**Exercise 5.4.1, Su Gao, p 132.**

Show that for any equivalence relation on a Polish space , iff there are infinitely many -equivalence classes.

Suppose . So there exists Borel such that iff . Since for each , the element lies in a different class, there are infinitely many classes.

Suppose there are infinitely many -classes; enumerate a countable subset: . By the Axiom of Choice, for each pick a , and define by . Since has the discrete topology, is cts, and so $latex \text{id}(\omega) \le_B E$.

**Exercise 5.4.2, Su Gao, p 132**

Show that if an equivalence relation is smooth then either , or , or for some finite , .

By Theorem 5.4.2, if is smooth, either (1) or (2) .

So assume (2). Then either has infinitely many classes or it does not.

If it does, by Exercise 5.4.1, , hence .

If it does not, suppose has classes. List the classes: . Since is smooth, each is Borel.

Then define by iff . Then is Borel. Also , since has discrete topology and we can pick

a point in each equivalence class to map to .

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I’m taking a class on Descriptive Set Theory now. Here is the proof of an interesting theorem from class: Assuming the Axiom of Choice, there exists an undetermined game.

Here, a game consists of two players, a pruned tree, and a payoff set . The players move alternately by picking an immediate extension of the last move. Player I wins if, after infinitely many moves, they have created an element of . Player II wins otherwise.

For more information about these kinds of games, you may refer to this wikipedia entry: http://en.wikipedia.org/wiki/Determinacy.

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Let be a Polish space. Decompose uniquely into , where is perfect, is countable open. This decomposition is possible by Cantor-Bendixson Theorem.

Exercise (6.6): is the largest perfect subset of .

*Proof*.

Suppose is perfect (so closed and perfect in its relative topology). From the proof of Theorem (6.4), and in the notation defined there, we have , hence . Now suppose , and further that ; notice that , being a set, is also a Polish space.

Claim: is a perfect space.

Let . Let be an open nbhd of . That is, there exists open in such that . Then is also open in , since is open in . Since is perfect, there exists . Therefore, is perfect and nonempty, hence uncountable by Theorem (6.2).

But this contradicts that is countable. Hence, , and thus .

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(B) Exercise (4.21).

Show that the Hausdorff metric is compatible with the Vietoris topology.

Proof.

Let be a metric space with . Let denote the topology from the Hausdorff metric on , with basis , denote the Vietoris topology on , with basis

.

If , .

(i) .

Proof.

Let . Suppose . Then, since and , we’re done.

So, suppose now that . Cover with finitely many balls , , of radius , such that, for all , . Let .

I will show that satisfies .

Let . That is , and for all , . To see that , I check that and . Let . Then , so for some , so . Similarly, if , .

So .

(ii) .

Proof.

Let , . If , then . So suppose that . I have that , and , .

Claim 1: .

Proof.

Suppose . Then, I can pick such that . Let such that . Then, has a convergent subsequence, . But then , which is a contradiction since is closed.

Let . Let . Then such that . Then let .

Claim 2: .

Proof.

Let . That is, . Then, , . Then, for each , . So there exists such that . Hence , and .

Finally, let . Then , so (see Claim 1). So , and .

So .

And hence .

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Kechris, p. 26

(B) Exercise (4.23). Let be metric space with . Show that for nonempty ,

(i) .

Proof.

Suppose .

Thus, for all , there exists , such that implies .

That is, .

Let . Now, choose such that implies .

Now, choose a sequence in , such that , in the following way: For ,

, for some . Pick any such that .

Then , so .

(ii) .

Proof.

Assume . Let . Since

and since is compact, hence closed, .

.

some such that , and .

Therefore, . Therefore .

Therefore , which is a contradiction.

Thus, .

(i) and (ii) show that .

To show that the converse fails, consider the real numbers with the trivial metric.

Let , which is clearly compact. Then

And .

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Let be a topological space. We denote by the space of all compact subsets of equipped with the **Vietoris topology**.

Let be a metric space with . We define the **Hausdorff metric** on , , as follows:

, if ,

, if exactly one of is ,

, if ,

where .

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(B)

Let be a compact metrizable space and a metrizable space. We denote by the space of continuous functions from into with the topology induced by the uniform metric: where is a compatible metric for . A simple compactness argument shows that this topology is independent of the choice of .

*Proof.*

Let be two compatible metrics for . I will show that the two corresponding uniform metrics, give the same topology.

Suppose that did not give the same topology. That is, without loss of generality, there exists , , such that for all , . (Here .) That is ; that is, , which is the same as

and there exists such that.

Now, let . Let . Then there exists such that and . Since is compact, the sequence must have a convergent subsequence . So .

Let . Choose such that implies and . Then implies . So .

Therefore, converges to under . Thus, it must also do so under . But for all , . So this is a contradiction.

Hence, both metrics give the same topology.

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1. Exercise 4.9: Let be separable Banach spaces. The **weak topology** on is the one generated by the functions (from into the scalar field) . Show that if is reflexive, with the weak topology is compact metrizable. Find a compatible metric.

2. Exercise 4.10:

A **topological vector space** is a vector space (over , where ), equipped with a topology in which addition and scalar multiplication are continuous. So Banach spaces and their duals with the weak*-topology are topological vector spaces. A point in a convex set is **extreme** (in ) if , with , , implies . Denote by the **extreme boundary** of , i.e., the set of extreme points of .

Show that if is a compact metrizable (in the relative topology) convex subset of a topological vector space, the the set is in .

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(Kechris exercise 4.11)

*(a) Let be a pruned tree on . It follows that is compact iff is finite splitting.*

* (b) In particular, if is compact, there is such that for all** , for every . Conclude that is not a countable union of** compact sets.*

*Proof.*

(a) “”. Suppose is not finite splitting. So there exists , of length , such that infinitely many satisfy . For each here, there exists such that . Pick sequence . Then, , for any . Hence, there is no convergent subsequence. Hence, is not compact.

“”. Suppose is finite splitting. Let be a sequence in . If , , let be the following statement: is an infinite

set. Then, it is clear that there exists such that holds, since there are only finitely many sequences of length in (since is finite splitting).

Now, assume inductively that there exists such that holds. Then, again since is finite splitting, there are only finitely many choices for . Hence, since is infinite, there must exist such that

holds.

Hence we have inductively defined a sequence such that for some . That is, contains a convergent subsequence, so is compact.

(b) Let be compact. Then is closed, and hence corresponds to a pruned tree , such that . So, for each , is finite (by the above). So define by . This satisfies the above property.

Now, suppose , where each is compact. Then for each we have such that implies for all . Now define by . Then, for any . Therefore, .

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First, we define . Let . There exists , such that . Define . We need to check that is well-defined. Let , . Now, let . Then, since is Cauchy, there exists such that implies . Therefore, so converges in . Now, there exists such that implies . Therefore, , by the triangle inequality. So . Hence

Since were arbitrary we see that is well-defined on .

*(b) *

Now, we check that is linear.

Let . Then, , , where . Then, so is additive. The demonstration that , for is quite similar.

*(c)*

Now, we need to show that is bounded. Thus is bounded with .

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