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I am new to the ultrafilters, so I apologise if the question is too elementary.

Let S be a collection of sets with the finite intersection property, in a non-compact Hausdorff space. S can be extended to an ultrafilter but can it be extended to a convergent ultrafilter, i.e., one containing the neighbourhood filter of some point?

This becomes simple if S has non-empty intersection (but this is not known) or when the space is compact (but it is not). If this can not be shown in general then, perhaps, under some more conditions on the space or the sets in S?

I would appreciate references to relevant techniques/literature as much as a direct answer.

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It cannot in general be extended to a convergent ultrafilter. Take the cofinite filter $\mathscr{F}$ on $\Bbb N$, where $\Bbb N$ has the discrete topology; then every non-principal ultrafilter on $\Bbb N$ extends $\mathscr{F}$, but none of them converges in $\Bbb N$.

Suppose that $\mathscr{F}$ is a filter on a space $X$. If $\mathscr{F}$ can be extended to an ultrafilter $\mathscr{U}$ that converges to some $x\in X$, then $x$ is a cluster point of $\mathscr{F}$, i.e., $x\in\bigcap_{F\in\mathscr{F}}\operatorname{cl}F$. Conversely, suppose that $x\in\bigcap_{F\in\mathscr{F}}\operatorname{cl}F$; and let $\mathscr{N}$ be the nbhd filter at $x$. Then $\mathscr{F}\cup\mathscr{N}$ can be extended to an ultrafilter $\mathscr{U}$, which clearly converges to $x$. Thus, a necessary and sufficient condition that $\mathscr{F}$ be extensible to a convergent ultrafilter is that $\mathscr{F}$ have a cluster point.

If you start with an arbitrary centred family $\mathscr{S}$ of subsets of $X$ (i.e., a family with the finite intersection property), just let $\mathscr{F}$ be the filter generated by $\mathscr{S}$, i.e., the set of all subsets $F$ of $X$ that contain the intersection of some finite subset of $\mathscr{S}$.

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  • $\begingroup$ Ok, not in general. But under any specific conditions? Note that the staring point is a collection S, not an arbitrary filter. $\endgroup$ Dec 4 '15 at 15:41
  • $\begingroup$ @Michal: $S$ generates a filter, so you might as well start with a filter. I’m in the process of extending my answer to say something about when it can be done. $\endgroup$ Dec 4 '15 at 15:42
  • $\begingroup$ @Michal: See if the added material is clear enough; I can expand it with a bit more detail if necessary. $\endgroup$ Dec 4 '15 at 15:49
  • $\begingroup$ This is perfectly clear. Thanks a lot. $\endgroup$ Dec 4 '15 at 16:01
  • $\begingroup$ @Michal: You’re very welcome. $\endgroup$ Dec 4 '15 at 16:02
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The statement "every filter extends to a convergent ultrafilter" is equivalent to compactness. First, suppose $X$ is not compact, and let $\{U_i\}$ be an open cover with no finite subcover. Then the collection $\{X\setminus U_i\}$ has the finite intersection property; let $F$ be the filter it generates and let $G$ be any ultrafilter containing $F$. For any $x\in X$, there is some $U_i$ such that $x\in U_i$, and hence some neighborhood of $x$ which is not in $G$. So $G$ cannot converge to any point. Thus the filter $F$ cannot be extended to a convergent ultrafilter.

Conversely, suppose $X$ is compact, and let $F$ be any filter. Choose any ultrafilter $G$ extending $F$, and take the intersection of all the closed sets in $G$. These sets have the finite intersetion property, so by compactness the intersection is nonempty. Take a point $x$ in the intersection. Then if $U$ is an open neighborhood of $x$, $X\setminus U$ is a closed set not containing $x$, so it cannot be in $G$. Since $G$ is an ultrafilter, that means $U\in G$. Thus $G$ contains every neighborhood of $x$ and thus converges to $x$.

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  • $\begingroup$ And in a Hausdorff space the intersection is to unique point, as well. $\endgroup$
    – Asaf Karagila
    Dec 4 '15 at 16:00
  • $\begingroup$ This I have seen, but the problem is that the space is not compact, while I need a convergent ultrafilter. The condition of filter generated by S having a cluster point, as described by Brian, is sufficient and necessary, so I can hardly expect anything more specific. $\endgroup$ Dec 4 '15 at 16:05

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