Alexander subbase theorem

Remarks

Suppose $X$ is not compact. By Zorn's lemma, there is an open cover $\mathcal C$ without finite subcover that is maximal for inclusion among such covers. For $x \in U \in \mathcal C$, choose $S_1, \ldots, S_n \in \mathcal S$ with $x \in S_1 \cap \cdots \cap S_n \subseteq U$. If no $S_i$ were in $\mathcal C$, maximality would give finite sets $\mathcal C_i \subseteq \mathcal C$ with $S_i \cup \bigcup \mathcal C_i = X$, and then $U$ with all $\mathcal C_i$ would be a finite subcover. So $\mathcal C \cap \mathcal S$ covers $X$, and by hypothesis it has a finite subcover, a contradiction.

Example: the sets $[0, c)$ and $(c, 1]$ for $0 < c < 1$ form a subbasis of $[0, 1]$. A cover by such sets contains some $[0, c)$ and some $(c', 1]$ with $c' < c$, by a supremum argument, so $[0, 1]$ is compact. The theorem also gives a direct proof of Tychonoff's theorem.

The subbasis hypothesis is needed. On $\R$, the family $\set{\R}$ has the finite subcover property, but it is not a subbasis of the usual topology, and $\R$ is not compact.

Going deeper

For a student

The product topology has a natural subbasis: the preimages of open sets under single projections. With the theorem, compactness of a product of compact spaces reduces to covers by such sets, which is a short argument; this is one standard proof of Tychonoff's theorem. The proof uses Zorn's lemma, and the theorem is equivalent to the ultrafilter lemma over the axioms of set theory without choice.