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.
 A: 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.

A: I am learning about ultrafilters, and I found this question as I was searching for application and equivalents of the Ultrafilter Theorem. My answer to (UT)$\Rightarrow$(AS) is different: Let $\mathcal{B}$ be the subbasis in the hypothesis of (AS), and let $\mathcal{C}$ be a family of closed subsets of $X$ having the finite intersection property but empty intersection, i.e. $\bigcap\mathcal{C}=\emptyset$.
The idea is to consider the filter generated by $\mathcal{C}$ and use (UT) to work with an ultrafilter containing $\mathcal{C}$ to reach a contradiction.
Indeed, consider the collection
$$F=\{S\subseteq X:C_{1}\cap\ldots\cap C_{n}\subseteq S\:\mbox{for some $n\in\mathbb{N}$ and some $C_{1},\ldots,C_{n}\in\mathcal{C}$}\}.$$
It is easy to verify that $F$ is a filter (indeed it is the smallest filter containing $\mathcal{C}$). By (UT) we have that $F$ is contained in some ultrafilter $F^{*}$. As $F^{*}$ is an ultrafilter, for every set $E\subseteq X$ we have that $E\in F^{*}$ or $X-E\in F^{*}$. This property will be key to reach a contradiction.
Consider the collection $\mathcal{B}^{c}\cap F^{*}$, where $\mathcal{B}^{c}=\{X-B:B\in\mathcal{B}\}$ (note that $\mathcal{B}^{c}$ is a family of closed subsets of $X$ with the property that any collection of elements in $\mathcal{B}^{c}$ having empty intersection must have a finite subcollection also having empty intersection).
Note that the collection $\mathcal{B}^{c}\cap F^{*}$ cannot have empty intersection. Indeed, if it had empty intersection there would be a finite subcollection in $\mathcal{B}^{c}\cap F^{*}$ having empty intersection, which is not possible since that collection would contain a non-empty finite intersection of elements in $\mathcal{C}$, which has the finite intersection property.
Let $x\in\bigcap(\mathcal{B}^{c}\cap F^{*})$. Then, since $\mathcal{C}$ has empty intersection, the complement $\mathcal{C}^{c}=\{X-C:C\in\mathcal{C}\}$ is an open cover of $X$. Therefore, since $\mathcal{B}$ is a subbasis for the topology of $X$ there is some open set $X-C\in\mathcal{C}^{c}$ and $B_{1},\ldots,B_{n}\in\mathcal{B}$ such that $x\in B_{1}\cap\ldots\cap B_{n}\subseteq X-C$.
Now, by the choice of $x$ it is clear that $X-B_{i}\notin\mathcal{B}^{c}\cap F^{*}$ for all $i=1,\ldots, n$. This implies that $B_{i}\in F^{*}$ for all $i=1,\ldots, n$ (the key property of ultrafilters put to good use), which in turn implies that $B_{1}\cap\ldots\cap B_{n}\in F^{*}$. Therefore, there are $C_{1},\ldots, C_{m}\in\mathcal{C}$ such that $C_{1}\cap\ldots\cap C_{m}\subseteq B_{1}\cap\ldots\cap B_{n}$.
Then the collection $\{C,C_{1},\ldots, C_{m}\}$ is a finite subcollection in $\mathcal{C}$ having empty intersection, which is a contradiction since $\mathcal{C}$ has the finite intersection property.
