EasyProofOfTychonoff

(Not Done)

Tychonoff's theorem says that the product of compact topological spaces is compact. We wish to give a fairly easy proof of this fact.

Definition: A collection of sets G is said to cover a set X if given any x in X there is a g(x) in G such that x is in g(x). A cover is said to be a T-cover if G is a subset of T. When T is the collection of open sets of the topological space X, a T-cover is also called an open cover. A cover is said to be a finite subcover if there are only finitely many g in G. A subset of G is called a subcover if it is also a cover of X. A topological space is called compact (or often quasi-compact) if every open cover of X has a finite subcover.

Theorem (Tychonoff): Suppose that is a set and for every suppose is a compact topological space. Endow with the product topology. is compact.

To prove this, we will make use of the following lemma.

Lemma (Alexander): If X is a topological space and S is a subbasis of X, then X is compact if and only if every S-cover of X has a finite subcover.

Proof of Tychonoff: Let S be the standard subbasis of the product topology, . By Alexander's lemma it suffices to consider S-covers. Let G be an S-cover, that is let G consist of sets for various and open in . Define . Assume BWOC, that for every that . Using the axiom of choice, choose in for every . Since is in X, for some and U open in . This is a contradiction since and . Therefore there must exist an such that . Since is compact, choose a finite subcover, . Let . C is a finite subcover of . Since G was an arbitrary S-cover, the conditions of Alexander's subbase lemma are satisfied, and X is compact.

Proof of Alexander subbasis lemma: Let B be the collection of all open covers of X that do not have finite subcovers, B for bad covers. Assume, by way of contradiction, that B is nonempty. Partial order B by set inclusion. Let C be a chain in B. Let D be the union of all elements of C. If D has a finite subcover, then those finitely many open sets would be contained in finitely many elements of the chain C. Every finite subset of the chain has a maximum element, and so each of those finitely many open sets that together cover X are contained in that maximum element. This is impossible since every element of C is an element of B, and thus has no finite subcover. Thus D has no finite subcover, so is in B, and is an upperbound for C inside B. Since C was an arbitrary chain in B, the conditions of Zorn's lemma are satisfied, and thus B has a maximal element, call it H. Consider . We show that G covers X, so that by hypothesis it has a finite subcover, but that finite subcover is a subcover of H as well. This is a contradiction, and so the assumption that B is nonempty is untenable. Thus every open cover of X has a finite subcover, and X is compact.