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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.

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1 Answer 1

up vote 8 down vote accepted

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 $X$.

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I just thought it is good to mention that *$p$ belongs to the closure of every element of $U$* is the definition of *cluster point* of $U$. For ultrafilters, the limit and cluster point are equivalent notins, for arbitrary filter they may differ. –  Martin Sleziak Apr 14 '12 at 6:50

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