Let $X$ denote a topological space and $O$ denote its poset of open subsets. Intuitively, $O$'s ultrafilters are kind of like generalized points of $X$. Is there a way to make these ultrafilters into a topological space $Y$ in their own right?

In a natural way, of course, so that the natural injection $f : X \rightarrow Y$ that maps a point of $X$ to its set of open neighbourhoods is continuous etc.

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    $\begingroup$ Are you familiar with the construction of the Stone–Čech compactification using ultrafilters? $\endgroup$ Dec 4, 2013 at 12:39
  • $\begingroup$ @HaraldHanche-Olsen, according to that link, $X$ has to be discrete... Unless that is a mistake in the article? $\endgroup$ Dec 4, 2013 at 12:44
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    $\begingroup$ In the Stone-Cech compactification, assuming sufficient regularity properties, one uses ultrafilters of closed sets. The space does not have to be discrete, but the Wikipedia article is not very clear. $\endgroup$ Dec 4, 2013 at 12:45
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    $\begingroup$ I know the construction which @CarlMummert mentions in his comment as Wallman compactification. $\endgroup$ Dec 22, 2013 at 15:01
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    $\begingroup$ @MartinSleziak, thanks. For non-normal spaces where they're different, how should we decide whether to use Stone-Cech, or whether to use Wallman? $\endgroup$ Dec 22, 2013 at 15:04

2 Answers 2


The standard topology for a set of ultrafilters is the "Stone topology". Let $X$ be the original space and let $U$ be the set of ultrafilters of the algebra of open sets of $X$. For each open set $O$ in $X$ we make an open set $N_O$ in $F$: $$ N_O = \{ F \in U : O \in F \}. $$ The Stone topology on $U$ is generated by the collection of these sets $N_O$. This topology is used in Stone's representation theorem, but in that context they only look at totally disconnected Hausdorff spaces ("Stone spaces").

The inclusion map from $X$ to $U$ sending each point to its neighborhood filter will be a homeomorphism onto its range, assuming $X$ has some mild separation properties ($T_1$ should be enough). This is in contrast to the Stone-Čech compactification, which requires that $X$ has to be a Tychonoff space for things to work out.

One way to see that your construction is not the Stone-Čech compactification: the compactification of $X = [0,1]$ is again $[0,1]$, but there are ultrafilters of open sets on $[0,1]$ that have empty intersection, and thus correspond to new points in $U$. For example, the filter of open sets $\{ (0,1/n) : n \in \mathbb{N}\}$ has empty intersection, and it extends to a maximal filter with the same property.

The idea of looking at filters of open sets, and coming up with conditions for when they will have nonempty intersection, is well-established. They usually use words like "topological completeness" for this; another closely related area is "domain representability". There are a lot of known results, and also many open questions.

  • $\begingroup$ Thank you this has been a very enlightening answer. $\endgroup$ Dec 4, 2013 at 13:02
  • $\begingroup$ This is called the soberification of $X$. $\endgroup$
    – Zhen Lin
    Dec 4, 2013 at 13:13
  • $\begingroup$ @ZhenLin I was wondering if you were pulling our collective leg, but apparently not. I see there is a related unanswered question on soberification. The notion of sobriety seems related, but different? $\endgroup$ Dec 4, 2013 at 14:21
  • $\begingroup$ A topological space is sober if and only if it is homeomorphic to its soberification. $\endgroup$
    – Zhen Lin
    Dec 4, 2013 at 15:44
  • $\begingroup$ @ZhenLin The soberification is related, but this can't be it: as Carl Mummert explains in this answer, there are ultrafilters of open sets on $[0,1]$ that don't correspond to points — whereas $[0,1]$ is Hausdorff, hence sober, so the soberification creates no new points. The confusion may come from the fact that the soberification can be described using “superfilters” of open sets: see Grayson, “Concepts of General Topology in Constructive Mathematics and in Sheaves II”, definition 1.3.2. $\endgroup$
    – Gro-Tsen
    Jan 19, 2022 at 14:29

This is an old question, but let me still explain how one can describe the topological space of ultrafilters of open sets, with the Stone topology.

So, let $X$ be a topological space (mild hypotheses on $X$ to be added later), $\mathscr{O}$ its set of open sets and $\mathscr{RO}$ its set of regular open sets, namely those that are equal to the interior of their closure. So $\mathscr{O}$ is a “frame” while $\mathscr{RO}$ is a complete Boolean algebra. The trivial inclusion $\mathscr{RO} \to \mathscr{O}$ preserves finite meets (which are just finite intersections) but not joins in general.

Recall that if $W$ is an open set, then $\operatorname{r.o.}(W) := \operatorname{int}(\operatorname{cl}(W))$, the interior of the closure of $W$, is the smallest regular open set containing $W$. Also recall that $\operatorname{r.o.}(W_1 \cap W_2) = \operatorname{r.o.}(W_1) \cap \operatorname{r.o.}(W_2)$; in particular, $W_1 \cap W_2 = \varnothing$ implies $\operatorname{r.o.}(W_1) \cap \operatorname{r.o.}(W_2) = \varnothing$.

Assertion: The map $\mathscr{U} \mapsto \mathscr{U} \cap \mathscr{RO}$ defines a bijection from the set of ultrafilters of $\mathscr{O}$ to that of ultrafilters of $\mathscr{RO}$, whose inverse is $\mathscr{U} \mapsto \{V \in \mathscr{O} : \operatorname{r.o.}(V) \in \mathscr{U}\}$.

Proof: First observe the following key facts:

  1. A filter $\mathscr{U}$ in $\mathscr{O}$ is an ultrafilter iff for every $V \in \mathscr{O}$ that does not belong to $\mathscr{U}$, there is $U \in \mathscr{U}$ such that $U\cap V = \varnothing$. (Indeed, the set of $W \in \mathscr{O}$ such that $W \supseteq U \cap V$ for some $U \in \mathscr{U}$, is a filter containing $\mathscr{U}$ and containing $V$, so it is the filter generated by $\mathscr{U}$ and $\{V\}$, and to say that $\mathscr{U}$ is an ultrafilter means precisely that this filter is improper for every $V \not\in \mathscr{U}$.) ❧ And exactly the same statement holds for $\mathscr{RO}$ (or indeed, suitably worded, for any distributive lattice).

  2. If $\mathscr{U}$ is an ultrafilter in $\mathscr{O}$ and $V \in \mathscr{O}$ is such that $\operatorname{r.o.}(V) \in \mathscr{U}$, then in fact $V \in \mathscr{U}$. (Indeed, if $V \not\in \mathscr{U}$, then by the previous point there is $U \in \mathscr{U}$ such that $U \cap V = \varnothing$, but then $U \cap \operatorname{r.o.}(V) = \varnothing$, contradicting the fact that $U \cap \operatorname{r.o.}(V) \in \mathscr{U}$.) ❧ The converse, of course, is trivial.

Now if $\mathscr{U}$ is an ultrafilter in $\mathscr{O}$, clearly $\mathscr{U} \cap \mathscr{RO}$ is a filter in $\mathscr{O}$. But it must even be an ultrafilter because if $V \in \mathscr{RO}$ is not in $\mathscr{U}$, by point 1 above (for $\mathscr{O}$), there is $U \in \mathscr{U}$ such that $U \cap V = \varnothing$, but then $\operatorname{r.o.}(U) \cap V = \varnothing$, while $\operatorname{r.o.}(U) \in \mathscr{U} \cap \mathscr{RO}$, and by point 1 again (for $\mathscr{RO}$ this time) this implies that $\mathscr{U} \cap \mathscr{RO}$ is an ultrafilter in $\mathscr{O}$.

Furthermore, the ultrafilter $\mathscr{U}$ in $\mathscr{O}$ can be recovered from the ultrafilter $\mathscr{U}_\flat := \mathscr{U} \cap \mathscr{RO}$ in $\mathscr{RO}$ as $\{V \in \mathscr{O} : \operatorname{r.o.}(V) \in \mathscr{U}_\flat\}$ by point 2 above. And conversely, if $\mathscr{U}$ is an ultrafilter in $\mathscr{RO}$ then $\mathscr{U}_\sharp := \{V \in \mathscr{O} : \operatorname{r.o.}(V) \in \mathscr{U}\}$ is an ultrafilter in $\mathscr{O}$ (it's clearly a filter, and the “ultra” part comes, e.g., from two applications of point 1 as above but easier), and we have $\mathscr{U}_\sharp \cap \mathscr{RO} = \mathscr{U}$ trivially.

To summarize, $\mathscr{U} \mapsto \mathscr{U}_\flat := \mathscr{U} \cap \mathscr{RO}$ and $\mathscr{U} \mapsto \mathscr{U}_\sharp := \{V \in \mathscr{O} : \operatorname{r.o.}(V) \in \mathscr{U}\}$ are inverse bijections between the set of ultrafilters in $\mathscr{O}$ and that in $\mathscr{RO}$. ∎

If on either set of ultrafilters we place the “Stone topology” whose open sets are generated by the $\mathfrak{N}(V) := \{\mathscr{U} : V \in \mathscr{U}\}$ for $V$ in the corresponding lattice ($\mathscr{O}$ or $\mathscr{RO}$), which are indeed the basis for a topology on account of $\mathfrak{N}(V_1 \cap V_2) = \mathfrak{N}(V_1) \cap \mathfrak{N}(V_2)$ (which is clear), then the two maps are easily seen to be continuous, so the two spaces are homeomorphic.

To summarize: ultrafilters in the frame $\mathscr{O}$ of open sets of $X$ can be identified (even with their Stone topology) with ultrafilters in the complete Boolean algebra $\mathscr{RO}$ of regular open sets of $X$, aka the Stone space of the complete Boolean algebra $\mathscr{RO}$.

Why is this any better? Because the Stone spaces of Boolean algebras have been extensively studied, and, in particular, the Stone space of the Boolean algebra of regular open sets of $X$, so, the space this question was about, is known as the Gleason space $\Theta X$ of $X$: see Porter & Woods, Extensions and Absolutes of Hausdorff Spaces (Springer 1988), §6.6(a) (p. 458). Of particular importance is (for $X$ Hausdorff) the subspace $EX$ of $\Theta X$, known as the Iliadis absolute, consisting of those ultrafilters which converge to a point $x$ in $X$ (in the sense that they are finer than the filter of regular open neighborhoods of $x$).

To summarize: the space ultrafilters of open sets of $X$ is Gleason space $\Theta X$ of $X$, which has been well studied.

This is in general not the Stone-Čech compactification of $X$, but it can be related to it: in the particular case where $X$ is compact (Hausdorff), the space $\Theta X = EX$ is described in this MathOverflow question as the projective limit of the Stone-Čech compactifications of the open dense subsets of $X$ (yet another description is provided in the answer).

  • $\begingroup$ Can you provide a reference for your claim that if $W_1,W_2$ are open sets then $\operatorname{r.o.}(W_1\cap W_2)=\operatorname{r.o.}(W_1)\cap \operatorname{r.o.}(W_2)$? According to this, $\operatorname{r.o.}(W_1\cap W_2)=(W_1\cap W_2)^{\perp\perp}\subseteq (W_1^\perp\cup W_2^\perp)^\perp = W_1^{\perp\perp}\cap W_2^{\perp\perp}=\operatorname{r.o.}(W_1)\cap \operatorname{r.o.}(W_2)$. $\endgroup$
    – Chad K
    Feb 28 at 8:25
  • $\begingroup$ If $\mathcal U$ is an ultrafilter in $\mathscr{RO}$ and $W_1,W_2\in\mathcal{U}_\sharp := \{V \in \mathscr{O} : \operatorname{r.o.}(V) \in \mathcal{U}\}$, why is $W_1\cap W_2\in\mathcal{U}_\sharp$? This is weaker than claiming $\operatorname{r.o.}(W_1 \cap W_2) = \operatorname{r.o.}(W_1) \cap \operatorname{r.o.}(W_2)$. I don't see why it's true though. Can this theorem be found somewhere? $\endgroup$
    – Chad K
    Feb 28 at 11:47
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    $\begingroup$ @ChadK My quick'n'dirty explanation why $r.o.(W_1\cap W_2)=r.o.(W_1)\cap r.o.(W_2)$ is to note that $\neg\neg(P_1\land P_2) \Leftrightarrow (\neg\neg P_1)\land(\neg\neg P_2)$ in intuitionistic propositional logic (IPC) and note that by a result of Tarski, open sets in a topological space provide a sound semantics for IPC, and furthermore $\neg\neg$ gets interpreted as $r.o.$. But there's probably a simpler (and less roundabout) way to do it. I'll give this some thought. $\endgroup$
    – Gro-Tsen
    Feb 28 at 14:46
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    $\begingroup$ @ChadK The operations $\land,\lor,\Rightarrow$ correspond respectively on open sets to intersection, union and $\operatorname{int}((X\setminus U)\cup V)$ (for $U\Rightarrow V$, where $X$ is the whole space; this is the largest open $W$ such that $U\cap W \subseteq V$); in particular, $\neg$ corresponds to the interior of the complement. The keyword “Heyting algebra” [of open sets] should yield additional clarifications. (But again, there certainly is a less roundabout answer to your question.) $\endgroup$
    – Gro-Tsen
    Feb 28 at 19:23
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    $\begingroup$ I succeeded in reconstructing it from Tarski's definition 4.1 and the proof of the identity you mentioned in intuitionistic logic. For open $A,B,C$, 1) $(A\setminus B^\perp)^\perp=(A\cap\overline{B})^\perp=(A\cap B)^\perp$ and 2) $(A\setminus C)^{\perp\perp}=A^{\perp\perp}\cap C^\perp$. Number 2 is particularly tricky although maybe there's a shortcut. Then setting $C=B^\perp$ gives $A^{\perp\perp}\cap B^{\perp\perp}=(A\setminus B^\perp)^{\perp\perp}=(A\cap B)^{\perp\perp}$. $\endgroup$
    – Chad K
    Feb 28 at 20:30

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