Math 5345 Fall semester 2000
Selected solutions to exercises due Wednesday, October 11
Posted October 24, 2000
Page 42, #1
One way to approach this problem systematically is to divide it into cases, depending on how many subsets of the form {point} there are in the topology. Let's call the basic set {a,b,c}.Page 42, #2
We're given that A = (0,1){2}[3,4].Page 43, #6
(ii) Recall that Ai is always a subset of A. So, if Ab is also a subset of A, then A contains its closure. Since the closure always contains A, we conclude that A is equal to its closure under the given hypothesis. Since the closure is always a closed set, we conclude that A is closed.
Page 43, #8
We're given that A = {rational numbers}Proof that Ai is empty. This time, let a be a rational number. If e is a [small] positive number, we know that there is an integer n such that 0 < 1/n < e. So, a - 1/n and a + 1/n are rational numbers. One thing to say is that we can mimic what was done in that previous exercise to show that there is an irrational number between a - 1/n and a + 1/n, so that a cannot be a point of Ai. Somewhat more explicitly, we can observe that a + 1/(n) is such an irrational number. So, it follows in the same way that a cannot be a point of Ai.
Page 47 #2
The key observation is that if a1, , an are points of X, then:Page 47 #3
.The hypothesis is that every set of the form {a} is closed in the given topology. So, it follows that X - {a} is open in the topology, and therefore (by the equality in problem #2 above) that every complement of a finite set is open in the given topology. We conclude that the topology of finite complements is coarser than [or equivalent to] any topology that satisfies this hypothesis.Page 47, #4
The first step is to show that the set of open rectangles is a basis, i.e. that the intersection of finitely many open rectangles is either empty or an open rectangle. . Thus, suppose that the kth rectangle Rk is described by the inequalities ak < x < bk and ck < y < dk for k = 1, ,n. Then the intersection of the rectangles R1, , Rn is described by the inequalities inequalities a < x < b and c < y < d, where, a = max(a1, ,an); b = min(b1, ,bn); c = max(c1, ,cn); and d = min(d1, ,dn).Next, we show that every open rectangle is open in the metric topology: given a point p in an open rectangle R, an open disk centered at p with radius r is contained in R, if we take r to be the minimum of the perpendicular distances from p to the sides of R.
Finally, we show that every open disk is open in the "rectangle topology". Namely, if p is contained in the open disk D, we can find a rectangle containing p and contained in D
as follows: (1) draw a radius of the disk, passing
through p, (2) draw a rectangle with a corner at
the point where this radius meets the boundary of D, and the
opposite corner located on the same radius, just large enough so
that p is in the interior of the rectangle.
Comments and questions to: roberts@math.umn.edu
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