The Hausdorff maximal theorem is one of several convenient versions of the axiom of choice. For a discussion of this, see the appendix on the subject. There is this one, the well ordering principal, and Zorn’s lemma. They are all equivalent to the axiom of choice and which one you use is a matter of taste.
Theorem 21.2.1 (Hausdorff maximal principle) Let ℱ be a nonempty partially ordered set. Then there exists a maximal chain.
The main tool in the study of products of compact topological spaces is the Alexander subbasis theorem which is presented next. Recall a set is compact if every basic open cover admits a finite subcover. This was pretty easy to prove. However, there is a much smaller set of open sets called a subbasis which has this property. The proof of this result is much harder.
Definition 21.2.2 S⊆ τ is called a subbasis for the topology τ if the set ℬ of finite intersections of sets of S is a basis for the topology, τ.
Proof: The only if part is obvious because the subasic sets are themselves open.
If every basic open cover admits a finite subcover then the set in question is compact. Suppose then that H is a subset of X having the property that subbasic open covers admit finite subcovers. Is H compact? Assume this is not so. Then what was just observed about basic covers implies there exists a basic open cover of H, O, which admits no finite subcover. Let ℱ be defined as

The assumption is that ℱ is nonempty. Partially order ℱ by set inclusion and use the Hausdorff maximal principle to obtain a maximal chain, C, of such open covers and let

If D admits a finite subcover, then since C is a chain and the finite subcover has only finitely many sets, some element of C would also admit a finite subcover, contrary to the definition of ℱ. Therefore, D admits no finite subcover. If D^{′} properly contains D and D^{′} is a basic open cover of H, then D^{′} has a finite subcover of H since otherwise, C would fail to be a maximal chain, being properly contained in C∪

because they are all basic open sets. If it is the case that for all U ∈D one of the B_{i} is found in D, then replace each such U with the subbasic set from D containing it. But then this would be a subbasic open cover of H which by assumption would admit a finite subcover contrary to the properties of D. Therefore, one of the sets of D, denoted by U, has the property that

and no B_{i} is in D. Thus D∪

Consider

If p ∈ H ∖∪{V _{j}^{i}}, then p ∈ B_{i} for each i and sop ∈ U. This is therefore a finite subcover of D contradicting the properties of D. Therefore, ℱ must be empty. ■
This profound result will be used a little later and is an essential part of the proof of Tychonoff’s theorem which involves the notion of compactness of the product of compact sets.