7
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $F$.

${}$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.