If $(X,\mathcal T)$ is a topological space then $\mathcal F\subseteq\mathcal T$ is called an open filter on $X$ is (i) $X\in\mathcal F$, $\emptyset\notin\mathcal F$; (ii) $A,B\in\mathcal F$ $\Rightarrow$ $A\cap B\in\mathcal F$; $A\in\mathcal F$, $A\subseteq B\in\mathcal T$ $\Rightarrow$ $B\in\mathcal F$. (The definition of an open filter is similar to the definition of a filter, the only difference is that we are working only with open subsets of $X$.)

A maximal open filter is called an open ultrafilter. We can show, using Zorn lemma, that every open filter is contained in an open ultrafilter. In fact, we can show using Zorn lemma that every system of open sets, which has finite intersection property, is contained in an open ultrafilter.

If I am not missing something, then we can show that for an open ultrafilter $\mathcal F$ we have $\bigcap \mathcal F\ne\emptyset$ if and only if $\mathcal F$ converges.

Let us denote by $\mathcal N_x$ the system of all neighborhoods of $X$.

Suppose that $x\in\bigcap \mathcal F$. Clearly the system $\mathcal N_x\cup\mathcal F$ than has finite intersection property. (Any set from this system contains $x$.) Maximality of $\mathcal F$ then implies $\mathcal N_x\subseteq\mathcal F$, which means that $\mathcal F$ converges to $x$.

On the other hand, if $\mathcal F$ converges, then there exists a point $x\in X$ such that $\mathcal N_x\subseteq\mathcal F$. This clearly implies $x\in\bigcap\mathcal F$. EDIT: As the counterexample posted in the answer shows, this is a place where my argument contained a mistake. In this way I can only show: An open ultrafilter converges if and only if it has a cluster point. (Or: An open ultrafilter $\mathcal F$ converges to $x$ if and only if $x$ is a cluster point of $\mathcal F$.)

The reason I have some doubts about this that I have seen mentioned in some papers that an open ultrafilter converges if and only if it has a cluster point.1 I thought that it would be quite logical to mention also the observation that it converges if and only if it has a non-empty intersection.

1 For example: C. Liu: Absolutely Closed Spaces; Transactions of the American Mathematical Society Vol. 130, No. 1 (Jan., 1968), pp. 86-104 AMS, JSTOR.

EDIT 2: My original claim is not true for Hausdorff spaces, either. It suffices to take any H-closed space (for example, any compact Hausdorff space) and choose any system of open subsets of $X$, which has finite intersection property, but has empty intersection. Using Zorn Lemma we can show that there is an open ultrafilter $\mathcal F$ containing this system. For this open ultrafilter we have $\bigcap \mathcal F = \emptyset$. But, since we are working in a H-closed space, the open ultrafilter $\mathcal F$ converges.

(A Hausdorff space $X$ is called H-closed if or absolutely closed if it is closed in any Hausdorff space, which contains $X$ as a subspace. It is known that a Hausdorff space is H-closed if and only if every open ultrafilter on $X$ converges.2)

More concrete example was given in Daniel Fischer's comment below.

2See, for example, Berri, M. P. ; Porter, J. R. ; Stephenson, R. M., Jr. A survey of minimal topological spaces, dml.cz or elsewhere

  • $\begingroup$ One last thing, A open ultrafilter in $X$ converges if and only if it has a cluster point, and the space $X$ is Hausdorff. The counterexample, actually has every point in the space as a cluster point. $\endgroup$ – Bill Trok Mar 30 '14 at 9:40
  • $\begingroup$ In the other words: In your example every point of $X$ is a limit of that open ultrafilter. Also, every point of $X$ is a cluster point of that open ultrafilter. (Which agrees with the claim: An open ultrafilter $\mathcal F$ converges to $x$ if and only if $x$ is a cluster point of $\mathcal F$.) By the way, thanks for your help! $\endgroup$ – Martin Sleziak Mar 30 '14 at 9:44

Look at the finite complement topology on $\mathbb{R}$. Any open set in this topology contains all but finitely many points, so for any two open sets $U$ and $V$, $U \cap V$ is nonempty. This means that the collection of all nonempty open sets form an open ultrafilter. However, the intersection of all of these sets is empty.

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    $\begingroup$ But I do not think that this open ultrafilter converges. $\endgroup$ – Martin Sleziak Mar 30 '14 at 9:12
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    $\begingroup$ en.wikipedia.org/wiki/… Going off of this definition of converge, I believe it should. I mean it should contain every open neighborhood of every point, right? $\endgroup$ – Bill Trok Mar 30 '14 at 9:15
  • $\begingroup$ Of course, you are right, the open ultrafilter you suggested converges to every point of $\mathbb R$. $\endgroup$ – Martin Sleziak Mar 30 '14 at 9:36
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    $\begingroup$ In $\mathbb{R}$ with the standard topology, an open ultrafilter containing the open filter generated by $\{(0,\varepsilon) : \varepsilon > 0\}$ has empty intersection and converges to $0$. (Also @MartinSleziak) $\endgroup$ – Daniel Fischer Mar 30 '14 at 10:01
  • $\begingroup$ Yes, this shows that my original claim is not true for Hausdorff space either @DanielFischer. (I have added remark about this in my recent edit.) $\endgroup$ – Martin Sleziak Mar 30 '14 at 10:03

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