$ \quad and \quad $g_c=y$.\\
Then $g_x=y=g_c$, as desired.\\
{\it End of Case 1.}\\
\\
{\it Case 2:}\\
Since $\{p,q\}\not\subseteq\{a,b\}$, \quad
choose $c\in\{p,q\}$ s.t.~$c\ne\{a,b\}$.\\
We have \quad $c\in\{p,q\}\subseteq K=[a;b]$ \quad and \quad $c\notin\{a,b\}$.\\
Then: \quad $c\in[a;b]\backslash\{a,b\}=(a;b)$.\\
Want: $g$ has a global semi-extremum at $c$.\\
Since $g_p\le g\le g_q$ on $K$, we get:\\
\phx\qquad $(\,\,g\le g_q\hbox{ on }K\,\,)\,\,\&\,\,(\,\,g\ge g_p\hbox{ on }K\,\,)$.\\
So, since $c\in\{p,q\}$, we get:\\
\phx\qquad $(\,\,g\le g_c\hbox{ on }K\,\,)\,\,\vee\,\,(\,\,g\ge g_c\hbox{ on }K\,\,)$.\\
So, since $K=\D_g$, we get:\\
\phx\qquad $(\,\,g\le g_c\hbox{ on }\D_g\,\,)\,\,\vee\,\,(\,\,g\ge g_c\hbox{ on }\D_g\,\,)$.\\
That is, \qquad either \qquad $\,g$ has a global semi-maximum at $c$\\
\phx\qquad\qquad\qquad or \qquad\quad $g$ has a global semi-minimum at $c$.\\
Then $g$ has a global semi-extremum at $c$, as desired.\\
{\it End of Case 2.}
\end{proof}

\begin{thm}\wrlab{thm-restr-has-globmax-in-interior-implies-fn-has-locmax}
Let $X$ be a metric space and $f:X\drta\R$.\\
Let $K\subseteq\D_f$ and $g:=f|K$ and $c\in\Int_XK$.\\
Assume: \quad $g$ has a global strict-maximum at $c$.\\
Then: \qquad $f$ has a local strict-maximum at $c$ in $X$.
\end{thm}

\begin{proof}
Want: $\exists B\in\scrb_X(c)$ s.t.~$f<f_c$ on $B_c^\times$.\\
Since $c\in\Int_XK$, choose $B\in\scrb_X(c)$ s.t.~$B\subseteq K$.\\
Then $B\in\scrb_X(c)$. \qquad Want: $f<f_c$ on $B_c^\times$.\\
Since $K\subseteq\D_f$ and $g=f|K$, we see that: \quad $\D_g=K$.\\
\phx\qquad Since $g=f|K$, it follows that $\,\,g=f$ on $K$.\\
\phx\quad So, since $\,K_c^\times\subseteq K$, we get: \qquad $\,\,\,g=f$ on $K_c^\times$.\\
We have $c\in\Int_XK\subseteq K$, so $(f|K)_c=f_c$. \qquad
Then $g_c=(f|K)_c=f_c$.\\
Since $B\subseteq K$, it follows that $B_c^\times\subseteq K_c^\times$.\\
Since $g$ has a global strict-maximum at $c$ and since $\D_g=K$,\\
\phx\qquad\qquad\qquad\qquad\qquad we conclude that: \quad\qquad $\,g<g_c$ on $K_c^\times$.\\
So, since $g=f$ on $K_c^\times$ and since $g_c=f_c$, we get: $f<f_c$ on $K_c^\times$.\\
\phx\qquad\qquad $\,$ So, since $B_c^\times\subseteq K_c^\times$, we get: \qquad\qquad $f<f_c$ on $B_c^\times$.
\end{proof}

\begin{thm}
Let $X$ be a metric space and $f:X\drta\R$.\\
Let $K\subseteq\D_f$ and $g:=f|K$ and $c\in\Int_XK$.\\
Assume: \quad $g$ has a global strict-minimum at $c$.\\
Then: \qquad $f$ has a local strict-minimum at $c$ in $X$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}
Let $X$ be a metric space and $f:X\drta\R$.\\
Let $K\subseteq\D_f$ and $g:=f|K$ and $c\in\Int_XK$.\\
Assume: \quad $g$ has a global semi-maximum at $c$.\\
Then: \qquad $f$ has a local semi-maximum at $c$ in $X$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}
Let $X$ be a metric space and $f:X\drta\R$.\\
Let $K\subseteq\D_f$ and $g:=f|K$ and $c\in\Int_XK$.\\
Assume: \quad $g$ has a global semi-minimum at $c$.\\
Then: \qquad $f$ has a local semi-minimum at $c$ in $X$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}\wrlab{thm-globsemiextr-at-inter-gives-locsemiextr-for-ext}
Let $X$ be a metric space and $f:X\drta\R$.\\
Let $K\subseteq\D_f$ and $g:=f|K$ and $c\in\Int_XK$.\\
Assume: \quad $g$ has a global semi-extremum at $c$.\\
Then: \qquad $f$ has a local semi-extremum at $c$ in $X$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}\wrlab{thm-globmax-at-interiorpt-is-locmax}
Let $X$ be a metric space and $f:X\drta\R$.\\
Let $K\subseteq\D_f$ and $c\in\Int_XK$.\\
Assume that $f$ has a global strict-maximum at $c$.\\
Then $f$ has a local strict-maximum at $c$ in $X$.
\end{thm}

\begin{proof}
Let $K:=\D_f$ and let $g:=f|K$. \quad Then $g=f$.\\
\phx\qquad\qquad So, since \qquad\quad $f$ has a global strict-maximum at $c$,\\
\phx\qquad we conclude that: $\,\,\,\,$ $g$ has a global strict-maximum at $c$.\\
Then, by \tref{thm-restr-has-globmax-in-interior-implies-fn-has-locmax},
$f$ has a local strict-maximum at $c$ in $X$.
\end{proof}

In class, we explained why, in \tref{thm-globmax-at-interiorpt-is-locmax},
it is insufficient to assume only that $c\in K$; we need $c\in\Int_XK$.

Recall: \qquad $\forall a,b\in\R$, \quad $[a;b]\in\scrt'_\R$ \quad and \quad $\Int_\R[a;b]=(a;b)$.\\

\begin{defn}
Let $f:\R\drta\R$ and let $S\subseteq\R$.\\
By $f$ is \bfindex[c/d] on $S$, we mean:\\
\phx\qquad $(\,\,f\hbox{ is continuous on }S\hbox{ from }\R\hbox{ to }\R\,\,)\,\,\&\,\,(\,\,\Int_\R S\subseteq\D_{f'}\,\,)$.
\end{defn}

That is, $\forall S\subseteq\R$, \quad c/d on $S$ \quad means:\\
\phx\qquad\qquad continuous on $S$ \quad and \quad differentiable on $\Int_\R S$.\\
In particular, $\forall a,b\in\R$, \quad c/d on $[a;b]$ \quad means:\\
\phx\qquad\qquad continuous on $[a;b]$ \quad and \quad differentiable on $(a;b)$.\\

The next theorem is \bfindex[Rolle's Theorem]:

\begin{thm}\wrlab{thm-Rolle}
Let $f:\R\drta\R$, \quad let $b\in\R$, \quad and let $a<b$.\\
Assume: \quad $f$ is c/d on $[a;b]$ \quad and \quad $f_a=f_b$.\\
Then: \qquad $\exists c\in(a;b)$ s.t.~$f'_c=0$.
\end{thm}

\begin{proof}
Since $f$ is c/d on $[a;b]$ we get:\\
\phx\qquad both \quad $f$ is continuous on $[a;b]$ from $\R$ to $\R$ \quad and \quad $(a;b)\subseteq\D_{f'}$.\\
Let $K:=[a;b]$. \quad
Then $f$ is continuous on $K$ from $\R$ to $\R$,\\
\phx\qquad so, by \tref{thm-restr-contin}, $\,\,\,f|K$ is continuous on $K$ from $\R$ to $\R$.\\
Let $g:=f|K$. \qquad\qquad\qquad Then $g$ is continuous on $K$ from $\R$ to $\R$.\\
Since $a,b\in[a;b]=K$ and $g=f|K$, we get: \quad $g_a=f_a$ \quad and \quad $g_b=f_b$.
By hypothesis, $f_a=f_b$. \quad Then $g_a=f_a=f_b=g_b$, \quad so $g_a=g_b$.\\
Then, by \tref{thm-endpt-equal-gives-global-extr-in-interior}, choose $c\in(a;b)$ s.t.\\
\phx\qquad\qquad\qquad\qquad $g$ has a global semi-extremum at $c$.\\
Then $c\in(a;b)$. \qquad Want: $f'_c=0$.\\
We have $c\in(a;b)=\Int_\R[a;b]=\Int_\R K$.\\
Then, by \tref{thm-globsemiextr-at-inter-gives-locsemiextr-for-ext},
$f$ has a local semi-extremum at $c$ in $\R$.\\
We have: \quad $c\in(a;b)\subseteq\D_{f'}$.\\
Then by Fermat's Theorem (\tref{thm-fermat}), we get $f'_c=0$.
\end{proof}

\subsection{Mean Value Theorem}

\begin{defn}
Let \quad $f:\R\to\R$ \quad and \quad $a,b\in\R$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: \qquad $\ds{\DQ_{ab}^f\,\,:=\,\,\frac{f_b-f_a}{b-a}}$.
\end{defn}

\begin{thm}
Let $f:\R\drta\R$, \quad $a,h\in\R$, \quad $b:=a+h$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: \qquad $\ds{\DQ_{ab}^f\,\,=\,\,\frac{(f_a^\T)_h}{h}}$.
\end{thm}

\begin{thm}\wrlab{thm-fn-minus-secslopeline}
Let $f:\R\drta\R$, \quad $a,b\in\D_f$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,$ Assume $a\ne b$. \qquad Let $m:=\DQ_{ab}^f$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,$ Let \qquad $L:=m\cdot(\bullet)$, \qquad $g:=f-L$.\\
\phx\qquad\qquad\qquad\qquad $\,\,\,$ Then: \qquad $g_a=g_b$.
\end{thm}

\begin{proof}
We have $\ds{\left(\frac{f_b-f_a}{b-a}\right)\cdot(b-a)=f_b-f_a}$.\\
So, since $\ds{m=\DQ_{ab}^f=\frac{f_b-f_a}{b-a}}$, we get $m\cdot(b-a)=f_b-f_a$.\\
Then $L_b-L_a=m\cdot b-m\cdot a=m\cdot(b-a)=f_b-f_a$.\\
Since $L_b-L_a=f_b-f_a$, we get: $f_a-L_a=f_b-L_b$.\\
Then $g_a=(f-L)_a=f_a-L_a=f_b-L_b=(f-L)_b=g_b$, as desired.
\end{proof}

\begin{defn}
Let $L:\R\to\R$.\\
\phx\qquad\qquad By $L$ is \bfindex[algebraically linear], we mean:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\forall s,t\in\R$, $\,$ $L_{s+t}=L_s+L_t$\\
\phx\qquad\qquad\qquad and \qquad\qquad $\,\forall c\in\R$, $\forall t\in\R$, $\,$ $L_{c\cdot t}=c\cdot L_t$.\\
\end{defn}

\begin{thm}
Let $L:\R\to\R$.\\
\phx\qquad\qquad Assume $L$ is algebraically linear.\\
\phx\qquad\qquad Let $m:=L_1$. \qquad Then $L=m\cdot(\bullet)$.
\end{thm}

\begin{proof}
Want: $\forall x\in\R$, $L_x=(m\cdot(\bullet))_x$.\\
Given $x\in\R$. \qquad Want: $L_x=(m\cdot(\bullet))_x$.\\
We have $L_x=L_{x\cdot1}=x\cdot L_1=x\cdot m=m\cdot x=(m\cdot(\bullet))_x$, as desired.
\end{proof}

\begin{thm}
Let $L:\R\to\R$.\\
\phx\qquad\qquad Then: \quad $(\,\,L\in\scrl\,\,)\,\,\Leftrightarrow\,\,(\,\,L\hbox{ is algebraically linear}\,\,)$.\\
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}\wrlab{thm-dbltranslate-linear}
Let $L\in\scrl$ and $x\in\R$. \qquad Then $L_x^\T=L$.
\end{thm}

\begin{proof}
Want: \quad $\forall h\in\R$, $\,\,(L_x^\T)_h=L_h$.\\
Given $h\in\R$. \qquad Want: $(L_x^\T)_h=L_h$.\\
Since $L$ is algebraically linear, we get: \quad $L_{x+h}=L_x+L_h$.\\
We have $(L_x^\T)_h=L_{x+h}-L_x=L_x+L_h-L_x=L_h$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-Dderiv-linear}
Let $L\in\scrl$ and $x\in\R$. \qquad Then $D_xL=L$.
\end{thm}

\begin{proof}
By \tref{thm-dbltranslate-linear}, $L_x^\T=L$. \qquad
Then $L_x^\T-L=\bfzero$.\\
Since $L_x^\T-L=\bfzero\in\scrlito_1$, we conclude: \quad $L\in\LINS_xL$.\\
Then, by \tref{thm-L-linearization-imlies-L-D-deriv}, we get $D_xL=L$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-primederiv-linear}
Let $m\in\R$ and $L:=m\cdot(\bullet)$. \qquad Then $L'=C_\R^m$.
\end{thm}

\begin{proof}
Want: $\,$ $\forall x\in\R$, $\,L'_x=(C_\R^m)_x$.\\
Given $x\in\R$. \quad Want: $\,\,L'_x=(C_\R^m)_x$. \quad
By \tref{thm-Dderiv-linear}, $D_xL=L$.\\
Then $L'_x=(D_xL)_1=L_1=m\cdot1=m=(C_\R^m)_x$, as desired.
\end{proof}

The next theorem is the \bfindex[Mean Value Theorem]:

\begin{thm}\wrlab{thm-MVT}
Let $f:\R\drta\R$, \quad let $b\in\R$, \quad let $a<b$.\\
Assume $f$ is c/d on $[a;b]$. \qquad
Then $\exists c\in(a;b)$ s.t.~$f'_c=\DQ_{ab}^f$.
\end{thm}

\begin{proof}
Let $m:=\DQ_{ab}^f$, \quad $L:=m\cdot(\bullet)$, \quad $B:=C_\R^m$.\\
By \tref{thm-primederiv-linear}, we have: \quad $L'=B$.\\
Then $\D_{L'}=\D_B=\R$. \quad
Then $[a;b]\subseteq\D_{L'}$. \quad
Then $L$ is c/d on $[a;b]$.\\
So, since $f$ is c/d on $[a;b]$,
it follows that $f-L$ is c/d on $[a;b]$.\\
Let $g:=f-L$. \qquad Then $g$ is c/d on $[a;b]$.\\
By \tref{thm-fn-minus-secslopeline}, we get: \quad $g_a=g_b$.\\
By Rolle's Theorem (\tref{thm-Rolle}), choose $c\in(a;b)$ s.t.~$g'_c=0$.\\
Then $c\in(a;b)$. \qquad Want: $f'_c=\DQ_{ab}^f$.\\
Since $f$ is c/d on $[a;b]$, we get $\Int_\R[a;b]\subseteq\D_{f'}$.\\
We have $c\in(a;b)=\Int_\R[a;b]\subseteq\D_{f'}$, \qquad so $c\in\D_{f'}$.\\
Since $\D_{f'}\subseteq\R$, we get: $\D_{f'}\bigcap\R=\D_{f'}$.\\
Then $c\in\D_{f'}=\D_{f'}\bigcap\R=\D_{f'}\bigcap\D_{L'}$.\\
Then, by linearity of differentiation, we see that $(f-L)'_c=f'_c-L'_c$.\\
So, since $f-L=g$ and $L'=B$, we get $g'_c=f'_c-B_c$.\\
So, since $g'_c=0$ and $B_c=(C_\R^m)_c=m$, we get $0=f'_c-m$, so $m=f'_c$.\\
Then $f'_c=m=\DQ_{ab}^f$, as desired.
\end{proof}

\begin{defn}
Let $f:\R\drta\R$ and let $S\subseteq\D_f$.\\
\phx\qquad\qquad\qquad Then \qquad $\DQ_S^f\,\,:=\,\,\{\DQ_{ab}^f\,|\,(\,a,b\in S\,)\,\&\,(\,a\ne b\,)\}$.
\end{defn}

\begin{defn}
Let $f$ be a function.\\
By $f$ is \bfindex[constant], we mean: \quad $\exists y$ s.t., $\forall x\in\D_f$, $\,$ $f_x=y$.
\end{defn}

\begin{thm}\wrlab{thm-DQ-and-precalc-conditions}
Let $f:\R\drta\R$ and let $S\subseteq\D_f$.\\
Then: \qquad\qquad (1) $\,$ $\,\,\,\,(\,\,0\notin\DQ_S^f\,\,)\,\,\,\,\Leftrightarrow\,\,\,\,(\,\,f|S\hbox{ is one-to-one}\,\,)$\\
\phx\qquad $\,$ and \qquad (2) $\,(\,\,\DQ_S^f\subseteq\{0\}\,\,)\,\,\,\,\Leftrightarrow\,\,\,\,(\,\,f|S\hbox{ is constant}\,\,)$\\
\phx\qquad $\,$ and \qquad (3) $\,$ $\,\,\,\,(\,\,\DQ_S^f>0\,\,)\,\,\,\,\Leftrightarrow\,\,\,\,(\,\,f|S\hbox{ is strictly-increasing}\,\,)$\\
\phx\qquad $\,$ and \qquad (4) $\,$ $\,\,\,\,(\,\,\DQ_S^f\ge0\,\,)\,\,\,\,\Leftrightarrow\,\,\,\,(\,\,f|S\hbox{ is semi-increasing}\,\,)$\\
\phx\qquad $\,$ and \qquad (5) $\,$ $\,\,\,\,(\,\,\DQ_S^f<0\,\,)\,\,\,\,\Leftrightarrow\,\,\,\,(\,\,f|S\hbox{ is strictly-decreasing}\,\,)$\\
\phx\qquad $\,$ and \qquad (6) $\,$ $\,\,\,\,(\,\,\DQ_S^f\le0\,\,)\,\,\,\,\Leftrightarrow\,\,\,\,(\,\,f|S\hbox{ is semi-decreasing}\,\,)$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{defn}
Let $S\subseteq\R$.\\
\phx\qquad By $S$ is an \bfindex[interval], we mean: \qquad $\forall a,b\in S$, $\,$ $[a;b]\subseteq S$.
\end{defn}

\noindent
Let $p,q\in\R$.
Then $(p;q)$, $[p;q)$, $(p;q]$ and $[p,q]$ are all intervals.\\

\noindent
Let $p\in\R$.
Then $(-\infty;p)$, $(-\infty,p]$, $(p,\infty)$ and $[p,\infty)$ are all intervals.\\

\noindent
Also, $(-\infty;\infty)$ is an interval. Note that $\R=(-\infty;\infty)$.

\begin{thm}\wrlab{thm-MVT-applications-prep}
Let $f:\R\drta\R$ and let $S\subseteq\D_{f'}$.\\
\phx\qquad\qquad\qquad Assume that $S$ is an interval. \qquad
Then $\DQ_S^f\subseteq f'_*S$.
\end{thm}

\begin{proof}
Want: $\forall m\in\DQ_s^f$, $m\in f'_*S$.\\
Given $m\in\DQ_s^f$. \qquad Want: $m\in f'_*S$.\\
Since $m\in\DQ_s^f$, choose $a,b\in S$ s.t.~$(\,\,a\ne b\,\,)\,\,\&\,\,(\,\,m=\DQ_{ab}^f\,\,)$.\\
Since $S$ is an interval and $a,b\in S$, we get: $[a;b]\in S$.\\
By hypothesis, $S\subseteq\D_{f'}$. \qquad Then $[a;b]\subseteq S\subseteq\D_{f'}$.\\
Since $[a;b]\subseteq\D_{f'}$, \qquad we conclude: \quad $f$ is c/d on $[a;b]$.\\
By the Mean Value Theorem (\tref{thm-MVT}),\\
\phx\qquad\qquad\qquad choose $c\in(a;b)$ s.t.~$f'_c=\DQ_{ab}^f$.\\
We have $c\in(a;b)\subseteq[a;b]\subseteq S$, \qquad so $c\in S$.\\
So, since $c\in S\subseteq\D_{f'}$, we get $c\in S\cap\D_{f'}$. \qquad
Then $f'_c\in f'_*S$.\\
Then $m=\DQ_{ab}^f=f'_c\in f'_*S$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-MVT-applications}
Let $f:\R\drta\R$ and let $S\subseteq\D_{f'}$.\\
\phx\qquad\qquad\qquad\qquad\qquad Assume that $S$ is an interval.\\
Then: \qquad\qquad (1) $\,$ $\,\,\,\,(\,\,0\notin f'_*S\,\,)\,\,\,\,\Rightarrow\,\,\,\,(\,\,f|S\hbox{ is one-to-one}\,\,)$\\
\phx\qquad $\,$ and \qquad (2) $\,(\,\,f'_*S\subseteq\{0\}\,\,)\,\,\,\,\Rightarrow\,\,\,\,(\,\,f|S\hbox{ is constant}\,\,)$\\
\phx\qquad $\,$ and \qquad (3) $\,$ $\,\,\,\,(\,\,f'_*S>0\,\,)\,\,\,\,\Rightarrow\,\,\,\,(\,\,f|S\hbox{ is strictly-increasing}\,\,)$\\
\phx\qquad $\,$ and \qquad (4) $\,$ $\,\,\,\,(\,\,f'_*S\ge0\,\,)\,\,\,\,\Rightarrow\,\,\,\,(\,\,f|S\hbox{ is semi-increasing}\,\,)$\\
\phx\qquad $\,$ and \qquad (5) $\,$ $\,\,\,\,(\,\,f'_*S<0\,\,)\,\,\,\,\Rightarrow\,\,\,\,(\,\,f|S\hbox{ is strictly-decreasing}\,\,)$\\
\phx\qquad $\,$ and \qquad (6) $\,$ $\,\,\,\,(\,\,f'_*S\le0\,\,)\,\,\,\,\Rightarrow\,\,\,\,(\,\,f|S\hbox{ is semi-decreasing}\,\,)$.
\end{thm}

\begin{proof}
Unassigned HW.\\
{\it Hint:} Combine \tref{thm-MVT-applications-prep} with \tref{thm-DQ-and-precalc-conditions}.
\end{proof}

\noindent
The converses of (1) and (3) of \tref{thm-MVT-applications} both fail because of:\\
\phx\qquad Let $f:=(\bullet)^3$ and let $S:=\R$.\\
Here, $f$ is 1-1 and strictly-increasing, but $0\in f'_*S$ and $\neg(f'_*S>0)$.\\

\noindent
The converse of (5) of \tref{thm-MVT-applications} fails because of:\\
\phx\qquad Let $f:=-(\bullet)^3$ and let $S:=\R$.\\
Here $f$ is strictly-decreasing, but $\neg(f'_*S<0)$.

Converses of (2), (4) and (6) of \tref{thm-MVT-applications}
happen only when $\#S=1$. See \tref{thm-converse-MVT-applications}, below.

\begin{defn}
We define $\sgn:\R\to\{-1,0,1\}$ by:\\
\phx\qquad\qquad $\forall x\in\R$, \qquad $\ds{\sgn_x=\begin{cases}1,&\hbox{if }x>0\\0,&\hbox{if }x=0\\-1,&\hbox{if }x<0.\end{cases}}$
\end{defn}

\begin{thm}\wrlab{thm-b-mimics-a}
Let $a,b\in\R$.\\
\phx\qquad Assume: \quad $|b-a|\le|a|/2$. \qquad Then: \quad $\sgn_b=\sgn_a$.
\end{thm}

\begin{proof}
Let $\vep:=|a|/2$. \quad Then $|b-a|\le\vep$, \quad so $a-\vep\le b\le a+\vep$.\\
Exactly one of the following holds:\\
\phx\qquad\quad (1) $a>0$ \qquad or \qquad (2) $a=0$ \qquad or \qquad (3) $a<0$.\\
\\
{\it Case (1):}
Since $a>0$, we get: \qquad $|a|=a$ \quad and \quad $a/2>0$.\\
Then $b\ge a-\vep=a-(|a|/2)=a-(a/2)=a/2>0$, so $b>0$.\\
Since $a>0$, we get $\sgn_a=1$. \quad Since $b>0$, we get $\sgn_b=1$.\\
Then $\sgn_b=1=\sgn_a$, as desired.\\
{\it End of Case (1).}\\
\\
{\it Case (2):}
Since $a=0$ and $\vep=|a|$, we get: $\vep=0$.\\
Then $|b-a|\le\vep=0$, so, since $|b-a|\ge0$, we get $|b-a|=0$.\\
Then $b=a$. \quad Then $\sgn_b=\sgn_a$, as desired.
{\it End of Case (2).}\\
\\
{\it Case (3):}
Since $a<0$, we get: \qquad $|a|=-a$ \quad and \quad $a/2<0$.\\
Then $b\le a+\vep=a+(|a|/2)=a+(-a/2)=a/2<0$, so $b<0$.\\
Since $a<0$, we get $\sgn_a=-1$. \quad Since $b<0$, we get $\sgn_b=-1$.\\
Then $\sgn_b=-1=\sgn_a$, as desired.\\
{\it End of Case (3).}
\end{proof}

\noindent
We express \tref{thm-b-mimics-a} by saying:\\
\phx\quad if $|b-a|\le|a|/2$, \quad then $b$ ``mimics'' $a$.

\begin{thm}\wrlab{thm-approx-mimics-fn-near-pt}
Let $\phi:\R\drta\R$, $k\in\N_0$, $H\in\scrh_k\backslash\{\bfzero\}$.\\
Assume: \quad $\phi-H\in\scrlito_k$.\\
Then: \quad $\exists B\in\scrb_\R(0)$ \quad s.t., \quad $\forall x\in B$, \qquad $\sgn_{\phi_x}=\sgn_{H_x}$.
\end{thm}

\begin{proof}
Since $H\in\scrh_k$, we have $H:\R\to\R$, so \quad $\forall x\in\R$, $H_x\in\R$.\\
Then, by \tref{thm-b-mimics-a}, it suffices to show:\\
\phx\qquad\qquad $\exists\delta>0$ s.t., $\forall x\in B_\R(0,\delta)$, \quad $|\phi_x-H_x|\le|H_x|/2$.\\
Since $H\in\scrh_k$, choose $c\in\R$ s.t.~$H=c\cdot(\bullet)^k$.\\
As $H\ne\bfzero$, we get $c\ne0$, so $|c|>0$. \quad
Let $\vep:=|c|/2$. \quad Then $\vep>0$.\\
By hypothesis, $\phi-H\in\scrlito_k$. \quad Let $R:=\phi-H$. \quad Then $R\in\scrlito_k$.\\
Choose $\delta>0$ s.t., $\forall x\in\R$, \qquad  $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|R_x|<\vep\cdot|x|^k\,\,)$.\\
Then $\delta>0$. \qquad\qquad
Want: $\,$ $\forall x\in B_\R(0,\delta)$, $\,\,$ $|\phi_x-H_x|\le|H_x|/2$.\\
\phx\qquad\qquad\qquad Given $x\in B_\R(0,\delta)$. \qquad Want: $|\phi_x-H_x|\le|H_x|/2$.\\
Since $x\in B_\R(0,\delta)$, we get: $d_\R(x,0)<\delta$.\\
Then $|x|=|x-0|=d_\R(x,0)<\delta$, \quad so,\\
\phx\qquad\qquad  by choice of $\delta$, \qquad $|R_x|<\vep\cdot|x|^k$.\\
We have $H_x=(c\cdot(\bullet)^k)_x=c\cdot x^k$, \quad so \quad $|H_x|=|c|\cdot|x|^k$.\\
\phx\qquad Then: \quad $|H_x|/2=(|c|/2)\cdot|x|^k=\vep\cdot|x|^k$.\\
Then: $\,\,|\phi_x-H_x|=|(\phi-H)_x|=|R_x|\le\vep\cdot|x|^k=|H_x|/2$, as desired.
\end{proof}

\noindent
We express \tref{thm-approx-mimics-fn-near-pt} by saying:\\
\phx\quad if $H\in\scrh_k$ and $\phi-H\in\scrlito_k$, \quad then $\phi$ ``mimics'' $H$ near $0$.\\
That is,\\
\phx\qquad if a function is approximated by a homogeneous $k$-polynomial,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $\,$with sub-$k$ remainder,\\
\phx\qquad then the function ``mimics'' the polynomial near $0$.\\
In class, we drew some graphs to indicate this geometrically.

\begin{thm}\wrlab{thm-converse-MVT-applications}
Let $f:\R\drta\R$, $S\subseteq\R$. Assume: $\#S\ne1$.\\
Then: \qquad\qquad (1) \qquad\quad $(\,\,f|S\hbox{ is one-to-one}\,\,)\,\,\,\,\not\Rightarrow\,\,\,\,(\,\,0\notin f'_*S\,\,)$\\
\phx\qquad $\,$ and \qquad (2) \qquad\qquad $(\,\,f|S\hbox{ is constant}\,\,)\,\,\,\,\Rightarrow\,\,\,\,(\,\,f'_*S\subseteq\{0\}\,\,)$\\
\phx\qquad $\,$ and \qquad (3) $(\,\,f|S\hbox{ is strictly-increasing}\,\,)\,\,\,\,\not\Rightarrow\,\,\,\,(\,\,f'_*S>0\,\,)$\\
\phx\qquad $\,$ and \qquad (4) \quad $(\,\,f|S\hbox{ is semi-increasing}\,\,)\,\,\,\,\Rightarrow\,\,\,\,(\,\,f'_*S\ge0\,\,)$\\
\phx\qquad $\,$ and \qquad (5) $(\,\,f|S\hbox{ is strictly-decreasing}\,\,)\,\,\,\,\not\Rightarrow\,\,\,\,(\,\,f'_*S<0\,\,)$\\
\phx\qquad $\,$ and \qquad (6) \quad $(\,\,f|S\hbox{ is semi-decreasing}\,\,)\,\,\,\,\Rightarrow\,\,\,\,(\,\,f'_*S\le0\,\,)$.
\end{thm}

\begin{proof}
{\it Proof of (1)-(5):} Unassigned HW. {\it End of proof of (1)-(5).}\\
\\
{\it Proof of (6):}
Want: \quad $\forall m\in f'_*S$, $\,$ $m\le0$.\\
Given $m\in f'_*S$. \qquad Want: $m\le0$.\\
Assume $m>0$. \qquad Want: Contradiction.\\
Since $m\in f'_*S$, choose $x\in S\cap\D_{f'}$ s.t.~$m=f'_x$.\\
Let $L:=D_xf$. \quad
Then $L_1=f'_x=m$, so $L_1=m$.\\
Since $m\in f'_*S$, we get $m\ne\frownie$, so $L_1\ne\frownie$, so $L\ne\frownie$.\\
Then $L\in\LINS_xf$. \quad
Then $L\in\scrl$ and $f_x^\T-L\in\scrlito_1$.\\
Let $\phi:=f_x^\T$. \qquad Then $\phi-L\in\scrlito_1$.\\
Since $L\in\scrl=\scrh_1$ and since $\phi-L\in\scrlito_1$,\\
\phx\qquad by \tref{thm-approx-mimics-fn-near-pt}, choose $\delta>0$ s.t., $\forall h\in B_\R(0,\delta)$,\\
\phx\qquad\qquad\qquad\qquad\qquad $[\,\,(\,\,L_h>0\,\,)\,\,\Rightarrow\,\,(\,\,\phi_h>0\,\,)\,\,]$\\
\phx\qquad\qquad\qquad and \qquad $[\,\,(\,\,L_h<0\,\,)\,\,\Rightarrow\,\,(\,\,\phi_h<0\,\,)\,\,]$.\\
Since $(\,\,x\in S\cap\D_{f'}\subseteq S\,\,)\,\,\&\,\,(\,\,\#S\ne1\,\,)$,\\
\phx\qquad we get $S_x^\times\ne\emptyset$, so choose $z\in S_x^\times$.\\
Since $z\ne x$, exactly one of the following must hold:\\
\phx\qquad\qquad (A) $z>x$ \qquad\qquad or \qquad\quad (B) $z<x$.\\
\\
{\it Case (A):}\\
Let $\gamma:=z-x$. \quad Then $\gamma>0$ and $x+\gamma=z$.\\
Then $x+\gamma=z\in S_x^\times\subseteq S$. \qquad Recall: $x\in S$.\\
Since $x,x+\gamma\in S$ and since $S$ is an interval,\\
\phx\qquad it follows that $[x;x+\gamma]\subseteq S$.\\
Let $h:=\min\{\gamma,\delta/2\}$. \quad
Then: $(\,\,h>0\,\,)\,\,\&\,\,(\,\,h\le\gamma\,\,)\,\,\&\,\,(\,\,h<\delta\,\,)$.\\
Recall: $L_1=m$. \quad Then $L_h=L_{h\cdot1}=h\cdot L_1=h\cdot m$.\\
So, since $h,m>0$, we conclude that $L_h>0$.\\
We have $h>0>-\delta$ and $h<\delta$, so $-\delta<h<\delta$, so $h\in(-\delta;\delta)$.\\
Since $h\in(-\delta;\delta)=B_\R(0,\delta)$ and since $L_h>0$,\\
\phx\qquad\qquad by choice of $\delta$, we get: \qquad $\,\,\phi_h>0$.\\
Since $0<h\le\gamma$, we get $x<x+h\le x+\gamma$, so $x+h\in(x;x+\gamma]$.\\
Then $x+h\in(x;x+\gamma]\subseteq[x;x+\gamma]\subseteq S$.\\
Since $x,x+h\in S$, since $x\le x+h$ and since $f|S$ is semi-decreasing,\\
\phx\qquad we conclude that: $f_x\ge f_{x+h}$. \qquad Then $f_{x+h}-f_x\le0$.\\
Then $0<\phi_h=(f_x^\T)_h=f_{x+h}-f_x\le0$, so $0<0$. \quad Contradiction.\\
{\it End of Case (A).}\\
\\
{\it Case (B):}\\
Let $\gamma:=x-z$. \quad Then $\gamma>0$ and $x-\gamma=z$.\\
Then $x-\gamma=z\in S_x^\times\subseteq S$. \qquad Recall: $x\in S$.\\
Since $x-\gamma,x\in S$ and since $S$ is an interval,\\
\phx\qquad it follows that $[x-\gamma;x]\subseteq S$.\\
Let $h:=\min\{\gamma,\delta/2\}$. \quad
Then: $(\,\,h>0\,\,)\,\,\&\,\,(\,\,h\le\gamma\,\,)\,\,\&\,\,(\,\,h<\delta\,\,)$.\\
Recall: $L_1=m$. \quad Then $L_{-h}=L_{-h\cdot1}=-h\cdot L_1=-h\cdot m$.\\
So, since $h,m>0$, we conclude that $L_h<0$.\\
We have $-h<0<\delta$ and $-h>-\delta$, so $-\delta<-h<\delta$, so $-h\in(-\delta;\delta)$.\\
Since $-h\in(-\delta;\delta)=B_\R(0,\delta)$ and since $L_{-h}<0$,\\
\phx\qquad\qquad by choice of $\delta$, we get: \qquad\quad $\phi_{-h}<0$.\\
Since $-\gamma\le-h<0$, we get $x-\gamma\le x-h<x$, so $x-h\in[x-\gamma;x)$.\\
Then $x-h\in[x-\gamma;x)\subseteq[x-\gamma;x]\subseteq S$.\\
Since $x-h,x\in S$, since $x-h\le x$ and since $f|S$ is semi-decreasing,\\
\phx\qquad $\,$ we conclude that: $\,f_{x-h}\ge f_x$. \qquad Then $f_{x-h}-f_x\ge0$.\\
Then $0>\phi_{-h}=(f_x^\T)_{-h}=f_{x-h}-f_x\ge0$, so $0>0$. \quad Contradiction.\\
{\it End of Case (B).}\\
\\
{\it End of proof of (6).}
\end{proof}

\subsection{Taylor's Theorem, second order}

\begin{thm}\wrlab{thm-thru-origin-dbltranslate}
Let $f:\R\drta\R$. \qquad Assume $f_0=0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $f_0^\T=f$.
\end{thm}

\begin{proof}
Want: $\,\,\,$ $\forall h\in\R$, $\,$ $(f_0^\T)_h=f_h$.\\
Given $h\in\R$. \qquad Want: $(f_0^\T)_h=f_h$.\\
We have $(f_0^\T)_h=f_{0+h}-f_0=f_h-0=f_h$.
\end{proof}

\begin{thm}\wrlab{thm-slopezero-lin-is-zero}
Let $L\in\scrl$. $\,$ Assume $L_1=0$. $\,$ Then $L=\bfzero$.
\end{thm}

\begin{proof}
$\forall x\in\R$, $\,\,$ $L_x=L_{x\cdot1}=x\cdot L_1=x\cdot0=0=\bfzero_x$. \quad
Then $L=\bfzero$.
\end{proof}

\begin{thm}\wrlab{thm-van-orderone-atzero-implies-litoone}
Let $f:\R\drta\R$. \qquad Assume: $f_0=f'_0=0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $f\in\scrlito_1$.
\end{thm}

\begin{proof}
Since $f_0=0$, by \tref{thm-thru-origin-dbltranslate}, we conclude that $f_0^\T=f$.\\
Let $L:=D_0f$. \quad Then $L=\UE(\LINS_0f)$ and $L_1=f'_0=0$.\\
Then $L_1=0\ne\frownie$, so $L\ne\frownie$, so $L\in\LINS_0f$.\\
Then \quad $L\in\scrl$ \quad and \quad $f_0^\T-L\in\scrlito_1$.\\
Since $L\in\scrl$ and $L_1=0$, by \tref{thm-slopezero-lin-is-zero}, we see that $L=\bfzero$.\\
Then $f=f-\bfzero=f_0^\T-L\in\scrlito_1$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-DQ-symmetric}
$\forall f:\R\drta\R$, $\forall a,b\in\R$, \quad $\DQ_{ab}^f=\DQ_{ba}^f$.\\
\end{thm}

\begin{thm}\wrlab{thm-DQ-symmetric-consequence}
Let $f:\R\drta\R$ and let $a,b\in\R$.\\
Let $\alpha:=\min\{a,b\}$ and let $\beta:=\max\{a,b\}$. \quad
Then $\DQ_{ab}^f=\DQ_{\alpha\beta}^f$.\\
\end{thm}

\begin{thm}\wrlab{thm-mvt-corollary}
Let $f:\R\drta\R$ and let $a,b\in\R$.\\
Assume $[a|b]\subseteq\D_{f'}$. \qquad
Then: $\exists c\in[a|b]$ s.t.~$f'_c\cdot(b-a)=f_b-f_a$.
\end{thm}

\begin{proof}
We know: \quad (1) $a=b$ \quad or \quad (2) $a\ne b$.\\
\\
{\it Case (1):}\\
Let $c:=a$. \quad Then $c=a\in[a|b]$. \quad Want: $f'_c\cdot(b-a)=f_b-f_a$.\\
Since $a,b\in[a|b]\subseteq\D_{f'}\subseteq\D_f$, we get $f_a,f_b\in\I_f$.\\
Since $c=a\in[a|b]\subseteq\D_{f'}$, we get $f'_c\in\I_{f'}$.\\
Then $f_a,f_b\in\I_f\subseteq\R$ and $f'_c\in\I_{f'}\subseteq\R$, so $f_a,f_b,f'_c\in\R$.\\
Since $a=b$, we get $b-a=0$.\\
Since $a=b$, we get $f_a=f_b$, so $f_b-f_a=0$.\\
Then $f'_c\cdot(b-a)=f'_c\cdot0=0=f_b-f_a$.\\
{\it End of Case (1).}\\
\\
{\it Case (2):}\\
Let $\alpha:=\min\{a,b\}$ and let $\beta:=\max\{a,b\}$. \quad Then $\alpha\le\beta$.\\
Since $a\ne b$, we get $\alpha\ne\beta$. \quad So, as $\alpha\le\beta$, we get: $\alpha<\beta$.\\
Since $[\alpha;\beta]=[a|b]\subseteq\D_{f'}$, we get: $f$ is c/d on $[\alpha;\beta]$.\\
By the Mean Value Theorem, choose $c\in(\alpha;\beta)$ s.t.~$f'_c=\DQ_{\alpha\beta}^f$.\\
Then $c\in(\alpha;\beta)\subseteq[\alpha;\beta]=[a|b]$. \quad Want: $f'_c\cdot(b-a)=f_b-f_a$.\\
By \tref{thm-DQ-symmetric-consequence}, we conclude: \qquad $\DQ_{ab}^f=\DQ_{\alpha\beta}^f$.\\
Since $a,b\in\R$ and $a\ne b$, we get $b-a\in\R_0^\times$, so $\ds{\frac{b-a}{b-a}=1}$.\\
We have $\ds{f'_c=\DQ_{\alpha\beta}^f=\DQ_{\alpha\beta}^f=\DQ_{ab}^f=\frac{f_b-f_a}{b-a}}$.\\
Then $\ds{f'_c\cdot(b-a)=(f_b-f_a)\cdot\frac{b-a}{b-a}=(f_b-f_a)\cdot1=f_b-f_a}$.\\
{\it End of Case (2).}
\end{proof}

\begin{thm}\wrlab{thm-ok-deriv-implies-okplusone}
Let $f:\R\drta\R$ and let $k\in\N_0$.\\
Assume: $(\,\,f'\in\scrlito_k\,\,)\,\,\&\,\,(\,\,f_0=0\,\,)$. \qquad
Then: $f\in\scrlito_{k+1}$.
\end{thm}

\begin{proof}
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|f_x|\le\vep\cdot|x|^{k+1}\,\,)$.\\
Given $\vep>0$. \qquad Want: $\exists\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|f_x|\le\vep\cdot|x|^{k+1}\,\,)$.\\
Since $f'\in\scrlito_k$, choose $\delta>0$ s.t., $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|f'_x|\le\vep\cdot|x|^k\,\,)$.\\
Then $\delta>0$. \qquad Want: $\forall x\in\R$,\\
\phx\qquad\qquad\qquad\qquad $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|f_x|\le\vep\cdot|x|^{k+1}\,\,)$.\\
Given $x\in\R$. \quad Want: $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|f_x|\le\vep\cdot|x|^{k+1}\,\,)$.\\
Assume: $|x|<\delta$. \qquad Want: $|f_x|\le\vep\cdot|x|^{k+1}$.\\
Since $f_0=0$, \quad it follows that: \quad $f_x-f_0=0$.\\
By \tref{thm-mvt-corollary}, choose $c\in[0|x]$ s.t.~$f'_c\cdot(x-0)=f_x-f_0$.\\
So, since $x-0=x$ and since $f_x-f_0=0$, we get: $f'_c\cdot x=f_x$.\\
Since $c\in[0|x]$, it follows that $|c|\le|x|$.\\
Since $|c|\le|x|<\delta$, by choice of $\delta$, we get: $|f'_c|\le\vep\cdot|c|^k$.\\
Then $|f_x|=|f'_c\cdot x|=|f'_c|\cdot|x|\le\vep\cdot|c|^k\cdot|x|\le\vep\cdot|x|^k\cdot|x|=\vep\cdot|x|^{k+1}$.
\end{proof}

\begin{thm}\wrlab{thm-van-ordertwo-atzero-implies-litotwo}
Let $f:\R\drta\R$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Assume: $f_0=f'_0=f''_0=0$. \qquad Then $f\in\scrlito_2$.
\end{thm}

\begin{proof}
Let $g:=f'$. \quad Then $g_0=f'_0=0$ and $g'_0=f''_0=0$.\\
Since $g_0=g'_0=0$, by \tref{thm-van-orderone-atzero-implies-litoone},
we get: $g\in\scrlito_1$.\\
Since $f'=g\in\scrlito_1$ and $f_0=0$, by \tref{thm-ok-deriv-implies-okplusone}, we get $f\in\scrlito_2$.
\end{proof}

\begin{thm}\wrlab{thm-dbltrans-of-dbltrans}
Let \quad $f:\R\drta\R$, \quad $x\in\D_f$ \quad and \quad $h\in\R$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \qquad $(f_x^\T)_h^\T=f_{x+h}^\T$.
\end{thm}

\begin{proof}
Let $g:=f_x^\T$. \qquad Want: $g_h^\T=f_{x+h}^\T$.\\
Want: $\forall s\in\R$, $(g_h^\T)_s=(f_{x+h}^\T)_s$.\\
Given $s\in\R$. \qquad Want: $(g_h^\T)_s=(f_{x+h}^\T)_s$.\\
Since $x\in\D_f$, we get $f_x\in\I_f$.\\
Since $f_x\in\I_f\subseteq\R$, we get: $f_x\,\,-\,\,f_x\,\,\,=\,\,\,0$.\\
Then: \qquad\quad\quad $(f_{x+h+s}\,-\,f_x)\,\,-\,\,(f_{x+h}\,-\,f_x)\,\,\,=\,\,\,f_{x+h+s}\,\,-\,\,f_{x+h}$.\\
Then $(g_h^\T)_s\,\,\,=\,\,\,g_{h+s}-g_h\,\,\,=\,\,\,(f_x^\T)_{h+s}-(f_x^\T)_h$\\
\phx\qquad\qquad\quad $=\,\,\,(f_{x+h+s}\,-\,f_x)\,\,-\,\,(f_{x+h}\,-\,f_x)\,\,\,=\,\,\,f_{x+h+s}\,\,-\,\,f_{x+h}$\\
\phx\qquad\qquad\quad $=\,\,\,(f_{x+h}^\T)_s$, \qquad as desired.
\end{proof}

\begin{thm}\wrlab{thm-lins-of-dbltransl}
Let $\,$ $f:\R\drta\R$, $\,$ $x\in\D_f$, $\,$ $g:=f_x^\T$, $\,$ $h\in\R$.\\
Then: \quad $\LINS_xg\,\,\,=\,\,\,\LINS_{x+h}f$.
\end{thm}

\begin{proof}
By \tref{thm-dbltrans-of-dbltrans}, we have $(f_x^\T)_h^\T=f_{x+h}^\T$. \quad
Then $g_h^\T=f_{x+h}^\T$.\\
Then $\LINS_hg\,\,\,=\,\,\,\{L\in\scrl\,|\,g_h^\T-L\in\scrlito_1\}$\\
\phx\qquad\qquad\qquad $\,=\,\,\,\{L\in\scrl\,|\,f_{x+h}^\T-L\in\scrlito_1\}\,\,\,=\,\,\,\LINS_{x+h}f$.
\end{proof}

\begin{thm}\wrlab{thm-deriv-of-dbltransl}
Let $\,$ $f:\R\drta\R$, $\,$ $x\in\D_f$, $\,$ $g:=f_x^\T$, $\,$ $h\in\R$.\\
Then: \quad $D_hg\,\,=\,\,D_{x+h}f$ \quad and \quad $g'_h=f'_{x+h}$.
\end{thm}

\begin{proof}
By \tref{thm-lins-of-dbltransl}, $\LINS_xg\,\,\,=\,\,\,\LINS_{x+h}f$.\\
Then $D_hg=\UE(\LINS_xg)=\UE(\LINS_{x+h}f)=D_{x+h}f$.\\
It remains to show that: \qquad $g'_h=f'_{x+h}$.\\
We have $g'_h=(D_hg)_1=(D_{x+h}f)_1=f'_{x+h}$, as desired.
\end{proof}

\begin{defn}
Let $a$ and $b$ be objects.\\
\phx\qquad\qquad\quad Then \quad $a\,\,\,=^*\,\,b$ \quad means: \quad $(\,\,b\,\ne\,\frownie\,\,)\,\,\Rightarrow\,\,(\,\,a\,=\,b\,\,)$.\\
\phx\qquad\qquad\quad Also, \quad $a\,\,\,{}^*\!\!=\,\,\,\,b$ \quad means: \quad $(\,\,a\,\ne\,\frownie\,\,)\,\,\Rightarrow\,\,(\,\,a\,=\,b\,\,)$.
\end{defn}

\noindent
We read \qquad ``$a\,\,\,=^*\,\,b$'' \qquad as\\
\phx\qquad ``$a$ is contingent equal to $b$, with the contingency on $b$''.\\
We read \qquad ``$a\,\,\,{}^*\!\!=\,\,\,\,b$'' \qquad as\\
\phx\qquad ``$a$ is contingent equal to $b$, with the contingency on $a$''.\\

The next two theorems restate linearity of derivatives, the
product rule and the chain rule, all in the language of contingent equalities.

\begin{thm}
Let \quad $f,g:\R\drta\R$ \quad and \quad $x\in\R$.\\
\phx\qquad\qquad Then: \qquad\qquad\qquad $(f\,+\,g)'_x\,\,=^*\,\,f'_x\,+\,g'_x$\\
\phx\qquad\qquad\qquad\qquad and \quad\qquad $\,\,\,\,\,(f\cdot g)'_x\,\,=^*\,\,f'_x\cdot g_x\,+\,f_x\cdot g'_x$\\
\phx\qquad\qquad\qquad\qquad and \quad\qquad $\,\,\,\,(g\circ f)'_x\,\,=^*\,\,g'_{f_x}\cdot f'_x$.
\end{thm}

\begin{thm}
Let \quad $c\in\R$, \quad $f:\R\drta\R$ \quad and \quad $x\in\R$.\\
\phx\qquad\qquad Then: \qquad\qquad\qquad $(c\cdot f)'_x\,\,=^*\,\,c\cdot f'_x$.
\end{thm}

We can sometimes ``remove the contingency'':

\begin{thm}
$\forall a,b$, $\,\,\,\,[\,\,\,(\,\,\,a\,\,=^*\,\,b\,\,\ne\,\,\frownie\,\,\,)\,\,\,\Rightarrow\,\,\,(\,\,\,a=b\,\,\,)\,\,\,]$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\&\,\,\,[\,\,\,(\,\,\,a\,\,=^*\,\,b\,\,\ne\,\,\frownie\,\,\,)\,\,\,\Rightarrow\,\,\,(\,\,\,a=b\,\,\,)\,\,\,]$.
\end{thm}

If we have contingencies simlutaneously on both sides, then the contingency is removable:

\begin{thm}
Let $a$ and $b$ be objects.\\
\phx\qquad\qquad Then: \qquad $[\,\,\,(\,\,a\,\,=^*\,\,b\,\,)\,\,\,\,\&\,\,\,(\,\,a\,\,\,\,{}^*\!\!=\,\,b\,\,)\,\,]\,\,\,\Rightarrow\,\,\,[\,\,\,a\,\,=\,\,b\,\,\,]$.
\end{thm}

\noindent
Idea of proof: at least one of the following must hold:\\
\phx\qquad $a=\frownie=b$ \quad or \quad $a\ne\frownie$ \quad or \quad $b\ne\frownie$.\\
In the first case, $a=b$ by contraction.\\
In the second, removing the contingency from $a\,\,\,\,{}^*\!\!=\,\,b$, we get $a=b$.\\
In the third, removing the contingency from $a\,\,=^*\,\,b$, we get $a=b$.\\
QED

\vskip.1in

The following captures that the derivative commutes with scalar multiplication, in the language of contingent equality:

\begin{thm}\wrlab{thm-deriv-comm-scmult-contingent}
Let $f:\R\drta\R$ and let $a,x\in\R$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then \quad $\,(a\cdot f)'_x\,\,=^*\,\,a\cdot f'_x$.
\end{thm}

\tref{thm-deriv-comm-scmult-contingent} follows from \tref{thm-prime-scalar-mult}.

The following captures additivity of the derivative, along with the product and chain rules,
all in the language of contingent equality:

\begin{thm}\wrlab{thm-add-prod-chrule-contingent}
Let $f,g:\R\drta\R$ and let $x\in\R$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then \quad $(f+g)'_x\,\,=^*\,\,f'_x+g'_x$\\
\phx\qquad\qquad\qquad\qquad\qquad and \qquad $\,(f\cdot g)'_x\,\,=^*\,\,f'_x\cdot g_x+f_x\cdot g'_x$\\
\phx\qquad\qquad\qquad\qquad\qquad and \qquad $(g\circ f)'_x\,\,=^*\,\,g'_{f_x}\cdot f'_x$.
\end{thm}

\tref{thm-add-prod-chrule-contingent} follows from
\tref{thm-prime-add-prime-prod-rule} and \tref{thm-prime-chain-rule}.

\begin{thm}
Let \quad $a\in\R_0^\times$, \quad $f:\R\drta\R$, \quad $x\in\R$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then \quad $\,(a\cdot f)'_x\,\,=\,\,a\cdot f'_x$
\end{thm}

\begin{proof}
We have $(a\cdot f)'_x\,\,=^*\,\,a\cdot f'_x$.\\
It therefore suffices to show: $a\cdot f'_x\,\,=^*\,\,(a\cdot f)'_x$.\\
Let $g:=a\cdot f$. \qquad\quad Want: $\,a\cdot f'_x\,\,=^*\,\,g'_x$.\\
Let $b:=1/a$. \qquad Then: \quad $b\cdot g=b\cdot a\cdot f=1\cdot f=f$.\\
Then $f=b\cdot g$. \quad Also, \quad $(b\cdot g)'_x\,\,=^*\,\,b\cdot g'_x$.\\
Then $a\cdot f'_x\,\,=\,\,a\cdot(b\cdot g)'_x\,\,=^*\,\,a\cdot b\cdot g'_x\,\,=\,\,1\cdot g'_x\,\,=\,\,g'_x$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-dbl-deriv-additive-contingent}
Let $f,g:\R\to\R$.\\
\phx\qquad\qquad\qquad Then: \qquad $\forall x\in\R$, \quad $(f+g)''_x\,\,=^*\,\,f''_x+g''_x$.
\end{thm}

\begin{proof}
By \tref{thm-add-prod-chrule-contingent}, we have:\\
\phx\qquad\qquad\qquad $\forall x\in\R$, \quad $(f+g)'_x\,\,=^*\,\,f'_x+g'_x$.\\
Then:\qquad\qquad $\forall x\in\R$, \quad $(f+g)'_x\,\,=^*\,\,(f'+g')_x$.\\
Let $\alpha:=f'+g'$ and $\beta:=(f+g)'$. \quad Then: \qquad $\forall x\in\R$, \quad $\beta_x\,\,=^*\,\,\alpha_x$.\\
\phx\qquad\qquad Then, by HW\#2-4, we get: \qquad\qquad $\,\,$ $\forall x\in\R$, \quad $\beta'_x\,\,=^*\,\,\alpha'_x$.\\
Therefore: \quad $\,\,\,\forall x\in\R$, \quad $\,\,\,(f+g)''_x\,\,=^*\,\,(f'+g')'_x$.\\
By \tref{thm-add-prod-chrule-contingent} (with $f$ replaced by $f'$ and $g$ by $g'$), we have:\\
\phx\qquad\qquad\qquad $\forall x\in\R$, \qquad\qquad\qquad\quad $\,\,\,(f'+g')'_x\,\,=^*\,\,f''_x+g''_x$.\\
Then: \qquad\quad $\,\,\,\forall x\in\R$, \quad $\,\,\,(f+g)''_x\,\,=^*\,\,(f'+g')'_x\,\,=^*\,\,f''_x+g''_x$.\\
Then: \qquad\quad $\,\,\,\forall x\in\R$, \quad $\,\,\,(f+g)''_x\,\,=^*\,\,f''_x+g''_x$, \qquad as desired.
\end{proof}

\begin{thm}\wrlab{thm-diff-mxpluscxsquared}
Let $m,c\in\R$ and let $\psi:\R\to\R$.\\
Assume: $\forall x\in\R$, $\psi_x=mx+cx^2$.\\
Then: $\forall x\in\R$, $[\,\,(\,\,\psi'_x=m+2cx\,\,)\,\,\&\,\,(\,\,\psi''_x=2c\,\,)\,\,]$.\\
\end{thm}

\begin{proof}
Define $L,Q:\R\to\R$ by: $\forall x\in\R$, $(\,\,L_x=mx\,\,)\,\,\&\,\,(\,\,Q_x=cx^2\,\,)$.\\
We have: \quad $\forall x\in\R$, $\,$ $\psi_x=mx+cx^2=L_x+Q_x=(L+Q)_x$.\\
Then $\forall x\in\R$, $\,$ $\psi_x=(L+Q)_x$. \qquad Then $\psi=L+Q$.\\
By HW\#2-2, we know: \quad $\forall x\in\R$, $\,$ $(\,\,L'_x=m\,\,)\,\,\&\,\,(\,\,L''_x=0\,\,)$.\\
By HW\#2-3, we know: \quad $\forall x\in\R$, $\,$ $(\,\,Q'_x=2cx\,\,)\,\,\&\,\,(\,\,Q''_x=2c\,\,)$.\\
Given $x\in\R$. \quad Want: \quad $(\,\,\psi'_x=m+2cx\,\,)\quad\,\,\,\,\,\&\,\,\,\,\,\quad(\,\,\psi''_x=2c\,\,)$.\\
We have: $\psi'_x\,\,=\,\,(L+Q)'_x\,\,=^*\,\,L'_x+Q'_x\,\,=\,\,m+2cx\,\,\ne\,\,\frownie$,\\
\phx\qquad\qquad\qquad and so \qquad $\,\psi'_x=m+2cx$. \quad $\,\,\,$ Want: $\,\psi''_x=2c$.\\
We have: $\psi''_x\,\,=\,\,(L+Q)''_x\,\,=^*\,\,L''_x+Q''_x\,\,=\,\,2c\,\,\ne\,\,\frownie$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad and so: $\,\,\,\,\psi''_x=2c$.
\end{proof}

\begin{thm}
Let $\ell\in\N$, $f\in\scrh_\ell$. \qquad
Then $f'\in\scrh_{\ell-1}$.
\end{thm}

\begin{proof}
Since $f\ni\scrh_\ell$, choose $c\in\R$ s.t.~$f=c\cdot(\bullet)^\ell$.\\
Since $c\cdot\ell\cdot(\bullet)^{\ell-1}\in\scrh_{\ell-1}$, it suffices to show:
$f'=c\cdot\ell\cdot(\bullet)^{\ell-1}$.\\
Want: $\forall x\in\R$, $f'_x=(c\cdot\ell\cdot(\bullet)^{\ell-1})_x$.\\
Given $x\in\R$. \qquad Want: $f'_x=(c\cdot\ell\cdot(\bullet)^{\ell-1})_x$.\\
By HW\#2-1, we have: \quad $\,\,\,\,\,((\bullet)^\ell)'\,\,=\,\,\,\,\,\,\,\,\,\ell\cdot(\bullet)^{\ell-1}$.\\
Then $f'_x\,=\,(c\cdot(\bullet)^\ell)'_x\,=^*\,c\cdot((\bullet)^\ell)'_x\,=\,c\cdot(\ell\cdot(\bullet)^{\ell-1})_x\,=\,c\cdot\ell\cdot x^{\ell-1}\,\ne\,\frownie$.\\
Then \quad\quad $\,\,f'_x\,=\,c\cdot\ell\cdot x^{\ell-1}\,=\,(c\cdot\ell\cdot(\bullet)^{\ell-1})_x$, \qquad as desired.
\end{proof}

\begin{thm}\wrlab{thm-deriv-diff-with-onediffeverywhere}
Let $\phi:\R\drta\R$ and $\psi:\R\to\R$.\\
\phx\qquad\qquad Assume: $\D_{\psi'}=\R$. \qquad Then: $(\phi-\psi)'=\phi'-\psi'$.
\end{thm}

\begin{proof}
Let $\rho:=\phi-\psi$. \qquad Want: $\rho'=\phi'-\psi'$.\\
\phx\qquad\qquad\qquad Want: \quad $\forall x\in\R$, $\,\,\,\,$ $\rho'_x=(\phi'-\psi')_x$.\\
Given $x\in\R$. \qquad Want: $\rho'_x=(\phi'-\psi')_x$. \qquad $\,\,\,\,$ Want: $\rho'_x\,\,=\,\,\phi'_x-\psi'_x$.\\
Since $\rho'_x\,\,=\,\,(\phi-\psi)'_x\,\,=^*\,\,\phi'_x-\psi'_x$,
it suffices to show: $\rho'_x\,\,\,{}^*\!\!=\,\,\phi'_x-\psi'_x$.\\
Want: $\phi'_x-\psi'_x\,\,=^*\,\,\rho'_x$. \qquad\qquad
Since $\phi:\R\to\R$, we get: $\psi-\psi=\bfzero$.\\
Then $\rho+\psi=\phi+\psi-\psi=\phi+\bfzero=\phi$, \quad so $\phi=\rho+\psi$.\\
We have $x\in\R=\D_{\psi'}$, \qquad so \qquad $\psi'_x\in\I_{\psi'}$.\\
Since $\psi'_x\in\I_{\psi'}\subseteq\R$, \qquad we get: \qquad $\psi'_x-\psi'_x=0$.\\
We have \quad $\phi'_x\,\,=\,\,(\rho+\psi)'_x\,\,=^*\,\,\rho'_x+\psi'_x$, \quad so \quad $\phi'_x\,\,=^*\,\,\rho'_x+\psi'_x$.\\
Then $\phi'_x-\psi'_x\,\,=^*\,\,\rho'_x+\psi'_x-\psi'_x\,\,=\,\,\rho'_x+0\,\,=\,\,\rho'_x$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-deriv-dbldiff-with-onedbldiffeverywhere}
Let $\phi:\R\drta\R$ and $\psi:\R\to\R$.\\
\phx\qquad\qquad Assume: $\D_{\psi''}=\R$. \qquad Then: $(\phi-\psi)''=\phi''-\psi''$.
\end{thm}

\begin{proof}
Since $\R=\D_{\psi''}\subseteq\D_{\psi'}$ and since $\D_{\psi'}\subseteq\R$, we get: $\D_{\psi'}=\R$.\\
Then, by \tref{thm-deriv-diff-with-onediffeverywhere}, we get: $\,\,\,(\phi-\psi)'=\phi'-\psi'$.\\
Let $\alpha:=\phi'$ and $\beta:=\psi'$. \quad $\,$ Then: $\,\,\,\,(\phi-\psi)'=\alpha-\beta$.\\
Then $\,$ $\alpha'=\phi''$ $\,$ and $\,$ $\beta'=\psi''$ $\,$ and $\,$ $(\phi-\psi)''=(\alpha-\beta)'$.\\
We have $\D_\beta=\D_{\psi'}=\R$ and $\I_\beta=\I_{\psi'}\subseteq\R$, \quad so \quad $\beta:\R\to\R$.\\
So, since $\D_{\beta'}=\D_{\psi''}=\R$, it follows,\\
\phx\qquad\qquad from \tref{thm-deriv-diff-with-onediffeverywhere}, that: \quad $(\alpha-\beta)'=\alpha'-\beta'$.\\
Then $(\phi-\psi)''=(\alpha-\beta)'=\alpha'-\beta'=\phi''-\psi''$.
\end{proof}

\begin{thm}\wrlab{thm-diff-fn-minus-mx-minus-cxsquared}
Let $m,c\in\R$ and let $\rho,g:\R\drta\R$.\\
Assume: $\,\,\forall x\in\R$, $\rho_x=g_x-mx-cx^2$.\\
Then: $\forall x\in\R$, $[\,\,(\,\,\rho'_x=g'_x-m-2cx\,\,)\,\,\&\,\,(\,\,\rho''_x=g''_x-2c\,\,)\,\,]$.\\
\end{thm}

\begin{proof}
Define $\psi:\R\to\R$ by: $\forall x\in\R$, $\psi_x=mx+cx^2$.\\
By \tref{thm-diff-mxpluscxsquared}, $\forall x\in\R$, $\psi''_x=2c$.\\
Since $\forall x\in\R$, $\psi''_x=2c\ne\frownie$, we get $\forall x\in\R$, $x\in\D_{\psi''}$, and so $\R\subseteq\D_{\psi''}$.\\
Since $\R\subseteq\D_{\psi''}\subseteq\D_{\psi'}\subseteq\R$, we get $\D_{\psi'}=\R$ and $\D_{\psi''}=\R$.\\
Then, by \tref{thm-deriv-dbldiff-with-onedbldiffeverywhere} and \tref{thm-deriv-diff-with-onediffeverywhere},\\
\phx\qquad we get: \qquad $(g-\psi)'=g'-\psi'$ \qquad and \qquad $(g-\psi)''=g''-\psi''$.\\
We have: \quad $\forall x\in\R$, $\,\,\rho_x=g_x-(mx+cx^2)=g_x-\psi_x=(g-\psi)_x$.\\
\phx\quad Then: \quad $\forall x\in\R$, $\,\,\rho_x=(g-\psi)_x$. \qquad\qquad Then: $\rho=g-\psi$.\\
Then: \qquad $\rho'=(g-\psi)'=g'-\psi'$ \qquad and \qquad $\rho''=(g-\psi)''=g''-\psi''$.\\
Given $x\in\R$. \qquad Want: $[\,\,(\,\,\rho'_x=g'_x-m-2cx\,\,)\,\,\&\,\,(\,\,\rho''_x=g''_x-2c\,\,)\,\,]$.\\
By \tref{thm-diff-mxpluscxsquared}, we have: $(\,\,\psi'_x=m+2cx\,\,)\,\,\&\,\,(\,\,\psi''_x=2c\,\,)$.\\
Then $\rho'_x=(g'-\psi')_x=g'_x-\psi'_x=g'_x-m-2cx$. \qquad Want: $\psi''_x=2c$.\\
We have $\rho''_x=(g''-\psi'')_x=g''_x-\psi''_x=g''_x-2c$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-translate-fn-implies-translate-deriv}
Let $\lambda,\mu:\R\drta\R$ and let $a\in\R$.\\
\phx\qquad Assume: $\forall h\in\R$, $\mu_h=\lambda_{h+a}$. \qquad
Then: $\forall h\in\R$, $\mu'_h=\lambda'_{h+a}$.
\end{thm}

\begin{proof}
Define $S,T:\R\to\R$ by: $\forall h\in\R$, $(\,S_h=h-a\,)\,\&\,(\,T_h=h+a\,)$.\\
By HW\#2-5, $S'=C_\R^1=T'$. \qquad
Also, $S\circ T=\id_\R=T\circ S$.\\
We have: $\forall h\in\R$, $\mu_h=\lambda_{h+a}=\lambda_{T_h}=(\lambda\circ T)_h$. \quad Then $\mu=\lambda\circ T$.\\
Then $\mu\circ S=\lambda\circ T\circ S=\lambda\circ\id_\R=\lambda$, \quad so $\mu\circ S=\lambda$.\\
Given $h\in\R$. \quad Want: $\mu'_h=\lambda'_{h+a}$. \quad Let $k:=h+a$. \quad Want: $\mu'_h=\lambda'_k$.\\
We have: \qquad $T_h=h+a=k$ \qquad and \qquad $S_k=k-a=h$.\\
Then $\mu'_h\,\,=\,\,(\lambda\circ T)'_h\,\,=^*\,\,\lambda'_{T_h}\cdot T'_h\,\,=\,\,\lambda'_k\cdot1\,\,=\,\,\lambda'_k$,\\
\phx\qquad\qquad\qquad\qquad so $\,\mu'_h\,\,\,\,=^*\,\,\lambda'_k$.\\
It suffices to show: \quad $\,\,\mu'_h\,\,\,\,{}^*\!\!=\,\,\,\,\lambda'_k$. \qquad\qquad
Want: $\lambda'_k\,\,=^*\,\,\mu'_h$.\\
We have $\lambda'_k\,\,=\,\,(\mu\circ S)'_k\,\,=^*\,\,\mu'_{S_k}\cdot S'_k\,\,=\,\,\mu'_h\cdot 1\,\,=\,\,\mu'_h$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-secondderiv-dbltranslate}
Let $f:\R\drta\R$ and $a\in\R$ and $g:=f_a^\T$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: \quad $\forall h\in\R$, $\,$ $g''_h=f''_{h+a}$.
\end{thm}

\begin{proof}
By \tref{thm-deriv-of-dbltransl}, we have: \quad $\forall h\in\R$, $\,$ $g'_h=f'_{h+a}$.\\
Let $\mu:=g'$ and $\lambda:=f'$. \quad Then: \quad $\forall h\in\R$, $\,$ $\mu_h=\lambda_{h+a}$.\\
Then, by \tref{thm-translate-fn-implies-translate-deriv}, \quad $\forall h\in\R$, $\,$ $\mu'_h=\lambda'_{h+a}$.\\
Then: \quad $\forall h\in\R$, $\,$ $g''_h=\mu'_h=\lambda'_{h+a}=f''_{h+a}$, \quad as desired.
\end{proof}

The next result is called the \bfindex[Second Order Taylor Theorem]\index[terms]{Taylor Theorem, Second Order}:

\begin{thm}\wrlab{thm-second-order-taylor}
Let $f:\R\drta\R$ and $a\in\D_{f''}$.\\
Let $m:=f'_a$ and $c:=f''_a/2$ and $L:=m\cdot(\bullet)$ and $Q:=c\cdot(\bullet)^2$.\\
\phx\qquad\qquad\qquad Then: \quad $f_a^\T\,-\,L\,-\,Q\,\,\,\in\,\,\,\scrlito_2$.
\end{thm}

\begin{proof}
Since $a\in\D_f$, it follows that $(f_a^\T)_0=0$.\\
Let $g:=f_a^\T$. \qquad Then $g_0=0$. \qquad Want: $g-L-Q\in\scrlito_2$.\\
Let $\rho:=g-L-Q$. \qquad\qquad\qquad\qquad Want: $\rho\in\scrlito_2$.\\
By \tref{thm-van-ordertwo-atzero-implies-litotwo}, it suffices to show: $\rho_0=\rho'_0=\rho''_0=0$.\\
By \tref{thm-deriv-of-dbltransl}, we have: \quad $\forall h\in\R$, $\,$ $g'_h=f'_{h+a}$.\\
Then $g'_0=f'_{0+a}=f'_a=m$.\\
By \tref{thm-secondderiv-dbltranslate}, we have: \quad $\forall h\in\R$, $\,$ $g''_h=f''_{h+a}$.\\
Since $c=f''_a/2$, we get: $2c=f''_a$.\\
Then $g''_0=f''_{0+a}=f''_a=2c$.\\
By \tref{thm-diff-fn-minus-mx-minus-cxsquared} (with $x$ replaced by $h$), we conclude:\\
\phx\qquad $\forall h\in\R$, $[\,\,(\,\,\rho'_h=g'_h-m-2ch\,\,)\,\,\&\,\,(\,\,\rho''_h=g''_h-2c\,\,)\,\,]$.\\
We have: \quad $g_0=0$ \quad and \quad $g'_0=m$ \quad and \quad $g''_0=2c$.\\
Then \qquad\qquad $\rho_0=g_0-m\cdot0-c\cdot0^2=0-0-0=0$\\
\phx\qquad and \qquad $\,\rho'_0=g'_0-m-2c\cdot 0=m-m-0=0$\\
\phx\qquad and \qquad $\,\rho''_0=g''_0-2c=2c-2c=0$.\\
We therefore have: \qquad $g_0=g'_0=g''_0=0$, \qquad as desired.
\end{proof}

The next result is called the \bfindex[Second Derivative Test for Maxima]:

\begin{thm}
Let $f:\R\drta\R$ and let $a\in\R$.\\
\phx\qquad\qquad\qquad Assume: \quad $(\,\,f'_a=0\,\,)\,\,\&\,\,(\,\,f''_a<0\,\,)$.\\
\phx\qquad\qquad\qquad Then: \quad $f$ has a local strict-maximum at $a$ in $\R$.
\end{thm}

\begin{proof}
Want: $f_a^\T$ has a local strict-maximum at $0$ in $\R$.\\
We have $f'_a=0\ne\frownie$, so $a\in\D_{f'}$. \quad Then $a\in\D_{f'}\subseteq\D_f$, so $(f_a^\T)_0=0$.\\
Let $g:=f_a^\T$. \qquad Want: $g$ has a local strict-maximum at $0$ in $\R$.\\
\phx\qquad\qquad\qquad\qquad\qquad\quad $\,$ Want: $\exists B\in\scrb_\R(0)$ s.t.~$g<g_0$ on $B_0^\times$.\\
We have $g_0=(f_a^\T)_0=0$. \quad
Want: $\exists B\in\scrb_\R(0)$ s.t.~$g<0$ on $B_0^\times$.\\
Let $m:=f'_a$ and $c:=f''_a/2$ and $L:=m\cdot(\bullet)$ and $Q:=c\cdot(\bullet)^2$.\\
By \tref{thm-second-order-taylor}, we have: $f_a^\T\,-\,L\,-\,Q\,\,\,\in\,\,\,\scrlito_2$.\\
Since $m=f'_a=0$, we get $m\cdot(\bullet)=\bfzero$. \quad
Then: $L=m\cdot(\bullet)=\bfzero$.\\
Then: \quad $g\,-\,Q\,\,=\,\,g\,-\,\bfzero\,-\,Q\,\,=\,\,f_a^\T\,-\,L\,-\,Q\,\,\in\,\,\scrlito_2$.\\
So, by \tref{thm-approx-mimics-fn-near-pt}, choose $B\in\scrb_\R(0)$ s.t., $\,$ $\forall h\in B$, $\sgn_{g_h}=\sgn_{Q_h}$.\\
Then $B\in\scrb_\R(0)$. \qquad Want: $g<0$ on $B_0^\times$.\\
Want: $\forall h\in B_0^\times$, $g_h<0$. \qquad Given $h\in B_0^\times$. \quad Want: $g_h<0$.\\
Since $f''_a<0$ and $c=f''_a/2$, we get: \quad $c<0$.\\
Since $h\in\B_0^\times\subseteq\R_0^\times$, we get: \quad $h^2>0$.\\
So, since $c<0$, we get $ch^2<0$. \quad Then $Q_h=(c\cdot(\bullet)^2)_h=ch^2<0$.\\
Since $Q_h<0$, we get: $\sgn_{Q_h}=-1$. \qquad Then $\sgn_{g_h}=\sgn_{Q_h}=-1$.\\
Since $\sgn_{g_h}=-1$, we conclude that $g_h<0$, as desired.
\end{proof}

\subsection{Some basic limit theorems}

\begin{thm}\wrlab{thm-powers-to-infty}
Let $u>1$, $a\in\R$. \quad Then $\exists j\in\N$ s.t.~$u^j>a$.
\end{thm}

\begin{proof}
Assume: $\forall j\in\N$, $u^j\le a$. \qquad\qquad Want: Contradiction.\\
Define $s\in\R^\N$ by: \quad $\forall j\in\N$, $s_j=u^j$.\\
Then: $(\,\,\forall j\in\N,\,s_{j+1}=u\cdot s_j\,\,)\,\,\&\,\,(\,\,s_j>1\,\,)$.\\
Since $u>1$, we know: \qquad\qquad $\forall j\in\N$, \quad $s_j\cdot u>s_j\cdot 1$.\\
Then: \quad $\forall j\in\N$, \quad $s_{j+1}=u\cdot s_j=s_j\cdot u>s_j\cdot 1=s_j$.\\
Then: \quad $\forall j\in\N$, \quad $s_{j+1}>s_j$, \qquad\qquad so $s$ is strictly-increasing.\\
We have: $\forall j\in\N$,\qquad $0<1<s_j$ and $s_j=u^j\le a$, \quad so \quad $0<s_j\le a$.\\
Then $\I_s\subseteq(0;a]$, \qquad so $\I_s$ is bounded in $\R$.\\
Since \quad $s$ is strictly-increasing \quad and since \quad $\I_s$ is bounded in $\R$,\\
\phx\qquad\qquad it follows (from \tref{thm-conv-of-bdd-semimonotone}) that $s$ is convergent in $\R$.\\
Choose $q\in\R$ s.t.~$s\to q$ in $\R$. \qquad\qquad
Define $\ell\in\N^\N$ by $\ell_j=j+1$.\\
We have: \qquad $\forall j\in\N$, \quad $\ell_{j+1}=j+2>j+1=\ell_j$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad so $\ell$ is strictly-increasing.\\
Then $s\circ\ell$ is a subsequence of $s$.\\
So, since $s\to q$ in $\R$, we get: \quad $s\circ\ell\to q$ in $\R$.\\
We have: \qquad $\forall j\in\N$, \quad $(u\cdot s)_j=u\cdot s_j=s_{j+1}=s_{\ell_j}=(s\circ\ell)_j$,\\
\phx\qquad\qquad\qquad\qquad so \quad $\,\,\,u\cdot s\,\,=\,\,u\circ\ell$.\\
Since \quad $u\cdot s=u\circ\ell$ \quad and since \quad $u\circ\ell\to q$ in $\R$\\
\phx\qquad\qquad\qquad\qquad\qquad we get: \quad $\,\,\,u\cdot s\to q$ in $\R$.\\
On the other hand, since $s\to q$ in $\R$, we conclude $u\cdot s\to u\cdot q$ in $\R$.\\
Since \quad both $u\cdot s\to q$ in $\R$ \quad and \quad $u\cdot s\to u\cdot q$ in $\R$,\\
\phx\qquad\qquad we get: $q=u\cdot q$. \qquad Then $(u-1)\cdot q=0$.\\
Since $u>1$, we get $u-1>0$, so $u-1\ne0$.\\
So, since $(u-1)\cdot q=0$, it follows that $q=0$.\\
Since \quad $\forall j\in\N$, $s_j>1$ \quad and since \quad $s\to q$ in $\R$,\\
\phx\qquad\qquad we conclude: \quad $q\ge1$.\\
Then $0=q\ge1>0$, so $0>0$. \quad Contradiction.
\end{proof}

\begin{thm}\wrlab{thm-powers-to-zero}
Let $c\in(0;1)$.\\
\phx\qquad\qquad Define $z\in\R^{\N_0}$ by: \qquad $\forall j\in\N_0$, \quad $z_j=c^j$.\\
Then $z\to0$ in $\R$.
\end{thm}

\begin{proof}
Want: $\forall\vep>0$, $\exists K\in\N_0$ s.t., $\forall j\in[K..\infty)$, \quad $|z_j|<\vep$.\\
Given $\vep>0$. \qquad Want: $\exists K\in\N_0$ s.t., $\forall j\in[K..\infty)$, \quad $|z_j|<\vep$.\\
Let $u:=1/c$. \qquad By \tref{thm-powers-to-infty}, choose $K\in\N$ s.t.~$u^K>1/\vep$.\\
Then $K\in\N\subseteq\N_0$. \qquad Want: $\forall j\in[K..\infty)$, \quad $|z_j|<\vep$.\\
Given $j\in[K..\infty)$. \qquad Want: $|z_j|<\vep$.\\
Since \quad $u^K>1/\vep>0$, \quad we conclude: \quad $1/(u^K)<1/(1/\vep)$.\\
Then \quad $c^K\,\,=\,\,(1/u)^K\,\,=\,\,1/(u^K)\,\,<\,\,1/(1/\vep)\,\,=\,\,\vep$, \quad so $c^K<\vep$.\\
Given $j\in[K..\infty)$. \qquad Want: $|z_j|<\vep$.\\
Since $c>0$, we get $c^K>0$. \qquad Since $0<c<1$, we get $c^{j-K}<1$.\\
Since \quad $c^K>0$ \quad and since \quad $c^{j-K}<1$,\\
\phx\qquad\qquad we conclude: \quad $c^K\cdot c^{j-K}<c^K\cdot 1$.\\
Since $c>0$, it follows that $c^j>0$. \quad Then $z_j=c^j>0$, so $|z_j|=z_j$.\\
Then: \quad $|z_j|\,\,=\,\,z_j\,\,=\,\,c^j\,\,=\,\,c^K\cdot c^{j-K}\,\,<\,\,c^K\cdot 1\,\,=\,\,c^K\,\,<\,\,\vep$.
\end{proof}

\begin{thm}\wrlab{thm-ratios-below-const-below-one}
Let $b\in(0;\infty)^{\N_0}$ and $c\in(0;1)$ and $k\in\N_0$.\\
Assume: \quad $\forall j\in[k..\infty)$, $\,$ $b_{j+1}/b_j\le c$. \qquad\qquad Then: \quad $b\to0$ in $\R$.
\end{thm}

\begin{proof}
Want: $\forall\vep>0$, $\exists K\in\N_0$ s.t., $\forall j\in[K..\infty)$, $|b_j|<\vep$.\\
Given $\vep>0$. \quad Want: $\exists L\in\N_0$ s.t., $\forall j\in[L..\infty)$, $|b_j|<\vep$.\\
Define $r\in(0;\infty)^\N$ by: \qquad $\forall j\in\N$, \quad $r_j=b_{k+j}/b_{k+j-1}$.\\
Then, $\forall j\in\N$, we have: \quad $r_j\le c$.\\
Then, $\forall i\in\N$, we have: \quad $r_i\,\,\,\,\,\cdot\,\,\,\,r_{i-1}\,\,\,\cdot\,\,\,r_{i-2}\,\,\,\,\cdots\,\,\,\,r_1\quad\le\quad c^i$.\\
Also, $\,\,\forall i\in\N$, we have: \quad $r_i\,\,\,\,\,\cdot\,\,\,\,r_{i-1}\,\,\,\cdot\,\,\,r_{i-2}\,\,\,\,\cdots\,\,\,\,r_1$\\
\phx\qquad\qquad\qquad\qquad $\ds{=\,\,\,\,\,\frac{b_{k+i}}{b_{k+i-1}}\cdot\frac{b_{k+i-1}}{b_{k+i-2}}\cdot\frac{b_{k+i-2}}{b_{k+i-3}}\cdots\frac{b_{k+1}}{b_k}\,\,\,=\,\,\,\frac{b_{k+i}}{b_k}}$.\\
\phx\qquad We conclude: \qquad\quad $\forall i\in\N$, \quad $b_{k+i}/b_{k}\le c^i$.\\
Let $a:=b_k$. \qquad Then: \quad $\,\forall i\in\N$, \quad $b_{k+i}/a\,\,\le\,c^i$.\\
Also, since $b\in(0;\infty)^{\N_0}$, we get $a\in(0;\infty)$. \quad Then $a>0$.\\
Define $z\in\R^{\N_0}$ by: \qquad $\forall j\in\N_0$, \quad $z_j=c^j$.\\
By \tref{thm-powers-to-zero}, we have: \quad $z\to0$ in $\R$.\\
Choose $R\in\N_0$ s.t., $\forall i\in[R..\infty)$, \quad $|z_i|<\vep/a$.\\
Let $M:=R+1$. \quad Then $M\in\N$. \quad Also, $[M..\infty)\,\,\subseteq\,\,[R..\infty)$.\\
Let $L:=M+k$. \quad Then $L\in\N_0$. \quad Want: $\forall j\in[L..\infty)$, $|b_j|<\vep$.\\
Given $j\in[L..\infty)$. \qquad Want: $|b_j|<\vep$.\\
Since \quad $j\in[L..\infty)$ and \quad since $L-k=M$, \quad we get $j-k\in[M..\infty)$.\\
Recall: $\forall i\in\N$, $b_{k+i}/a\,\le\,c^i$. \qquad Let $i:=j-k$. \quad Then: $i\in[M..\infty)$.\\
Also, $k+i=j$ \qquad Since $i\in[M..\infty)$ and since $M\in\N$, we get $i\in\N$.\\
\phx\quad Then: $b_j=b_{k+i}/a\,\le\,c^i$. \qquad So, since $a>0$, we get: $b_j\,\le\,a\cdot c^i$.\\
Since $i\in[M..\infty)\subseteq[R..\infty)$, by choice of $R$, we have $|z_i|<\vep/a$.\\
Since $c\in(0;1)>0$, we get $c^i>0$. \quad Then $|c^i|=c^i$.\\
By definition of $z$, we have $z_i=c^i$. \quad Then $c^i=|c^i|=|z_i|<\vep/a$.\\
Since \quad $c^i<\vep/a$ \quad and since \quad $a>0$, \quad we get: \quad $a\cdot c^i<\vep$.\\
Then $b_j\le a\cdot c^i<\vep$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-make-some-room}
Let $\alpha,\beta\in\R$. \quad Then:\\
\phx\qquad\qquad $(\,\,\alpha\le\beta\,\,)\,\,\Leftrightarrow\,\,(\,\,\forall\eta>0,\,\alpha\le\beta+\eta\,\,)$.
\end{thm}

\begin{proof}
{\it Proof of $\Rightarrow$:}\\
Assume: $\alpha\le\beta$. \qquad $\,$ Want: $\forall\eta>0,\,\alpha\le\beta+\eta$.\\
\phx\qquad\qquad\qquad Given $\eta>0$. \quad Want: $\alpha\le\beta+\eta$.\\
Since $\eta>0$, it follows that $\eta\ge0$. \quad Then $\beta+\eta\ge\beta+0$.\\
Then $\alpha\le\beta=\beta+0\le\beta+\eta$, as desired.\\
{\it End of proof of $\Rightarrow$.}\\
\\
{\it Proof of $\Leftarrow$:}\\
Assume: $\forall\eta>0,\,\alpha\le\beta+\eta$. \qquad $\,$ Want: $\alpha\le\beta$.\\
Assume $\alpha>\beta$. \qquad Want: Contradiction.\\
We have $\alpha-\beta>0$. \quad Let $\eta:=(\alpha-\beta)/2$. \quad Then $\eta>0$.\\
Then, by our assumption, $\alpha\le\beta+\eta$. \qquad Then $\alpha-\beta\le\eta$.\\
Since $\eta=(\alpha-\beta)/2$, we get $2\eta=\alpha-\beta$.\\
Then $2\eta=\alpha-\beta\le\eta$, so $2\eta\le\eta$, so $2\eta-\eta\le\eta-\eta$.\\
Then $\eta=2\eta-\eta\le\eta-\eta=0$, so $\eta\le0$.\\
Then $0\ge\eta>0$, so $0>0$. \quad Contradiction.\\
{\it End of proof of $\Leftarrow$.}
\end{proof}

\begin{thm}\wrlab{thm-quantequiv-for-derivcomputation}
Let $f:\R\drta\R$ and let $x,m\in\R$.\\
Then: \qquad $(\,\,f'_x=m\,\,)\,\,\Leftrightarrow\,\,(\,\,\forall\vep>0,\,\exists\delta>0~\hbox{s.t.},\forall h\in(-\delta;\delta),$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $|f_{x+h}-f_x-m\cdot h|\le\vep\cdot|h|\,\,)$.
\end{thm}

\begin{proof}
Define $L\in\scrl$ by: \quad $\forall h\in\R$, $L_h=m\cdot h$.\\
We have: $\forall h\in\R$, $\,$ $(f_x^\T)_h=f_{x+h}-f_x$. \qquad\quad Let $\rho:=f_x^\T-L$.\\
Then: $\forall h\in\R$, $\,$ $\rho_h\,=\,(f_x^T-L)_h\,=\,(f_x^\T)_h-L_h\,=\,f_{x+h}-f_x-m\cdot h$.\\
Then: $\forall h\in\R$, $\,$ $\rho_h\,=\,f_{x+h}-f_x-m\cdot h$.\\
Want: $(\,\,f'_x=m\,\,)\,\,\Leftrightarrow\,\,(\,\,\forall\vep>0,\,\exists\delta>0~\hbox{s.t.},\forall h\in(-\delta;\delta),\,|\rho_h|\le\vep\cdot|h|\,\,)$.\\
Know: $(\,\,\rho\in\scrlito_1\,\,\,)\,\,\,\Leftrightarrow\,\,(\,\,\forall\vep>0,\,\exists\delta>0~\hbox{s.t.},\forall h\in(-\delta;\delta),\,|\rho_h|\le\vep\cdot|h|\,\,)$.\\
Want: $(\,\,f'_x=m\,\,)\,\,\Leftrightarrow\,\,(\,\,\rho\in\scrlito_1\,\,\,)$.\\
By definition of $L$, we conclude: $(\,\,f'_x=m\,\,)\,\,\Leftrightarrow\,\,(\,\,D_xf=L\,\,)$.\\
By \tref{thm-L-linearization-imlies-L-D-deriv}, we conclude: $(\,\,D_xf=L\,\,)\,\,\,\,\Leftrightarrow\,\,(\,\,f_x^\T-L\in\scrlito_1\,\,)$.\\
Then: $\,\,(\,\,f'_x=m\,\,)\,\,\Leftrightarrow\,\,(\,\,D_xf=L\,\,)$\\
\phx\qquad\qquad\qquad\qquad $\Leftrightarrow\,\,(\,\,f_x^\T-L\in\scrlito_1\,\,)\,\,\Leftrightarrow\,\,(\,\,\rho\in\scrlito_1\,\,\,)$.
\end{proof}

\begin{thm}\wrlab{thm-consecdiffs-of-conv-series-bdd}
Let $s\in\R^{\N_0}$.
Assume $s$ is convergent in $\R$.\\
\phx\qquad\qquad\qquad\qquad\quad$\,$ Then $\{s_{j+1}-s_j\,|\,j\in\N_0\}$ is bounded in $\R$.
\end{thm}

\begin{proof}
Define $t\in\R^{\N_0}$ by: $\forall j\in\N_0$, $t_j=s_{j+1}$.\\
Let $u:=t-s$. \quad
Then: $\forall j\in\N_0$, $u_j=(t-s)_j=t_j-s_j=s_{j+1}-s_j$,\\
\phx\qquad so $\I_u=\{s_{j+1}-s_j\,|\,j\in\N_0\}$. \qquad Want: $\I_u$ is bounded in $\R$.\\
Since $s$ is convergent in $\R$, choose $q\in\R$ s.t.~$s\to q$ in $\R$.\\
By HW\#3-1, $t\to q$ in $\R$. \qquad
Then $t-s\to q-q$ in $\R$.\\
So, since $t-s=u$ and $q-q=0$, we get: $u\to0$ in $\R$.\\
Then $u$ is convergent in $\R$, so $\I_u$ is bounded in $\R$, as desired.
\end{proof}

\subsection{Differentiation commutes with uniform limit}
\quad
\vskip.1in\noindent
We illustrated in class that: \quad $\exists\phi,\psi\in(\R^\R)^\N$, \quad $\exists f,g\in\R^\R$\\
\phx\qquad such that \qquad\quad $\,\,\,\phi\to f$ pointwise from $\R$ to $\R$\\
\phx\qquad\qquad\qquad\quad and \quad $\psi\to g$ pointwise from $\R$ to $\R$\\
\phx\qquad\qquad\qquad\quad and \quad $\forall j\in\N$, \quad $(\phi_j)'\,\,=\,\,\,\psi_j$\\
\phx\qquad\qquad\qquad\quad and \qquad\qquad\qquad $f'\,\,\,\,\,\,\ne\,\,\,g$.\\
That is, differentiation does not commute with pointwise limit.\\
In this section, we will show:\\
\phx\qquad\quad differentiation {\it DOES} commute with {\it UNIFORM} limit.

\begin{defn}
Let $X$ be a set and let $S\subseteq X$.\\
Let $Y$ be a metric space. Let $f\in(\PF^X_Y)^\N$ and let $g\in\PF^X_Y$.\\
By $f\to g$ pointwise\index[terms]{pointwise limit} on $S$ from $X$ to $Y$, we mean:\\
\phx\qquad\qquad $\forall p\in S$, \quad $f_\bullet(p)\to g(p)$ in $Y$.\\
By $f\to g$ uniformly\index[terms]{uniform limit} on $S$ from $X$ to $Y$, we mean:\\
\phx\qquad\qquad $\forall\vep>0$, $\exists K\in\N$ s.t., $\forall j\in\N$, $\forall p\in S$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(\,f_j(p)\,,\,g(p)\,)\,<\,\vep\,\,)$.
\end{defn}

\noindent
Let $X$ be a set and let $S\subseteq X$. Let $Y$ be a metric space.\\
\phx\qquad\qquad Let \quad $f\in(\PF^X_Y)^\N$ \quad and let \quad $g\in\PF^X_Y$.\\
Note that, \qquad $\,\,$if \quad $f\to g$ $\,$ uniformly $\,$ on $S$ from $X$ to $Y$,\\
\phx\qquad\qquad\quad $\,\,$then \quad $f\to g$ $\,$ pointwise $\,$ on $S$ from $X$ to $Y$.\\
Also, note that, if $\,\,\,\,\,\,f\to g$ $\,$ pointwise $\,$ on $S$ from $X$ to $Y$,\\
\phx\qquad\qquad\quad $\,\,$then \quad\quad $S\,\,\subseteq\,\,\D_g\,\,\bigcap\,\,\D_{f_1}\,\,\bigcap\,\,\D_{f_2}\,\,\bigcap\,\,\D_{f_3}\,\,\bigcap\,\,\cdots$.

\begin{thm}\wrlab{thm-diff-commutes-with-uniflimit}
Let $\phi,\psi\in(\PF^\R_\R)^\N$ and let $f,g\in\PF^\R_\R$.\\
Let $U$ be an open subset of $\R$.\\
Assume: \qquad\qquad $\,\,\phi\to f$ pointwise on $U$ from $\R$ to $\R$\\
\phx\qquad\qquad and \qquad $\psi\to g$ uniformly on $U$ from $\R$ to $\R$\\
\phx\qquad\qquad and \qquad $\forall j\in\N$, \quad $(\phi_j)'\,\,=\,\,\,\psi_j$.\\
\phx Then: \qquad\qquad\qquad\qquad\quad $\,\,\,\,\,f'\,\,\,\,\,\,=\,\,\,g$.\\
\end{thm}

\begin{proof}
Want: $\forall x\in U$, $f'_x=g_x$. \qquad
Given $x\in U$. \quad Want: $f'_x=g_x$.\\
By \tref{thm-quantequiv-for-derivcomputation}, we wish to show:\\
\phx\qquad\quad $\,$ $\forall\vep>0$, \quad $\,\,\exists\delta>0$ s.t., $\forall h\in(-\delta;\delta)$, $|f_{x+h}-f_x-g_x\cdot h|\le\vep\cdot|h|$.\\
Given $\vep>0$. Want: $\exists\delta>0$ s.t., $\forall h\in(-\delta;\delta)$, $|f_{x+h}-f_x-m\cdot h|\le\vep\cdot|h|$.\\
Since $U$ is an open subset of $\R$, we know that $U=\Int_\R U$.\\
Since $x\in U=\Int_\R U$, choose $S\in\scrb_\R(U)$ s.t.~$S\subseteq U$.\\
Since $S\in\scrb_\R(U)$, choose $\alpha>0$ s.t.~$S=B_\R(x,\alpha)$.\\
Then $(x-\alpha;x+\alpha)=B_\R(x,\alpha)=S\subseteq U$.\\
Since $\psi\to g$ uniformly on $U$ from $\R$ to $\R$,\\
\phx\qquad choose $k\in\N$ s.t., \quad $\forall j\in[k..\infty)$, \quad $|g-\psi_j|<\vep/8$ on $U$.\\
Since \quad $(\phi_k)'=\psi_k$ on $U$ \quad and since $x\in U$,\\
\phx\qquad\qquad\qquad we conclude: \qquad $\,\,\,(\phi_k)'_x=(\psi_k)_x$.\\
Let $\Phi:=\phi_k$ and $\Psi:=\psi_k$. \quad $\,$ Then: $\Phi'_x=\Psi$.\\
Since $\Phi'_x=\Psi$, by \tref{thm-quantequiv-for-derivcomputation}, choose $\beta>0$ s.t.,\\
\phx\qquad\quad $\forall h\in(-\beta;\beta)$, \qquad $|\Phi_{x+h}-\Phi_x-\Psi_x\cdot h|\le(\vep/2)\cdot|h|$.\\
Let $\delta:=\min\{\alpha,\beta\}$. \qquad Then $\delta>0$.\\
\phx\quad Want: \quad $\forall h\in(-\delta;\delta)$, $\,\,\,$ $|f_{x+h}-f_x-m\cdot h|\le\vep\cdot|h|$.\\
Given $h\in(-\delta;\delta)$. \quad Want: $|f_{x+h}-f_x-m\cdot h|\le\vep\cdot|h|$.\\
Since $0<\delta\le\alpha$ and $0<\delta\le\beta$, we get:\\
\phx\qquad both \quad $(-\delta;\delta)\subseteq(-\alpha;\alpha)$ \quad and \quad $(-\delta;\delta)\subseteq(-\beta;\beta)$.\\
Then \quad $h\in(-\delta;\delta)\subseteq(-\alpha;\alpha)$ \quad and \quad $h\in(-\delta;\delta)\subseteq(-\beta;\beta)$.\\
Since $h\in(-\beta;\beta)$, by choice of $\beta$, we get: $|\Phi_{x+h}-\Phi_x-\Psi_x\cdot h|\le(\vep/2)\cdot|h|$.\\
Let \quad $A:=\Phi_{x+h}-\Phi_x-\Psi_x\cdot h$ \quad and let \quad $B:=f_{x+h}-f_x-g_x\cdot h$.\\
Then $|A|\le(\vep/2)\cdot|h|$. \qquad Want: $|B|\le\vep\cdot|h|$.\\
Since $|B|\le|B-A|+|A|\le|B-A|+(\vep/2)\cdot|h|$,\\
\phx\quad it suffices to show: \quad $|B-A|\le(\vep/2)\cdot|h|$.\\
By \tref{thm-make-some-room}, it suffices to prove: $\forall\eta>0$,\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,\,|B-A|\le(\vep/2)\cdot|h|+\eta$.\\
\phx$\,\,\,$ Given $\eta>0$. $\,$ Want: $|B-A|\le(\vep/2)\cdot|h|+\eta$.\\
Since $h\in(-\alpha;\alpha)$, we get: $x+h\in(x-\alpha;x+\alpha)$.\\
Then $x+h\in(x-\alpha;x+\alpha)\subseteq U$, so $x+h\in U$.\\
Since $\,$ $x,x+h\in U$ $\,$ and $\,$ $\phi\to f$ pointwise on $U$ from $\R$ to $\R$, $\,$ we get:\\
\phx\quad both \quad $\phi_\bullet(x)\to f(x)$ in $\R$ \quad and \quad $\phi_\bullet(x+h)\to f(x+h)$ in $\R$.\\
Since \qquad\qquad\qquad\qquad\qquad $\psi\to g$ uniformly on $U$ from $\R$ to $\R$,\\
\phx\quad and since uniform convergence implies pointwise convergence,\\
\phx\qquad\qquad we conclude: \qquad $\,\psi\to g$ pointwise on $U$ from $\R$ to $\R$.\\
So, since $x\in U$, we get: $\,\,\psi_\bullet(x)\to g(x)$ in $\R$.\\
We have proved: \qquad\qquad\quad $\phi_\bullet(x+h)\to f(x+h)\,$ \quad in $\R$\\
\phx\qquad\qquad\qquad\qquad and \qquad\qquad $\phi_\bullet(x)\to f(x)$ \qquad\quad in $\R$\\
\phx\qquad\qquad\qquad\qquad and \qquad\qquad $\psi_\bullet(x)\to g(x)$ \qquad\quad in $\R$.\\
So, since $B=f_{x+h}-f_x-g_x\cdot h=[f(x+h)]-[f(x)]-[g(x)]\cdot h$,
\phx\qquad we get: \quad $\,\,\,\,[\phi_\bullet(x+h)]-[\phi_\bullet(x)]-[\psi_\bullet(x)]\cdot h\to B$ \quad in $\R$.\\
Choose $\ell\in\N$ such that, $\forall j\in[\ell..\infty)$,\\
\phx\qquad $\,\,\,\,|\,\,\,\,([\phi_\bullet(x+h)]-[\phi_\bullet(x)]-[\psi_\bullet(x)]\cdot h)_j\,\,\,\,-\,\,\,\,B\,\,\,\,|\quad<\quad\eta$.\\
Let $m:=\max\{k,\ell\}$. \qquad Then \quad $m\in[k..\infty)$ \quad and \quad $m\in[\ell..\infty)$.\\
Since $m\in[\ell..\infty)$, by choice of $\ell$, we conclude:\\
\phx\qquad $\,\,\,\,|\,\,\,\,([\phi_\bullet(x+h)]-[\phi_\bullet(x)]-[\psi_\bullet(x)]\cdot h)_m\,\,\,\,-\,\,\,\,B\,\,\,\,|\quad<\quad\eta$.\\
Let $C:=([\phi_\bullet(x+h)]-[\phi_\bullet(x)]-[\psi_\bullet(x)]\cdot h)_m$. \quad Then $|C-B|<\eta$.\\
Then $|B-A|\le|B-C|+|C-A|=|C-A|+|C-B|\le|C-A|+\eta$.\\
\phx\qquad\quad $\,$ Then: \quad $\,\,|B-A|\le|C-A|+\eta$.\\
It therefore suffices to prove: $|C-A|\le(\vep/2)\cdot|h|$.\\
Let \quad $\sigma:=\phi_m-\phi_k$ \quad and let \quad $\tau:=\psi_m-\psi_k$.\\
We know: \quad $\forall j\in\N$, $(\phi_j)'=\psi_j$ on $U$.\\
We conclude: \qquad\qquad $\,\sigma'=\tau$ on $U$. \qquad
Then: \quad $U\subseteq\D_{\sigma'}$.\\
Recall both that \quad $\Phi=\phi_k$ \quad and that \quad $\Psi=\psi_k$.\\
We have \quad $\,\,A\,\,=\,\,\Phi_{x+h}-\Phi_x-\Psi_x\cdot h$\\
\phx\qquad\qquad\qquad $\,=\,\,(\phi_k)_{x+h}\,-\,(\phi_k)_x\,-\,(\psi_k)_x\cdot h$.\\
We have \quad $\,\,C\,\,=\,\,([\phi_\bullet(x+h)]-[\phi_\bullet(x)]-[\psi_\bullet(x)]\cdot h)_m$\\
\phx\qquad\qquad\qquad $\,=\,\,[\phi_m(x+h)]-[\phi_m(x)]-[\psi_m(x)]\cdot h$\\
\phx\qquad\qquad\qquad $\,=\,\,(\phi_m)_{x+h}-(\phi_m)_x-(\psi_m)_x\cdot h$.\\
So, since $\,\,\,\,\,\,\,A\,\,=\,\,(\phi_k)_{x+h}\,-\,(\phi_k)_x\,-\,(\psi_k)_x\cdot h$.\\
we get $\,C-A\,\,=\,\,\,\,\,\sigma_{x+h}\,\,\,\,\,-\,\,\,\,\sigma_x\,\,\,\,\,-\,\,\,\,\tau_x\,\,\,\,\,\,\cdot h$.\\
Since $(x-\alpha;x+\alpha)$ is an interval, and since $x,x+h\in(x-\alpha;x+\alpha)$,\\
\phx$\,$ we get $[x|x+h]\subseteq(x-\alpha;x+\alpha)$.\\
So, \qquad\qquad $\,\,\,\,\,$ since $(x-\alpha;x+\alpha)\subseteq U$, \quad we get: \quad $[x|x+h]\subseteq U$.\\
So, \qquad\qquad \qquad\qquad\qquad since $U\subseteq\D_{\sigma'}$, \quad we get: \quad $[x|x+h]\subseteq\D_{\sigma'}$.\\
Then, by \tref{thm-mvt-corollary}, choose $q\in[x|x+h]$ s.t.~$\sigma_{x+h}-\sigma_x=\sigma'_q\cdot h$.\\
Since \quad $q\in[x|x+h]\subseteq U$, \quad we conclude: \quad $q\in U$.\\
So, \quad since $\sigma'=\tau$ on $U$, \qquad we conclude: \quad $\sigma'_q=\tau'_q$.\\
We have: \qquad\quad $\,\sigma_{x+h}-\sigma_x\,\,=\,\,\sigma'_q\cdot h\,\,=\,\,\tau_q\cdot h$.\\
Then $\,\,\,\,\,C-A=\sigma_{x+h}-\sigma_x-\tau_x\cdot h\,=\tau_q\cdot h-\tau_x\cdot h$.\\
Then $\,\,\,\,\,C-A\,=\,\tau_q\cdot h\,\,+(\,-\,\tau_x\cdot h\,)$.\\
Then $\,\,\,|C-A|\le|\tau_q\cdot h|+|-\tau_x\cdot h|=|\tau_q|\cdot|h|+|-\tau_x|\cdot|h|$.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $\,\,\,=|\tau_q|\cdot|h|+\,\,\,\,|\tau_x|\,\,\,\cdot|h|$.\\
Then $\,\,\,|C-A|\le\,\,|\tau_q|\,\,\cdot|h|+\,\,\,|\tau_x|\,\,\cdot|h|$.\\
Want: $|C-A|\le(\vep/4)\cdot|h|+(\vep/4)\cdot|h|$.\\
Want: \quad $|\tau_q|\le\vep/4$ \quad and \quad $|\tau_x|\le\vep/4$.\\
Since \qquad\qquad $\,\,\,\,\,\,\forall j\in[k..\infty)$, \quad $|g-\psi_j|<\vep/8$ on $U$,\\
\phx\qquad snd since $k,m\in[k..\infty)$,\\
we conclude: \quad $|g-\psi_k|<\vep/8$ on $U$ \quad and \quad $|g-\psi_m|<\vep/8$ on $U$.\\
By hypothesis, $\psi\to g$ uniformly on $U$ from $\R$ to $\R$.\\
Then $U\subseteq\D_g$. \quad
Then $|\psi_m-\psi_k|\le|\psi_m-g|+|g-\psi_k|$ on $U$,\\
Since \quad $\,\,\,\,\,\,\,\tau=\psi_m-\psi_k$,\\
and since \quad $\,\,\,\,\,\,|\psi_m-\psi_k|\le|\psi_m-g|+|g-\psi_k|$ \qquad\qquad on $U$,\\
and since \qquad\qquad\qquad\quad $\,\,\,|\psi_m-g|=|g-\psi_m|$,\\
we get \quad $|\tau|\le|g-\psi_m|+|g-\psi_k|$ \qquad\quad $\,\,\,$ on $U$.\\
Then \quad $\,\,\,|\tau|\le\,\,\,\,(\vep/8)\,\,\,\,\,+\,\,\,(\vep/8)$ \qquad\qquad $\,$ on $U$,\\
and so \quad $|\tau|\le\qquad\quad\,\,\,\,\vep/4$ \qquad\qquad\qquad\quad on $U$.\\
So, since $q,x\in U$, we conclude: \qquad $|\tau_q|\le\vep/4$ and $|\tau_x|\le\vep/4$.
\end{proof}

\subsection{Power series}
\quad
\vskip.1in\noindent
A function with domain $\N_0$ will be called a \bfindex[zero-sequence].\\
The theory of $0$-sequences is completely parallel to that of sequences.\\
For example:

\begin{defn}
Let $X$ be a metric space and let $s\in X^{\N_0}$.\\
Then \quad $s$ is \bfindex[Cauchy] in $X$ \quad means:\\
\phx\qquad $\forall\vep>0$, $\exists K\in\N_0$ s.t., $\forall i,j\in[K..\infty)$, $d(s_i,s_j)<\vep$.\\
Also, $\forall q\in X$, \quad $s\to q$ in $X$ \quad means:\\
\phx\qquad $\forall\vep>0$, $\exists K\in\N_0$ s.t., $\forall j\in[K..\infty)$, $d(s_j,q)<\vep$.\\
Also, $s$ is \bfindex[convergent] in $X$ means: \quad $\exists q\in X$ s.t.~$s\to q$ in $X$.\\
Also, \quad $X\hbox{-}\lim s\,\,:=\,\,\UE\{q\in X\,|\,s\to q\hbox{ in }X\}$.
\end{defn}

\begin{defn}
Let $s$ be a zero-sequence.\\
Then $s_{\bullet-1}$ is the sequence defined by: $\forall j\in\N$, $(s_{\bullet-1})_j=s_{j-1}$.
\end{defn}

\begin{thm}
Let $X$ be a complete metric space, $s\in X^{\N_0}$.\\
Assume: $s$ is Cauchy in $X$. \qquad Then: $s$ is convergent in $X$.
\end{thm}

\noindent
Idea of proof:\\
Since $s$ is Cauchy in $X$, $s_{\bullet-1}$ is Cauchy in $X$.\\
Then, as $X$ is complete, $s_{\bullet-1}$ is convergent in $X$.\\
Then $s$ is convergent in $X$. QED\\

\vskip.in\noindent
We have the \bfindex[Principle of Zero-Induction]:

\begin{thm}
Let $S\subseteq\N_0$.\\
\phx\qquad Assume: \quad $(\,0\in S\,)\quad\&\quad(\,\forall j\in S,\,j+1\in S\,)$.\\
\phx\qquad $\,\,\,\,\,$ Then: \qquad\qquad $\,\,S=\N_0$.
\end{thm}

\noindent
Idea of proof:\\
Since $S\subseteq\N_0$, $S+1\subseteq\N$.\\
Since $0\in S$, $1\in S+1$.\\
Since \quad $\forall j\in S$, $j+1\in S$\\
\phx\qquad we conclude: \qquad $\forall j\in S+1$, \quad $j+1\in S+1$.\\
Then, by the Principle of Mathematical Induction, $S+1=\N$.\\
Then $S=\N-1=\N_0$. QED\\

\noindent
The following can be proved by Zero-Induction:

\begin{thm}
Let $u\in\R$. \quad Then: $\forall k\in\N_0$,\\
\phx\qquad\qquad $(1-u)(1+u+u^2+\cdots+u^k)=1-u^{k+1}$.
\end{thm}

\begin{thm}
Let $u\in[0;1)$, $k\in\N_0$.\\
\phx\qquad\qquad Then: \quad $1+u+u^2+\cdots+u^k\,\,\le\,\,1/(1-u)$.
\end{thm}

\begin{thm}
Let $u\in[0;1)$, $k\in\N_0$, $M\ge0$.\\
\phx\qquad\qquad Then: \quad $M+Mu+Mu^2+\cdots+Mu^k\,\,\le\,\,M/(1-u)$.
\end{thm}

\begin{defn}
Let $S\subseteq\R$.
Then $|S|:=\{|x|~\hbox{s.t.}~x\in S\}$.
\end{defn}

\begin{thm}\wrlab{thm-setbdd-iff-absvalset-bdd}
Let $S\subseteq\R$. Then:\\
\phx\qquad\qquad\qquad $(\,\,S\hbox{ is bounded in }\R\,\,)\,\,\Rightarrow\,\,(\,\,|S|\hbox{is bounded in }\R\,\,)$.
\end{thm}

\begin{proof}
We have: $\forall x\in\R$, $\big|\,|x|\,\big|=|x|$. \quad It follows that:\\
\phx\quad $\forall x\in\R$, $\forall K\ge0$, \quad $(\,\,x\,\in\,B_\R(0,K)\,\,)\,\,\Leftrightarrow\,\,(\,\,|x|\,\in\,B_\R(0,K)\,\,)$.\\
Then: \qquad $\,\forall K\ge0$, $\,\,$ $(\,\,S\,\subseteq\,B_\R(0,K)\,\,)\,\,\Leftrightarrow\,\,(\,\,|S|\,\subseteq\,B_\R(0,K)\,\,)$.\\
Then: \quad $(\,\,S\hbox{ is bounded in }\R\,\,)\,\,\Leftrightarrow\,\,(\,\,\exists K\ge0~\hbox{s.t.}~S\subseteq B_\R(0,K)\,\,)$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $\,\,\,\Leftrightarrow\,\,(\,\,\exists K\ge0~\hbox{s.t.}~|S|\subseteq B_\R(0,K)\,\,)$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $\,\,\,\Leftrightarrow\,\,(\,\,|S|\hbox{ is bounded in }\R\,\,)$.
\end{proof}

\begin{defn}
Let $a\in\R^{\N_0}$, $x\in\R$.\\
Then: \quad $\TS_x^a\,\,:=\,\,\{\,a_0\,,\,a_1x\,,\,a_2x^2\,,\,\ldots\,\}$.
\end{defn}

``TS'' stands for ``Terms of Series''.

\begin{defn}
Let $a\in\R^{\N_0}$, $j\in\N_0$.\\
Then $P_jSS^a:\R\to\R$ is defined by: $\forall x\in\R$,\\
\phx\qquad $P_jSS_x^a\,\,=\,\,a_0\,+\,a_1x\,+\,a_2x^2\,+\,\cdots\,+\,a_jx^j$.
\end{defn}

\noindent
``PSS'' stands for ``Partial Sum of Series''.

\noindent
Note that $P_jSS^a\,\,=\,\,a_0\,+\,a_1(\bullet)\,+\,a_2(\bullet)^2\,+\,\cdots\,+\,a_j(\bullet)^j$;\\
this is a (not necessarily homogeneous) polynomial.

\begin{defn}
Let $a\in\R^{\N_0}$.\\
Then $P_\bullet SS^a\in(\R^\R)^{\N_0}$ is defined by: $\forall j\in\N_0$, $\,$ $(P_\bullet SS^a)_j\,\,=\,\,P_jSS^a$.
\end{defn}

\begin{defn}
Let $a\in\R^{\N_0}$, $x\in\R$.\\
Then $P_\bullet SS_x^a\in\R^{\N_0}$ is defined by: $\forall j\in\N_0$, $\,$ $(P_\bullet SS_x^a)_j\,\,=\,\,P_jSS_x^a$.
\end{defn}

\begin{defn}
Let $a\in\R^{\N_0}$.\\
\phx\qquad\qquad\qquad Then $\RC_a:=\sup\{s\ge0\,|\,\TS_s^a\hbox{ is bounded in }\R\}$.
\end{defn}

\noindent
``RC'' stands for ``Radius of Convergence''.

\begin{defn}
Let $a\in\R^{\N_0}$, $\rho:=\RC_a$.\\
\phx\qquad\qquad\qquad\qquad\qquad\quad Then $\IC_a:=(-\rho;\rho)$.
\end{defn}

\noindent
``IC'' stands for ``Interval of Convergence''.

\begin{defn}
Let $X$ and $Y$ be sets, $f\in(Y^X)^\N$, $p\in X$.\\
\phx\qquad\qquad Then $f_\bullet(p)\in Y^\N$ is defined by: \quad $\forall j\in\N$, $(f_\bullet(p))_j=f_j(p)$.
\end{defn}

\begin{defn}
Let $a\in\R^{\N_0}$. Then $a$ is denoted\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad ${}_0(a_0,a_1,a_2,\ldots)$.
\end{defn}

\begin{thm}
Let $a\in\R^{\N_0}$ and let $r<\RC_a$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: $\exists s>r$ s.t.~$\TS_s^a$ is bounded in $\R$.
\end{thm}

\begin{proof}
Let $T:=\{s\ge0\,|\,TS_s^a\hbox{ is bounded in }\R\}$.\\
Since $r<\RC_a=\sup T$, we conclude: $\neg(\sup T\le r)$.\\
Then $\neg(T\le r)$, so choose $s\in T$ s.t.~$\neg(s\le r)$.\\
Then $s>r$. Want: $TS_s^a$ is bounded in $\R$.\\
Since $s\in T$, we get: $TS_s^a$ is bounded in $\R$, \qquad as desired.
\end{proof}

\begin{thm}
Let $a\in\R^{\N_0}$ and let $x\in\R$.\\
\phx\qquad Assume: $|x|>\RC_a$. \qquad Then: $P_\bullet SS_x^a$ is not convergent in $\R$.
\end{thm}

\begin{proof}
Assume $P_\bullet SS_x^a$ is convergent in $\R$. \quad Want: Contradiction.\\
Let $u:=P_\bullet SS_x^a$. \quad Then $u$ is convergent in $\R$.\\
By \tref{thm-consecdiffs-of-conv-series-bdd}, $\{u_{j+1}-u_j\,|\,j\in\N_0\}$ is bounded in $\R$.\\
Let $D:=\{u_{j+1}-u_j\,|\,j\in\N_0\}$. \qquad Then $D$ is bounded in $\R$.\\
We have: $\forall j\in\N_0$,\\
\phx\quad $u_{j+1}-u_j\quad\,\,=\,\,\quad P_{j+1}SS_x^a-P_jSS_x^a$\\
\phx\qquad\qquad\quad $\quad\,\,\,=\,\,\quad(a_0+a_1x+\cdots+a_jx^j+a_{j+1}x^{j+1})$\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,\,-\,\,(a_0+a_1x+\cdots+a_jx^j)\qquad\quad\qquad=\quad a_{j+1}x^{j+1}$.\\
Then $D=\{u_{j+1}-u_j\,|\,j\in\N_0\}=\{\,a_1x\,,\,a_2x^2\,,\,a_3x^3\,,\,\ldots\,\}$.\\
Since $D$ is bounded in $\R$ and $\{a_0\}$ is bounded in $\R$,\\
\phx\qquad we see that $\{a_0\}\,\bigcup\,D$ is bounded in $\R$.\\
Then, by \tref{thm-setbdd-iff-absvalset-bdd}, $|\,\{a_0\}\,\bigcup\,D\,|$ is bounded in $\R$.\\
We have $|\,\{a_0\}\,\bigcup\,D\,|\,\,=\,\,\{\,|a_0|\,,\,|a_1|\cdot|x|\,,\,|a_2|\cdot|x|^2\,,\,|a_3|\cdot|x|^3\,,\,\ldots\,\}$.\\
Let $E:=\TS_{|x|}^a$. \quad Then $E=\{\,a_0\,,\,a_1\cdot|x|\,,\,a_2\cdot|x|^2\,,\,a_3\cdot|x|^3\,,\,\ldots\,\}$.\\
Then $|E|\,\,=\,\,\{\,|a_0|\,,\,|a_1|\cdot|x|\,,\,|a_2|\cdot|x|^2\,,\,|a_3|\cdot|x|^3\,,\,\ldots\,\}\,\,=\,\,|\,\{a_0\}\,\bigcup\,D\,|$.\\
So, since $|\,\{a_0\}\,\bigcup\,D\,|$ is bounded in $\R$, we get: $|E|$ is bounded in $\R$.\\
Then, by \tref{thm-setbdd-iff-absvalset-bdd}, $E$ is bounded in $\R$.\\
Let $T:=\{s\ge0\,|\,TS_s^a\hbox{ is bounded in }\R\}$. \quad Then $\RC_a=\sup\,T$.\\
Since $E=\TS_{|x|}^a$ and since $E$ is bounded in $\R$, we get: $|x|\in T$.\\
By hypothesis, $|x|>\RC_a$. \quad Then $|x|\in T\le\sup\,T=\RC_a<|x|$.\\
Then $|x|<|x|$. \qquad Contradiction.
\end{proof}

\begin{thm}
Let $a\in\R^{\N_0}$ and let $x\in\IC_a$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then $P_\bullet SS_x^a$ is convergent in $\R$.
\end{thm}

\begin{proof}
Let $b:=P_\bullet SS_x^a$. \qquad Want: $b$ is convergent in $\R$.\\
Want: $b$ is Cauchy in $\R$.\\
\phx\qquad\qquad Want: $\forall\vep>0$, $\exists K\in\N_0$ s.t., $\forall i,j\in[K..\infty)$, $\,$ $|b_i-b_j|<\vep$.\\
Given $\vep>0$. \qquad Want: $\,\exists K\in\N_0$ s.t., $\forall i,j\in[K..\infty)$, $\,$ $|b_i-b_j|<\vep$.\\
Let $\alpha:=|a|$ and let $\rho:=\RC_a$. \quad By HW\#4-4, $\rho=\RC_\alpha$.\\
Since $x\in\IC_a=(-\rho;\rho)$, we conclude: \quad $|x|<\rho$.\\
Let $r:=|x|$. \qquad
Since $r=|x|$ and $0\le|x|<\rho$, we get $r\in[0;\rho)$.\\
Let $c:=P_\bullet SS_r^\alpha$. \qquad
Since $r\in[0;\rho)=[0;\RC_\alpha)$, \quad by HW\#4-3,\\
\phx\qquad\qquad\qquad\qquad\qquad we conclude: \quad $c$ is convergent in $\R$.\\
Then $c$ is Cauchy in $\R$, so choose $K\in\N_0$ s.t.,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\forall i,j\in[K..\infty)$, $\,$ $|c_i-c_j|<\vep$.\\
Then $K\in\N_0$. \qquad\quad Want: $\forall i,j\in[K..\infty)$, $\,$ $|b_i-b_j|<\vep$.\\
Given $i,j\in[K..\infty)$. \qquad\qquad\qquad\quad $\,\,$ Want: $|b_i-b_j|<\vep$.\\
Since $i,j\in[K..\infty)$, by choice of $K$, we have: $|c_i-c_j|<\vep$.\\
By HW\#4-2, we conclude that: \quad $|b_i-b_j|\le|c_i-c_j|$.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,\,$ Then: \quad $|b_i-b_j|\le|c_i-c_j|<\vep$.
\end{proof}

\begin{defn}
Let $a\in\R^{\N_0}$.\\
\phx\qquad\qquad\qquad\qquad\qquad\quad Then $SS^a:\IC_a\to\R$ is defined by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\forall x\in\IC_a$, \quad $SS_x^a\,\,=\,\,\R\hbox{-}\lim\,P_\bullet SS_x^a$.
\end{defn}

\begin{defn}
Let $\ds{a:={}_0\left(\,\frac{1}{0!}\,,\,\frac{1}{1!}\,,\,\frac{1}{2!}\,,\,\frac{1}{3!}\,,\,\frac{1}{4!},\ldots\right)}$.\\
\phx\qquad\qquad\qquad\qquad\qquad\quad Then we define: \qquad $\exp\,\,:=\,\,SS^a$.
\end{defn}

\noindent
By HW\#4-1, we have $\D_\exp=\R$.\\
More colloquially, we would say:\\
\phx\qquad ``Let $\exp:\R\to\R$ be defined by:\\
\phx\qquad\qquad\qquad $\forall x\in\R$, \quad $\ds{\exp_x\,\,=\,\,\frac{1}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^3}{4!}+\cdots}$.''\\
\\
Our main remaining goal in this section is to show that\\
\phx\qquad differentiation of power series\\
\phx\qquad\qquad\qquad\qquad works via \qquad\qquad term-by-term differentiation.\\
In particular, we will show that $\exp'=\exp$.

\begin{defn}
Let $a\in\R^{\N_0}$.
Then $a^*\in\R^{\N_0}$ is defined by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\forall j\in\N_0$, \quad $a^*_j\,\,:=\,\,(j+1)\cdot a_{j+1}$.
\end{defn}

\noindent
We have: \qquad $\forall a\in\R^{\N_0}$, \qquad $a^*\,\,\,=\,\,\,{}_0(\,a_1\,,\,2\cdot a_2\,,\,3\cdot a_3\,,\,4\cdot a_4\,,\,\ldots)$.

\begin{thm}
Let $\ds{a:={}_0\left(\,\frac{1}{0!}\,,\,\frac{1}{1!}\,,\,\frac{1}{2!}\,,\,\frac{1}{3!}\,,\,\frac{1}{4!},\ldots\right)}$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: \qquad\quad $a^*\,\,=\,\,a$.
\end{thm}

\noindent
By the preceding theorem, to show $\exp'=\exp$, it suffices to show:\\
\phx\qquad\qquad $\forall a\in\R^{\N_0}$, \qquad $(SS^a)'\,\,=\,\,SS^{a^*}$.\\
We begin with the partial sum version:

\begin{thm}\wrlab{thm-deriv-partsum-powerseries}
Let $a\in\R^{\N_0}$ and let $j\in\N$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: \quad $(P_jSS^a)'\,\,=\,\,P_{j-1}SS^{a^*}$.
\end{thm}

\begin{proof}
Since $P_jSS^a=a_0+a_1\cdot(\bullet)+a_2\cdot(\bullet)^2+\cdots+a_j\cdot(\bullet)^j$,\\
we get: $(P_jSS^a)'=a_1+2\cdot a_2\cdot(\bullet)+\cdots+j\cdot a_j\cdot(\bullet)^{j-1}$.\\
So, since $P_{j-1}SS^{a^*}=a_1+2\cdot a_2\cdot(\bullet)+\cdots+j\cdot a_j\cdot(\bullet)^{j-1}$,\\
we get: \quad $(P_jSS^a)'\,\,=\,\,P_{j-1}SS^{a^*}$, \quad as desired.
\end{proof}

\begin{defn}
Let $X$ be a set and let $S\subseteq X$.\\
Let $Y$ be a metric space. Let $f\in(\PF^X_Y)^{\N_0}$ and let $g\in\PF^X_Y$.\\
By $f\to g$ pointwise\index[terms]{pointwise limit} on $S$ from $X$ to $Y$, we mean:\\
\phx\qquad\qquad $\forall p\in S$, \quad $f_\bullet(p)\to g(p)$ in $Y$.\\
By $f\to g$ uniformly\index[terms]{uniform limit} on $S$ from $X$ to $Y$, we mean:\\
\phx\qquad\qquad $\forall\vep>0$, $\exists K\in\N_0$ s.t., $\forall j\in\N_0$, $\forall p\in S$,\\
\phx\qquad\qquad\qquad\qquad\qquad $(\,\,j\ge K\,\,)\,\,\Rightarrow\,\,(\,\,d(\,f_j(p)\,,\,g(p)\,)\,<\,\vep\,\,)$.
\end{defn}

\noindent
Let $X$ be a set and let $S\subseteq X$. Let $Y$ be a metric space.\\
\phx\qquad\qquad Let \quad $f\in(\PF^X_Y)^{\N_0}$ \quad and let \quad $g\in\PF^X_Y$.\\
Note that, \qquad $\,\,$if \quad $f\to g$ $\,$ uniformly $\,$ on $S$ from $X$ to $Y$,\\
\phx\qquad\qquad\quad $\,\,$then \quad $f\to g$ $\,$ pointwise $\,$ on $S$ from $X$ to $Y$.\\
Also, note that, if $\,\,\,\,\,\,f\to g$ $\,$ pointwise $\,$ on $S$ from $X$ to $Y$,\\
\phx\qquad\qquad\quad $\,\,$then \quad\quad $S\,\,\subseteq\,\,\D_g\,\,\bigcap\,\,\D_{f_0}\,\,\bigcap\,\,\D_{f_1}\,\,\bigcap\,\,\D_{f_2}\,\,\bigcap\,\,\cdots$.

\begin{defn}
Let $X$ and $Y$ be sets, $f\in(Y^X)^{\N_0}$, $p\in X$.\\
\phx\qquad\qquad Then $f_\bullet(p)\in Y^{\N_0}$ is defined by: \quad $\forall j\in\N_0$, $(f_\bullet(p))_j=f_j(p)$.
\end{defn}

\begin{thm}\wrlab{thm-powerseries-converges-unifoncompact}
Let $a\in\R^{\N_0}$ and let $r\in[0,\RC_a)$.\\
Then: \qquad $P_\bullet SS^a\,\,to\,\,SS^a$ \quad uniformly on $[-r;r]$ from $\R$ to $\R$.
\end{thm}

\begin{proof}
Let $f:=P_\bullet SS^a$, $g:=SS^a$, $T:=[-r;r]$.\\
Then $T=[-r;r]\subseteq(-\RC_a,\RC_a)=\IC_a$, \quad so \quad $T\subseteq\IC_a$.\\
We wish to show: $f\to g$ uniformly on $T$ from $\R$ to $\R$.\\
\phx\qquad Want: $\,\,\,\,\,\,$ $\forall\vep>0$, $\exists K\in\N_0$ s.t., $\forall i\in[K..\infty)$, $\,$ $|g-f_i|<\vep$ on $T$.\\
Given $\vep>0$. \quad Want: $\,$ $\exists K\in\N_0$ s.t., $\forall i\in[K..\infty)$, $\,$ $|g-f_i|<\vep$ on $T$.\\
By HW\#4-4, we have: $\,\IC_a=\IC_{|a|}$.\\
Since $r\in[-r;r]=T\subseteq\IC_a=\IC_{|a|}$, \quad we get $r\in\IC_{|a|}$.\\
Then $P_\bullet SS_r^{|a|}$ is convergent in $\R$, \quad and $P_\bullet SS_r^{|a|}\to SS_r^{|a|}$ in $\R$.\\
Let $c:=P_\bullet SS_r^{|a|}$ and let $z:=SS_r^{|a|}$. \qquad Then $c\to z$ \quad $\,\,$ in $\R$.\\
Choose $\,K\in\N_0$ s.t., $\forall i\in[K..\infty)$, $\,$ $|z-c_i|<\vep$.\\
\phx$\,$ Then $K\in\N_0$. \qquad Want: $\forall i\in[K..\infty)$, $\,$ $|g-f_i|<\vep$ on $T$.\\
\phx\qquad\qquad\qquad Given $i\in[K..\infty)$. \quad Want: $|g-f_i|<\vep$ on $T$.\\
\phx\qquad\qquad\qquad\qquad\qquad\quad Want: $\forall x\in T$, $\,$ $|g-f_i|_x<\vep$.\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,\,$ Given $x\in T$. \quad Want: $|g-f_i|_x<\vep$.\\
Since $i\in[K..\infty)$, by choice of $K$, we get: $\,\,|z-c_i|<\vep$.\\
It therefore suffices to show: \quad $|g-f_i|_x\le|z-c_i|$.\\
\phx\qquad\qquad\qquad Want: $\forall\eta>0$, $\,\,$ $|g-f_i|_x\le|z-c_i|+\eta$.\\
\phx\qquad\qquad Given $\eta>0$. Want: $\,\,|g-f_i|_x\le|z-c_i|+\eta$.\\
We have $x\in T\subseteq\IC_a$.\\
Then $P_\bullet SS_x^a$ is convergent in $\R$, \quad and $P_\bullet SS_x^a\to SS_x^a$ in $\R$.\\
Let $b:=P_\bullet SS_x^a$. \quad Recall: $g=SS^a$. \quad Then $b\to g_x$ $\,\,\,$ in $\R$.\\
Choose $L\in\N_0$ s.t., $\,\,\forall j\in[L..\infty)$, $\,\,\,$ $|g_x-b_j|<\eta/2$.\\
Recall: $c\to z$ in $\R$.\\
Choose $M\in\N_0$ s.t., $\forall j\in[M..\infty)$, $\,\,$ $|z-c_j|<\eta/2$.\\
Let $j:=\max\{L,M\}$. \quad Then $j\in[L..\infty)$ and $j\in[M..\infty)$.\\
Since $j\in[L..\infty)$, $\,\,$by choice of $L$, $\,\,$we get: $|g_x-b_j|<\eta/2$.\\
Since $j\in[M..\infty)$, by choice of $M$, we get: $\,\,|z-c_j|<\eta/2$.\\
Since $f=P_\bullet SS^a$, we get $f_i=P_iSS^a$, and so $(f_i)_x=P_iSS_x^a$.\\
Then $b_i=(P_\bullet SS_x^a)_i=P_iSS_x^a=(f_i)_x$.\\
By HW\#4-2, $|b_j-b_i|\le|c_j-c_i|$.\\
Then: $|g-f_i|_x\,\,=\,\,|g_x-(f_i)_x|$\\
\phx\qquad\qquad\qquad $\,\,\,=\,\,|g_x-b_i|$\\
\phx\qquad\qquad\qquad $\,\,\,\le\,\,|g_x-b_j|+|b_j-b_i|$\\
\phx\qquad\qquad\qquad $\,\,\,<\,\,\,\,\,\,(\eta/2)\,\,\,\,+|b_j-b_i|$\\
\phx\qquad\qquad\qquad $\,\,\,\le\,\,\,\,\,\,(\eta/2)\,\,\,\,+|c_j-c_i|$\\
\phx\qquad\qquad\qquad $\,\,\,\le\,\,\,\,\,\,(\eta/2)\,\,\,\,+|c_j-z\,|+|z-c_i|$\\
\phx\qquad\qquad\qquad $\,\,\,\le\,\,\,\,\,\,(\eta/2)\,\,\,\,+|z\,-c_j|+|z-c_i|$\\
\phx\qquad\qquad\qquad $\,\,\,<\,\,\,\,\,\,(\eta/2)\,\,\,\,+\,\,\,(\eta/2)\,\,\,+|z-c_i|$\\
\phx\qquad\qquad\qquad $\,\,\,=\,\,|z-c_i|+\eta$, \qquad as desired.
\end{proof}

Unassigned HW:\\
Let $X$ be a set, let $S\subseteq X$ and let $Y$ be a metric space.\\
Let $\alpha\in(\PF^X_Y)^{\N_0}$ and let $\omega\in\PF^X_Y$.\\
Assume that: \quad $\alpha\to\omega$ \quad uniformly on $S$ from $X$ to $Y$.\\
Define $\beta\in(\PF^X_Y)^\N$ by: \qquad $\forall j\in\N$, $\,$ $\beta_j=\alpha_{j-1}$.\\
Show that: \quad $\,\,\,\,\,\,\beta\to\omega$ \quad uniformly on $S$ from $X$ to $Y$.\\

Unassigned HW:\\
Let $X$ be a set, let $S\subseteq X$ and let $Y$ be a metric space.\\
Let $\alpha\in(\PF^X_Y)^{\N_0}$ and let $\omega\in\PF^X_Y$.\\
Assume that: \quad $\alpha\to\omega$ \quad uniformly on $S$ from $X$ to $Y$.\\
Define $\beta\in(\PF^X_Y)^\N$ by: \qquad $\forall j\in\N$, $\,$ $\beta_j=\alpha_j$.\\
Show that: \quad $\,\,\,\,\,\,\beta\to\omega$ \quad uniformly on $S$ from $X$ to $Y$.

\begin{thm}
Let $a\in\R^{\N_0}$. \qquad
Then: $\,$ $(SS^a)'\,\,=\,\,SS^{a^*}$.
\end{thm}

\begin{proof}
Let $b:=a^*$ and $f:=SS^a$ and $g:=SS^b$. \qquad Want: $\,$ $f'=g$.\\
\phx\qquad Let $\,\,\,U:=\IC_a$. \qquad\qquad By HW\#4-5, $\RC_a=\RC_b$.\\
\phx\qquad Then $U=\IC_b$. \qquad\qquad Then: $\,$ $\D_{SS^a}\,=\,U\,=\,\D_{SS^b}$.\\
Want: \quad $\forall x\in U$, $\,$ $(SS^a)'_x\,=\,SS^b_x$.\\
Given $x\in U$. \qquad Want: $(SS^a)'_x\,=\,SS^b_x$.\\
Let $\rho:=\RC_a$. \qquad Then $x\in\IC_a=(-\rho;\rho)$, \quad so $|x|<\rho$.\\
Let $\ds{r:=\frac{|x|+\rho}{2}}$. \qquad Then $|x|<r<\rho$.\\
Since \quad $|x|<r$, \quad we get: \quad $x\in(-r;r)$.\\
By \tref{thm-powerseries-converges-unifoncompact}, we have:\\
\phx\qquad\qquad\quad $\,\,\,\,P_\bullet SS^a\,\,\to\,\,SS^a$ \quad uniformly on $[-r;r]$ from $\R$ to $\R$\\
\phx\qquad\quad and $\,\,\,$ $P_\bullet SS^b\,\,\to\,\,SS^b$ \quad uniformly on $[-r;r]$ from $\R$ to $\R$.\\
Define $\phi,\psi\in(\PF^\R_\R)^\N$ by: \quad $\forall j\in\N$,\\
\phx\qquad\qquad\qquad $\phi_j\,\,=\,\,P_jSS^a$ \qquad and \qquad $\psi_j=P_{j-1}SS^b$.\\
Then\qquad\qquad\qquad $\,\phi\,\,\to\,\,SS^a$ \quad uniformly on $[-r;r]$ from $\R$ to $\R$\\
\phx\qquad\quad and \qquad $\,\,\,\,\,\psi\,\,\to\,\,SS^b$ \quad uniformly on $[-r;r]$ from $\R$ to $\R$.\\
Recall that \quad $f=SS^a$ \quad and that \quad $g=SS^b$.\\
Then\qquad\qquad\qquad $\,\phi\,\,\to\,\,f$ \quad uniformly on $[-r;r]$ from $\R$ to $\R$\\
\phx\qquad\quad and \qquad $\,\,\,\,\,\psi\,\,\to\,\,g$ \quad uniformly on $[-r;r]$ from $\R$ to $\R$.\\
So, as $(-r;r)\subseteq[-r;r]$, we conclude:\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,\,\phi\,\,\to\,\,f$ \quad uniformly on $(-r;r)$ from $\R$ to $\R$\\
\phx\qquad\quad and \qquad $\,\,\,\,\,\psi\,\,\to\,\,g$ \quad uniformly on $(-r;r)$ from $\R$ to $\R$.\\
Since uniform convergence implies pointwise convergence, we get:\\
\phx\qquad\qquad\qquad\quad $\,\,\,\,\,\phi\,\,\to\,\,f$ \quad pointwise on $(-r;r)$ from $\R$ to $\R$.\\
By \tref{thm-deriv-partsum-powerseries}, $\forall j\in\N$, \quad $(P_jSS^a)'\,\,=\,\,P_{j-1}SS^{a^*}$.\\
Then, $\forall j\in\N$, \quad $(\phi_j)'\,\,=\,\,(P_jSS^a)'\,\,=\,\,P_{j-1}SS^{a^*}\,\,=\,\,P_{j-1}SS^b\,\,=\,\,\psi_j$.\\
Then, by \tref{thm-diff-commutes-with-uniflimit}, $f'=g$, as desired.
\end{proof}

\section{Multivariable Differential Calculus}

\subsection{Euclidean spaces}

\begin{thm}
$(5)=(1\mapsto5)\in\R^1$ and $5\in\R$ and $(5)\ne5$.
\end{thm}

\begin{thm}
We have: \quad $\R^1\,\bigcap\,\R\,\,=\,\,\emptyset$.
\end{thm}

\begin{defn}
$\ES\,\,:=\,\,\{\R\}\,\bigcup\,\{\R^1\,,\,\R^2\,,\,\R^3\,,\,\ldots\}$.
\end{defn}

\noindent
An element of $\ES$ is called a \bfindex[Euclidean space].\\
We will call $\R$ the \quad \bfindex[scalar space] \quad or \quad \bfindex[scalar Euclidean space].\\
An element of $\{\R^1,\R^2,\R^3,\ldots\}$ will be called a\\
\phx\qquad\qquad \bfindex[vector space] \quad or \quad \bfindex[vector Euclidean space].\\
Recall that any element of $\R$ is called a scalar.\\
Elements of $\R^1\bigcup\R^2\bigcup\R^3\bigcup\cdots$ will be called \bfindex[Euclidean vectors]
or, simply, \bfindex[vectors] in this course.\\
\\
Note that $\bigcup\ES=\R\cup\R^1\cup\R^2\cup\R^3\cup\cdots$.\\
Then $\bigcup\ES$ is the set of all objects that are either scalar or vector.\\
In the next definition, ``v/s'' is read ``vector or scalar'':

\begin{defn}
Let $x$ be an object.\\
\phx\qquad By $x$ is a \bfindex[v/s], we mean: \quad $x\in\bigcup\ES$.
\end{defn}

\begin{thm}
$(2,3)$ is a v/s \qquad and \qquad $5$ is a v/s\\
\phx\qquad and\qquad $(-3,6,5)$ is a v/s \qquad and \qquad $\{-3,6,5\}$ is NOT a v/s.
\end{thm}

\begin{thm}
$\forall m\in\N$, $\forall x\in\R^m$, $\forall j\in[1..m]$, $x_j\in\R$.
\end{thm}

\begin{defn}
Let $m\in\N$, $V:=\R^m$. \quad Then $\scri_V:=[1..m]$.
\end{defn}

\begin{thm}
$\forall V\in\ES\backslash\{\R\}$, $\forall j\in\scri_V$, $x_j\in\R$.
\end{thm}

\begin{defn}
$\forall x\in\R$, $x_0:=x$.
\end{defn}

\begin{thm}
$7_0=7\in\R$.
\end{thm}

\begin{thm}
$\forall x\in\R$, $x_0=x\in\R$.
\end{thm}

\begin{defn}
$\scri_\R:=\{0\}$.
\end{defn}

\begin{thm}
$\forall x\in\R$, $\forall j\in\scri_\R$, $x_j=x\in\R$.
\end{thm}

\begin{thm}
$\forall V\in\ES$, $\forall j\in\scri_V$, $x_j\in\R$.
\end{thm}

\noindent
Let $V\in\ES$. Then $\scri_V$ is called the \bfindex[indexing set] of $V$.

\begin{thm}
Let $V\in\ES$ and $x,y\in V$.\\
Then: \qquad $(\,\,x=y\,\,)\,\,\Leftrightarrow\,\,(\,\,\forall j\in\scri_V,\,x_j=y_j\,\,)$.
\end{thm}

\begin{defn}
Let $V\in\ES$.\\
Then $0_V\in V$ is defined by: $\forall j\in\scri_V$, $(0_V)_j=0$.
\end{defn}

\begin{thm}
$(\,\,0_\R=0\,\,)\,\,\&(\,\,0_{\R^1}=(0)\,\,)\,\,\&\,\,(\,\,0_{\R^2}=(0,0)\,\,)$\\
\phx\qquad $\&\,\,(\,\,0_{\R^3}=(0,0,0)\,\,)\,\,\&\,\,(\,\,0_{\R^4}=(0,0,0,0)\,\,)\,\,\&\,\,(\,\,0_{\R^5}=(0,0,0,0,0)\,\,)$.
\end{thm}

\begin{defn}
Let $V\in\ES$ and let $x,y\in V$.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ Then \qquad $\ds{x\sbullet y\,\,:=\,\,\sum_{j\in\scri_V}\,x_j\cdot y_j}$.
\end{defn}

\begin{thm}
$4\sbullet2=4_0\cdot2_0=4\cdot2=8$.
\end{thm}

\begin{thm}
$(4)\sbullet(2)=(4)_1\cdot(2)_1=4\cdot2=8$.
\end{thm}

\begin{thm}
$(3,6)\sbullet(8,7)=(3,6)_1\cdot(8,7)_1+(3,6)_2\cdot(8,7)_2$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $=3\cdot8+6\cdot7=24+42=66$.
\end{thm}

\begin{defn}
Let $V,W\in\ES$ and $x\in V$ and $y\in W$.\\
Assume $V\ne W$. \qquad Then $x\sbullet y:=\frownie$ and $x+y:=\frownie$.
\end{defn}

\noindent
Let $x:=(5,6)$ and $y:=(7,8,9)$.
If we think of $x$ and $y$ as functionals,\\
\phx\qquad $\ds{x=\left(\begin{tabular}{c}$1\mapsto5$\\$2\mapsto6$\end{tabular}\right)}$ \qquad and \qquad $\ds{y=\left(\begin{tabular}{c}$1\mapsto7$\\$2\mapsto8$\\$3\mapsto9$\end{tabular}\right)}$,\\
then, as $5+7=12$ and $6+8=14$, we compute\\
\phx\qquad\qquad\qquad\qquad $\ds{x+y=\left(\begin{tabular}{c}$1\mapsto12$\\$2\mapsto14$\end{tabular}\right)=(12,14)}$.\\
On the other hand, thinking of $x$ and $y$ as vectors,\\
\phx\qquad since $x\in\R^2$ and since $y\in\R^3$ and since $\R^2\ne\R^3$,\\
\phx\qquad by the preceding definition, we compuate $x+y=\frownie$.\\
Each object, has a {\it type} and the meaning of $x+y$ depends on\\
\phx\qquad whether we type $x$ and $y$ \quad as functionals \quad or \quad as vectors.\\
We could avoid this confusion\\
\phx\qquad by introducing, say, ${}_{{\rm vector}}+_{{\rm vector}}$ to indicate vector addition.\\
Then \quad $x+y=(12,14)$ \qquad and \qquad $x\,\,\,\,{}_{{\rm vector}}\!+_{{\rm vector}}\,\,\,\,y\quad=\quad\frownie$.\\
However, ${}_{{\rm vector}}+_{{\rm vector}}$ is simply hard to read.\\
We have decided, instead, to use $+$ to denote\\
\phx\qquad both \quad addition of functionals \quad and \quad addition of vectors,\\
and the reader has to determine, by context, which is meant.

Frownie is infective:

\begin{defn}
$\forall x$, $x\sbullet\frownie:=\frownie$ and $\frownie\sbullet x:=\frownie$.
\end{defn}

\begin{thm}
$[\,\,4\sbullet(2)=\frownie\,\,]\,\,\&\,\,[\,\,(3,6)\sbullet(2,4,7)=\frownie\,\,]$\\
\phx\qquad\qquad\qquad $\&\,\,[\,\,(2,8)\sbullet\frownie=\frownie=\frownie\sbullet6\,\,]$.
\end{thm}

\begin{thm}
$[\,\,3\cdot(5,8)=(15,24)\,\,]\,\,\&\,\,[\,\,3\sbullet(5,8)=\frownie\,\,]$.
\end{thm}

The next two theorems can be summarized as: ``$\sbullet$ is {\it bilinear}''.

\begin{thm}\wrlab{thm-dotprod-biadditive}
Let $V\in\ES$, $x,y,z\in V$.\\
Then $(x+y)\sbullet z=(x\sbullet z)+(y\sbullet z)$ and $x\sbullet(y+z)=(x\sbullet y)+(x\sbullet z)$.
\end{thm}

\begin{thm}\wrlab{thm-dotprod-commutes-with-scalarmult-on-LandR}
Let $V\in\ES$, $c\in\R$, $x,y\in V$.\\
Then $(c\cdot x)\sbullet y=c\cdot(x\sbullet y)$ and $x\sbullet(c\cdot y)=c\cdot(x\sbullet y)$.
\end{thm}

The next theorem can be summarized as:\\
\phx\qquad\qquad\qquad ``$\sbullet$ is {\it commutative} (a.k.a.~{\it symmetric})''.

\begin{thm}\wrlab{thm-dotprod-symmetric}
Let $V\in\ES$, $x,y\in V$.
Then $x\sbullet y=y\sbullet x$.
\end{thm}

\begin{defn}
$\forall a,b$, \quad $\ds{\delta_b^a\,\,:=\,\,\begin{cases}1,&\hbox{if }a=b\\0,&\hbox{if }a\ne b.\end{cases}}$
\end{defn}

\begin{defn}
Let $V\in\ES$ and $j\in\scri_V$.\\
\phx\qquad\qquad Then $\vep_j^V\in V$ is defined by: \quad $\forall i\in\scri_V$, $\,$ $(\vep_j^V)_i=\delta_j^i$.
\end{defn}

\begin{thm}
Let $V:=\R^3$. \qquad Then:\\
\phx\quad $[\,\,\vep_1^V\,=\,(1,0,0)\,\,]\,\,\&\,\,[\,\,\vep_2^V\,=\,(0,1,0)\,\,]\,\,\&\,\,[\,\,\vep_3^V\,=\,(0,0,1)\,\,]$.
\end{thm}

\begin{thm}
$\vep_0^\R=1$.
\end{thm}

\begin{thm}
Let $V:=\R^3$ and $x:=(-3,6,5)$.\\
\phx\qquad Then: \quad $x\sbullet\vep_2^V=(3,6,-5)\sbullet(0,1,0)=6=x_2$.
\end{thm}

\begin{thm}
$\forall V\in\ES,\,\forall x\in V,\,\forall j\in\scri_V,\quad x\sbullet\vep_j^V=x_j$.
\end{thm}

\begin{thm}
Let $V:=\R^3$ and $x:=(-3,6,5)$. \quad Then:\\
\phx\quad $x=(-3,6,5)=-3\vep_1^V+6\vep_2^V+5\vep_3^V=x_1\vep_1^V+x_2\vep_2^V+x_3\vep_3^V$.
\end{thm}

\begin{thm}
Let $V\in\ES$ and $x\in V$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \qquad $\ds{x\,\,=\,\,\sum_{j\in\scri_V}\,x_j\vep_j^V}$.
\end{thm}

\begin{defn}
Let $V\in\ES$ and $j\in\scri_V$.\\
\phx\qquad\qquad Then $\pi_j^V:V\to\R$ is defined by: \quad $\forall x\in V$, $\,$ $\pi_j^V(x)=x_j$.
\end{defn}

\begin{thm}
Let $V:=\R^3$ and $x:=(-3,6,5)$. \quad Then:\\
\phx\qquad\qquad $[\,\,\pi_1^V(x)\,=\,-3\,\,]\,\,\&\,\,[\,\,\pi_2^V(x)\,=\,6\,\,]\,\,\&\,\,[\,\,\pi_3^V(x)\,=\,5\,\,]$.
\end{thm}

\begin{thm}
$\pi_0^\R(4)=4_0=4$.
\end{thm}

\begin{thm}
$\pi_0^\R=\id_\R$.
\end{thm}

\begin{thm}
$\forall V\in\ES,\forall x\in V,\forall j\in\scri_V,\,\pi_j^V(x)=x_j=x\sbullet\vep_j^V$.
\end{thm}

\begin{defn}
$\forall V\in\ES$, \quad $\Pi_V\,\,:=\,\,\{\pi_j^V\,|\,j\in\scri_V\}$.
\end{defn}

\begin{thm}
Let $V:=\R^3$. $\,$ Then: $\Pi_V\,\,=\,\,\{\,\pi_1^V\,,\,\pi_2^V\,,\,\pi_3^V\,\}$.
\end{thm}

\begin{thm}
$\Pi_\R\,\,=\,\,\{\,\id_\R\,\}$.
\end{thm}

\begin{defn}
Let $X$ and $Y$ be sets.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ Then: \qquad $\PF_Y^X\,\,:=\,\,\{f\,|\,f:X\drta Y\}$.
\end{defn}

\noindent
Let $X$ and $Y$ be sets.
Then $\ds{\PF_Y^X=\bigcup_{W\subseteq X}\,Y^W}$.

\begin{defn}
Let $Y$ be a set and let $f$ be an object.\\
By $f$ is a {\bf$Y$-function}, we mean: $(\,\,f\hbox{ is a function}\,\,)\,\,\&\,\,(\,\,\I_f\subseteq Y\,\,)$.
\end{defn}

\begin{defn}
Let $V\in\ES$, $f$ a $V$-function and $j\in\scri_V$.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ Then: \qquad $\pi_jf\,\,:=\,\,\pi_j^V\circ f$.
\end{defn}

\begin{thm}
Let $f:\R^2\to\R^3$ be defined by:\\
\phx\qquad\qquad\qquad $\forall x\in\R^2$, \quad $f_x\,\,=\,\,(\,x_1x_2\,,\,x_1^2+x_2^2\,,\,x_1^3-6x_1x_2\,)$.\\
Then, $\forall x\in\R^2$, we have:\\
\phx\quad $[\,\,(\pi_1f)_x=x_1x_2\,\,]\,\,\&\,\,[\,\,(\pi_2f)_x=x_1^2+x_2^2\,\,]\,\,\&\,\,[\,\,(\pi_3f)_x=x_1^3-6x_1x_2\,\,]$.
\end{thm}

\begin{defn}
Let $A,B\in\bigcup\ES$.\\
Then $A+B:=\{x+y\,|\,x\in A,\,y\in B,\,x+y\ne\frownie\}$.\\
Also, $A-B:=\{x-y\,|\,x\in A,\,y\in B,\,x-y\ne\frownie\}$.\\
Also, $A\sbullet B:=\{x\sbullet y\,|\,x\in A,\,y\in B,\,x\sbullet y\ne\frownie\}$.
\end{defn}

\begin{defn}
Let $x\in\bigcup\ES$, $A\subseteq\bigcup\ES$.\\
Then $x+A:=\{x\}+A$. \qquad
Also, $A+x:=A+\{x\}$.\\
Also, $x-A:=\{x\}-A$. \qquad
Also, $A-x:=A-\{x\}$.\\
Also, $\,\,x\sbullet A:=\{x\}\sbullet A$. \qquad
$\,$ Also, $\,\,A\sbullet x:=A\sbullet\{x\}$.
\end{defn}

\begin{defn}
Let $A\subseteq\R$ and $B\subseteq\bigcup\ES$.\\
Then $A\cdot B:=\{c\cdot y\,|\,c\in A,\,y\in B\}$. \quad
Also, $B\cdot A:=\{y\cdot c\,|\,c\in A,\,y\in B\}$.
\end{defn}

\begin{defn}
Let $c\in\R$, $B\subseteq\ES$.\\
Then $c\cdot B:=\{c\}\cdot B$. \qquad
Also, $B\cdot c:=B\cdot\{c\}$.
\end{defn}

\begin{defn}
Let $A\subseteq\R$, $y\in\bigcup\ES$.\\
Then $A\cdot y:=A\cdot\{y\}$. \qquad
Also, $y\cdot A:=\{y\}\cdot A$.
\end{defn}

In the last three definitions,
``$\cdot$'' is sometimes omitted.

\begin{defn}
Let $f$ be an object.\\
By $f$ is a \bfindex[v/s-function], we mean: \quad $f$ is a $(\bigcup\ES)$-function.
\end{defn}

That is, by a ``v/s-function'', we mean a function whose image consists of vectors and scalars.

\begin{thm}
Let $V\in\ES$ and let $f$ be a $V$-function.\\
\phx\quad\qquad\qquad\qquad\qquad $\,$ Then $f$ is a v/s-function.
\end{thm}

Recall that a ``functional'' is a scalar-valued function, or a real-valued function, or an $\R$-function,
or an $\R$-valued function; in other words, a function whose image consists of scalars.

\begin{thm}
Let $f$ be a functional. Then $f$ is a v/s-function.
\end{thm}

\begin{defn}
Let $c\in\R$, $f$ a v/s-function.\\
Then $c\cdot f$ and $f\cdot c$ are the v/s-functions defined by:\\
\phx\qquad\qquad $\forall x$, \qquad $[\,\,(c\cdot f)_x=c\cdot f_x\,\,]\,\,\&\,\,[\,\,(f\cdot c)_x=f_x\cdot c\,\,]$.
\end{defn}

Again, $\cdot$ is often omitted, but no $\sbullet$ or $+$ or $-$.

\begin{defn}
Let $f$ and $g$ be v/s-functions.\\
Then $f+g$ is the v/s-function defined by:\\
\phx\qquad\qquad $\forall x$, \qquad $(f+g)_x=f_x+g_x$.\\
Also, $f-g$ is the v/s-function defined by:\\
\phx\qquad\qquad $\forall x$, \qquad $(f-g)_x=f_x-g_x$.\\
Also, $f\sbullet g$ is the v/s-function defined by:\\
\phx\qquad\qquad $\forall x$, \qquad $(f\sbullet g)_x=f_x\sbullet g_x$.
\end{defn}

\begin{thm}
Define $f,g:\R^2\to\R^3$ by: \quad $\forall x\in\R^2$,\\
\phx\qquad $f_x\,=\,(\,x_1x_2\,,\,x_1^2+x_2^2\,,\,x_1^3-6x_1x_2\,)$ \quad and \quad $g_x\,=\,(\,0\,,\,1\,,\,8x_2\,)$.\\
Then, $\forall x\in\R^2$, \quad $\,\,\,\,\,(f+g)_x\,=\,(\,x_1x_2\,,\,x_2^2+x_2^2+1\,,\,x_1^3-6x_1x_2+8x_2\,)$\\
\phx\qquad\qquad\qquad $\,\,\,\&$ \quad $(f\sbullet g)_x\,=\,x_1^2+x_2^2+[x_1^3+6x_1x_2]\cdot[8x_2]$.
\end{thm}

\begin{defn}
Let $f$ be a functional, $g$ a v/s-function.\\
Then $f\cdot g$ and $g\cdot f$ are the v/s-functions defined by:\\
\phx\qquad\qquad $\forall x$, \qquad $[\,\,(f\cdot g)_x=f_x\cdot g_x\,\,]\,\,\&\,\,[\,\,(g\cdot f)_x=g_x\cdot f_x\,\,]$.
\end{defn}

Again, $\cdot$ is often omitted.

\begin{thm}
Define $f:\R^2\to\R$ and $g:\R^2\to\R^3$ by:\\
\phx\quad $\forall x\in\R^2$, \quad $[\,f_x\,=\,x_1^5-9x_2^4\,]\,\&\,[\,g_x\,=\,(\,0\,,\,1\,,\,8x_1x_2\,)\,]$.\\
Then, $\forall x\in\R^2$,\\
\phx\qquad $(f\cdot g)_x\,=\,(\,0\,,\,x_1^5-9x_2^4\,,\,[x_1^5-9x_2^4]\cdot[8x_1x_2]\,)\,=\,(g\cdot f)_x$.
\end{thm}

\begin{defn}
Let $f$ be a v/s-function, $v\in\bigcup\ES$.\\
Then $f+v$ and $v+f$ are the v/s-functions defined by: $\forall x$,\\
\phx\qquad\qquad\qquad $(f+v)_x=f_x+v$ \qquad and \qquad $(v+f)_x=v+f_x$.\\
Also, $f-v$ and $v-f$ are the v/s-functions defined by: $\forall x$,\\
\phx\qquad\qquad\qquad $(f-v)_x=f_x-v$ \qquad and \qquad $(v-f)_x=v-f_x$.\\
Also, $f\sbullet v$ and $v\sbullet f$ are the v/s-functions defined by: $\forall x$,\\
\phx\qquad\qquad\qquad $(f\sbullet v)_x=f_x\sbullet v$ \qquad and \qquad $(v\sbullet f)_x=v\sbullet f_x$.
\end{defn}

Again, no $\sbullet$ or $+$ or $-$ is ever omitted.

\begin{defn}
Let $F$ and $G$ be sets of v/s-functions.\\
\phx\qquad Then $F+G\,\,:=\,\,\{f+g\,|\,f\in F,\,g\in G\}$.\\
\phx\qquad Also, $F-G\,\,:=\,\,\{f-g\,|\,f\in F,\,g\in G\}$.\\
\phx\qquad Also, $F\sbullet G\,\,:=\,\,\{f\sbullet g\,|\,f\in F,\,g\in G\}$.
\end{defn}

In the preceding definition, $F+G$, $F-G$ and $F\sbullet G$ are all sets of v/s-functions.

\begin{defn}
Let $f$ be a v/s-function, $G$ a set of v/s-functions.\\
\phx\quad Then $f+G\,\,:=\,\,\{f\}+G$. \quad Also, $G+f\,\,:=\,\,G+\{f\}$.\\
\phx\quad Also, $f-G\,\,:=\,\,\{f\}-G$. \quad Also, $G-f\,\,:=\,\,G-\{f\}$.\\
\phx\quad Also, $f\sbullet G\,\,:=\,\,\{f\}\sbullet G$. \quad Also, $G\sbullet f\,\,:=\,\,G\sbullet\{f\}$.
\end{defn}

\begin{defn}
Let $F$ be a set of v/s-functions, $B\subseteq\bigcup\ES$.\\
Then $F+B\,\,:=\,\,\{f+v\,|\,f\in F,\,v\in B\}$.\\
Also, $B+F\,\,:=\,\,\{v+f\,|\,f\in F,\,v\in B\}$.\\
Also, $F-B\,\,:=\,\,\{f-v\,|\,f\in F,\,v\in B\}$.\\
Also, $B-F\,\,:=\,\,\{v-f\,|\,f\in F,\,v\in B\}$.\\
Also, $F\sbullet B\,\,:=\,\,\{f\sbullet v\,|\,f\in F,\,v\in B\}$.\\
Also, $B\sbullet F\,\,:=\,\,\{v\sbullet f\,|\,f\in F,\,v\in B\}$.
\end{defn}

\begin{defn}
Let $f$ be a v/s-function, $B\subseteq\bigcup\ES$.\\
Then $f+B\,\,:=\,\,\{f\}+B$. \quad
Also, $B+f\,\,:=\,\,B+\{f\}$.\\
Also, $f-B\,\,:=\,\,\{f\}-B$. \quad
Also, $B-f\,\,:=\,\,B-\{f\}$.\\
Also, $f\sbullet B\,\,:=\,\,\{f\}\sbullet B$. \quad
Also, $B\sbullet f\,\,:=\,\,B\sbullet\{f\}$.
\end{defn}

\begin{defn}
Let $F$ be a set of v/s-functions, $v\in\bigcup\ES$.\\
Then $F+v\,\,:=\,\,F+\{v\}$. \quad
Also, $v+F\,\,:=\,\,\{v\}+F$.\\
Also, $F-v\,\,:=\,\,F-\{v\}$. \quad
Also, $v-F\,\,:=\,\,\{v\}-F$.\\
Also, $F\sbullet v\,\,:=\,\,F\sbullet\{v\}$. \quad
Also, $v\sbullet F\,\,:=\,\,\{v\}\sbullet F$.
\end{defn}

\begin{defn}
Let $F$ be a set of functionals.\\
\phx\quad\qquad\qquad\qquad\qquad\quad $\,$ Let $G$ be a set of v/s-functions.\\
\phx\qquad Then $F\cdot G\,\,:=\,\,\{f\cdot g\,|\,f\in A,\,g\in B\}$.\\
\phx\qquad Also $G\cdot F\,\,:=\,\,\{g\cdot f\,|\,f\in A,\,g\in B\}$.
\end{defn}

\begin{defn}
Let $f$ be a functional.\\
\phx\quad\qquad\qquad\qquad\qquad\quad $\,$ Let $G$ be a set of v/s-functions.\\
\phx\qquad Then $f\cdot G\,\,:=\,\,\{f\}\cdot G$. \quad
Also $G\cdot f\,\,:=\,\,G\cdot\{f\}$.
\end{defn}

\begin{defn}
Let $F$ be a set of v/s-functions.\\
\phx\quad\qquad\qquad\qquad\qquad\quad $\,$ Let $g$ be a functional.\\
\phx\qquad Then $F\cdot g\,\,:=\,\,F\cdot\{g\}$. \quad
Also $g\cdot F\,\,:=\,\,\{g\}\cdot F$.
\end{defn}

\begin{defn}
Let $A\subseteq\R$, $G$ a set of v/s-functions.\\
\phx\qquad Then $A\cdot G\,\,:=\,\,\{c\cdot g\,|\,c\in A,\,g\in G\}$.\\
\phx\qquad Also, $G\cdot A\,\,:=\,\,\{g\cdot c\,|\,c\in A,\,g\in G\}$.
\end{defn}

\begin{defn}
Let $c\in\R$, $G$ a set of v/s-functions.\\
\phx\qquad Then $c\cdot G\,\,:=\,\,\{c\}\cdot G$. \quad
Also, $G\cdot c\,\,:=\,\,G\cdot\{c\}$.
\end{defn}

\begin{defn}
Let $A\subseteq\R$, $g$ a v/s-function.\\
\phx\qquad Then $A\cdot g\,\,:=\,\,A\cdot\{g\}$. \quad
Also, $g\cdot A\,\,:=\,\,\{g\}\cdot A$.
\end{defn}

\noindent
In all of the preceding defitions, keep in mind that\\
\phx\qquad $\cdot$, $\bullet$ and $\sbullet$ are all commutative, \quad but $-$ is not.\\
So, for example, in the last definition, $g\cdot A=A\cdot g$.

We also have a dot product on any matrix space:

\begin{defn}
Let $p,q\in\N$ and let $A,B\in\R^{p\times q}$.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ Then $\ds{A\sbullet B\,\,:=\,\,\sum_{i=1}^p\,\sum_{j=1}^q\,A_{ij}\cdot B_{ij}}$.
\end{defn}

\noindent
Keep in mind that this dot prouct is scalar-valued.\\
It is {\it NOT} the same as matrix multiplication.

\subsection{Basics of dot product and the standard norm}

\begin{thm}
$\forall V\in\ES$, $\forall x\in V$, \quad $\ds{x\sbullet x=\sum_{j\in\scri_V}\,x_j^2}$.
\end{thm}

\begin{thm}
Let $V\in\ES$ and let $x\in V$.\\
Then: $[\,\,x\sbullet x\ge0\,\,]\,\,\&\,\,[\,\,(x\sbullet x=0)\,\,\Leftrightarrow\,\,(\,\,x=0_V\,\,)\,\,]$.
\end{thm}

\begin{defn}
$\forall x\in\bigcup\ES$, $|x|:=\sqrt{x\sbullet x}$.
\end{defn}

\noindent
The notation ``$|x|$'' is read ``the norm of $x$''.

\begin{thm}
We have: $|(5,-2,1)|=\sqrt{25+4+1}=\sqrt{30}$.
\end{thm}

\noindent
The next theorem is expressed by saying:\\
\phx\qquad\qquad $|\bullet|$ separates zero.

\begin{thm}
Let $V\in\ES$, $x\in V$.\\
\phx\qquad\qquad Then: $(\,\,|x|=0\,\,)\,\,\Leftrightarrow\,\,(\,\,x=0_V\,\,)$.
\end{thm}

\begin{thm}
Let $x\in\bigcup\ES$ and let $c\in\R$.\\
\phx\qquad\qquad Then: \quad $(c\cdot x)\sbullet(c\cdot x)=c^2\cdot(x\sbullet x)$.\\
\phx\qquad\qquad Also: \qquad\qquad $|c\cdot x|=|c|\cdot|x|$.
\end{thm}

\noindent
The formula $|c\cdot x|=|c|\cdot|x|$ is expressed by saying:\\
\phx\qquad\qquad $|\bullet|$ is absolute homogeneous.

\begin{thm}
Let $V\in\ES$ and let $j\in\scri_V$.\\
\phx\qquad\qquad Then: \quad $(\,\,\vep_j^V\sbullet\vep_j^V=1\,\,)\,\,\&\,\,(\,\,|\vep_j^V|=1\,\,)$.
\end{thm}

\begin{defn}
Let $V\in\ES$. Then $S_V:=\{x\in V~\hbox{s.t.}~|x|=1\}$.
\end{defn}

\noindent
In the last definition, $S_V$ is called the \bfindex[unit sphere] in $V$.\\
Also, the elements of $S_V$ are called \bfindex[unit vectors].\\
Every standard basis vector is a unit vector:

\begin{thm}
Let $V\in\ES$, $j\in\scri_V$. Then $\vep_j^V\in S_V$.
\end{thm}

\begin{thm}
Let $V\in\ES$. Then $S_V\ne\emptyset$.
\end{thm}

The next result asserts that every Euclidean vector can be written
as its norm times a unit vector; such a form for the vector is called a ``polar form''.
Unassigned HW: Show that every nonzero Euclidean vector has a unique polar form,
but, for zero, the form is not unique.

\begin{thm}
Let $V\in\ES$ and let $x\in V$.\\
\phx\quad\qquad\qquad\qquad\qquad\quad $\,$ Then $\exists u\in S_V$ s.t.~$x\,=\,|x|\cdot u$.
\end{thm}

\begin{proof}
Either \quad (1) $x=0_V$ \quad or \quad (2) $x\ne0_V$.\\
\\
{\it Case (1):}
Since $x=0_V$, we get: $|x|=0$.\\
Since $S_V\ne\emptyset$, choose $u\in S_V$.\\
Then $u\in S_V$ and we wish to show: $x\,=\,|x|\cdot u$.\\
We have $x=0_V=0\cdot u=|x|\cdot u$, as desired.\\
{\it End of Case (1).}\\
\\
{\it Case (2):}
Since $x\ne0_V$, we get: $|x|\ne0$. \quad Then $(1/|x|)\cdot|x|=1$.\\
Since $|x|\ge0$ and $|x|\ne0$, we get $|x|>0$. \quad Then $1/|x|>0$.\\
Let $a:=1/|x|$. \quad Then $a>0$. \quad Then $|a|=a$.\quad Let $u:=a\cdot x$.\\
We have $a\cdot|x|=(1/|x|)\cdot|x|=1$. Then $|u|=|a\cdot x|=|a|\cdot|x|=a\cdot|x|=1$.\\
Then $u\in S_V$. \qquad Want: $x=|x|\cdot u$.\\
We have $x=1\cdot x=(a\cdot|x|)\cdot x=|x|\cdot(a\cdot x)=|x|\cdot u$, as desired.\\
{\it End of Case (2).}
\end{proof}

We also have a standard norm on any matrix space:

\begin{defn}
Let $p,q\in\N$ and let $A\in\R^{p\times q}$.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ Then $|A|:=\sqrt{A\sbullet A}$.
\end{defn}

\subsection{Cauchy-Schwarz}

The next theorem is \bfindex[Weak Cauchy-Schwarz]:

\begin{thm}
Let $V\in\ES$, $x,y\in V$. \quad Then $x\sbullet y\le|x|\cdot|y|$.
\end{thm}

\begin{proof}
Choose $t,u\in S_V$ s.t.~$(\,x=|x|\cdot t\,)\,\&\,(\,y=|y|\cdot u\,)$.\\
Want: $(|x|\cdot t)\sbullet(|y|\cdot u)\le|x|\cdot|y|$.\\
By bilinearity, we have: $(|x|\cdot t)\sbullet(|y|\cdot u)=|x|\cdot|y|\cdot(t\sbullet u)$.\\
Want: $|x|\cdot|y|\cdot(t\sbullet u)\le|x|\cdot|y|$. \qquad Want: $t\sbullet u\le1$.\\
Since $t,u\in S_V$, we get: $|t|=1=|u|$.\\
Then: $0\,\,\le\,\,|t-u|^2\,\,=(t-u)\sbullet(t-u)\,\,=\,\,(t\sbullet t)-2\cdot(t\sbullet u)+(u\sbullet u)$\\
\phx\qquad\quad $\,\,\,\,=\,\,|t|^2-2\cdot(t\sbullet u)+|u|^2\,\,=\,\,1^2-2\cdot(t\sbullet u)+1^2\,\,=\,\,2-2\cdot(t\sbullet u)$.\\
Then $0\le2-2\cdot(t\sbullet u)$, \quad so \quad $2\cdot(t\sbullet u)\le2$, \quad so \quad $t\sbullet u\le1$.
\end{proof}

The next theorem is \bfindex[Cauchy-Schwarz]:

\begin{thm}
Let $V\in\ES$, $x,y\in V$. \quad Then $|x\sbullet y|\le|x|\cdot|y|$.
\end{thm}

\begin{proof}
Since $|x\sbullet y|\,\,=\,\,\max\{\,x\sbullet y\,,\,-(x\sbullet y)\,\}$,\\
\phx\qquad\qquad it suffices to show $(\,\,x\sbullet y\le|x|\cdot|y|\,\,)\,\,\&\,\,(\,\,-(x\sbullet y)\le|x|\cdot|y|\,\,)$.\\
By Weak Cauchy-Schwarz, $x\sbullet y\le|x|\cdot|y|$. $\,$ Want: $-(x\sbullet y)\le|x|\cdot|y|$.\\
By Weak Cauchy-Schwarz, $(-x)\sbullet y\le|-x|\cdot|y|$.\\
We have $(-x)\sbullet y=((-1)\cdot x)\sbullet y=(-1)\cdot(x\sbullet y)=-(x\sbullet y)$.\\
We have $|-x|=|(-1)\cdot x|=|-1|\cdot|x|=1\cdot|x|=|x|$.\\
Then $-(x\sbullet y)=(-x)\sbullet y\le|-x|\cdot|y|=|x|\cdot|y|$, as desired.
\end{proof}

\subsection{The $1$-norm}

\begin{defn}
$\forall V\in\ES$, $\forall x\in V$, \quad $\|x\|:=\sum_{j\in\scri_V}\,|x_j|$.
\end{defn}

The quantity $\|x\|$ is called the \bfindex[one-norm] of $x$.

\begin{thm}
Let $x:=(5,-2,1)$.\\
\phx\qquad Then $|x|=\sqrt{25+4+1}=\sqrt{30}$ and $\|x\|=5+2+1=8$.
\end{thm}

\begin{thm}
$\forall$finite set $I$, $\forall a\in[0;\infty)^I$, we have:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\ds{\left(\sum_{j\in I}\,a_j\right)^2\,\,\,\ge\,\,\,\sum_{i\in I}\,a_i^2}$.
\end{thm}

\begin{proof}
We have:\\
\phx $\,$ $\ds{\left(\sum_{j\in I}a_j\right)^2=\left[\sum_{j\in I}a_j\right]\cdot\left[\sum_{k\in I}a_k\right]
=\sum_{j\in I}\sum_{k\in I}a_ja_k\ge\sum_{i\in I}a_ia_i=\sum_{i\in I}a_i^2}$.
\end{proof}

\begin{thm}\wrlab{thm-1-2-norms-comparable}
Let $V\in\ES$, $m:=\#\scri_V$, $x\in V$.\\
\phx\qquad\qquad Then: \quad $|x|\,\,\le\,\,\|x\|\,\,\le\,\,\sqrt{m}\cdot|x|$.
\end{thm}

\begin{proof}
We have:\\
\phx\qquad $\ds{\|x\|^2\,\,=\,\,\left(\sum_{j\in\scri_V}\,|x_j|\right)^2\,\,\ge\,\,\sum_{j\in\scri_V}\,|x_j|^2\,\,=\,\,\sum_{j\in\scri_V}\,x_j^2\,\,=\,\,|x|^2}$.\\
Then \quad $0\le|x|^2\le\|x\|^2$, \quad so \quad $\sqrt{|x|^2}\le\sqrt{\|x\|^2}$, \quad so \quad $|x|\le\|x\|$.\\
It remains to show: \qquad $\|x\|\le\sqrt{m}\cdot|x|$.\\
We have: $\forall j\in\scri_V$, $\exists a\in\{-1,1\}$ \quad s.t. \quad $a\cdot x_j=|x_j|$.\\
For all $j\in\scri_V$, let $A_j:=\{a\in\{-1,1\}~\hbox{s.t.}~a\cdot x_j=|x_j|\}$.\\
Then we have: \qquad\quad $\forall j\in\scri_V$, \quad $A_j\ne\emptyset$.\\
Choose $s\in V$ s.t., \quad $\,\,\,\,\forall j\in\scri_V$, \quad $s_j\in A_j$.\\
Then \quad $\forall j\in\scri_V$, \quad we have \quad $s_j\cdot x_j=|x_j|$.\\
Also, \quad $\forall j\in\scri_V$, \quad we have \quad $s_j\in A_j\subseteq\{-1,1\}$, \quad and so \quad $s_j^2=1$.\\
Then $\ds{|s|^2=\sum_{j\in\scri_V}\,s_j^2=\sum_{j\in\scri_V}\,1=\#\scri_V=m}$. \quad
Then $|s|=\sqrt{|s|^2}=\sqrt{m}$.\\
By Weak Cauchy-Schwarz, \qquad\qquad $\,\,s\sbullet x\le|s|\cdot|x|$.\\
Then: $\ds{\|x\|=\sum_{j\in\scri_V}\,|x_j|=\sum_{j\in\scri_V}\,s_j\cdot x_j=s\sbullet x\le|s|\cdot|x|=\sqrt{m}\cdot|x|}$.
\end{proof}

\begin{thm}\wrlab{thm-1-2-norms-in-R2-comparable}
Let $a,b\in\R$.\\
\phx\qquad Then: \quad $\sqrt{a^2+b^2}\,\,\le\,\,|a|+|b|\,\,\le\,\,\sqrt{2}\cdot\sqrt{a^2+b^2}$.
\end{thm}

\begin{proof}
By \tref{thm-1-2-norms-comparable},\\
\phx\qquad\qquad\qquad $\,\,\,\,|(a,b)|\,\,\,\,\,\le\,\,\|(a,b)\|\,\,\,\le\,\,\sqrt{2}\cdot\,\,\,\,|(a,b)|$.\\
\phx\qquad Then: \quad $\sqrt{a^2+b^2}\,\,\le\,\,|a|+|b|\,\,\le\,\,\sqrt{2}\cdot\sqrt{a^2+b^2}$.
\end{proof}

We also have a one-norm on any matrix space:

\begin{defn}
Let $p,q\in\N$ and let $A\in\R^{p\times q}$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,$ Then $\ds{\|A\|:=\sum_{i=1}^p\,\sum_{j=1}^q\,|A_{ij}|}$.
\end{defn}

\subsection{The standard metrics on a Euclidean space}

\begin{thm}
Let $V\in\ES$ and let $x,y\in V$.\\
\phx\quad\qquad\qquad\qquad\qquad Then $|x+y|\le|x|+|y|$.
\end{thm}

\begin{proof}
By weak Cauchy-Schwarz, \qquad\quad $\,\,\,\,x\sbullet y\le|x|\cdot|y|$.\\
Then \qquad\qquad $\,\,\,2\cdot(x\sbullet y)\qquad\qquad\,\le\qquad\qquad2\cdot|x|\cdot|y|$.\\
Then \quad $\,|x|^2\,\,+\,\,2\cdot(x\sbullet y)\,\,+\,\,|y|^2\,\,\,\,\le\,\,\,\,|x|^2\,\,+\,\,2\cdot|x|\cdot|y|\,\,+\,\,|y|^2$.\\
Then \quad $|x+y|^2\,\,=\,\,(x+y)\sbullet(x+y)$\\
\phx\quad\qquad\qquad\qquad $=\,\,(x\sbullet x)+2\cdot(x\sbullet y)+(y\sbullet y)$\\
\phx\quad\qquad\qquad\qquad $=\,\,|x|^2+2\cdot(x\sbullet y)+|y|^2$\\
\phx\quad\qquad\qquad\qquad $\le\,\,|x|^2+2\cdot|x|\cdot|y|+|y|^2\,\,=\,\,(|x|+|y|)^2$.\\
Then $0\le|x+y|^2\le(|x|+|y|)^2$, so $\sqrt{|x+y|^2}\le\sqrt{(|x|+|y|)^2}$.\\
Since $|x+y|\ge0$ and $|x|+|y|\ge0$, we conclude:\\
\phx\qquad\qquad $\sqrt{|x+y|^2}=|x+y|\qquad\hbox{and}\qquad\sqrt{(|x|+|y|)^2}=|x|+|y|$.\\
Then $|x+y|=\sqrt{|x+y|^2}\le\sqrt{(|x|+|y|)^2}=|x|+|y|$, as desired.
\end{proof}

\begin{thm}
Let $V\in\ES$.\\
Define $d:V\times V\to[0;\infty)$ by: \quad $\forall x,y\in V$, $\,$ $d(x,y)=|y-x|$.\\
\phx\quad\qquad\qquad\qquad\qquad Then: \quad $d\in\scrm(V)$.
\end{thm}

\begin{proof}
We wish to show:\\
\phx\qquad\qquad\qquad (1) $\forall x,y\in V$, $\,$ $(\,d(x,y)=0\,)\,\Leftrightarrow\,(\,x=y\,)$.\\
\phx\qquad and \qquad (2) $\forall x,y\in V$, $\,$ $d(x,y)\,=\,d(y,x)$.\\
\phx\qquad and \qquad (3) $\forall x,y,z\in V$, $\,$ $d(x,z)\,\le\,[\,d(x,y)\,]\,+\,[\,d(y,z)\,]$.\\
\\
{\it Proof of (1):}
Given $x,y\in V$. \qquad Want: $(\,d(x,y)=0\,)\,\Leftrightarrow\,(\,x=y\,)$.\\
We have: $(\,d(x,y)=0\,)\,\Leftrightarrow\,(\,|y-x|=0\,)\,\Leftrightarrow\,(\,y-x=0_V\,)\,\Leftrightarrow\,(\,x=y\,)$.\\
{\it End of proof of (1).}\\
\\
{\it Proof of (2):}
Given $x,y\in V$. \qquad Want: $d(x,y)\,=\,d(y,x)$.\\
We have: $d(x,y)\,=\,|y-x|\,=\,|(-1)\cdot(x-y)|$\\
\phx\qquad\qquad\qquad\qquad\quad $=\,|-1|\cdot|x-y|\,=\,1\cdot[d(y,x)]\,=\,d(y,x)$.\\
{\it End of proof of (2).}\\
\\
{\it Proof of (3):}
Given $x,y,z\in V$. Want: $d(x,z)\,\le\,[\,d(x,y)\,]\,+\,[\,d(y,z)\,]$.\\
We have $d(x,z)\,=\,|z-x|\,=\,|\,(z-y)\,+\,(y-x)\,|$\\
\phx\qquad\qquad\qquad $\,\,\,\,\,\le\,|z-y|\,+\,|y-x|\,=\,[\,d(y,z)\,]\,+\,[\,d(x,y)\,]$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad $\,\,\,\,\,=\,[\,d(x,y)\,]\,+\,[\,d(y,z)\,]$.\\
{\it End of proof of (3).}
\end{proof}

\noindent
Notation: \qquad
Let $V\in\ES$ and define $d:V\times V\to[0;\infty)$ by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\forall x,y\in V$, \quad $d(x,y)=|y-x|$.\\
Then the metric space $(V,d)$ is denoted $V$.

\begin{thm}
Let $V\in\ES$ and let $x,y\in V$.\\
\phx\qquad\qquad\qquad\qquad $\,\,$ Then $d_V(x,y)=|y-x|$.
\end{thm}

\subsection{Multivariable polynomials}

\begin{defn}
Let $X$ and $Y$ be sets. Then $\scrc^X_Y:=\{C_X^z\,|\,z\in Y\}$.
\end{defn}

Then $\scrc^X_Y$ is the set of constant functions from $X$ to $Y$.

\begin{thm}
Let $V:=\R^2$, $W:=\R^3$, $\alpha:=\vep_1^W$, $C:=C_V^{8\alpha}$.\\
\phx\qquad\qquad\qquad\qquad $\,\,$ Then $C\in\scrc^V_W$.
\end{thm}

\begin{defn}
Let $V,W\in\ES$ and let $k\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,\,\,$ Then $\scrm_k^{VW}\,\,:=\,\,\scrc^V_W\,\cdot\,\Pi_V\,\cdot\,\Pi_V\,\cdot\,\cdots\,\cdot\,\Pi_V$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad with $k$ copies of $\Pi_V$.
\end{defn}

The logic purist would dislike the preceding definition, preferring:

\begin{defn}
Let $V,W\in\ES$ and let $\scrf:=W^V$.\\
\phx\qquad Define $\Phi:2^\scrf\to2^\scrf$ by: \quad $\forall S\subseteq\scrf$, $\,$ $\Phi_S=S\cdot\Pi_V$.\\
\phx\qquad Then, for all $k\in\N_0$, we define: $\scrm_k^{VW}\,\,:=\,\,\Phi_\circ^k(\scrc^V_W)$.
\end{defn}

Elements of $\scrm_k^{VW}$ are called $k$-\bfindex[monomials] from $V$ to $W$.

\begin{thm}
Let $V:=\R^2$ and let $W:=\R^3$.\\
Define $f:V\to W$ by: \quad $\forall x,y\in\R$, $\,$ $f(x,y)\,=\,(\,8x^3y\,,\,0\,,\,0\,)$.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad Then $f\in\scrm_4^{VW}$.
\end{thm}

\begin{proof}
Let \quad $\alpha:=\vep_1^W$, \quad $C:=C_V^{8\alpha}$, \quad $X:=\pi_1^V$, \quad $Y:=\pi_2^V$.\\
Then $f=C\cdot X^3\cdot Y=C\cdot X\cdot X\cdot X\cdot Y\in\scrc^V_W\cdot\pi_V\cdot\pi_V\cdot\pi_V\cdot\pi_V=\scrm_4^{VW}$.
\end{proof}

\begin{defn}
Let $V$ be a set, let $W\in\ES$ and let $S\subseteq W^V$.\\
\phx\qquad\qquad\qquad\qquad\qquad By $S$ is \bfindex[v.op.-closed], we mean:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\quad $S+S\subseteq S$ \qquad and \qquad $\R\cdot S\subseteq S$.
\end{defn}

\noindent
That is, $S$ is closed under vector addition and scalar multiplication.\\
\\
We leave it to the reader to verify that,\\
\phx\qquad $\forall$set $V$, $\forall W\in\ES$, \qquad\quad $\scrc^V_W$ is v.op.-closed.\\
In particular, we get closure under scalar multiplication:\\
\phx\qquad $\forall$set $V$, $\forall W\in\ES$, \qquad\quad $\R\cdot\scrc^V_W\subseteq\scrc^V_W$.\\
Let $X:=\R^2$. We also leave it to the reader to verify that,\\
\phx\qquad\qquad\qquad\quad $\scrm_1^{X\R}+\scrm_1^{X\R}\not\subseteq\scrm_1^{X\R}$,\\
so $\scrm_1^{X\R}$ is {\it NOT} v.op.-closed.\\
However, $\scrm_1^{X\R}$ {\it IS} closed under scalar multiplication:\\

\begin{thm}
Let $V,W\in\ES$ and let $k\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\quad Then $\R\cdot\scrm_k^{VW}\subseteq\scrm_k^{VW}$.
\end{thm}

\begin{proof}
Let $\scra:=\Pi_V\,\cdot\,\Pi_V\,\cdot\,\cdots\,\cdot\,\Pi_V$, with $k$ copies of $\Pi_V$.\\
Then $\scrm_k^{VW}=\scrc^V_W\cdot\scra$.
Then $\R\cdot\scrm_k^{VW}=\R\cdot\scrc^V_W\cdot\scra\subseteq\scrc^V_W\cdot\scra=\scrm_k^{VW}$.
\end{proof}

\begin{defn}
Let $V,W\in\ES$ and let $k\in\N_0$.\\
Then $\scrh_k^{VW}\,\,:=\,\,\{\phi_1+\cdots+\phi_m\,|\,m\in\N,\,\phi_1,\ldots,\phi_m\in\scrm_k^{VW}\}$.
\end{defn}

\noindent
Elements of $\scrh_k^{VW}$ are called $k$-\bfindex[polynomials] from $V$ to $W$.\\
So a $k$-polynomial is, by definition, a sum of $k$-monomials.\\
\\
Another perspective: The function space $\scrh_k^{VW}$ is exactly\\
\phx\qquad\qquad the closure of $\scrm_k^{VW}$ under vector operations.\\
Since $\scrm_k^{VW}$ was already closed under scalar multiplication,\\
\phx\qquad we only needed to close it under vector addition for v.op.-closure.\\
In particular, we record:

\begin{thm}
Let $V,W\in\ES$ and let $k\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\quad Then $\scrh_k^{VW}$ is v.op.-closed.
\end{thm}

\begin{thm}
Let $V:=\R^2$ and $W:=\R^3$.\\
Define $g:V\to W$ by: $\forall x,y\in\R$,\\
\phx\qquad $g(x,y)\quad=\quad(\,\,x^4+6x^3y\,\,,\,\,7x^2y^2+\sqrt{2}xy^3\,\,,\,\,9x^4+y^4\,\,)$.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad Then: \quad $g\in\scrh_4^{VW}$.
\end{thm}

\begin{proof}
Let \quad $\alpha:=\vep_1^W$, \quad$\beta:=\vep_2^W$, \quad $\gamma:=\vep_3^W$, \quad $X:=\pi_1^V$, \quad $Y:=\pi_2^V$.\\
Then \qquad $g\qquad=\qquad\quad\,\,\,\,\,C_V^\alpha\cdot X^4\quad\qquad\,+\quad C_V^{3\alpha}\cdot X^3\cdot Y$\\
\phx\qquad\qquad\qquad\qquad\qquad $+\quad C_V^{7\beta}\cdot X^2\cdot Y^2\quad+\quad C_V^{\sqrt{2}\beta}\cdot X\cdot Y^3$\\
\phx\qquad\qquad\qquad\qquad\qquad $+\quad C_V^{9\gamma}\cdot X^4\qquad\quad+\quad C_V^{\gamma}\cdot Y^4$,\\
so $g\in\scrm_4^{VW}+\scrm_4^{VW}+\scrm_4^{VW}+\scrm_4^{VW}+\scrm_4^{VW}+\scrm_4^{VW}\subseteq\scrh_4^{VW}$.
\end{proof}

\begin{thm}
Let $V,W\in\ES$. Then $\scrc^V_W=\scrm_0^{VW}=\scrh_0^{VW}$.
\end{thm}

\begin{defn}
Let $V,W\in\ES$.\\
Then: \quad $\scrl^V_W:=\scrh_1^{VW}$ \quad and \quad $\scrq^V_W:=\scrh_2^{VW}$ \quad and \quad $\scrk^V_W:=\scrh_3^{VW}$.
\end{defn}

\noindent
Let $V,W\in\ES$. Then:\\
\phx\quad element of $\scrc^V_W$ are called constant functions from $V$ to $W$ \qquad and\\
\phx\quad element of $\scrl^V_W$ are called linear functions from $V$ to $W$ \qquad and\\
\phx\quad element of $\scrk^V_W$ are called cubic functions from $V$ to $W$.\\
We next argue that linear is the same as ``algebraically linear'':

\begin{thm}
Let $V,W\in\ES$ and let $f:V\to W$.\\
Then: \qquad\qquad\quad $\,\,\,(\,\,\,\,f\in\scrl^V_W\,\,\,\,)\qquad\qquad\Leftrightarrow$\\
\phx\qquad\qquad\qquad\qquad $(\qquad\qquad[\,\,\forall c\in\R,\,\forall x\in V,\quad f_{c\cdot x}=c\cdot f_x\,\,]$\\
\phx\qquad\qquad\qquad\qquad\qquad $\&\quad\,\,\,\,[\,\,\forall x,y\in V,\qquad\quad f_{x+y}=f_x+f_y\,\,]\quad)$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\noindent
The next two theorems show that constants and linear functions are all Lipschitz.\\
Consequently, they are all uniformly continuous, and, therefore, continuous.

\begin{thm}
Let $V$ and $W$ be metric spaces, and let $C\in\scrc^V_W$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then $C$ is Lipschitz-$0$ from $V$ to $W$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}\wrlab{thm-lin-lip}
Let $V,W\in\ES$ and let $L\in\scrl^V_W$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then $L$ is Lipschitz from $V$ to $W$.
\end{thm}

\begin{proof}
Want: $\exists K\ge0$ s.t.~$L$ is Lipschitz-$K$ from $V$ to $W$.\\
Let \quad $A\,:=\,\max\,\{\,|L_{\vep_j^V}|~\hbox{s.t.}~j\in\scri_V\,\}$, \quad $m:=\#\scri_V$, \quad $K:=A\cdot\sqrt{m}$.\\
Then $K\ge0$. \quad Want: $L$ is Lipschitz-$K$ from $V$ to $W$.\\
Want: $\forall x,y\in V$, $d(L_x,L_y)\le K\cdot[d(x,y)]$.\\
Given $x,y\in V$. \qquad Want: $d(L_x,L_y)\le K\cdot[d(x,y)]$.\\
Let $I:=\scri_V$. $\,$
For all $j\in I$, let $e_j:=\vep_j^V$. $\,$ For all $j\in I$, let $q_j:=L_{e_j}$.\\
Since $\ds{x=\sum_{j\in I}\,x_j\cdot e_j}$ and $L$ is algebraically linear, we get: $\ds{L_x=\sum_{j\in I}\,x_j\cdot q_j}$.\\
Since $\ds{y=\sum_{j\in I}\,y_j\cdot e_j}$ and $L$ is algebraically linear, we get: $\ds{L_y=\sum_{j\in I}\,y_j\cdot q_j}$.\\
By definition of $A$, we have:\\
\phx\qquad\qquad $\forall j\in I$, \qquad $|q_j|\,\,\,\,\le\,\,\,\,A$.\\
By Cauchy-Schwarz, we have:\\
\phx\qquad\qquad $\forall j\in I$, \qquad $|(y_j-x_j)\cdot q_j|\,\,\,\,\le\,\,\,\,|y_j-x_j|\cdot|q_j|$.\\
We have $\ds{\|y-x\|=\sum_{j\in I}\,|y_i-x_i|}$ \quad and \quad $\|y-x\|\le\sqrt{m}\cdot|y-x|$.\\
Then $\ds{d(L_x,L_y)\,\,=\,\,|L_y-L_x|\,\,=\,\,\left|\left(\sum_{j\in I}\,y_jq_j\right)-\left(\sum_{j\in I}\,x_jq_j\right)\right|}$\\
\phx\qquad\qquad\qquad $\,\,$ $\ds{=\,\,\left|\sum_{j\in I}\,(y_jq_j-x_j\cdot q_j)\right|}$\\
\phx\qquad\qquad\qquad $\,\,$ $\ds{=\,\,\left|\sum_{j\in I}\,\bigg((y_j-x_j)\cdot q_j\bigg)\right|}$\\
\phx\qquad\qquad\qquad $\,\,$ $\ds{\le\,\,\sum_{j\in I}\,|(y_j-x_j)\cdot q_j|}$\\
\phx\qquad\qquad\qquad $\,\,$ $\ds{\le\,\,\sum_{j\in I}\,\bigg(|y_j-x_j|\cdot|q_j|\bigg)}$\\
\phx\qquad\qquad\qquad $\,\,$ $\ds{\le\,\,\sum_{j\in I}\,\bigg(|y_j-x_j|\cdot A\bigg)\,\,\,\qquad=\,\,\quad\bigg(\sum_{j\in I}\,|y_j-x_j|\bigg)\cdot A}$\\
\phx\qquad\qquad\qquad $\,\,$ $=\qquad\|y-x\|\cdot A\qquad\qquad\,\le\qquad\sqrt{m}\cdot|y-x|\cdot A$\\
\phx\qquad\qquad\qquad $\,\,$ $=\qquad A\cdot\sqrt{m}\cdot|y-x|\qquad=\qquad K\cdot[d(x,y)]$.
\end{proof}

\noindent
MATERIAL ABOVE IS COVERED ON MIDTERM 1 SPRING 2020

\subsection{Bilinear multiplications}

\begin{defn}
Let $U$, $V$ and $W$ be sets and let $*:U\times V\to W$.\\
Then, $\forall x\in U$, $\forall y\in V$, we denote $*(x,y)$ by $x*y$.\\
Also, $\forall x\in U$, the function $x*\bullet:V\to W$ is defined by:\\
\phx\qquad\qquad\qquad $\forall y\in V$, \quad $(x*\bullet)_y=x*y$.\\
Also, $\forall y\in V$, the function $\bullet*y:U\to W$ is defined by:\\
\phx\qquad\qquad\qquad $\forall x\in U$, \quad $(\bullet*y)_x=x*y$.
\end{defn}

\begin{defn}
Let $U,V,W\in\ES$. Then:\\
\phx\quad $\scrb^{UV}_W\quad:=\quad\{\quad*:U\times V\to W\,\,\,\,|\,\,\,\,(\forall x\in U,\,\,x*\bullet\in\scrl^V_W)\qquad\&$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\,(\forall y\in V,\,\,\bullet*y\in\scrl^U_W)\qquad\qquad\}$.
\end{defn}

\begin{thm}
Define $*:\R\times\R\to\R$ by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\forall a,b\in\R$, $a*b=a\cdot b$.\\
\phx\qquad\qquad\qquad\qquad\quad Then $*\in\scrb^{\R\R}_\R$.
\end{thm}

\begin{thm}
Let $V\in\ES$. Define $*:\R\times V\to V$ by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\forall a\in\R$, $\forall x\in V$, $a*x=a\cdot x$.\\
\phx\qquad\qquad\qquad\qquad\quad Then $*\in\scrb^{V\R}_V$.
\end{thm}

\begin{thm}
Let $V\in\ES$. Define $*:V\times V\to\R$ by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $\forall x,y\in V$, $x*y=x\sbullet y$.\\
\phx\qquad\qquad\qquad\qquad\quad Then $*\in\scrb^{VV}_\R$.
\end{thm}

\begin{thm}
Let $U:=\R^3$, $V:=\R^4$, $W:=\R$ and\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\ds{A:=\left[\begin{matrix}0&2\pi&4&6\\-1&3&-5&\sqrt{7}\\-2&7&8&9\end{matrix}\right]}$.\\
Define $*:U\times V\to W$ by:\\
\phx\qquad $\forall x\in U$, $\forall y\in V$, \quad $\ds{x*y\,\,=\,\,\sum_{j=1}^3\,\sum_{k=1}^4\,x_j\cdot A_{jk}\cdot y_k}$.\\
Then $*\in\scrb^{UV}_W$.
\end{thm}

Continuing in the theme of the last theorem, there is a map\\
\phx\qquad\qquad $\R^{3\times4}\,\,\to\,\,\scrb^{\R^3,\R^4}_\R$.\\
It is not hard to show that this map is\\
\phx\quad\qquad\qquad\quad algebraically linear \quad and \quad bijective onto $\scrb^{\R^3,\R^4}_\R$.\\
More generally, $\forall p,q\in\N$, there is an algebraically linear bijection\\
\phx\qquad\qquad $\R^{p\times q}\,\,\hookrightarrow>\,\,\scrb^{\R^p,\R^q}_\R$.\\
Also, $\forall p,q,m\in\N$, there is an algebraically linear bijection\\
\phx\qquad\qquad $\R^{p\times q\times m}\,\,\hookrightarrow>\,\,\scrb^{\R^p,\R^q}_{\R^m}$.\\

In this course, we will content ourselves with quadratic approximations (the Second Order Taylor Theorem),
which means that we will use many bilinar maps.
Were we to focus on cubic approximations (Third Order Taylor), we would need to develop
the bookkeeping of trilinear maps.
Quartic and Higher order polynomial approximations would require quadrilinear and higher order multilinear maps.

Just as we showed that linear maps are Lipschitz (\tref{thm-lin-lip}),\\
we now show that bilinear maps are ``doubly Lipschitz'':

\begin{thm}\wrlab{thm-bilin-dbllip}
Let $U,V,W\in\ES$ and let $*\in\scrb^{UV}_W$.\\
Then: $\exists L\ge0$ s.t., $\forall x\in U$, $\forall y\in V$, \quad $|x*y|\le L\cdot|x|\cdot|y|$.
\end{thm}

\begin{proof}
Let $\,$ $\ell:=\#\scri_U$, $\,$ $m:=\#\scri_V$, $\,\,\,$ $J:=\scri_V$, $\,$ $K:=\scri_W$.\\
Let $A\,:=\,\max\,\{\,|\vep_j^U*\vep_k^V|~\hbox{s.t.}~j\in J,\,k\in K,\}$.\\
Let $L:=A\cdot\sqrt{\ell}\cdot\sqrt{m}$. \qquad Then $L\ge0$.\\
\phx\qquad Want: $\forall x\in U$, $\forall y\in V$, \quad $\,\,|x*y|\le L\cdot|x|\cdot|y|$.\\
Given $x\in U$, $y\in V$. \qquad Want: $|x*y|\le L\cdot|x|\cdot|y|$.\\
By the reproducing formula, $\ds{x=\sum_{j\in J}\,x_j\vep_j^U}$ and $\ds{y=\sum_{k\in K}\,y_k\vep_k^V}$.\\
Then: $\,\,\ds{x*y=\left(\sum_{j\in J}\,x_j\vep_j^U\right)*\left(\sum_{j\in J}\,y_k\vep_k^V\right)}$.\\
By bilinearity, $\ds{\left(\sum_{j\in J}\,x_j\vep_j^U\right)*\left(\sum_{j\in J}\,y_k\vep_k^V\right)=\sum_{j\in J}\,\sum_{k\in K}\,\left[x_j\cdot y_k\cdot(\vep_j^U*\vep_k^V)\right]}$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: \qquad $\ds{\,x*y=\sum_{j\in J}\,\sum_{k\in K}\,\left[x_j\cdot y_k\cdot(\vep_j^U*\vep_k^V)\right]}$.\\
Then, by subadditivity, $\ds{|x*y|\le\sum_{j\in J}\,\sum_{k\in K}\,|x_j\cdot y_k\cdot(\vep_j^U*\vep_k^V)|}$.\\
Then, by absolute homogeneity, $\ds{|x*y|\le\sum_{j\in J}\,\sum_{k\in K}\,(\,|x_j|\cdot|y_k|\cdot|\vep_j^U*\vep_k^V|\,)}$.\\
Then, by choice of $A$, $\ds{|x*y|\le\sum_{j\in J}\,\sum_{k\in K}\,(\,|x_j|\cdot|y_k|\cdot A\,)}$.\\
By the distributive law, $\ds{\left(\sum_{j\in J}\,|x_j|\right)\cdot\left(\sum_{k\in K}\,|y_k|\right)=\sum_{j\in J}\,\sum_{k\in K}\,(\,|x_j|\cdot|y_k|\,)}$.\\
So, by commutativity of multiplication and the distributive law,\\
\phx\qquad\qquad $\ds{A\cdot\left(\sum_{j\in J}\,|x_j|\right)\cdot\left(\sum_{k\in K}\,|y_k|\right)=\sum_{j\in J}\,\sum_{k\in K}\,(\,|x_j|\cdot|y_k|\cdot A\,)}$.\\
So, since $\ds{\|x\|=\sum_{j\in J}|x_j|}$ and $\ds{\|y\|=\sum_{k\in K}|y_k|}$, we get\\
\phx\qquad\qquad\qquad\qquad\qquad $\ds{\,\,\,A\cdot\|x\|\cdot\|y\|=\sum_{j\in J}\,\sum_{k\in K}\,(\,|x_j|\cdot|y_k|\cdot A\,)}$.\\
By \tref{thm-1-2-norms-comparable}, \qquad $\|x\|\le\sqrt{\ell}\cdot|x|$ \quad and \quad $\|y\|\le\sqrt{m}\cdot|y|$.\\
Then \quad $\ds{|x*y|\quad\le\quad\sum_{j\in J}\,\sum_{k\in K}\,(\,|x_j|\cdot|y_k|\cdot A\,)}$\\
\phx\qquad\qquad\qquad\quad $=\quad A\cdot\|x\|\cdot\|y\|\quad\le\quad A\cdot\sqrt{\ell}\cdot|x|\cdot\sqrt{m}\cdot|y|$\\
\phx\qquad\qquad\qquad\quad $=\quad L\cdot|x|\cdot|y|$, \qquad as desired.
\end{proof}

\begin{defn}
Let $U,V,W\in\ES$ and let $*\in\scrb^{UV}_W$.\\
Let $f$ be a $U$-function \quad and \quad let $g$ be a $V$-function.\\
Then $f*g$ is the $W$-function defined by: \quad $\forall t$, $\,$ $(f*g)_t=f_t*g_t$.
\end{defn}

\begin{defn}
Let $U,V,W\in\ES$ and let $*\in\scrb^{UV}_W$.\\
Let $F$ be a set of $U$-functions \quad and \quad let $G$ be a set of $V$-functions.\\
Then \quad $F*G\,\,:=\,\,\{\,f*g\,|\,f\in F,\,g\in G\,\}$.
\end{defn}

\begin{defn}
Let $U,V,W\in\ES$ and let $*\in\scrb^{UV}_W$.\\
Let $f$ be a $U$-function \quad and \quad let $G$ be a set of $V$-functions.\\
Then \quad $f*G\,\,:=\,\,\{\,f*g\,|\,g\in G\,\}$.
\end{defn}

\begin{defn}
Let $U,V,W\in\ES$ and let $*\in\scrb^{UV}_W$.\\
Let $F$ be a set of $U$-functions \quad and \quad let $g$ be a $V$-function.\\
Then \quad $F*g\,\,:=\,\,\{\,f*g\,|\,f\in F\,\}$.
\end{defn}

\begin{defn}
Let $U,V,W\in\ES$ and let $*\in\scrb^{UV}_W$.\\
Let $X\subseteq U$-function \quad and \quad let $G$ be a set of $V$-functions.\\
Then \quad $X*G\,\,:=\,\,\{\,x*g\,|\,x\in X,\,g\in G\,\}$.
\end{defn}

\begin{defn}
Let $U,V,W\in\ES$ and let $*\in\scrb^{UV}_W$.\\
Let $x\in U$ \quad and \quad let $g$ be a $V$-function.\\
Then $x*g$ is the $W$-function defined by: \quad $\forall t$, $\,$ $(x*g)_t=x*g_t$.
\end{defn}

\begin{defn}
Let $U,V,W\in\ES$ and let $*\in\scrb^{UV}_W$.\\
Let $f$ be a $U$-function \quad and \quad let $y\in V$.\\
Then $f*y$ is the $W$-function defined by: \quad $\forall t$, $\,$ $(f*y)_t=f_t*y$.
\end{defn}

\begin{defn}
Let $U,V,W\in\ES$ and let $*\in\scrb^{UV}_W$.\\
Let $F$ be a set of $U$-functions \quad and \quad let $Y\subseteq V$.\\
Then \quad $F*Y\,\,:=\,\,\{\,f*y\,|\,f\in F,\,y\in Y\,\}$.
\end{defn}

\begin{defn}
Let $U,V,W\in\ES$ and let $*\in\scrb^{UV}_W$.\\
Let $X\subseteq U$ \quad and \quad let $Y\subseteq V$.\\
Then \quad $X*Y\,\,:=\,\,\{\,x*y\,|\,x\in X,\,y\in Y\,\}$.
\end{defn}

\begin{defn}
Let $U,V,W\in\ES$ and let $*\in\scrb^{UV}_W$.\\
Let $x\in U$ \quad and \quad let $Y\subseteq V$.\\
Then \quad $x*Y\,\,:=\,\,\{\,x*y\,|\,y\in Y\,\}$.
\end{defn}

\begin{defn}
Let $U,V,W\in\ES$ and let $*\in\scrb^{UV}_W$.\\
Let $X\subseteq U$ \quad and \quad let $y\in V$.\\
Then \quad $X*y\,\,:=\,\,\{\,x*y\,|\,x\in X\,\}$.
\end{defn}

\begin{thm}\wrlab{thm-prod-monoms-is-monom}
Let $S,U,V,W\in\ES$, $k,\ell\in\N_0$ and $*\in\scrb^{UV}_W$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $\scrm_k^{SU}\,*\,\scrm_\ell^{SV}\,\,\subseteq\,\,\scrm_{k+\ell}^{SW}$.
\end{thm}

\begin{proof}
Want: $\forall h\in\scrm_k^{SU}\,*\,\scrm_\ell^{SV}$, $h\in\scrm_{k+\ell}^{SW}$.\\
Given $h\in\scrm_k^{SU}\,*\,\scrm_\ell^{SV}$. \qquad Want: $h\in\scrm_{k+\ell}^{SW}$.\\
Since $h\in\scrm_k^{SU}\,*\,\scrm_\ell^{SV}$, choose $f\in\scrm_k^{SU}$ and $g\in\scrm_\ell^{SV}$ s.t. $h=f*g$.\\
As $f\in\scrm_k^{SU}$, choose $C\in\scrc^S_U$ and $\alpha_1,\ldots,\alpha_k\in\Pi_S$ s.t.~$f=C\cdot\alpha_1\cdots\alpha_k$.\\
As $g\in\scrm_\ell^{SV}$, choose $D\in\scrc^S_V$ and $\beta_1,\ldots,\beta_\ell\in\Pi_S$ s.t.~$g=D\cdot\beta_1\cdots\beta_\ell$.\\
By bilinearity of $*$, as $\alpha_1,\ldots,\alpha_k,\beta_1,\ldots,\beta_\ell$ are all functionals, we get:\\
\phx\qquad $\forall t\in S$, \qquad $(f*g)_t\,\,=\,\,((C*D)\cdot\alpha_1\cdots\alpha_k\cdot\beta_1\cdots\beta_\ell)_t$.\\
Then $f*g\,\,=\,\,(C*D)\cdot\alpha_1\cdots\alpha_k\cdot\beta_1\cdots\beta_\ell$.\\
Since $C\in\scrc^S_U$ and $D\in\scrc^S_V$ and $*:U\times V\to W$, we conclude: $C*D\in\scrc^S_W$.\\
So, since $\alpha_1,\ldots,\alpha_k,\beta_1,\ldots,\beta_k\in\Pi_S$, we see that:\\
\phx\qquad $(C*D)\cdot\alpha_1\cdots\alpha_k\cdot\beta_1\cdots\beta_\ell\quad\in\quad\scrm_{k+\ell}^{SW}$.\\
Then: $h\,=\,f*g\,=\,(C*D)\cdot\alpha_1\cdots\alpha_k\cdot\beta_1\cdots\beta_\ell\,\in\,\scrm_{k+\ell}^{SW}$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-prod-polys-is-poly}
Let $S,U,V,W\in\ES$, $k,\ell\in\N_0$ and $*\in\scrb^{UV}_W$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $\scrh_k^{SU}\,*\,\scrh_\ell^{SV}\,\,\subseteq\,\,\scrh_{k+\ell}^{SW}$.
\end{thm}

\begin{proof}
Want: $\forall h\in\scrh_k^{SU}\,*\,\scrh_\ell^{SV}$, $h\in\scrh_{k+\ell}^{SW}$.\\
Given $h\in\scrh_k^{SU}\,*\,\scrh_\ell^{SV}$. \qquad Want: $h\in\scrh_{k+\ell}^{SW}$.\\
Since $h\in\scrh_k^{SU}\,*\,\scrh_\ell^{SV}$, choose $f\in\scrh_k^{SU}$ and $g\in\scrh_\ell^{SV}$ s.t. $h=f*g$.\\
As $f\in\scrm_k^{SU}$, choose $m\in\N$, $\phi_1,\ldots,\phi_m\in\scrm_k^{SU}$ s.t.~$f=\phi_1+\cdots+\phi_m$.\\
As $g\in\scrm_\ell^{SV}$, choose $n\in\N$, $\psi_1,\ldots,\psi_n\in\scrm_k^{SV}$ s.t.~$g=\psi_1+\cdots+\psi_n$.\\
By bilinearity, we have: \quad $\ds{\left(\sum_{j=1}^m\,\phi_j\right)*\left(\sum_{k=1}^n\,\psi_k\right)=\sum_{j=1}^m\,\sum_{k=1}^n\,(\phi_j*\psi_k)}$,\\
Then: $\ds{h=f*g=\left(\sum_{j=1}^m\,\phi_j\right)*\left(\sum_{k=1}^n\,\psi_k\right)=\sum_{j=1}^m\,\sum_{k=1}^n\,(\phi_j*\psi_k)}$.\\
It therefore suffices to show: $\forall j\in[1..m]$, $\forall k\in[1..n]$, $\phi_j*\psi_k\,\,\in\,\,\scrm_{k+\ell}^{SW}$.\\
Given $j\in[1..m]$, $k\in[1..n]$. \qquad\qquad\qquad\qquad Want: $\phi_j*\psi_k\,\,\in\,\,\scrm_{k+\ell}^{SW}$.\\
By \tref{thm-prod-monoms-is-monom}, \quad $\scrm_k^{SU}\,*\,\scrm_\ell^{SV}\,\,\subseteq\,\,\scrm_{k+\ell}^{SW}$.\\
Then: \qquad $\,\,\phi_j*\psi_k\,\,\in\,\,\scrm_k^{SU}\,*\,\scrm_\ell^{SV}\,\,\subseteq\,\,\scrm_{k+\ell}^{SW}$, \qquad as desired.
\end{proof}

\begin{thm}
Let $S,U\in\ES$, $k,\ell\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $\scrh_k^{SU}\,\sbullet\,\scrh_\ell^{SU}\,\,\subseteq\,\,\scrh_{k+\ell}^{S\R}$.
\end{thm}

\begin{proof}
Define $*\in\scrb^{UU}_\R$ by: \quad $\forall x,y\in U$, \quad $x*y=x\sbullet y$.\\
Let $V:=U$ and $W:=\R$. \qquad Then $*\in\scrb^{UV}_W$.\\
By \tref{thm-prod-polys-is-poly}, we have: \quad $\scrh_k^{SU}\,*\,\scrh_\ell^{SV}\,\,\subseteq\,\,\scrh_{k+\ell}^{SW}$.\\
Then $\scrh_k^{SU}\,\sbullet\,\scrh_\ell^{SV}\,\,=\,\,\scrh_k^{SU}\,*\,\scrh_\ell^{SV}\,\,\subseteq\,\,\scrh_{k+\ell}^{SW}\,\,=\,\,\scrh_{k+\ell}^{S\R}$, as desired.
\end{proof}

\begin{thm}
Let $S\in\ES$, $k,\ell\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $\scrh_k^{S\R}\,\cdot\,\scrh_\ell^{S\R}\,\,\subseteq\,\,\scrh_{k+\ell}^{S\R}$.
\end{thm}

\begin{proof}
Define $*\in\scrb^{\R\R}_\R$ by: \quad $\forall a,b\in\R$, \quad $a*b=a\cdot b$.\\
Let $V:=\R$ and $W:=\R$. \qquad Then $*\in\scrb^{UV}_W$.\\
By \tref{thm-prod-polys-is-poly}, we have: \quad $\scrh_k^{SU}\,*\,\scrh_\ell^{SV}\,\,\subseteq\,\,\scrh_{k+\ell}^{SW}$.\\
Then $\scrh_k^{S\R}\,\cdot\,\scrh_\ell^{S\R}\,\,=\,\,\scrh_k^{SU}\,*\,\scrh_\ell^{SU}\,\,\subseteq\,\,\scrh_{k+\ell}^{SW}\,\,=\,\,\scrh_{k+\ell}^{W\R}$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-prod-realvalued-polys-is-realvalued-poly}
Let $S\in\ES$, $k,\ell\in\N_0$. \quad Then: $\forall m\in\N$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\forall f_1,\ldots,f_m\in\scrh_k^{S\R}$, \quad $f_1\cdots f_m\in\scrh_{km}^{S\R}$.
\end{thm}

\begin{proof}
Unassigned HW. Hint: Use induction, starting with:\\
\phx\qquad\qquad Let $A:=\{m\in\N\,|\,\forall f_1,\ldots,f_m\in\scrh_k^{S\R},\,f_1\cdots f_m\in\scrh_{km}^{S\R}\}$.\\
\phx\qquad\qquad\qquad\qquad\qquad Want: $A=\N$.
\end{proof}

\begin{thm}\wrlab{thm-prod-vectorpoly-scalarpoly-is-vectorpoly}
Let $S,U\in\ES$, $k,\ell\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $\scrh_k^{SU}\,\cdot\,\scrh_\ell^{S\R}\,\,\subseteq\,\,\scrh_{k+\ell}^{SU}$.
\end{thm}

\begin{proof}
Define $*\in\scrb^{U\R}_U$ by: \quad $\forall x\in U$, $\forall a\in\R$, \quad $x*a=x\cdot a$.\\
Let $V:=\R$ and $W:=U$. \qquad Then $*\in\scrb^{UV}_W$.\\
By \tref{thm-prod-polys-is-poly}, we have: \quad $\scrh_k^{SU}\,*\,\scrh_\ell^{SV}\,\,\subseteq\,\,\scrh_{k+\ell}^{SW}$.\\
Then $\scrh_k^{SU}\,\cdot\,\scrh_\ell^{S\R}\,\,=\,\,\scrh_k^{SU}\,*\,\scrh_\ell^{SV}\,\,\subseteq\,\,\scrh_{k+\ell}^{SW}\,\,=\,\,\scrh_{k+\ell}^{SU}$, as desired.
\end{proof}

\begin{thm}\wrlab{thm-monom-then-lin-is-monom}
Let $V,W,X\in\ES$, $k\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $\scrl^W_X\,\circ\,\scrm_k^{VW}\,\,\subseteq\,\,\scrm_k^{VX}$.
\end{thm}

\begin{proof}
Want: $\forall h\in\scrl^W_X\,\circ\,\scrm_k^{VW}$, $h\in\scrm_k^{VX}$.\\
Given $h\in\scrl^W_X\,\circ\,\scrm_k^{VW}$. \qquad Want: $h\in\scrm_k^{VX}$.\\
Since $h\in\scrl^W_X\,\circ\,\scrm_k^{VW}$, choose $g\in\scrl^W_X$ and $f\in\scrm_k^{VW}$ s.t. $h=g\circ f$.\\
As $f\in\scrm_k^{VW}$, choose $C\in\scrc^V_W$ and $\alpha_1,\ldots,\alpha_k\in\Pi_V$ s.t.~$f=C\cdot\alpha_1\cdots\alpha_k$.\\
Since $g\in\scrl^W_X$, we know that $g$ is algebrically linear.\\
Then: \qquad $\,\forall t\in S$, \qquad $(g\circ f)_t\,\,=\,\,((g\circ C)\cdot\alpha_1\cdots\alpha_k)_t$.\\
\phx\qquad\qquad Then: \qquad\qquad $g\circ f\,\,=\,\,(g\circ C)\cdot\alpha_1\cdots\alpha_k$.\\
Since $C\in\scrc^V_W$ and $g:W\to X$, we get: $g\circ C\in\scrc^V_X$.\\
So, since $\alpha_1,\ldots,\alpha_k\in\Pi_V$, we get: $(g\circ C)\cdot\alpha_1\cdots\alpha_k\,\,\in\,\,\scrm_k^{VX}$.\\
Then \quad $h\,\,=\,\,g\circ f\,\,=\,\,(g\circ C)\cdot\alpha_1\cdots\alpha_k\,\,\in\,\,\scrm_k^{VX}$, \quad as desired.
\end{proof}

\begin{thm}\wrlab{thm-poly-then-lin-is-poly}
Let $V,W,X\in\ES$, $k\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $\scrl^W_X\,\circ\,\scrh_k^{VW}\,\,\subseteq\,\,\scrh_k^{VX}$.
\end{thm}

\begin{proof}
Want: $\forall h\in\scrl^W_X\,\circ\,\scrh_k^{VW}$, $h\in\scrh_k^{VX}$.\\
Given $h\in\scrl^W_X\,\circ\,\scrh_k^{VW}$. \qquad Want: $h\in\scrh_k^{VX}$.\\
Since $h\in\scrl^W_X\,\circ\,\scrh_k^{VW}$, choose $g\in\scrl^W_X$ and $f\in\scrh_k^{VW}$ s.t. $h=g\circ f$.\\
As $f\in\scrh_k^{VW}$, choose $m\in\N$, $\phi_1,\ldots,\phi_m\in\scrm_k^{VW}$ s.t.~$f=\phi_1+\cdots+\phi_m$.\\
Since $g\in\scrl^W_X$, we know that $g$ is algebrically linear.\\
Then: \qquad $\,\forall t\in S$, \qquad $(g\circ f)_t\,\,=\,\,((g\circ\phi_1)+\cdots+(g\circ\phi_m))_t$.\\
\phx\qquad\qquad Then: \qquad\qquad $g\circ f\,\,=\,\,(g\circ\phi_1)+\cdots+(g\circ\phi_m)$.\\
By \tref{thm-monom-then-lin-is-monom}, $\scrl^W_X\,\circ\,\scrm_k^{VW}\,\,\subseteq\,\,\scrm_k^{VX}$.\\
Then, \quad $\forall j\in[1..m]$, \quad $g\circ\phi_j\,\,\in\,\,\scrl^W_X\,\circ\,\scrm_k^{VW}\,\,\subseteq\,\,\scrm_k^{VX}$.\\
Then: \quad $(g\circ\phi_1)+\cdots+(g\circ\phi_m)\,\,\in\,\,\scrh_k^{VX}$.\\
Then \quad $h\,\,=\,\,g\circ f\,\,=\,\,(g\circ\phi_1)+\cdots+(g\circ\phi_m)\,\,\in\,\,\scrh_k^{VX}$, \quad as desired.
\end{proof}

\begin{thm}\wrlab{thm-poly-then-monom-is-poly}
Let $V,W,X\in\ES$, $k\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $\scrm_\ell^{WX}\,\circ\,\scrh_k^{VW}\,\,\subseteq\,\,\scrh_{k\ell}^{VX}$.
\end{thm}

\begin{proof}
Want: $\forall h\in\scrm_\ell^{WX}\,\circ\,\scrh_k^{VW}$, $h\in\scrh_k^{VX}$.\\
Given $h\in\scrm_\ell^{WX}\,\circ\,\scrh_k^{VW}$. \qquad Want: $h\in\scrh_k^{VX}$.\\
Since $h\in\scrm_\ell^{WX}\,\circ\,\scrh_k^{VW}$, choose $g\in\scrm_\ell^{WX}$ and $f\in\scrh_k^{VW}$ s.t. $h=g\circ f$.\\
As $g\in\scrm_\ell^{WX}$, choose $C\in\scrc^W_X$ and $\alpha_1,\ldots,\alpha_\ell\in\Pi_W$ s.t.~$g=C\cdot\alpha_1\cdots\alpha_\ell$.\\
Then: \qquad $\,\forall t\in S$, \qquad $(g\circ f)_t\,\,=\,\,((C\circ f)\cdot(\alpha_1\circ f)\cdots(\alpha_\ell\circ f))_t$.\\
\phx\qquad\qquad Then: \qquad\qquad $g\circ f\,\,=\,\,(C\circ f)\cdot(\alpha_1\circ f)\cdots(\alpha_\ell\circ f)$.\\
Let $\beta:=(\alpha_1\circ f)\cdots(\alpha_\ell\circ f)$. \quad Then $g\circ f\,\,=\,\,(C\circ f)\cdot\beta$.\\
Since $f\in\scrh_k^{VW}$, we conclude that $f:V\to W$.\\
So, since  $C\in\scrc^W_X$, we get: $C\circ f\in\scrc^V_X$. \qquad Then $C\circ f\in\scrc^V_X=\scrh_0^{VX}$.\\
We have: \quad $\alpha_1\,,\,\ldots\,,\,\alpha_\ell\,\,\in\,\,\Pi_W\,\,\subseteq\,\,\scrl^W_\R$.\\
By \tref{thm-poly-then-lin-is-poly}, we have: $\scrl^W_\R\,\circ\,\scrh_k^{VW}\,\,\subseteq\,\,\scrh_k^{V\R}$.\\
Then: \quad $\alpha_1\circ f,\ldots,\alpha_\ell\circ f\,\,\in\,\,\scrl^W_\R\,\circ\,\scrh_k^{VW}\,\,\subseteq\,\,\scrh_k^{V\R}$.\\
Then, by \tref{thm-prod-realvalued-polys-is-realvalued-poly}, $(\alpha_1\circ f)\cdots(\alpha_\ell\circ f)\,\in\,\scrh_{k\ell}^{V\R}$.\\
\phx\qquad Then: \qquad\qquad $\,\beta=(\alpha_1\circ f)\cdots(\alpha_\ell\circ f)\,\in\,\scrh_{k\ell}^{V\R}$.\\
By \tref{thm-prod-vectorpoly-scalarpoly-is-vectorpoly}, we have: \quad $\scrh_0^{VX}\cdot\scrh_{k\ell}^{V\R}\,\subseteq\,\scrh_{k\ell}^{VX}$.\\
Then: \quad $h=g\circ f=(C\circ f)\cdot\beta\in\scrh_0^{VX}\cdot\scrh_{k\ell}^{V\R}\,\subseteq\,\scrh_{k\ell}^{VX}$, \quad as desired.
\end{proof}

\begin{thm}\wrlab{thm-poly-then-poly-is-poly}
Let $V,W,X\in\ES$, $k,\ell\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $\scrh_\ell^{WX}\,\circ\,\scrh_k^{VW}\,\,\subseteq\,\,\scrh_{k\ell}^{VX}$.
\end{thm}

\begin{proof}
Want: $\forall h\in\scrh_\ell^{WX}\,\circ\,\scrh_k^{VW}$, $h\in\scrh_k^{VX}$.\\
Given $h\in\scrh_\ell^{WX}\,\circ\,\scrh_k^{VW}$. \qquad Want: $h\in\scrh_k^{VX}$.\\
Since $h\in\scrh_\ell^{WX}\,\circ\,\scrh_k^{VW}$, choose $g\in\scrh_\ell^{WX}$ and $f\in\scrh_k^{VW}$ s.t. $h=g\circ f$.\\
As $g\in\scrh_\ell^{WX}$, choose $m\in\N$, $\psi_1,\ldots,\psi_m\in\scrm_\ell^{WX}$ s.t.~$g=\psi_1+\cdots+\psi_m$.\\
Then: \qquad $\,\forall t\in S$, \qquad $(g\circ f)_t\,\,=\,\,((\psi_1\circ f)+\cdots+(\psi_m\circ f))_t$.\\
\phx\qquad\qquad Then: \qquad\qquad $g\circ f\,\,=\,\,(\psi_1\circ f)+\cdots+(\psi_m\circ f)$.\\
By \tref{thm-poly-then-monom-is-poly}, we have: $\scrm_\ell^{WX}\,\circ\,\scrh_k^{VW}\,\,\subseteq\,\,\scrh_{k\ell}^{VX}$.\\
Then: $\psi_1\circ f\,,\,\ldots\,,\,\psi_m\circ f\,\,\in\,\,\scrm_\ell^{WX}\,\circ\,\scrh_k^{VW}\,\,\subseteq\,\,\scrh_{k\ell}^{VX}$.\\
So, since $\scrh_{k\ell}^{VX}$ is v.op.-closed, $(\psi_1\circ f)+\cdots+(\psi_m\circ f)\,\,\in\,\,\scrh_{k\ell}^{VX}$.\\
Then $h\,\,=\,\,g\circ f\,\,=\,\,(\psi_1\circ f)+\cdots+(\psi_m\circ f)\,\,\in\,\,\scrh_{k\ell}^{VX}$, as desired.
\end{proof}

\subsection{Continuity of polynomials}

\begin{thm}\wrlab{thm-continuity-of-sum}
Let $V,W\in\ES$, $f,g:V\drta W$, $t\in V$.\\
Assume: $f$ and $g$ are both continuous at $t$ from $V$ to $W$.\\
Then $f+g$ is continuous at $t$ from $V$ to $W$.
\end{thm}

\begin{proof}
Let $h:=f+g$. \quad Want: $h$ is continuous at $t$ from $V$ to $W$.\\
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\D_h$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,d(x,t)<\delta\,\,]\,\,\Rightarrow\,\,[\,\,d(h_x,h_t)<\vep\,\,]$.\\
Given $\vep>0$. \quad Want: $\exists\delta>0$ s.t., $\forall x\in\D_h$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,d(x,t)<\delta\,\,]\,\,\Rightarrow\,\,[\,\,d(h_x,h_t)<\vep\,\,]$.\\
Let $\sigma:=\vep/2$. \qquad Then $\sigma>0$.\\
So, as $f$ is continuous at $t$ from $V$ to $W$, choose $\lambda>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,d(x,t)<\lambda\,\,]\,\,\Rightarrow\,\,[\,\,d(f_x,f_t)<\sigma\,\,]$.\\
Also, as $g$ is continuous at $t$ from $V$ to $W$, choose $\mu>0$ s.t., $\forall x\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,d(x,t)<\mu\,\,]\,\,\Rightarrow\,\,[\,\,d(g_x,g_t)<\sigma\,\,]$.\\
Let $\delta:=\min\{\lambda,\mu\}$. \quad Then $\delta>0$.\\
Want: $\forall x\in\D_h$, $\,$ \qquad\qquad $[\,\,d(x,t)<\delta\,\,]\,\,\Rightarrow\,\,[\,\,d(h_x,h_t)<\vep\,\,]$.\\
Given $x\in\D_h$. \qquad Want: $[\,\,d(x,t)<\delta\,\,]\,\,\Rightarrow\,\,[\,\,d(h_x,h_t)<\vep\,\,]$.\\
Assume: $d(x,t)<\delta$. \qquad Want: $d(h_x,h_t)<\vep$.\\
Since $x\in\D_h=\D_{f+g}=\D_f\bigcap\D_g$, we get: $x\in\D_f$ and $x\in\D_g$.\\
So, since $d(x,t)<\delta\le\lambda$, by choice of $\lambda$, we have: $d(f_x,f_t)<\sigma$.\\
Also, since $d(x,t)<\delta\le\mu$, by choice of $\mu$, we have: $d(g_x,g_t)<\sigma$.\\
Since $h=f+g$, we get both $h_x=f_x+g_x$ and $h_t=f_t+g_t$.\\
Since $\sigma=\vep/2$, we get $\sigma+\sigma=\vep$.\\
Then: \quad $d(h_x,h_t)\,\,=\,\,d(f_x+g_x,f_t+g_t)\,\,=\,\,|\,(f_x+g_x)\,-\,(f_t+g_t)\,|$\\
\phx\qquad\qquad\quad $=\,\,|\,(f_x-f_t)\,+\,(g_x-g_t)\,|\,\,\le\,\,|\,f_x\,-\,f_t\,|+|\,g_x\,-\,g_t\,|$\\
\phx\qquad\qquad\quad $=\,\,[\,d(f_x,f_t)\,]\,+\,[\,d(g_x,g_t)\,]\,\,\le\,\,\sigma+\sigma\,\,=\vep$.
\end{proof}

\begin{thm}\wrlab{thm-continuity-of-bigsums}
Let $V,W\in\ES$, $t\in V$. Then, $\forall\ell\in\N$,\\
\phx\quad $\forall\alpha_1,\ldots,\alpha_\ell:V\drta W$,\\
\phx\qquad\qquad\qquad\quad $\,\,[\,\alpha_1,\ldots,\alpha_\ell\hbox{ are all continuous at }t\hbox{ from }V\hbox{ to }W\,]$\\
\phx\qquad\qquad $\Rightarrow\qquad[\,\alpha_1+\cdots+\alpha_\ell\hbox{ is continuous at }t\hbox{ from }V\hbox{ to }W\,]$.
\end{thm}

\begin{proof}
Unassigned HW. Hint: Use induction, starting with:\\
\phx\quad Let $A:=\{\ell\in\N\,|\,\forall\alpha_1,\ldots,\alpha_\ell:V\drta W$,\\
\phx\qquad\qquad\qquad\quad $\,\,[\,\alpha_1,\ldots,\alpha_\ell\hbox{ are all continuous at }t\hbox{ from }V\hbox{ to }W\,]$\\
\phx\qquad\qquad $\Rightarrow\qquad[\,\alpha_1+\cdots+\alpha_\ell\hbox{ is continuous at }t\hbox{ from }V\hbox{ to }W\,]\quad\}$.\\
Want: $A=\N$.
\end{proof}

\begin{thm}\wrlab{thm-continuity-of-products}
Let $S,U,V,W\in\ES$, let $*\in\scrb^{UV}_W$ and let $t\in S$.\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad Let $f:S\drta U$ and let $g:S\drta V$.\\
\phx\qquad\qquad\qquad Assume that $f$ is continuous at $t$ from $S$ to $U$.\\
\phx\qquad\qquad\qquad Assume that $g$ is continuous at $t$ from $S$ to $V$.\\
\phx\qquad\qquad\qquad Then $f*g$ is continuous at $t$ from $S$ to $W$.
\end{thm}

\begin{proof}
Let $h:=f*g$. \quad Want: $h$ is continuous at $t$ from $S$ to $W$.\\
Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall x\in\D_h$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,d(x,t)<\delta\,\,]\,\,\Rightarrow\,\,[\,\,d(h_x,h_t)<\vep\,\,]$.\\
Given $\vep>0$. \quad Want: $\exists\delta>0$ s.t., $\forall x\in\D_h$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,d(x,t)<\delta\,\,]\,\,\Rightarrow\,\,[\,\,d(h_x,h_t)<\vep\,\,]$.\\
Choose $L\ge0$ s.t., \qquad $\forall p\in U$, $\forall q\in V$, \quad $|p*q|\le L\cdot|p|\cdot|q|$.\\
Let $a:=f_t$ and $b:=g_t$ and $\ds{\sigma:=\min\left\{\,\,1\,\,,\,\,\frac{\vep}{(L+1)\cdot(|b|+|a|+1)}\,\,\right\}}$.\\
Then: \qquad $\sigma\le1$ \quad and \quad $L\cdot\sigma\cdot(|b|+|a|+1)<\vep$.\\
Also, we have: \qquad $\sigma>0$.\\
So, as $f$ is continuous at $t$ from $S$ to $V$, choose $\lambda>0$ s.t., $\forall x\in\D_f$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,d(x,t)<\lambda\,\,]\,\,\Rightarrow\,\,[\,\,d(f_x,f_t)<\sigma\,\,]$.\\
Also, as $g$ is continuous at $t$ from $S$ to $W$, choose $\mu>0$ s.t., $\forall x\in\D_g$,\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $[\,\,d(x,t)<\mu\,\,]\,\,\Rightarrow\,\,[\,\,d(g_x,g_t)<\sigma\,\,]$.\\
Let $\delta:=\min\{\lambda,\mu\}$. \quad Then $\delta>0$.\\
Want: $\forall x\in\D_h$, $\,$ \qquad\qquad $[\,\,d(x,t)<\delta\,\,]\,\,\Rightarrow\,\,[\,\,d(h_x,h_t)<\vep\,\,]$.\\
Given $x\in\D_h$. \qquad Want: $[\,\,d(x,t)<\delta\,\,]\,\,\Rightarrow\,\,[\,\,d(h_x,h_t)<\vep\,\,]$.\\
Assume: $d(x,t)<\delta$. \qquad Want: $d(h_x,h_t)<\vep$.\\
Since $x\in\D_h=\D_{f*g}=\D_f\bigcap\D_g$, we get: $x\in\D_f$ and $x\in\D_g$.\\
So, since $d(x,t)<\delta\le\lambda$, by choice of $\lambda$, we have: $d(f_x,f_t)<\sigma$.\\
Also, since $d(x,t)<\delta\le\mu$, by choice of $\mu$, we have: $d(g_x,g_t)<\sigma$.\\
Recall that $a=f_t$ and $b=g_t$. \quad Then $h_t=(f*g)_t=f_t*g_t=a*b$.\\
Let $A:=f_x$ and $B:=g_x$. \quad Then $h_x=(f*g)_x=f_x*g_x=A*B$.\\
Also, we have: \quad $|A-a|=|f_x-f_t|=d(f_x,f_t)<\sigma$.\\
Also, we have: \quad $|B-b|=|g_x-g_t|=d(g_x,g_t)<\sigma$.\\
By bilinearity, we have:\\
\phx\quad $[A*B]\,-\,[a*b]\,\,=\,\,[\,(A-a)*b\,]\,+\,[\,a*(B-b)\,]\,+\,[\,(A-a)*(B-b)\,]$.\\
Recall: \qquad $\sigma\,\,\le\,\,1$ \quad and \quad $L\cdot\sigma\cdot(\,|\,b\,|\,+\,|\,a\,|\,+\,1\,)\,\,<\,\,\vep$.\\
Then:\\
\phx$\,$ $d(h_x,h_t)\,\,=\,\,d(A*B,a*b)\,\,=\,\,|\,[A*B]\,-\,[a*b]\,|=$\\
\phx\quad\quad $=\,\,|\,[\,(A-a)*b\,]\,\,\,\,\,+\,\,\,[\,a*(B-b)\,]\,\,\,\,\,\,\,+\,\,\,[\,(A-a)*(B-b)\,]\,|$\\
\phx\quad\quad $\le\,\,\,\,|\,(A-a)*b\,|\,\,\,\,\,\,+\,\,\,\,|\,a*(B-b)\,|\,\,\,\,\,\,\,+\,\,\,|\,(A-a)*(B-b)\,|$\\
\phx\quad\quad $\le\,L\cdot|\,A-a\,|\cdot|\,b\,|\,+\,L\cdot|\,a\,|\cdot|\,B-b\,|\,\,+\,L\cdot|\,A-a\,|\cdot|\,B-b\,|$\\
\phx\quad\quad $\le\quad\,\,\,L\cdot\sigma\cdot|\,b\,|\,\,\,\,\quad+\quad\,\,\,L\cdot|\,a\,|\cdot\sigma\,\,\quad\,\,\,+\qquad\quad\,L\cdot\sigma\cdot\sigma$\\
\phx\quad\quad $=\,\,L\cdot\sigma\cdot(\,|\,b\,|\,+\,|\,a\,|\,+\,\sigma\,)$\\
\phx\quad\quad $\le\,\,L\cdot\sigma\cdot(\,|\,b\,|\,+\,|\,a\,|\,+\,1\,)\,\,\,\,<\,\,\,\,\vep$, \qquad as desired.
\end{proof}

\begin{thm}
Let $S,U\in\ES$, let $f,g:S\drta U$ and let $t\in S$.\\
\phx\qquad\quad Assume that $f$ and $g$ are both continuous at $t$ from $S$ to $U$.\\
\phx\qquad\quad Then $f\sbullet g$ is continuous at $t$ from $S$ to $\R$.
\end{thm}

To use \tref{thm-continuity-of-products} to prove the preceeding theorem,\\
\phx\qquad define $*\in\scrb^{UU}_\R$ by: $\forall x,y\in U$, $x*y=x\sbullet y$.

\begin{thm}
Let $S\in\ES$, let $f,g:S\drta\R$ and let $t\in S$.\\
\phx\qquad\quad Assume that $f$ and $g$ are both continuous at $t$ from $S$ to $\R$.\\
\phx\qquad\quad Then $f\cdot g$ is continuous at $t$ from $S$ to $\R$.
\end{thm}

To use \tref{thm-continuity-of-products} to prove the preceeding theorem,\\
\phx\qquad define $*\in\scrb^{\R\R}_\R$ by: $\forall a,b\in\R$, $a*b=a\cdot b$.

The preceding theorem can be used the prove the following theorem,
by induction on $k$.

\begin{thm}
Let $S\in\ES$ and let $t\in S$. \quad Then, $\forall k\in\N$,\\
\phx\quad $\forall\alpha_1,\ldots,\alpha_k:S\drta\R$,\\
\phx\qquad\qquad\qquad $\,(\,\,\alpha_1,\ldots,\alpha_k\hbox{ are all continuous at }t\hbox{ from }S\hbox{ to }\R\,\,)$\\
\phx\qquad\qquad $\Leftrightarrow\quad(\,\,\alpha_1\cdots\alpha_k\hbox{ is continuous at }t\hbox{ from }S\hbox{ to }\R\,\,)$.
\end{thm}

\begin{thm}
Let $S,U\in\ES$, $f:S\drta U$, $g:S\drta\R$, $t\in S$.\\
\phx\qquad\quad Assume that $f$ is continuous at $t$ from $S$ to $U$.\\
\phx\qquad\quad Assume that $g$ is continuous at $t$ from $S$ to $\R$.\\
\phx\qquad\quad Then $f\cdot g$ is continuous at $t$ from $S$ to $U$.
\end{thm}

To use \tref{thm-continuity-of-products} to prove the preceeding theorem,\\
\phx\qquad define $*\in\scrb^{U\R}_U$ by: $\forall x\in U$, $\forall a\in\R$, $x*a=x\cdot a$.

\begin{thm}
Let $V,W\in\ES$, $k\in\N_0$, $f\in\scrm_k^{VW}$.\\
\phx\qquad\qquad\qquad\qquad\quad Then $f$ is continuous from $V$ to $W$.
\end{thm}

\begin{proof}
Since $f\in\scrm_k^{VW}$, choose $\phi\in\scrc^V_W$ and $\alpha_1,\ldots,\alpha_k\in\Pi_V$\\
\phx\qquad\qquad\qquad such that: \qquad $f=\phi\cdot\alpha_1\cdots\alpha_k$.\\
Since $\alpha_1,\ldots,\alpha_k\in\Pi_V\subseteq\scrl^V_W$,\\
\phx\qquad we conclude that $\alpha_1,\ldots,\alpha_k$ are all Lipschitz from $V$ to $\R$,\\
\phx\qquad and so $\alpha_1,\ldots,\alpha_k$ are all continuous from $V$ to $\R$.\\
Then $\alpha_1\cdots\alpha_k$ is continuous from $V$ to $\R$.\\
Since $\phi\in\scrc^V_W$, we conclude that $\phi$ is Lipschitz-$0$ from $V$ to $W$,\\
\phx\qquad and so $\phi$ is continuous from $V$ to $W$.\\
Then $\phi\cdot\alpha_1\cdots\alpha_k$ is continuous from $V$ to $W$.\\
So, since $f=\phi\cdot\alpha_1\cdots\alpha_k$, we conclude that:\\
\phx\qquad $f$ is continuous from $V$ to $W$, \qquad as desired.
\end{proof}

\begin{thm}
Let $V,W\in\ES$, $k\in\N_0$, $f\in\scrh_k^{VW}$.\\
\phx\qquad\qquad\qquad\qquad\quad Then $f$ is continuous from $V$ to $W$.
\end{thm}

\begin{proof}
Since $f\in\scrh_k^{VW}$, choose $\ell\in\N$ and $\alpha_1,\ldots,\alpha_\ell\in\scrm_k^{VW}$\\
\phx\qquad\qquad\qquad such that: \qquad $f=\alpha_1+\cdots+\alpha_k$.\\
By the preceding theorem,\\
\phx\qquad $\alpha_1,\ldots,\alpha_\ell$ are all continuous from $V$ to $\R$.\\
Then $\alpha_1+\cdots+\alpha_\ell$ is continuous from $V$ to $\R$.\\
So, since $f=\alpha_1+\cdots+\alpha_\ell$, we conclude that:\\
\phx\qquad $f$ is continuous from $V$ to $W$, \qquad as desired.
\end{proof}

\subsection{Homogeneity of (homogeneous) polynomials}

\begin{thm}
Let $V,W\in\ES$, $k\in\N_0$, $f\in\scrm_k^{VW}$, $x\in V$, $c\in\R$.\\
\phx\qquad\qquad\qquad\qquad\quad Then: $f_{c\cdot x}\,\,=\,\,c^k\cdot f_x$.
\end{thm}

\begin{proof}
Since $f\in\scrm_k^{VW}$, choose $\phi\in\scrc^V_W$ and $\alpha_1,\ldots,\alpha_k\in\Pi_V$\\
\phx\qquad\qquad\qquad\qquad s.t. \qquad $f=\phi\cdot\alpha_1\cdots\alpha_k$.\\
For all $j\in[1..k]$, as $\alpha_j\in\Pi_V\subseteq\scrl^V_\R$, we get: $(\alpha_j)_{c\cdot x}=c\cdot((\alpha_j)_x)$.\\
Since $\phi\in\scrc^V_W$, we conclude: $\phi_{c\cdot x}=\phi_x$.\\
Then $f_{c\cdot x}=(\phi\cdot\alpha_1\cdots\alpha_k)_{c\cdot x}$\\
\phx\qquad\quad $\,\,\,\,\,=\phi_{c\cdot x}\cdot[(\alpha_1)_{c\cdot x}]\cdots[(\alpha_k)_{c\cdot x}]$\\
\phx\qquad\quad $\,\,\,\,\,=\phi_x\cdot[c\cdot((\alpha_1)_x)]\cdots[c\cdot((\alpha_k)_x)]$\\
\phx\qquad\quad $\,\,\,\,\,=c^k\cdot\phi_x\cdot[(\alpha_1)_x]\cdots[(\alpha_k)_x]$\\
\phx\qquad\quad $\,\,\,\,\,=c^k\cdot(\phi\cdot\alpha_1\cdots\alpha_k)_x$\\
\phx\qquad\quad $\,\,\,\,\,=c^k\cdot f_x$, \qquad as desired.
\end{proof}

\begin{thm}
Let $V,W\in\ES$, $k\in\N_0$, $f\in\scrh_k^{VW}$, $x\in V$, $c\in\R$.\\
\phx\qquad\qquad\qquad\qquad\quad Then: $f_{c\cdot x}\,\,=\,\,c^k\cdot f_x$.
\end{thm}

\begin{proof}
Since $f\in\scrm_k^{VW}$, choose $\ell\in\N$ and $\alpha_1,\ldots,\alpha_\ell\in\scrm_k^{VW}$\\
\phx\qquad\qquad\qquad\qquad s.t. \qquad $f=\alpha_1+\cdots+\alpha_k$.\\
By the preceding theorem, $\forall j\in[1..\ell]$, we have: $(\alpha_j)_{c\cdot x}=c^k\cdot((\alpha_j)_x)$.\\
Then $f_{c\cdot x}=(\alpha_1+\cdots+\alpha_k)_{c\cdot x}$\\
\phx\qquad\quad $\,\,\,\,\,=[(\alpha_1)_{c\cdot x}]+\cdots+[(\alpha_k)_{c\cdot x}]$\\
\phx\qquad\quad $\,\,\,\,\,=[c^k\cdot((\alpha_1)_x)]+\cdots+[c^k\cdot((\alpha_k)_x)]$\\
\phx\qquad\quad $\,\,\,\,\,=c^k\cdot[((\alpha_1)_x)+\cdots+((\alpha_k)_x)]$\\
\phx\qquad\quad $\,\,\,\,\,=c^k\cdot[(\alpha_1+\cdots+\alpha_k)_x]$\\
\phx\qquad\quad $\,\,\,\,\,=c^k\cdot f_x$, \qquad as desired.
\end{proof}

\subsection{Tensors of linear and bilinear maps}

\begin{defn}
Let $V,W\in\ES$ and let $L:V\to W$.\\
By $L$ is \bfindex[algebraically linear], we mean:\\
\phx\qquad\qquad\qquad\qquad $\forall u,v\in V$, \quad $L_{u+v}=L_u+L_v$\\
\phx\qquad\qquad and \qquad $\forall c\in\R,\,\forall v\in V$, \quad $L_{c\cdot v}=c\cdot L_v$.\\
\end{defn}

\begin{defn}
Let $V,W\in\ES$. \quad Then:\\
\phx\qquad\qquad\qquad $\scral^V_W\,\,\,:=\,\,\,\{\,\,f:V\to W\,\,|\,\,f\hbox{ is algebraically linear}\,\,\}$.
\end{defn}

\begin{thm}\wrlab{thm-linearmonomials-are-alglin}
Let $V,W\in\ES$. Then $\scrm_1^{VW}\subseteq\scral^V_W$.
\end{thm}

\begin{proof}
Want: $\forall M\in\scrm_1^{VW}$, \quad $M\in\scral^V_W$.\\
Given $M\in\scrm_1^{VW}$. \qquad Want: $M\in\scral^V_W$.\\
Choose $C\in\scrc^V_W$ and $p\in\Pi_V$ s.t. $M=C\cdot p$.\\
Unassigned HW: Show that $\Pi_V\subseteq\scral^V_\R$. \qquad Then $p\in\scral^V_\R$.\\
Unassigned HW: Show that $\scrc^V_W\cdot\scral^V_\R\subseteq\scral^V_W$.\\
Then $M=C\cdot p\in\scrc^V_W\cdot\scral^V_\R\subseteq\scral^V_\R$.
\end{proof}

\begin{thm}\wrlab{thm-lin-are-alglin}
Let $V,W\in\ES$. \quad Then $\scrl^V_W\subseteq\scral^V_W$.
\end{thm}

\begin{proof}
Want: $\forall L\in\scrl^V_W$, \quad $L\in\scral^V_W$.\\
Given $L\in\scrl^V_W$. \qquad Want: $L\in\scral^V_W$.\\
Since $L\in\scrl^V_W=\scrh_1^{VW}$, choose $k\in\N$ and $\phi_1,\ldots,\phi_k\in\scrm_1^{VW}$\\
\phx\qquad such that \quad $L=\phi_1+\cdots+\phi_k$.\\
By \tref{thm-linearmonomials-are-alglin}, $\scrm_1^{VW}\subseteq\scral^V_W$.\\
Unassigned HW: Show $\scral^V_W+\scral^V_W\subseteq\scral^V_W$.\\
Then, as $\phi_1,\ldots,\phi_k\in\scrm_1^{VW}\subseteq\scral^V_W$, we get: $\phi_1+\cdots+\phi_k\in\scral^V_W$.\\
Then $L=\phi_1+\cdots+\phi_k\in\scral^V_W$.
\end{proof}

\begin{defn}
We define:\\
\phx\qquad\qquad\qquad $\forall\alpha,\beta\in\N$, \quad $\R^\alpha\otimes\R^\beta\,\,\,:=\,\,\,\R^{\alpha\times\beta}$\\
\phx\qquad and \qquad\quad $\,\forall\beta\in\N$, \quad $\,\,\,\R\otimes\R^\beta\,\,\,:=\,\,\,\R^\beta$\\
\phx\qquad and \qquad\quad $\,\forall\alpha\in\N$, \quad $\R^\alpha\otimes\R\,\,\,\,\,\,:=\,\,\,\R^\alpha$\\
\phx\qquad and \qquad\qquad\qquad\qquad $\,\,\R\otimes\R\,\,\,\,\,\,:=\,\,\,\R$.
\end{defn}

\begin{defn}\wrlab{defn-lin-tensor-bijection1}
Let $\alpha,\beta\in\N$, $V:=\R^\alpha$, $W:=\R^\beta$.\\
Then: \quad $\forall L\in\scral^V_W$, \quad $[L]\in W\otimes V$ is defined by:\\
\phx\qquad\qquad\qquad $\forall j\in[1..\beta]$, $\forall i\in[1..\alpha]$, \quad $[L]_{ji}\,\,=\,\,(L(\vep_i^V))\sbullet\vep_j^W$.\\
Also: \quad $\forall S\in W\otimes V$, \quad $\Lin_{VW}^S\in\scrl^V_W$ is defined by:\\
\phx\qquad\qquad\qquad $\forall v\in V$, \quad $\ds{\Lin_{VW}^S(v)\,\,=\,\,\sum_{j=1}^\beta\,\sum_{i=1}^\alpha\,S_{ji}\cdot v_i\cdot \vep_j^W}$.
\end{defn}

When the Euclidean spaces are obvious, we sometimes omit $V$ and $W$, and write ``$\Lin^S$'', instead of ``$\Lin_{VW}^S$''.

\begin{thm}
Define $L:\R^3\to\R^2$ by: $\forall x,y,z\in\R$,\\
\phx\qquad\qquad\quad $L(x,y,z)\,\,=\,\,(\,2x+3y-7z\,,\,4x-y-9z\,)$.\\
Then $L\in\scral^{\R^3}_{\R^2}$ and $\ds{[L]=\left[\begin{array}{ccc}2&3&-7\\4&-1&-9\end{array}\right]\in\R^{2\times3}=\R^3\otimes\R^2}$.
\end{thm}

\begin{defn}\wrlab{defn-lin-tensor-bijection2}
Let $\alpha\in\N$, $V:=\R^\alpha$, $W:=\R$.\\
Then: \quad $\forall L\in\scral^V_W$, \quad $[L]\in W\otimes V$ is defined by:\\
\phx\qquad\qquad\qquad $\forall i\in[1..\alpha]$, \quad $[L]_i\,\,=\,\,(L(\vep_i^V))\sbullet1$.\\
Also: \quad $\forall S\in W\otimes V$, \quad $\Lin_{VW}^S\in\scrl^V_W$ is defined by:\\
\phx\qquad\qquad\qquad $\forall v\in V$, \quad $\ds{\Lin_{VW}^S(v)\,\,=\,\,\sum_{i=1}^\alpha\,S_i\cdot v_i\cdot1}$.
\end{defn}

When the Euclidean spaces are obvious, we sometimes omit $V$ and $W$, and write ``$\Lin^S$'', instead of ``$\Lin_{VW}^S$''.

\begin{defn}\wrlab{defn-lin-tensor-bijection3}
Let $\beta\in\N$, $V:=\R$, $W:=\R^\beta$.\\
Then: \quad $\forall L\in\scral^V_W$, \quad $[L]\in W\otimes V$ is defined by:\\
\phx\qquad\qquad\qquad $\forall j\in[1..\beta]$, \quad $[L]_j\,\,=\,\,(L(1))\sbullet\vep_j^W$.\\
Also: \quad $\forall S\in W\otimes V$, \quad $\Lin_{VW}^S\in\scrl^V_W$ is defined by:\\
\phx\qquad\qquad\qquad $\forall v\in V$, \quad $\ds{\Lin_{VW}^S(v)\,\,=\,\,\sum_{j=1}^\beta\,S_j\cdot v\cdot \vep_j^W}$.
\end{defn}

When the Euclidean spaces are obvious, we sometimes omit $V$ and $W$, and write ``$\Lin^S$'', instead of ``$\Lin_{VW}^S$''.

\begin{defn}\wrlab{defn-lin-tensor-bijection4}
Let $\beta\in\N$, $V:=\R$, $W:=\R$.\\
Then: \quad $\forall L\in\scral^V_W$, \quad $[L]\in W\otimes V$ is defined by:\\
\phx\qquad\qquad\qquad $[L]\,\,=\,\,(L(1))\sbullet1$.\\
Also: \quad $\forall S\in W\otimes V$, \quad $\Lin_{VW}^S\in\scrl^V_W$ is defined by:\\
\phx\qquad\qquad\qquad $\forall v\in V$, \quad $\ds{\Lin_{VW}^S(v)\,\,=\,\,S\cdot v\cdot1}$.
\end{defn}

When the Euclidean spaces are obvious, we sometimes omit $V$ and $W$, and write ``$\Lin^S$'', instead of ``$\Lin_{VW}^S$''.

\begin{thm}
Let $V,W\in\ES$, $L\in\scral^V_W$.
Then $\Lin^{[L]}=L$.\
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}
Let $V,W\in\ES$. Then $\scral^V_W=\scrl^V_W$.
\end{thm}

\begin{proof}
By \tref{thm-lin-are-alglin}, we have: $\scrl^V_W\subseteq\scral^V_W$.\\
Want: $\scral^V_W\subseteq\scrl^V_W$. \quad Given $L\in\scral^V_W$. \quad Want: $L\in\scrl^V_W$.\\
We have $L=\Lin^{[L]}\in\scrl^V_W$, as desired.
\end{proof}

\begin{thm}
Let $V,W\in\ES$, $S\in W\otimes V$.
Then $[\Lin^S]=S$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\noindent
Let $V,W\in\ES$. \quad Then\\
\phx\qquad\qquad $L\mapsto[L]:\scrl^V_W\to W\otimes V$ \quad and \quad $S\mapsto\Lin^S:W\otimes V$\\
are bijections are are inverses of one another.

\begin{defn}
We define:\\
\phx\qquad\qquad\qquad $\,\forall i,j\in\N$, \quad $\,\,\,\,\,\,i\|j\,\,:=\,\,(i,j)$\\
\phx\qquad and \qquad\quad $\,\forall j\in\N$, \quad $\,\,\,\,0\|j\,\,:=\,\,j$\\
\phx\qquad and \qquad\quad $\,\,\forall i\in\N$, \quad $\,\,\,\,\,i\|0\,\,:=\,\,i$\\
\phx\qquad and \qquad\qquad\qquad\qquad $\,\,0\|0\,\,:=\,\,0$.
\end{defn}

\noindent
We can now fold the eight formulas in\\
\phx\qquad\qquad\qquad \dref{defn-lin-tensor-bijection1}, \qquad\quad $\,\,$ \dref{defn-lin-tensor-bijection2},\\
\phx\qquad\qquad\qquad \dref{defn-lin-tensor-bijection3} \quad and \quad \dref{defn-lin-tensor-bijection4}\\
into two:

\begin{thm}\wrlab{thm-lin-tensor-bijectionall}
Let $V,W\in\ES$.\\
Then: \quad $\forall L\in\scral^V_W$, $\forall j\in\scri_W$, $\forall i\in\scri_V$, \quad $[L]_{j\|i}\,\,=\,\,(L(\vep_i^V))\sbullet\vep_j^W$.\\
Also: \quad $\forall S\in W\otimes V$, $\forall v\in V$, \quad $\ds{\Lin^S(v)\,\,=\,\,\sum_{j\in\scri_W}^\beta\,\sum_{i\in\scri_V}^\alpha\,S_{j\|i}\cdot v_i\cdot \vep_j^W}$.
\end{thm}

\begin{defn}
We define:\\
\phx\qquad\qquad\qquad $\forall\alpha,\beta,\gamma\in\N$, \quad $\,\,\,\,\,\,\R^\alpha\otimes\R^{\beta\times\gamma}\,\,:=\,\,\,\R^{\alpha\times\beta\times\gamma}$\\
\phx\qquad\qquad\qquad $\forall\alpha,\beta,\gamma\in\N$, \quad $\,\,\R^{\alpha,\beta}\otimes\R^\gamma\,\,\,\,\,\,\,\,:=\,\,\,\R^{\alpha\times\beta\times\gamma}$\\
\phx\qquad and \qquad\quad $\,\forall\beta,\gamma\in\N$, \quad $\,\,\,\,\,\,\,\,\,\R\otimes\R^{\beta\times\gamma}\,\,:=\,\,\,\R^{\beta\times\gamma}$\\
\phx\qquad and \qquad\quad $\,\forall\alpha,\beta\in\N$, \quad $\R^{\alpha\times\beta}\otimes\R\,\,\,\,\,\,\,\,\,\,:=\,\,\,\R^{\alpha\times\beta}$.
\end{defn}

\begin{defn}
We define:\\
\phx\qquad\qquad\qquad $\,\forall i,j,k\in\N$, \quad $\,\,\,\,\,\,i\|(j,k)\,\,:=\,\,(i,j,k)$\\
\phx\qquad\qquad\qquad $\,\forall i,j,k\in\N$, \quad $\,\,\,\,\,\,(i,j)\|k\,\,:=\,\,(i,j,k)$\\
\phx\qquad and \qquad\quad $\,\forall j,k\in\N$, \quad $\,\,\,\,0\|(j,k)\,\,:=\,\,(j,k)$\\
\phx\qquad and \qquad\quad $\,\,\forall i,j\in\N$, \quad $\,\,\,\,\,(i,j)\|0\,\,:=\,\,(i,j)$.
\end{defn}

\begin{defn}
Let $V,W\in\ES$.\\
Then: \quad $\forall B\in\scrb^V_W$, \quad $[B]\in X\otimes V\otimes W$ is defined by:\\
\phx\quad $\forall k\in\scri_X$, $\forall i\in\scri_V$, $\forall j\in\scri_W$, $\,\,\,$ $[B]_{k\|i\|j}\,\,=\,\,(B(\vep_i^V,\vep_j^W))\sbullet\vep_k^X$.\\
Also: \quad $\forall T\in X\otimes V\otimes W$, \quad $\Bilin_{VWX}^T\in\scrb^{VW}_X$ is defined by:\\
\phx$\,$ $\forall v\in V$, $\forall w\in W$, $\,$ $\ds{\Bilin_{VWX}^T(v,w)=\sum_{k\in\scri_X}\,\sum_{i\in\scri_V}\,\sum_{j\in\scri_W}\,T_{k\|i\|j}\cdot v_i\cdot w_j\cdot \vep_k^X}$.
\end{defn}

When the Euclidean spaces are obvious, we sometimes omit $V$ and $W$ and $X$, and write ``$\Bilin^T$'', instead of ``$\Bilin_{VWX}^T$''.

\subsection{Polarization and diagonal restriction}

\begin{defn}
Let $V,W\in\ES$ and $*\in\scrb^{VV}_W$.\\
Then $\Qd^*:V\to W$ is defined by: $\forall v\in V$, $\Qd^*_v=v*v$.
\end{defn}

\noindent
Let $V,W\in\ES$. According to HW\#5-5, $\Qd^*\in\scrq^V_W$.\\
\\
Let $V,W\in\ES$. The next two results can be summarized as saying:\\
\phx\qquad $*\mapsto\Qd^*:\scrb^{VV}_W\to\scrq^V_W$ is algebraically linear.

\begin{thm}
let $V,W\in\ES$ and $B,C\in\scrb^{VV}_W$.\\
Then $\Qd^{B+C}=\Qd^B+\Qd^C$.
\end{thm}

\begin{thm}
let $V,W\in\ES$ and $a\in\R$ and $B\in\scrb^{VV}_W$.\\
Then $\Qd^{a\cdot B}=a\cdot\Qd^B$.
\end{thm}

\begin{thm}\wrlab{thm-monomials-polarizable}
Let $V,W\in\ES$ and $F\in\scrm_2^{VW}$.\\
Then $\exists*\in\scrb^{VV}_W$ s.t.~$\Qd^*=F$.
\end{thm}

\begin{proof}
Since $F\in\scrm_2^{VW}$,\\
\phx\qquad\qquad choose $C\in\scrc^V_W$ and $p,q\in\Pi_V$ s.t.~$F=C\cdot p\cdot q$.\\
Since $C\in\scrc^V_W$, choose $w\in W$ s.t.~$C=C_V^w$.\\
Define $*:V\times V\to W$ by $\forall u,v\in V$, $u*v=w\cdot p_u\cdot q_v$.\\
Since $p,q\in\Pi_V\subseteq\scrl^V_\R$, it follows that: $*\in\scrb^{VV}_W$. \quad Want: $\Qd^*=F$.\\
Want: $\forall v\in V$, $\Qd^*_v=F_v$. \quad
Given $v\in V$. \quad Want: $\Qd^*_v=F_v$.\\
We have $C_v=(C_V^w)_v=w$. \qquad Then $w=C_v$.\\
Then: $\Qd^*_v=v*v=w\cdot p_v\cdot q_v=C_v\cdot p_v\cdot q_v$\\
\phx\qquad\qquad $\,\,\,=(C\cdot p\cdot q)_v=F_v$, \qquad as desired.
\end{proof}

\noindent
Let $V,W\in\ES$. The next result can be summarized as saying:\\
\phx\qquad $*\mapsto\Qd^*:\scrb^{VV}_W\to\scrq^V_W$ is surjective.

\begin{thm}
Let $V,W\in\ES$ and $F\in\scrq^V_W$.\\
Then $\exists*\in\scrb^{VV}_W$ s.t.~$\Qd^*=F$.
\end{thm}

\begin{proof}
Since $F\in\scrq^V_W=\scrh_2^{VW}$, choose $k\in\N$ and $\psi_1,\ldots,\psi_k\in\scrm_2^{VW}$\\
\phx\qquad\qquad\qquad\qquad s.t. \qquad $F=\psi_1+\cdots+\psi_k$.\\
By \tref{thm-monomials-polarizable}, $\forall j\in[1..k]$, choose $B_j\in\scrb^{VV}_W$ s.t. $\psi_j=\Qd^{B_j}$.\\
Let $*:=B_1+\cdots+B_k$. \qquad Then $*\in\scrb^{VV}_W$. \qquad Want: $\Qd^*=F$.\\
We have $\Qd^*=\Qd^{B_1+\cdots+B_k}=\Qd^{B_1}+\cdots+\Qd^{B_k}$\\
\phx\qquad\qquad\quad $\,\,\,=\psi_1+\cdots+\psi_k=F$, \quad as desired.
\end{proof}

\subsection{Principal minors and positive definiteness}

\begin{defn}
Let $V,W\in\ES$. Then:\\
\phx\qquad\qquad\qquad $\scrsb^V_W\quad:=\quad\{\,\,*\in\scrb^{VV}_W\,\,|\,\,\forall u,v\in V,\,\,u*v=v*u\,\,\}$.
\end{defn}

\begin{defn}
Let $V,W\in\ES$ and let $*\in\scrb^{VV}_W$.\\
Then $\Sym^*:V\times V\to W$ is defined by:\\
\phx\qquad\quad $\forall u,v\in V$, \qquad $\ds{\Sym^*(u,v)\,\,=\,\,\frac{u*v\,\,\,\,+\,\,\,\,v*u}{2}}$.
\end{defn}

\noindent
Unassigned HW: Let $V,W\in\ES$ and let $*\in\scrsb^V_W$. Show: $\Sym^*=*$.\\
\\
Monastery and convent story about the joys of polarization \dots

\begin{defn}
Let $m\in\N$ and let $A\in\R^{m\times m}$.\\
\phx\qquad\qquad\qquad\qquad\qquad Assume that $A$ is symmetric.\\
\phx\qquad\qquad Then $\PM_A$ denotes the set of principal minors of $A$.
\end{defn}

\begin{thm}
Let $m\in\N$, $V:=\R^m$, $B\in\scrsb^V_\R$, $Q:=\Qd^B$.\\
Then: \quad $(\,\,Q>0\hbox{ on }V_{0_V}^\times\,\,)\,\,\Leftrightarrow\,\,(\,\,\PM_{[B]}>0\,\,)$.
\end{thm}

\begin{proof}
Omitted. Belongs in a linear algebra course.
\end{proof}

\noindent
We described the Multivariable Second Derivative for Minima.\\
We'll come back to this later.

\subsection{Multivariable $\DNZ$, $\BNZ$, $\CVZ$, big-O and little-o}

\begin{defn}
Let $V,W\in\ES$. Then $\bfzero^V_W:=C_V^{0_W}$.
\end{defn}

\begin{defn}
Let $f$ be a v/s-function.\\
\phx\qquad\qquad\qquad Then $|f|$ is the functional defined by:\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\forall x$, \quad $|f|_x=|f_x|$.
\end{defn}

\begin{defn}
Let $V,W\in\ES$. Then:\\
\phx $\DNZ^V_W\,\,:=\,\,\{\,\,f:V\drta W\,\,|\,\,f\hbox{ is defined near }0_V\hbox{ in }V\,\,\}$,\\
\phx $\BNZ^V_W\,\,:=\,\,\{\,\,f:V\drta W\,\,|\,\,f\hbox{ is bounded near }0_V\hbox{ from }V\hbox{ to }W\,\,\}$,\\
\phx $\CVZ^V_W\,\,:=\,\,\{\,\,f\in\DNZ^V_W\,\,|\,\,(\,f_{0_V}=0_W\,)\,\,\&$\\
\phx\quad\qquad\qquad\qquad\qquad\qquad $\,\,(\,f\hbox{ is continuous at }0_V\hbox{ from }V\hbox{ to }W\,)\,\,\}$.
\end{defn}

\begin{thm}
Let $V,W\in\ES$.\\
\phx\qquad\quad Then $\DNZ^V_W$ and $\BNZ^V_W$ and $\CVZ^V_W$ are all v.op-closed.
\end{thm}

\begin{thm}
Let $V,W\in\ES$ and let $f,g:V\drta W$.\\
\phx\qquad\qquad\qquad\qquad Assume: \quad $g=f$ near $0_V$ in $V$.\\
\phx\qquad\qquad Then: \quad $\,[\,\,(\,\,f\in\DNZ^V_W\,\,)\,\,\Leftrightarrow\,\,(\,\,g\in\DNZ^V_W\,\,)\,\,]$\\
\phx\qquad\qquad\qquad\quad $\&\,\,[\,\,(\,\,f\in\BNZ^V_W\,\,)\,\,\Leftrightarrow\,\,(\,\,g\in\BNZ^V_W\,\,)\,\,]$\\
\phx\qquad\qquad\qquad\quad $\&\,\,[\,\,(\,\,f\in\CVZ^V_W\,\,)\,\,\Leftrightarrow\,\,(\,\,g\in\CVZ^V_W\,\,)\,\,]$.
\end{thm}

\begin{thm}
Let $V,W\in\ES$ and let $f,g:V\drta W$. Then:\\
\phx $\,\,[f\in\BNZ^V_W]\,\Leftrightarrow\,[\exists\delta,L>0~\hbox{s.t.},\forall x\in V,\,(|x|<\delta)\Rightarrow(|f_x|\le L)]$\\
and\\
\phx $\,\,[f\in\CVZ^V_W]\,\Leftrightarrow\,[\forall\vep>0,\exists\delta>0~\hbox{s.t.},\forall x\in V,\,(|x|<\delta)\Rightarrow(|f_x|\le\vep)]$.
\end{thm}

\begin{defn}
Let $V,W\in\ES$ and let $k\in\N_0$.\\
\phx\qquad Then: \qquad\qquad\quad $\scrbigo_k^{VW}\,\,:=\,\,(\BNZ^V_W)\cdot(|\bullet|_V^k)$\\
\phx\qquad\qquad\qquad $\&$\qquad\quad $\,\,\scrlito_k^{VW}\,\,:=\,\,(\CVZ^V_W)\cdot(|\bullet|_V^k)$.
\end{defn}

\begin{thm}
Let $V,W\in\ES$ and let $k\in\N_0$.\\
\phx\qquad\qquad\qquad Then $\scrbigo_k^{VW}$ and $\scrlito_k^{VW}$ are both v.op.-closed.
\end{thm}

\begin{thm}
Let $V,W\in\ES$, $k\in\N_0$ and $f,g:V\drta W$.\\
\phx\qquad\qquad\qquad\qquad Assume: \quad $g=f$ near $0_V$ in $V$.\\
\phx\qquad\qquad Then: \quad $\,[\,\,(\,\,f\in\scrbigo_k^{VW}\,\,)\,\,\Leftrightarrow\,\,(\,\,g\in\scrbigo_k^{VW}\,\,)\,\,]$\\
\phx\qquad\qquad\qquad\quad $\&\,\,[\,\,(\,\,f\in\,\scrlito_k^{VW}\,\,\,)\,\,\Leftrightarrow\,\,(\,\,g\in\,\scrlito_k^{VW}\,\,\,)\,\,]$.
\end{thm}

\begin{thm}
Let $V,W\in\ES$, $k\in\N_0$, $f:V\drta W$. Then:\\
\phx $[f\in\scrbigo_k^{VW}]\Leftrightarrow[\exists\delta,L>0~\hbox{s.t.},\forall x\in V,\,(|x|<\delta)\Rightarrow(|f_x|\le L\cdot|x|^k)]$\\
and\\
\phx $\,[f\in\scrlito_k^{VW}]\Leftrightarrow[\forall\vep>0,\exists\delta>0~\hbox{s.t.},\forall x\in V,\,(|x|<\delta)\Rightarrow(|f_x|\le\vep\cdot|x|^k)]$.
\end{thm}

\noindent
MATERIAL ABOVE IS COVERED ON MIDTERM 2 SPRING 2020

\begin{thm}
Let $V\in\ES$. Then $S_V$ is compact.
\end{thm}

\begin{proof}
Since $S_V\in B_V(0_V,2)$, we see that $S_V$ is bounded in $V$.\\
So, since $V$ is proper, it suffices to show: $S_V$ is closed in $V$.\\
Want: $S_V\in\scrt'_V$. \qquad We have $S_V=(|\bullet|_V)^*(\{1\})$.\\
Because finite sets in metric spaces are closed,\\
\phx\qquad we conclude that $\{1\}\in\scrt'_\R$.\\
Since $|\bullet|_V$ is Lipschitz-1 from $V$ to $\R$,\\
\phx\qquad it follows that $|\bullet|_V$ is continuous from $V$ to $\R$.\\
Then, we have: \qquad $\forall C\in\scrt'_\R$, $\,$ $(|\bullet|_V)^*(C)\in\scrt'_V$.\\
Then $(|\bullet|_V)^*(\{1\})\in\scrt'_V$. \quad Then $S_V=(|\bullet|_V)^*(\{1\})\in\scrt'_V$.
\end{proof}

\begin{thm}
Let $V,W\in\ES$ and let $k\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: $\scrh_k^{VW}\subseteq\scrbigo_k^{VW}$.
\end{thm}

\begin{proof}
Want: $\forall f\in\scrh_k^{VW}$, $f\in\scrbigo_k^{VW}$.
Given $f\in\scrh_k^{VW}$. Want: $f\in\scrbigo_k^{VW}$.\\
Want: $\exists\delta,L>0$ s.t., $\forall x\in V$, \quad $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|f_x|\le L\cdot|x|^k\,\,)$.\\
We have $f\in\scrh_k^{VW}$, so $f$ is continuous from $V$ to $W$.\\
So, since $S_V$ is compact, we see that $f_*S_V$ is compact as well.\\
Then $f_*S_V$ is closed and bounded in $W$.\\
Since $f_*S_V$ is bounded in $W$,\\
\phx\qquad choose $L>0$ s.t. $f_*S_V\subseteq B_W(0_W,L)$.\\
Let $\delta:=1$. \qquad Then $\delta,L>0$.\\
Want: $\forall x\in V$, \quad $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|f_x|\le L\cdot|x|^k\,\,)$.\\
Given $x\in V$. \qquad Want: $(\,\,|x|<\delta\,\,)\,\,\Rightarrow\,\,(\,\,|f_x|\le L\cdot|x|^k\,\,)$.\\
Assume $|x|<\delta$. \qquad Want: $|f_x|\le L\cdot|x|^k$.\\
By polar decomposition, choose $u\in S_V$ s.t.~$x=|x|\cdot u$.\\
Let $c:=|x|$. \quad Then $x=c\cdot u$. \quad Also, as $c\ge0$, we get: $|c|=c$.\\
We have $f_x=f_{cu}=c^k\cdot f_u$, so $|f_x|=|c|^k\cdot|f_u|$.\\
Then, as $|c|=c$, this gives: $|f_x|=c^k\cdot|f_u|$.\\
Since $f\in\scrh_k^{VW}$, we conclude: $\D_f=V$.\\
Recall: $u\in S_V$. \quad Then $u\in S_V\subseteq V=\D_f$. \quad Then $u\in S_V\bigcap\D_f$.\\
Then $f_u\in f_*S_V\subseteq B_W(0_W,L)$, so $d_W(f_u,0_W)<L$.\\
Then $|f_u|=|f_u-0|=d_W(f_u,0_W)<L$, and so $|f_u|\le L$.\\
So, since $|x|^k\ge0$, we get: $|f_u|\cdot|x|^k\le L\cdot|x|^k$.\\
Recall that $|f_x|=c^k\cdot |f_u|$ and that $c=|x|$.\\
Then $|f_x|=c^k\cdot |f_u|=|f_u|\cdot c^k=|f_u|\cdot|x|^k\le L\cdot|x|^k$, as desired.
\end{proof}

\begin{thm}
Let $V,W\in\ES$ and $k\in\N_0$ and $f\in\scrlito_k^{VW}$.\\
\phx\qquad Let $u\in S_V$ and $\vep>0$. \qquad Then $\exists t>0$ s.t.~$|f_{t\cdot u}|<\vep\cdot t^k$.
\end{thm}

\begin{proof}
Choose $\delta>0$ s.t., $\forall x\in V$, \quad $[\,\,|x|<\delta\,\,]\,\,\Rightarrow\,\,[\,\,|f_x|\le(\vep/2)\cdot|x|^k\,\,]$.\\
Let $t:=\delta/2$. \quad Then $t>0$. \quad Want: $|f_{t\cdot u}|<\vep\cdot t^k$.\\
Let $x:=t\cdot u$. \qquad\qquad\qquad $\,\,\,\,\,$ Want: $\,|f_x|\,<\vep\cdot t^k$.\\
Since $u\in S_V$, we get: $|u|=1$. \qquad\qquad Since $t>0$, we get: $|t|=t$.\\
Then $|x|=|t\cdot u|=|t|\cdot|u|=t\cdot1=t$. \quad Then $(\vep/2)\cdot|x|^k=(\vep/2)\cdot t^k$.\\
Since $|x|=t=\delta/2<\delta$, by choice of $\delta$, we get: $|f_x|\le(\vep/2)\cdot|x|^k$.\\
Since $t>0$, it follows that $t^k>0$.\\
So, since $\vep/2<\vep$, we get: $\,(\vep/2)\cdot t^k<\vep\cdot t^k$.\\
Then: $|f_x|\le(\vep/2)\cdot|x|^k=(\vep/2)\cdot t^k<\vep\cdot t^k$, as desired.
\end{proof}

\begin{thm}
Let $V,W\in\ES$ and $k\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then: $\scrh_k^{VW}\bigcap\scrlito_k^{VW}=\{\bfzero^V_W\}$.
\end{thm}

\begin{proof}
Since $\bfzero^V_W\in\scrh_k^{VW}$ and $\bfzero^V_W\in\scrlito_k^{VW}$,
we get $\bfzero^V_W\in\scrh_k^{VW}\bigcap\scrlito_k^{VW}$.\\
Then $\{\bfzero^V_W\}\subseteq\scrh_k^{VW}\bigcap\scrlito_k^{VW}$. \qquad
Want: $\scrh_k^{VW}\bigcap\scrlito_k^{VW}\subseteq\{\bfzero^V_W\}$.\\
\phx\quad $\,$ Want: \quad $\forall f\in\scrh_k^{VW}\bigcap\scrlito_k^{VW}$, $\,$ $f\in\{\bfzero^V_W\}$.\\
Given $f\in\scrh_k^{VW}\bigcap\scrlito_k^{VW}$. \quad Want: $f\in\{\bfzero^V_W\}$. \quad Want: $f=\bfzero^V_W$.\\
Assume that $f\ne\bfzero^V_W$. \qquad Want: Contradiction.\\
Since $f\in\scrh_k^{VW}$, we get $\D_f=V$. \quad Also $\D_{\bfzero^V_W}=V$.\\
So, since $f\ne\bfzero^V_W$, choose $y\in V$ s.t.~$f_y\ne(\bfzero^V_W)_y$.\\
By polar decomposition, choose $u\in S_V$ s.t.~$y=|y|\cdot u$.\\
Let $c:=|y|$. \qquad Then $y=c\cdot u$.\\
Since $c^k\cdot f_u=f_{c\cdot u}=f_y\ne(\bfzero^V_W)_y=(C_V^{0_W})_y=0_W$, \quad we get $f_u\ne0_W$.\\
Then $|f_u|>0$. \qquad Let $\vep:=|f_u|$. \qquad Then $\vep>0$.\\
By the preceding theorem, choose $t>0$ s.t.~$|f_{t\cdot u}|<\vep\cdot t^k$.\\
Then $t^k\cdot\vep=\vep\cdot t^k>|f_{t\cdot u}|$, \quad so $t^k\cdot\vep>|f_{t\cdot u}|$.\\
Then $t^k\cdot\vep>|f_{t\cdot u}|=|t^k\cdot f_u|=|t|^k\cdot|f_u|=t^k\cdot\vep$,\\
\phx\qquad and so $t^k\cdot\vep>t^k\cdot\vep$. \qquad Contradiction.
\end{proof}

\begin{thm}
Let $V,W\in\ES$. \qquad Then $\CVZ^V_W\subseteq\BNZ^V_W$.
\end{thm}

\begin{proof}
Unassigned HW. {\it Hint:} Follow the proof of \tref{thm-cvz-circ-bnz-not-subseteq-dnz}.
\end{proof}

\begin{thm}
Let $V,W\in\ES$, $k\in\N_0$. Then $\scrlito_k^{VW}\subseteq\scrbigo_k^{VW}$.
\end{thm}

\begin{proof}
We have: $\scrlito_k^{VW}=\CVZ^V_W\cdot|\bullet|^k\subseteq\BNZ^V_W\cdot|\bullet|^k=\scrbigo_k^{VW}$.
\end{proof}

\subsection{Bilinear products of function spaces}

\begin{thm}
Let $S,U,V,W\in\ES$ and let $*\in\scrb^{VW}_W$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then $\BNZ^S_U\,*\,\CVZ^S_V\,\subseteq\,\CVZ^S_W$.
\end{thm}

\begin{proof}
Want: \quad $\forall h\in\BNZ^S_U\,*\,\CVZ^S_V$, $\,$ $h\in\CVZ^S_W$.\\
Given $h\in\BNZ^S_U\,*\,\CVZ^S_V$. \qquad Want: $h\in\CVZ^S_W$.\\
\phx\qquad $\,\,\,$ Want: $\forall\vep>0$, $\exists\delta>0$ s.t., $\forall t\in S$, $(|t|<\delta)\Rightarrow(|h_t|\le\vep)$.\\
Given $\vep>0$. \quad Want: $\exists\delta>0$ s.t., $\forall t\in S$, $(|t|<\delta)\Rightarrow(|h_t|\le\vep)$.\\
As $h\in\BNZ^S_U\,*\,\CVZ^S_V$, choose $f\in\BNZ^S_U$, $g\in \CVZ^S_V$ s.t. $h=f*g$.\\
Since $f\in\BNZ^S_U$, choose $K,\alpha>0$ s.t.,\\
\phx\qquad\qquad\qquad $\forall t\in S$, \qquad\qquad\quad $\,\,\,\,(|t|<\alpha)\Rightarrow(|f_t|\le K)$.\\
Since $*$ is bilinear, and therefore double-Lipschitz, choose $L\ge0$ s.t.,\\
\phx\qquad\qquad\qquad $\forall x\in U$, $\forall y\in V$, \qquad $\!|x*y|\le L\cdot|x|\cdot|y|$.\\
Since $g\in\CVZ^S_V$, choose $\beta>0$ s.t.,\\
\phx\qquad\qquad\qquad $\forall t\in S$, \qquad\qquad\quad $\ds{\,\,\,\,(|t|<\beta)\Rightarrow\left(|g_t|\le\frac{\vep}{(L+1)\cdot K}\right)}$.\\
Let $\delta:=\min\{\alpha,\beta\}$. \qquad\qquad Then $\delta>0$.\\
\phx\qquad Want: $\forall t\in S$, \quad $(|t|<\delta)\Rightarrow(|h_t|\le\vep)$.\\
Given $t\in S$. \quad Want: $\,(|t|<\delta)\Rightarrow(|h_t|\le\vep)$.\\
\phx\qquad $\,\,\,\,\,\,$ Assume: $|t|<\delta$. \quad Want: $|h_t|\le\vep$.\\
Since $|t|<\delta\le\alpha$, by choice of $\alpha$, we get: \quad $|f_t|\le K$.\\
Since $|t|<\delta\le\beta$, by choice of $\beta$, we get: \quad $\ds{|g_t|\le\frac{\vep}{(L+1)\cdot K}}$.\\
Since $|f_t|\le K$ and $\ds{|g_t|\le\frac{\vep}{(L+1)\cdot K}}$, we get: $f_t\ne\frownie\ne g_t$.\\
Then $t\in\D_f$ and $t\in\D_g$. \quad Then $f_t\in\I_f\subseteq U$ and $g_t\in\I_g\subseteq V$.\\
Let $x:=f_t$ and $y:=g_t$. \quad Then $x\in U$ and $y\in V$.\\
So, by choice of $L$, we get: \quad $|x*y|\le L\cdot|x|\cdot|y|$.\\
We have $|x|=|f_t|\le K$ and $\ds{|y|=|g_t|\le\frac{\vep}{(L+1)\cdot K}}$.\\
Since $0\le|x|\le K$ and $\ds{0\le|y|\le\frac{\vep}{(L+1)\cdot K}}$,\\
\phx\qquad we conclude that \quad $\ds{|x|\cdot|y|\le\frac{\vep}{L+1}}$,\\
\phx\qquad and so, as $L\ge0$, we get \quad $L\cdot|x|\cdot|y|<\vep$.\\
We have $h_t=(f*g)_t=f_t*g_t=x*y$.\\
Then $|h_t|=|x*y|\le L\cdot|x|\cdot|y|<\vep$, \quad and so $|h_t|\le\vep$, as desired.
\end{proof}

\begin{thm}
Let $S\in\ES$. Then $\BNZ^S_\R\cdot\CVZ^S_\R\subseteq\CVZ^S_\R$.
\end{thm}

\begin{proof}
Define $*\in\scrb^{\R\R}_\R$ by: \quad $\forall a\in\R$, $\forall b\in\R$, $\,$ $a*b=a\cdot b$.\\
Then $\BNZ^S_\R\cdot\CVZ^S_\R=\BNZ^S_\R*\CVZ^S_\R\subseteq\CVZ^S_\R$, as desired.
\end{proof}

\begin{thm}
Let $S,U\in\ES$. Then $\BNZ^S_U\cdot\CVZ^S_\R\subseteq\CVZ^S_U$.
\end{thm}

\begin{proof}
Define $*\in\scrb^{U\R}_U$ by: \quad $\forall x\in U$, $\forall c\in\R$, $\,$ $x*c=x\cdot c$.\\
Then $\BNZ^S_U\cdot\CVZ^S_\R=\BNZ^S_U*\CVZ^S_\R\subseteq\CVZ^S_U$, as desired.
\end{proof}

\begin{thm}
Let $S,U\in\ES$. Then $\BNZ^S_U\sbullet\CVZ^S_U\subseteq\CVZ^S_\R$.
\end{thm}

\begin{proof}
Define $*\in\scrb^{UU}_\R$ by: \quad $\forall x\in U$, $\forall y\in U$, $\,$ $x*y=x\sbullet y$.\\
Then $\BNZ^S_U\cdot\CVZ^S_U=\BNZ^S_U*\CVZ^S_U\subseteq\CVZ^S_\R$, as desired.
\end{proof}

\begin{thm}
Let $S,U,V,W\in\ES$ and let $*\in\scrb^{UV}_W$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then \qquad\qquad $\,\,\BNZ^S_U\,*\,\BNZ^S_V\,\subseteq\,\BNZ^S_W$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad and \quad $\CVZ^S_U\,*\,\BNZ^S_V\,\subseteq\,\CVZ^S_W$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad and \quad $\BNZ^S_U\,*\,\CVZ^S_V\,\subseteq\,\CVZ^S_W$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad and \quad $\CVZ^S_U\,*\,\CVZ^S_V\,\subseteq\,\CVZ^S_W$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}\wrlab{thm-prod-contin-fnals-contin}
Let $S\in\ES$. Then $\CVZ^S_\R\cdot\CVZ^S_\R\subseteq\CVZ^S_U$.
\end{thm}

\begin{proof}
Define $*\in\scrb^{\R\R}_\R$ by: \quad $\forall a\in\R$, $\forall b\in\R$, $\,$ $a*b=a\cdot b$.\\
Then $\CVZ^S_\R\cdot\CVZ^S_\R=\CVZ^S_\R*\CVZ^S_\R\subseteq\CVZ^S_\R$, as desired.
\end{proof}

\begin{thm}
Let $S,U,V,W\in\ES$ and $*\in\scrb^{UV}_W$ and $k,\ell\in\N_0$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then \qquad\qquad $\,\,\scrbigo_k^{SU}\,*\,\scrbigo_\ell^{SV}\,\subseteq\,\scrbigo_{k+\ell}^{SW}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad and \quad $\,\scrlito_k^{SU}\,*\,\scrbigo_\ell^{SV}\,\subseteq\,\,\scrlito_{k+\ell}^{SW}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad and \quad $\scrbigo_k^{SU}\,*\,\scrlito_\ell^{SV}\,\,\subseteq\,\,\scrlito_{k+\ell}^{SW}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad and \quad $\,\scrlito_k^{SU}\,*\,\,\scrlito_\ell^{SV}\,\,\subseteq\,\,\scrlito_{k+\ell}^{SW}$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\begin{thm}
Let $V\in\ES$. \quad Then $|\bullet|_V\in\CVZ^V_\R$.
\end{thm}

\begin{proof}
Since $\D_{|\bullet|_V}=V$ and $\I_{|\bullet|_V}\subseteq\R$, we get: $|\bullet|_V\in\DNZ^V_\R$.\\
Also, we have: \quad $(|\bullet|_V)_{0_V}=|0_V|=0=0_\R$.\\
It remains to show: $|\bullet|_V$ is continuous at $0_V$ from $V$ to $\R$.\\
By HW\#3-5(a), we know that $|\bullet|_V$ is Lipschitz-$1$ from $V$ to $\R$,\\
\phx\qquad and so $|\bullet|_V$ is continuous from $V$ to $\R$.\\
Then $|\bullet|_V$ is continuous at $0_V$ from $V$ to $\R$.
\end{proof}

\begin{thm}
Let $V,W\in\ES$, $k\in\N_0$. Then $\scrbigo_{k+1}^{VW}\subseteq\scrlito_k^{VW}$.
\end{thm}

\begin{proof}
We have: $\scrbigo_{k+1}^{VW}\,\,\,=\,\,\,\BNZ^V_W\cdot\,\,|\bullet|_V^{k+1}$\\
\phx\qquad\qquad\qquad\qquad\qquad $=\,\,\,\BNZ^V_W\cdot\,\,|\bullet|_V\,\,\cdot\,\,\,|\bullet|_V^k$\\
\phx\qquad\qquad\qquad\qquad\qquad $\subseteq\,\,\,\BNZ^V_W\cdot\CVZ^V_W\cdot\,\,\,|\bullet|_V^k$\\
\phx\qquad\qquad\qquad\qquad\qquad $\subseteq\,\,\,\qquad\CVZ^V_W\quad\,\,\,\,\cdot\,\,\,|\bullet|_V^k\quad=\quad\scrlito_k^{VW}$.
\end{proof}

\subsection{Compositions of function spaces}

\begin{thm}
Let $V,W,X\in\ES$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then $\BNZ^W_X\,\circ\,\CVZ^V_W\,\subseteq\,\BNZ^V_X$.
\end{thm}

\begin{proof}
Want: \quad $\forall h\in\BNZ^W_X\,\circ\,\CVZ^V_W$, $\,$ $h\in\BNZ^V_X$.\\
Given $h\in\BNZ^W_X\,\circ\,\CVZ^V_W$. \qquad Want: $h\in\BNZ^V_X$.\\
Want: $\exists\delta,L>0$ s.t., $\forall x\in V$, $(|x|<\delta)\Rightarrow(|h_x|\le L)$.\\
As $h\in\BNZ^W_X\,\circ\,\CVZ^V_W$, choose $g\in\BNZ^W_X$, $f\in \CVZ^V_W$ s.t. $h=g\circ f$.\\
Since $g\in\BNZ^W_X$, choose $\vep,L>0$ s.t.,\\
\phx\qquad\qquad\qquad $\forall y\in W$, \qquad\qquad\quad $\,\,\,\,(|y|<\vep)\Rightarrow(|g_y|\le L)$.\\
Since $f\in\CVZ^V_W$, choose $\delta>0$ s.t.,\\
\phx\qquad\qquad\qquad $\forall x\in V$, \qquad\qquad\quad $\,\,\,\,\,(|x|<\delta)\Rightarrow(|f_x|\le\vep/2)$.\\
Then $\delta,L>0$. \quad Want: $\forall x\in V$, $\,(|x|<\delta)\Rightarrow(|h_x|\le L)$.\\
\phx\qquad\qquad Given $x\in V$. \quad Want: $(|x|<\delta)\Rightarrow(|h_x|\le L)$.\\
\phx\qquad\qquad\qquad\qquad Assume: $|x|<\delta$. \quad Want: $|h_x|\le L$.\\
Since $\vep>0$, we get: $\vep/2<\vep$. \qquad\qquad Let $y:=f_x$.\\
Since $|x|<\delta$, by choice of $\delta$, we get:\\
\phx\qquad\qquad $\,\,\,|f_x|\le\vep/2$.\\
Since $|y|=|f_x|\le\vep/2<\vep$, by choice of $\vep$, we get: $|g_y|\le L$.\\
We have $h_x=(g\circ f)_x=g_{f_x}=g_y$. \quad Then $|h_x|=|g_y|\le L$.
\end{proof}

\begin{thm}
Let $V,W,X\in\ES$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then $\CVZ^W_X\,\circ\,\CVZ^V_W\,\subseteq\,\CVZ^V_X$.
\end{thm}

\begin{proof}
Unassigned HW.
\end{proof}

\noindent
Warning: As we observed earlier, we have:\\
\phx\qquad\qquad\qquad\qquad $\CVZ^\R_\R\,\circ\,\BNZ^\R_\R\,\,\,\not\subseteq\,\,\,\DNZ^\R_\R$.\\
So, if $\BNZ$ or (any larger function space) appears on the RHS of $\circ$,\\
\phx\qquad\qquad then all bets are off.

\begin{thm}
Let $V,W\in\ES$, $\phi\in\CVZ^V_W$.\\
\phx\qquad\qquad\qquad Then: $(\,\,|\phi|\in\CVZ^V_\R\,\,)\,\,\&\,\,(\,\,\forall\ell\in\N,\,\,|\phi|^\ell\in\CVZ^V_\R\,\,)$.
\end{thm}

\begin{proof}
We have $|\phi|=(|\bullet|_W)\circ\phi\in\CVZ^W_\R\circ\CVZ^V_W\subseteq\CVZ^V_\R$.\\
\phx\quad Want: \quad $\forall\ell\in\N,\,\,|\phi|^\ell\in\CVZ^V_\R$.\\
As $|\phi|\in\CVZ^V_\R$, by \tref{thm-prod-contin-fnals-contin} and induction on $\ell$, we get:\\
\phx\qquad\qquad\quad $\forall\ell\in\N,\,\,|\phi|^\ell\in\CVZ^V_\R$, \qquad\qquad\qquad as desired.
\end{proof}

\begin{thm}
Let $V,W,X\in\ES$ and $k,\ell\in\N$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Then: \quad $\scrbigo_\ell^{WX}\,\circ\,\scrlito_k^{VW}\,\subseteq\,\scrlito_{k\ell}^{VX}$.
\end{thm}

\begin{proof}
\quad Want: \quad $\forall h\in\scrbigo_\ell^{WX}\,\circ\,\scrlito_k^{VW}$, $\,$ $h\in\scrlito_{k\ell}^{VX}$.\\
\phx\quad $\,\,$ Given $h\in\scrbigo_\ell^{WX}\,\circ\,\scrlito_k^{VW}$. \quad Want: $h\in\scrlito_{k\ell}^{VX}$.\\
Since $h\in\scrbigo_\ell^{WX}\,\circ\,\scrlito_k^{VW}$, choose $g\in\scrbigo_\ell^{WX}$ and $f\in\scrlito_k^{VW}$ s.t.~$h=g\circ f$.\\
Since $g\in\scrbigo_\ell^{WX}$, choose $\psi\in\BNZ^W_X$ s.t.~$g=\psi\cdot|\bullet|_W^\ell$.\\
Since $f\in\scrlito_k^{VW}$, choose $\phi\in\CVZ^V_W$ s.t.~$f=\phi\cdot|\bullet|_V^k$.\\
We have: $\forall t\in V$, \qquad\qquad\qquad $\,f_t=(\phi\cdot|\bullet|_V^k)=\phi_t\cdot|t|^k$, \qquad so\\
\phx\quad $(g\circ f)_t=g(f_t)=[\psi(f_t)]\cdot|f_t|^\ell=[(\psi\circ f)_t]\cdot|\,\,[\phi_t\cdot|t|^k]\,\,|^\ell$\\
\phx\qquad\qquad $\,\,=[(\psi\circ f)_t]\cdot|\phi_t|^\ell\cdot|t|^{k\ell}=((\psi\circ f)\cdot|\phi|^\ell\cdot|\bullet|_V^{k\ell})_t$.\\
\phx\qquad\qquad\qquad\qquad Then: $\,\,\,$ $g\circ f\,\,=\,\,\,(\psi\circ f)\cdot|\phi|^\ell\cdot|\bullet|_V^{k\ell}$.\\
We have \quad $f\in\scrlito_k^{VW}\subseteq\scrlito_0^{VW}=\CVZ^V_W$,\\
\phx\qquad so \quad $\psi\circ f\in\BNZ^W_X\circ\CVZ^V_W\subseteq\BNZ^V_X$.\\
Since $|\phi|=|\bullet|_W\circ\phi\in\CVZ^W_\R\circ\CVZ^V_W\subseteq\CVZ^V_\R$, we get $|\phi|^\ell\in\CVZ^V_\R$.\\
Then $g\circ f=(\psi\circ f)\,\,\cdot\,\,|\phi|^\ell\,\,\,\,\cdot\,\,|\bullet|_V^k$\\
\phx\qquad\qquad $\,\,\,\in\,\,\,\,\BNZ^V_X\cdot\CVZ^V_\R\cdot\,|\bullet|_V^{k\ell}$\\
\phx\qquad\qquad $\,\,\,\subseteq\,\,\,\,\qquad\CVZ^V_\R\quad\,\,\,\cdot\,|\bullet|_V^{k\ell}\,\,=\,\,\scrlito_k^{VX}$, \quad as desired.
\end{proof}

\begin{thm}
Let $V,W,X\in\ES$, and $k,\ell\in\N$.\\
\phx\qquad\qquad\qquad\qquad\qquad Then \qquad\qquad $\,\,\scrbigo_k^{WX}\,\circ\,\scrbigo_\ell^{VW}\,\subseteq\,\scrbigo_{k\ell}^{VX}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad and \quad $\,\scrlito_k^{WX}\,\circ\,\scrbigo_\ell^{VW}\,\subseteq\,\,\scrlito_{k\ell}^{VX}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad and \quad $\scrbigo_k^{WX}\,\circ\,\scrlito_\ell^{VW}\,\,\subseteq\,\,\scrlito_{k\ell}^{VX}$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad\qquad and \quad $\,\scrlito_k^{WX}\,\circ\,\,\scrlito_\ell^{VW}\,\,\subseteq\,\,\scrlito_{k\ell}^{VX}$.
\end{thm}

\begin{proof}
One of these is the preceding theorem, and the rest are proved similarly and left as unassigned HW.
\end{proof}

\subsection{The multi-variable $D$-derivative}

\begin{defn}
Let $V,W\in\ES$, $f:V\drta W$, $q\in V$.\\
Then $f_q^\T:V\drta W$ is defined by: $\,$ $\forall h\in V$, $\,\,\,$ $(f_q^\T)_h\,\,=\,\,f_{q+h}\,-\,f_q$.
\end{defn}

\begin{defn}
Let $V,W\in\ES$, $f:V\drta W$, $q\in V$.\\
\phx\qquad\qquad\qquad Then: \quad $\LINS_qf\,\,:=\,\,\{\,\,L\in\scrl^V_W\,\,|\,\,f_q^\T-L\in\scrlito_1^{VW}\,\,\}$.
\end{defn}

\begin{thm}
Let $V,W\in\ES$, $f:V\drta W$, $q\in V$.\\
\phx\qquad Assume: $\LINS_qf\ne\emptyset$. \qquad Then: $f_q^\T\in\scrbigo_1^{VW}$.
\end{thm}

\begin{proof}
Choose $L\in\LINS_qf$. \quad Then $L\in\scrl^V_W$ and $f_q^\T-L\in\scrlito_1^{VW}$.\\
Since $L\in\scrl^V_W$, we get $L:V\to W$, and so $L-L=\bfzero^V_W$.\\
Then $f_q^\T-L+L=f_q^\T$. \qquad
We have $L\in\scrl^V_W=\scrh_1^{VW}\subseteq\scrbigo_1^{VW}$.\\
Then $f_q^\T=(f_q^\T-L)+L\in\scrlito_1^{VW}+L\subseteq\scrbigo_1^{VW}+\scrbigo_1^{VW}\subseteq\scrbigo_1^{VW}$.
\end{proof}

\begin{thm}
Let $V,W\in\ES$, $f:V\drta W$, $q\in V$.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ Then: $\#\LINS_qf\le1$.
\end{thm}

\begin{proof}
Want: \quad $\forall L,M\in\LINS_qf$, $\,$ $L=M$.\\
Given $L,M\in\LINS_qf$. \qquad Want: $L=M$.\\
We have $f_q^\T-L\in\scrlito_1^{VW}$ and $f_q^\T-M\in\scrlito_1^{VW}$,\\
\phx\qquad and so $(f_q^\T-L)-(f_q^\T-M)\in\scrlito_1^{VW}-\scrlito_1^{VW}\subseteq\scrlito_1^{VW}$.\\
We have $f_q^\T\in\scrbigo_1^{VW}\subseteq\DNZ^V_W$, so $f_q^\T-f_q^\T=\bfzero^V_W$ near $0_V$ in $V$.\\
Then $(f_q^\T-L)-(f_q^\T-M)=M-L$ near $0_V$ in $V$.\\
So, since $(f_q^\T-L)-(f_q^\T-M)\in\scrlito_1^{VW}$ and since $\scrlito_1^{VW}$ is zero-local,\\
\phx\quad we conclude: $M-L\in\scrlito_1^{VW}$.\\
Also, $M-L\in\scrl^V_W-\scrl^V_W\subseteq\scrl^V_W=\scrh_1^{VW}$.\\
Then $M-L\in\scrh_1^{VW}\bigcap\scrlito_1^{VW}=\{\bfzero^V_W\}$.\\
Then $M-L=\bfzero^V_W$, and so $L=M$, as desired.
\end{proof}

\begin{defn}
Let $V,W\in\ES$, $f:V\drta W$, $q\in V$.\\
\phx\qquad\qquad\qquad\qquad\qquad $\,$ Then: \quad $D_qf\,\,:=\,\,\UE(\LINS_qf)$.
\end{defn}

\begin{thm}
Let $V,W\in\ES$, $f:V\drta W$, $q\in V$.\\
\phx\qquad\qquad\qquad\qquad\quad $\,$ Assume: $D_qf\ne\frownie$.\\
\phx\qquad\qquad\qquad $\,$ Then:\qquad\quad $\,\,(\,\,f_q^\T\in\scrbigo_1^{VW}\,\,)$\\
\phx\qquad\qquad\qquad\qquad\quad\qquad $\,\,\&\,\,(\,\,f\hbox{ is defined near }q\hbox{ in }V\,\,)$\\
\phx\qquad\qquad\qquad\qquad\quad\qquad $\,\,\&\,\,(\,\,f\hbox{ is continuous at }q\hbox{ from }V\hbox{ to }W\,\,)$.
\end{thm}

\begin{proof}
Since $\UE(\LINS_qf)=D_qf\ne\frownie$, we get: $\LINS_qf\ne\emptyset$.\\
Then $f_q^\T\in\scrbigo_1^{VW}$. \quad
Want: \qquad $(\,\,f\hbox{ is defined near }q\hbox{ in }V\,\,)$\\
\phx\qquad\qquad\qquad\qquad\qquad\qquad $\&\,\,\,(\,\,f\hbox{ is continuous at }q\hbox{ from }V\hbox{ to }W\,\,)$.\\
We have $f_q^\T\in\scrbigo_1^{VW}\subseteq\scrlito_1^{VW}\subseteq\scrlito_0^{VW}=\CVZ^V_W$.\\
Then: \qquad $(\,\,f_q^\T\hbox{ is defined near }0_V\hbox{ in }V\,\,)$\\
\phx\qquad\quad $\&\,\,\,\,(\,\,f_q^\T\hbox{ is continuous at }0_V\hbox{ from }V\hbox{ to }W\,\,)$.\\
Then: \qquad $(\,\,f\hbox{ is defined near }q\hbox{ in }V\,\,)$\\
\phx\qquad\quad $\&\,\,\,\,(\,\,f\hbox{ is continuous at }q\hbox{ from }V\hbox{ to }W\,\,)$,\\
as desired.
\end{proof}

\begin{thm}
Let $V,W\in\ES$, $f:V\drta W$, $L\in\scrl^V_W$, $q\in V$.\\
\phx\qquad\qquad\qquad Assume: $L\in\LINS_qf$. \qquad Then: $D_qf=L$.
\end{thm}

\begin{proof}
Since $L\in\LINS_qf$ and $\#\LINS_qf\le1$, we get: $\LINS_qf=\{L\}$.\\
Then $D_qf=\UE(\LINS_qf)=\UE\{L\}=L$, as desired.
\end{proof}

\begin{thm}
Let $V:=\R^2$ and $q:=(1,-1)$.\\
Define $f:V\to\R$ by: $\forall x,y\in\R$, $f(x,y)=x^3+9xy+3y^2$.\\
Define $L\in\scrl^V_\R$  by: $\forall s,t\in\R$, $L(s,t)=-6s+3t$. \qquad Then: $D_qf=L$.
\end{thm}

\begin{proof}
Want: $L\in\LINS_qf$. \quad Want: $f_q^\T-L\in\scrlito_1^{V\R}$.\\
We have: $\forall s,t\in\R$,\\
\phx\qquad $f_q^\T(s,t)\quad=\quad f_{q+(s,t)}\,\,-\,\,f_q\quad=\quad f_{(s+1,t-1)}\,\,-\,\,f_{(1,-1)}$\\
\phx\qquad\qquad\qquad $\,\,\,=\quad[(s+1)^3+9\cdot(s+1)\cdot(t-1)+3\cdot(t-1)^2]$\\
\phx\qquad\qquad\qquad\quad $\,\,-\,\,[\,\,\,\,\,\,\,\,1^3\,\,\,\,\,\,\,\,+9\,\,\,\cdot\,\,\,1\,\,\,\cdot\,\,\,\,\,\,\,\,(-1)\,\,+3\cdot\,\,(-1)^2\,\,\,]$\\
\phx\qquad\quad $=\quad\,\,\,(s^3+3s^2+3s)\,\,+\,\,(9st-9s+9t)\,\,\,\,\,\,+\,\,(3t^2-6t)$\\
\phx\qquad\quad $=\quad\,\,\,(-6s+3t)\,\,+\,\,(3s^2+9st+3t^2)\,\,+\,\,s^3$\\
\phx\qquad\quad $=\quad\,\,\,\,\,\,(\,L(s,t)\,)\,\,\,\,+\,\,(3s^2+9st+3t^2)\,\,+\,\,s^3$.\\
Let $S:=\pi_1^V$ and $T:=\pi_2^V$. \quad Then: $\forall s,t\in\R$,\\
\phx\quad $\,\,\,\,\,(f_q^\T-L)(s,t)\quad=\quad(\,\,f_q^T(s,t)\,\,)\,\,-\,\,(\,\,L(s,t)\,\,)$\\
\phx\qquad\qquad\qquad\qquad\quad $=\quad\,\,\,\,(\,3s^2+\,\,9st\,\,+\,3t^2\,)\,+\,\,s^3$\\
\phx\qquad\qquad\qquad\qquad\quad $=\quad(\,\,(3S^2+9ST+3T^2)\,+\,S^3\,\,)(s,t)$.\\
Then $f_q^\T-L=(\,3S^2\,\,\,+\,9ST\,\,+\,\,\,3T^2\,)+\,\,\,S^3$.\\
Then $f_q^T-L\in\,(\scrm_2^{V\R}+\scrm_2^{V\R}+\scrm_2^{V\R})+\scrm_3^{V\R}$\\
\phx\qquad\qquad\quad $\,\,\subseteq\quad\scrh_2^{V\R}+\scrh_3^{V\R}\quad\subseteq\quad\scrbigo_2^{V\R}+\scrbigo_3^{V\R}$\\
\phx\qquad\qquad\quad $\,\,\subseteq\quad\scrlito_1^{V\R}\,\,+\,\scrlito_1^{V\R}\quad\,\subseteq\quad\scrlito_1^{V\R}$, \quad as desired.
\end{proof}

\subsection{Miscellaneous}

The following is the $D$-derivative \bfindex[chain rule].

\begin{thm}
Let $V,W,X\in\ES$.\\
\phx\qquad\qquad Let $f:V\drta W$, let $g:W\drta X$ and let $q\in X$.\\
\phx\qquad\qquad Then: \quad $D_q(g\circ f)\,\,\,=^*\,\,\,(D_{f_q}g)\,\circ\,(D_qf)$.
\end{thm}

\begin{proof}
I owe you.
\end{proof}

Here is a definition we needed in class:

\begin{defn}
Let $V,W\in\ES$.\\
\phx\qquad\qquad Let $f:V\drta W$ and let $j\in\scri_V$.\\
\phx\qquad\qquad Then: \quad $\partial_jf\,\,\,:=\,\,\,\nabla_{\vep_j^V}f$.
\end{defn}

The notation ``$\partial_jf$'' is read ``the $j$th \bfindex[partial derivative] of $f$''.

\printindex[terms]

FOR OFFICE USE ONLY: \tref{thm-deriv-of-dbltransl}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{document}

% The remainder of this file is GNU Emacs commands to set the mode and
% other aspects of the editing session.  These must come at the end of
% the file 

% Local Variables:
% TeX-command-default: "LaTeX"
% TeX-parse-self: t
% TeX-auto-regexp-list:LaTeX-auto-regexp-list
% End:
