College Math Teaching

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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1 Comment »

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

    Pingback by Compact Spaces and the Tychonoff Theorem IV: conclusion | College Math Teaching — March 17, 2015 @ 9:14 pm


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