508 CHAPTER 19. HAUSDORFF SPACES AND MEASURES

Definition 19.2.2 S ⊆ τ is called a subbasis for the topology τ if the set B offinite intersections of sets of S is a basis for the topology τ .

Theorem 19.2.3 Let (X ,τ) be a topological space and let S ⊆ τ be a subbasis forτ . Then if H ⊆ X , H is compact if and only if every open cover of H consisting entirely ofsets of S admits a finite subcover.

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 admitfinite subcovers. Is H compact? Assume this is not so. Then what was just observedabout basic covers implies there exists a basic open cover of H, O , which admits no finitesubcover. Let F be defined as

{O : O is a basic open cover of H which admits no finite subcover}.

The assumption is that F is nonempty. Partially order F by set inclusion and use theHausdorff maximal principle to obtain a maximal chain, C , of such open covers and letD = ∪C . If D admits a finite subcover, then since C is a chain and the finite subcover hasonly finitely many sets, some element of C would also admit a finite subcover, contrary tothe definition of F . Therefore, D admits no finite subcover. If D ′ properly contains D andD ′ is a basic open cover of H, then D ′ has a finite subcover of H since otherwise, C wouldfail to be a maximal chain, being properly contained in C∪{D ′}. Every set of D is of theform

U = ∩mi=1Bi, Bi ∈S

because they are all basic open sets. If it is the case that for all U ∈ D one of the Bi isfound in D , then replace each such U with the subbasic set from D containing it. Butthen this would be a subbasic open cover of H which by assumption would admit a finitesubcover contrary to the properties of D . Therefore, one of the sets of D , denoted byU , has the property that U = ∩m

i=1Bi, Bi ∈ S and no Bi is in D . Thus D ∪{Bi} admitsa finite subcover, for each of the above Bi because it is strictly larger than D . Let thisfinite subcover corresponding to Bi be denoted by V i

1, · · · ,V imi,Bi. Consider {U,V i

j , j =1, · · · ,mi, i = 1, · · · ,m}. If p ∈ H \∪{V i

j}, then p ∈ Bi for each i and so p ∈ U . This istherefore a finite subcover of D contradicting the properties of D . Therefore, F must beempty. ■

19.3 The Product Topology and CompactnessNow here is the definition of the product topological space.

Definition 19.3.1 Let (Xi,τ i) for i ∈ I be a topological space. By the axiom ofchoice, ∏i∈I Xi ̸= /0. Then by definition, a basis for a topology on ∏i∈I Xi consists of sets ofthe form ∏i∈I Ai where Ai = Xi except forfinitely many i and for these, Ai ∈ τ i. A sub-basisfor this topology consists of sets of the form ∏i∈I Ai where Ai = Xi for all but a single i andfor this one, Ai ∈ τ i. This product topology is denoted by ∏τ i. The resulting topologicalspace is (∏i∈I Xi,∏τ i). “Subbasic” sets will be those which are in the sub-basis.

It is important that the basic open sets have the ith entries not all of Xi in only finitelymany i. If you don’t insist on this, you will be dealing with something called the “box