Topological subbasis

Alexander subbase theorem

Let be a topological space. Then is compact iff has a subbasis such that every open subbasic cover of (i.e. by elements of ) has a finite subcover. topology

Proof

The forward direction is trivially proven. For the converse, suppose towards contradiction that is not compact but every subbasic open cover from has a finite subcover. Let denote the set of all open covers of lacking a finite subcover partially ordered by inclusion. By Zorn’s lemma, has a maximal element , which itself is an open cover of admitting no finite subcover, By maximality, if is an open set such that then has a finite subcover for some finite .

Now clearly does not cover , for if it did it would have a finite subcover. Thus there exists some not covered by . But since does cover , there exists some with . Since is a subbasis, there must exist some finite collection of subbasic open sets such that . Now , for otherwise would cover . Defining as above and , it follows is a finite subcover of for any . Let and . Now for any , covers iff , whence for each and thus , implying is a finite subcover of , contradicting . Therefore must be compact.

This proof requires Zorn’s lemma, and therefore depends on the Axiom of Choice, however it may be formulated to only require the weaker Ultrafilter lemma which is equivalent to the Boolean prime ideal theorem.

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