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It can be easily proven using the ultrafilter lemma that if every ultrafilter on a topological space converges to at most one point, then the space is Hausdorff. My question is is whether this statement implies the ultrafilter lemma.

It certainly implies some fragment of it, for instance it implies that every infinite set has a nonprincipal ultrafilter (the cofinite topology on an infinite set is not Hausdorff and also has the property that every principal ultrafilter converges to a unique point).

Apologies if this is in the consequences of AC book… I’ve temporarily lost access to my copy.

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  • $\begingroup$ I can't see it on the list of Form 14 in Consequences. $\endgroup$
    – Asaf Karagila
    Jan 16 at 21:07
  • $\begingroup$ Also didn't find anything remotely similar elsewhere in the book. Herrlich's also seem to focus more on compactness properties. $\endgroup$
    – Asaf Karagila
    Jan 16 at 21:28
  • $\begingroup$ @Asaf thanks for looking. Yeah, the analogous characterization of compactness is definitely equivalent to UL (don’t recall if Herrlich proves that one… probably). $\endgroup$ Jan 16 at 21:40

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Yes, this is equivalent to the ultrafilter lemma. Let $F$ be a proper filter on a set $X$, which we may assume to not be contained in any principal ultrafilter. Consider the space $Y=X\sqcup\{a,b\}$ with the topology that a set is open iff either it contains an element of $F$ or it contains neither $a$ nor $b$. (This is the topology "generated" by saying the filter $F$ converges to both $a$ and $b$, i.e. the finest topology with that property.) This space is not Hausdorff, since $a$ and $b$ do not have disjoint neighborhoods. However, any other two points of $Y$ do have disjoint neighborhoods (here we use the assumption that $F$ is not contained in any principal ultrafilter to ensure that $a$ and $b$ can be separated from the points of $X$).

Now suppose there is some ultrafilter $G$ on $Y$ that converges to two different points, which can only be $a$ and $b$. This means that $G$ contains every neighborhood of both $a$ and $b$, so it contains $A\cup\{a\}$ and $A\cup\{b\}$ for each $A\in F$. It follows that $G$ contains every element of $F$, so the restriction of $G$ to $X$ is an ultrafilter on $X$ that contains $F$.

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  • $\begingroup$ I wonder if there's a direct proof, without working towards a contradiction. It kind of feels like this proof is as direct as it gets. Very nice! $\endgroup$
    – Asaf Karagila
    Jan 16 at 23:21

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