College Math Teaching

March 17, 2015

Compact Spaces and the Tychonoff Theorem IV: conclusion

Filed under: topology — Tags: , , — collegemathteaching @ 9:14 pm

We are finishing up a discussion of the Tychonoff Theorem: an arbitrary product of compact spaces is compact (in the product topology, of course). The genesis of this discussion comes from this David Wright article.

In the first post in this series, we gave an introduction to “compactness”.

In the second post, we gave a proof that the finite product of compact spaces is compact.

In the third post, we gave come equivalent definitions of compactness

In particular, we showed that:

1. A space is compact if and only if the space has the following property: if A \subset X is an infinite union of open sets with no finite subcover, then A is a proper subset of X ; that is, X -A \neq \emptyset and

2. A space is compact if and only if the space has the following property: every infinite subset E has a perfect limit point. Note: a perfect limit point for a set E is a point x \in X such that, for every open U, x \in U , |U \cap E| = |E| (the intersection of every open neighborhood of a perfect limit point with E has the same cardinality as E .

Note the following about these two facts: each of these facts promises the existence of a specific point rather than the existence/non-existence of a cover of a particular type. Fact 1 promises the existence of an excluded point, and fact 2 promises the existence of a perfect limit point.

When it comes to a point in an infinite of topological spaces, constructing a point is really like constructing a sequence of points (in the case of countable products) or a net of points (in the case of uncountable products). That is, if one wants to construct a point in an infinite product of spaces, one can assume some well ordering of the index used in the product, then construct the first coordinate of the point from the first factor space, the second coordinate from the second factor space, and so on.

We’ll use fact one: the excluded point property to prove Tychonoff’s Theorem.

Proof. Assume that X = \Pi_{\alpha \in I} X_{\alpha} and that I is well ordered. We start out by showing that the product of two compact spaces is compact, and use recursion to get the general result.

Let \mathscr{O} be an infinite union of open sets in X_1 \times X_2 with no finite subcover. First, we show that there is some a \in X_1 such that for each open set U, a \in U , no finite subcollection of \mathscr{O} covers U \times X_2 . Now if there is some open U \subset X_1 where U  is disjoint from every \pi_1 (O), O \in \mathscr{O} we are done with this step. So assume not; assume that every U is a subset of the first factor of some O \in \mathscr{O} . If it isn’t the case that there is x \in X_1 where U_x \times X_2 has no finite subcover of elements of \mathscr{O} , for each such U_x \subset X_1, x \in X_1 there is a finite number of elements of \mathscr{O} that covers U_x \times X_2 . Now since X_1 is compact, a finite number of U_x covers X_1 , hence a finite subcover of \mathscr{O} covers ALL of X_1 \times X_2 . Hence some point a \in X_1 exists such that no finite subcover of \mathscr{O} covers U \times X_2 for any open U \subset X_1, a \in U .

Similarly, we can find b \in X_2 so that for all open V \subset X_2, b \in V , no finite subcollection of \mathscr{O} covers U \times V where U is a basic open set in X_1 that contains a . This shows that (a,b) \notin \cup_{O \in \mathscr{O}} because, if it were, this single point would lie in some basic open set U \times V which, by definition, is a finite subcover.

Now given an arbitrary product with a well ordered index set I we can now assume that there is some collection of open sets that lacks a finite subcover and inductively define a_{\gamma} \in X_{\gamma} so that, if U is any basic open set containing \Pi_{\alpha \leq \gamma} \{a_{\alpha} \} \times \Pi_{\alpha > \gamma} X_{\alpha} then no finite subcollection of \mathscr{O} covers U . The point (a_{\gamma}) thus constructed lies in no \mathscr{O} .

Note: if you are wondering why this “works”, note that we assumed NOTHING about the compactness of the remaining product space factors \Pi_{\alpha > \gamma} X_{\alpha} .
And remember that we are using the product topology: an open set in this topology has the entire space as factors for all but a finite number of indices. So we only exploit the compactness of the leading factors.

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March 16, 2015

Compact Spaces and Tychonoff’s Theorem II

Filed under: advanced mathematics, topology — Tags: , , — collegemathteaching @ 6:10 pm

Ok, now lets prove the following: If X, Y are compact spaces, then X \times Y is compact (in the usual product topology). Note: this effectively proves that the finite product of compact spaces is compact. One might call this a “junior” Tychonoff Theorem.

Proof. We will prove this theorem a couple of times; the first proof is the more elementary but less elegant proof. It can NOT be easily extended to show that the arbitrary product of compact spaces is compact (which is the full Tychonoff Theorem).

We will show that an open covering of X \times Y by basis elements of the form U \times V , U open in X and V open in Y has a finite subcover.

So let \mathscr{U} be an open cover of X \times Y . Now fix x_{\beta} \in X and consider the subset x_{\beta} \times Y . This subset is homeomorphic to Y and is therefore compact; therefore there is a finite subcollection of \mathscr{U} which overs x_{\beta} \times Y , say \cup^{\beta, k}_{i=1} U_{\beta, i} \times V_{\beta, i} Note that each U_{\beta, i} is an open set in X which contains x_{\beta} and there are only a finite number of these. Hence \cap^{\beta k}_{i=1} U_{\beta i} = U_{\beta} is also an open set which contains x_{\beta} . Also know that U_{\beta} \times Y \subset \cup^{\beta, k}_{i=1} U_{\beta, i} \times V_{\beta, i}

tychonoff

We can do this for each x_{\beta} \in X and so obtain an open cover of X by \cup_{x_{\beta} \in X} U_{\beta} and because X is compact, a finite subcollection of these covers X . Call these U_1, U_2, U_3....U_m . For each one of these, we have U_j \times Y \subset \cup^{j, k}_{i=1} U_{j, i} \times V_{j, i} .

So, our finite subcover of X \times Y is \cup^m_{j=1}\cup^{j, k}_{i=1} U_{j, i} \times V_{j, i} .

Now while this proof is elementary, it doesn’t extend to the arbitrary infinite product case.

So, to set up such an extension, we’ll give some “equivalent” definitions of compactness. Note: at some point, we’ll use some elementary cardinal arithmetic.

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