Alexander subbase theorem

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

Proof

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

Now clearly $C \cap S$ does not cover $𝑋$ , for if it did it would have a finite subcover. Thus there exists some $𝑥 \in 𝑋$ not covered by $C \cap S$ . But since $C$ does cover $𝑋$ , there exists some $𝑈 \in C$ with $𝑥 \in 𝑈$ . Since $S$ is a subbasis, there must exist some finite collection ${𝑆 𝑖} 𝑛 𝑖 = 1 \subseteq S$ of subbasic open sets such that $𝑥 \in ⋂ 𝑛 𝑖 = 1 𝑆 𝑖 \subseteq 𝑈$ . Now ${𝑆 𝑖} 𝑛 𝑖 = 1 ⊈ C$ , for otherwise $C \cap S$ would cover $𝑥$ . Defining $C 𝑆 𝑖$ as above and $˜ C = ⋂ 𝑛 𝑖 = 1 C 𝑆 𝑖$ , it follows ${𝑆 𝑖} \cup ˜ C$ is a finite subcover of ${𝑆 𝑖} \cup C$ for any $𝑖 \in ℕ 𝑛$ . Let $𝑌 = ⋃ ˜ C$ and $𝑍 = 𝑋 ∖ 𝑌$ . Now for any $𝐴 \subseteq 𝑋$ , ${𝐴} \cup ˜ C$ covers $𝑋$ iff $𝑍 \subseteq 𝐴$ , whence $𝑍 \subseteq 𝑆 𝑖$ for each $𝑖 \in ℕ 𝑛$ and thus $𝑍 \subseteq ⋂ 𝑛 𝑖 = 1 𝑆 𝑖 \subseteq 𝑈 \in C$ , implying ${𝑈} \cup ˜ C$ is a finite subcover of $C$ , contradicting $C \in 𝑃$ . Therefore $𝑋$ 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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Alexander subbase theorem

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