Ultrafilter Lemma and Alexander subbase theorem

I'm trying to prove the equivalence in ZF between the Ultrafilter Lemma (UF) and the Alexander subbase theorem (AS). Although I've found a way to prove that (AS) $\Rightarrow$ (UF), with the help of the intermediate step "$2^X$ (endowed with the product topology) is compact for any set $X$", I didn't find a way to prove the converse.

So, I would appreciate any hints or references to prove that (UF) $\Rightarrow$ (AS).

Thanks.

Solution 1:

Hint: Suppose a space $X$ has a subbasis that satisfies the hypotheses of (AS). Assuming (UF), to prove that $X$ is compact it suffices to show that any ultrafilter on $X$ has a limit. So to prove (AS), suppose an ultrafilter $F$ on $X$ has no limit, and show that the set of subbasic open sets whose complements are in $F$ would then be an open cover with no finite subcover.

A full proof is hidden below.

Let $F$ be an ultrafilter on $X$. The set $S$ of limits of $F$ is the intersection of all closed sets in $F$. Every closed set is an intersection of basic closed sets, so $S$ is also equal to the intersection of all basic closed sets in $F$. Now if $C\in F$ is a basic closed set, $C$ is a finite union $C_1\cup\dots\cup C_n$ of subbasic closed sets. Since $F$ is an ultrafilter, $F$ contains some $C_i$. It follows that in fact $S$ is the intersection of all subbasic closed sets in $F$.

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Now suppose $F$ has no limit, so $S=\emptyset$. The complements of the subbasic closed sets in $F$ are then an open cover with no finite subcover (they cover $X$ because $S=\emptyset$, and have no finite subcover because $F$ is a proper filter). But this open cover consists of subbasic open sets, and so this contradicts our assumption.