TITLE: Lindelöf in terms of filters
QUESTION [6 upvotes]: The following characterisation of compactness is well known.

Let $X$ be a space. Then $X$ is compact if and only if every filter on $X$ has a cluster point if and only if every ultrafilter on $X$ converges.

Now recall that a space is Lindelöf if every open covering of it has a countable subcovering.
My question is whether there is a characterisation of the Lindelöf property in terms of filters similar to the statement given above for compactness.

REPLY [7 votes]: A space $X$ is Lindelöf iff every filter on $X$ with the countable intersection property has a cluster point. (A filter $\mathscr{F}$ has the countable intersection property if $\bigcap\mathscr{C}\ne\varnothing$ whenever $\mathscr{C}\subseteq\mathscr{F}$ is countable.) I doubt that you can get an analogue for the ultrafilter part, though. An ultrafilter with the countable intersection property would be countably complete. Take $X$ to be $\omega_1$ with the discrete topology. The only countably complete ultrafilter on $\omega_1$ are the principal ultrafilters, which all converge, but $X$ is not Lindelöf.
Added: I should probably add that this is not a special property of $\omega_1$. There is a countably complete free (i.e., non-principal) ultrafilter on a set $X$ iff $|X|$ is a measurable cardinal, and the existence of any such cardinals cannot be proved in $\mathsf{ZFC}$. They are well up in the hierarchy of large cardinals.