HW4

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Homework Assignment 4

Required reading. Read, reread and rereread your notes to this point, and make sure that you really, really really, really really really understand everything in them. Do the same every week! Also, read all of Munkres sections 31, 32, 33, 35, 43, 44 (for fun) and 45.

Solve the following problems. (But submit only the underlined ones). In Munkres' book, page 199 problems 2, 7a, 7b, 7c and 7d (for fun), page 205 problems 3 and 5, pages 212-213 problems 3 and 7, and page 271 problems 8 and 10 (the last bit may be cancelled if we don't cover enough material. Stay tuned on Tuesday November 15th).

Due date. This assignment is due at the end of our class on Thursday, November 17, 2005.

On Term Exam 1. Monday November 7 from 6PM (UofT time) until 8PM at Sidney Smith 1083 (http://www.osm.utoronto.ca/cgi-bin/class_spec/spec03?bldg=SS&room=1083), choose 4 out of 5 questions, on everything covered in class until and including Thursday November 3 except the obvious "asides". Some questions may be straight class material (reproduce some proof or some definition, for example), some may be straight from homework, some may require some original thought and some may involve a combination of these three things. You may wish to take a look at exams I gave in 1993 (http://www.math.toronto.edu/~drorbn/classes/9293/131/midterm.pdf) and in 2004 (http://www.math.toronto.edu/~drorbn/classes/0405/Topology/TE1/Exam.html), but remember that the material for those exams was somewhat different.

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Just for fun. Participate in writing a page, Zorn from Choice, that will include a complete proof of Zorn's Lemma from the Axiom of Choice. We want the full details. But even more, we want understanding. Learn for yourself and tell others what happens in the proof and why. What is the global structure of the proof? Why is every step of the proof "natural"? Why is every step of the proof "the right thing to do"? Can you illustrate parts of the proof to make them more easily understandable?

Here are some hints and questions, following the proof in a 1982 Hebrew textbook on set theory by Shmockler (spell?), used in a class I (Dror) took when I was an undergraduate student.

Start with the following "climbing" (better name?) lemma, saying that if every up-going trail is convergent, every climber must come to a stop (can you say it better?). Precisely, if X is a poset in which every chain has a least upper bound and if f:X\to X climbs (i.e., \forall x\in X\, f(x)\geq x), then f has a fixed point - an x_0\in X with f(x0) = x0.

Instead of Zorn's Lemma, prove "The Hausdorff Maximality Principle", saying that in every partially ordered set (poset) P there is a maximal chain. Why is this equivalent to Zorn's Lemma? How does the climbing lemma imply the Hausdorff Maximality Principle? Idea: X will be the collection of all chains in P, ordered using inclusion. If there isn't a maximal chain in P, you can extend any given chain (climb). But every chain in X, a chain of chains in P, is least bounded by its union. Hence by the climbing lemma the extension process must come to a stop.

How about proving the climbing lemma? Start climbing at some point, any point, a_0\in X, and when you can't climb any more, you're at a good x0. But what's climbing? Simple iteration of f won't do, the mountain may be higher than just the integers. Instead, a "climber's path" will be a minimal subset A of X having the following three properties:

  1. a_0\in A.
  2. f(A)\subset A.
  3. For every chain C in A, \sup C\in A.

(Why is there such a minimal A? Is it reasonable to think of such an A as a climber's path? Do you have a good name for sets satisfying 1-3 without the minimality condition?)

If A is itself a chain, we are done (why?), so that's what we aim to show. We do this in steps, all using the same trick - if step k ain't true, there's a smaller set having the above three properties. The steps are:

  1. a0 is a lower bound for A.
  2. Let B:=\{x\in A: (y\in A)\wedge(y<x)\Rightarrow f(y)\leq x\} (what does this mean?). For any x\in B let B_x:=\{z\in B:(z\leq x)\vee(z\geq f(x))\} (in words?). Then Bx = A.
  3. B = A, hence A is a chain.

The axiom of choice is used here precisely once, and not in the hardest part of the proof. Where is it used?

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Table of contents
1 Problems for submission

1.1 31, Problem 7, p. 199
1.2 32, Problem 3, p. 205
1.3 33, Problem 7, p. 213
1.4 44, Problem 8, p. 271

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Problems for submission

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\S31, Problem 7, p. 199

Let p \colon X \to Y be a closed continuous surjective map such that p − 1({y}) is compact for each y \in Y. (Such a map is called a perfect map.)

  1. Show that if X is Hausdorff, then so is Y.
  2. Show that if X is regular, then so is Y.
  3. Show that if X is locally compact, then so is Y.
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\S32, Problem 3, p. 205

  1. Show that every locally compact Hausdorff space is regular.
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\S33, Problem 7, p. 213

  1. Show that every locally compact Hausdorff space is completely regular.
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\S44, Problem 8, p. 271

  1. If X and Y are spaces, define e \colon X \times \mathcal{C}(X,Y) \to Y by the equation e(x,f) = f(x); the map e is called the evaluation map. Show that if d is a metric for Y and \mathcal{C}(X,Y) has the corresponding uniform topology, then e is continuous. We shall generalize this result in \S46.
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