If a topological space $X$ enjoys the property that every ultrafilter $U$ on $X$ has a single limit, must $X$ be a Hausdorff space?
(Ultrafilters here consist of arbitrary subsets (so not necessarily, for example, $z$-sets or closed sets) but the limit of such a $U$ means the intersection of the closures of all its sets.)
I guess not, but I don't have an example.
An ultrafilter $U$ on the space $X$ is said to converge to the point $p$ if every neighborhood of $p$ belongs to $U$. (This is equivalent to saying that the complement of an element of $U$ is never a neighborhood of $p$. Thus it is also equivalent to saying that $p$ belongs to the closure of every element of $U$.)
If $U$ converges to both $p$ and $q$, then $p$ and $q$ cannot have disjoint neighborhoods, because this would mean that the empty set belongs to $U$.
Conversely, if $p$ and $q$ do not have disjoint neighborhoods then there is a proper filter containing all neighborhoods of $p$ and all neighborhoods of $q$ (namely the set of all subsets $S\subset X$ such that $S$ contains the intersection of a neighborhood of $p$ and a neighborhood of $q$), and therefore there is an ultrafilter converging to both $p$ and $q$.
Thus $X$ is a Hausdorff space if and only if no ultrafilter on the point set of $X$ converges to more than one point of $X$.
