# Ultrafilters of closed sets

The following definition should be standard, but let me state it just in case there is some ambiguity:

If $$\mathscr{L}$$ is a set of subsets of a set $$X$$ that is closed under finite unions and intersections and contains $$\varnothing,X$$ (or more generally, if $$\mathscr{L}$$ is a distributive lattice with top and bottom elements), let us say that $$\mathscr{F} \subseteq \mathscr{L}$$ is a filter in $$\mathscr{L}$$ provided it contains $$X$$ (the top element), is closed under finite intersections (meets), and is an up-set (meaning that if $$A \subseteq B$$ with $$A \in \mathscr{F}$$ and $$B \in \mathscr{L}$$ then in fact $$B \in \mathscr{F}$$). Let us say that such a filter is proper iff it does not contain $$\varnothing$$; and that it is an ultrafilter iff it is proper and maximal for inclusion among proper filters.

If $$\operatorname{Ult}(\mathscr{L})$$ denotes the set of ultrafilters in $$\mathscr{L}$$, then for every $$A\in\mathscr{L}$$ we define a set $$Z(A) := \{\mathscr{U}\in\operatorname{Ult}(\mathscr{L}) : \mathscr{U} \ni A\}$$ of ultrafilters containing $$A$$ as an element. It is almost trivial that $$Z(A\cap B) = Z(A) \cap Z(B)$$, and it is also true, though slightly less trivial, that $$Z(A\cup B) = Z(A) \cup Z(B)$$ (sketch of proof: the inclusion $$\supseteq$$ is clear, so let us see $$\subseteq$$: if $$\mathscr{U} \ni A\cup B$$ and $$\mathscr{U} \not\ni A$$ then the filter generated by $$\mathscr{U}$$ and $$\{A\}$$ contains $$\varnothing$$, so $$A\cap U = \varnothing$$ for some $$U\in\mathscr{U}$$, and then $$\mathscr{U} \ni (A\cup B)\cap U = B\cap U \subseteq B$$ so $$\mathscr{U}\ni B$$).

In particular, the $$Z(A)$$ form a basis of closed sets for a topology on $$\operatorname{Ult}(\mathscr{L})$$ called the Zariski topology. (Digression: note that they also form a basis of open sets for another topology, the Stone topology; I mention this in passing, because this confused the hell out of me: part of the confusion comes from the fact that, if $$\mathscr{L}$$ is actually a Boolean algebra, — e.g., if we consider all subsets of $$X$$, or more generally the clopen subsets of a topological space $$X$$, — then the complement of $$Z(A)$$ is $$Z(X\setminus A)$$ so the two topologies coincide.)

As an example, if $$X$$ is a topological space and $$\mathscr{Z}$$ is the lattice of zero-sets of continuous real-valued functions on $$X$$ (“z-sets”), then $$\operatorname{Ult}(\mathscr{Z})$$, with its Zariski topology, is the Stone-Čech compactification of $$X$$ (Gillman & Jerison, Rings of Continuous Functions (1960), points (a) and (b) in the proof of theorem 6.5).

This example has long bothered me because z-sets seem to make a fairly incongruous appearance, and I wondered why we don't take closed sets instead, which seem more “fundamental”, and what happens if we do. Let me ask precisely that:

Question: if $$X$$ is a topological space, $$\mathscr{C}$$ is the lattice of closed sets of $$X$$, can we better describe the space $$\operatorname{Ult}(\mathscr{C})$$ of ultrafilters in $$\mathscr{C}$$, endowed with its Zariski topology? Does it have a name? What is “wrong” with it that makes it less interesting than the Stone-Čech compactification?

(Note that if $$X$$ is metric, then closed sets and z-sets coincide, so the above space is the Stone-Čech compactification.)

Maybe assume that $$X$$ is $$T_1$$, which ensures that we have a continuous map $$X \to \operatorname{Ult}(\mathscr{C})$$ taking $$x\in X$$ to the set of $$F \subseteq X$$ closed such that $$F\ni x$$.

Remark: For the description of the space of ultrafilters of open sets (with the Stone topology), see this answer on MSE, where I prove that it is the “Gleason space” of $$X$$.

• This topology is briefly mentioned here en.m.wikipedia.org/wiki/… in addition to the Stone topology. Jan 20, 2022 at 1:08
• I've seen the term de Groot dual for this topology on Google which can be defined for slightly more general spaces. Jan 20, 2022 at 1:12
• This is just the Wallman-Shanin compactification, right? (try the Encyclopedia of General Topology pg.218) Jan 20, 2022 at 16:31
• @Tyrone: Indeed, that seems to be the term I was looking for (except that with the terminology in the book you cite, it's just the “Wallman compactification”: Wallman-Shanin compactifications are a bit more general). And the main “problem” with it seems to be that it fails to be Hausdorff in general. Do you want to post this as an answer? Jan 20, 2022 at 17:28

The construction you describe when $$\mathscr{C}$$ consists of all closed sets of $$X$$ is known as the Wallman compactification of $$X$$. I'll denote if $$\omega(X)$$. It is due to Wallman; Lattices and topological spaces, Ann. Math. 39 (1938) 112-126.

Of course some sort of techincal assumption is required.

Let $$X$$ be a $$T_1$$ space. Then $$\omega(X)$$ is a compact $$T_1$$ space containing $$X$$ as a dense subspace. Moreover it has the property that every continuous map $$X\rightarrow K$$ into a compact Hausdorff space $$K$$ extends over $$\omega(X)$$. The space $$\omega(X)$$ is Hausdorff if and only if $$X$$ is normal, and in this case $$\omega(X)\cong\beta(X)$$.

Shanin later generalised Wallman's construction; On special extensions of topological spaces, Dokl. SSSR 38 (1943) 6-9, On separation in topological spaces, Dokl. SSSR 38 (1943) 110-113, On the theory of bicompact extensions of topological spaces, Dokl. SSSR 38 (1943) 154-156. The compactifications that Shanin constructed allowed for the ultrafilters to come from more general lattices $$\mathscr{L}$$ of closed subsets of $$X$$. Of course at the expense of added assumptions: $$\mathscr{L}$$ is required to be a so-called $$T_1$$-base for the closed subsets of $$X$$. Denote by $$\omega(X;\mathscr{L})$$ the Wallman-Shanin compactification built using the $$T_1$$-base $$\mathscr{L}$$.

Here are some examples to convince you that these compactifications are interesting.

1. $$X$$ is locally compact $$T_2$$ and $$\mathscr{L}$$ consists of all $$(i)$$ compact subsets of $$X$$, and $$(ii)$$ all closed subsets $$A\subseteq X$$ for which there is a compact $$K\subseteq X$$ with $$A\cup K=X$$. Then $$\omega(X;\mathscr{L})$$ is the Alexandroff compactification of $$X$$.
1. $$X$$ is Tychonoff and $$\mathscr{L}=\mathscr{Z}(X)$$ is the collection of zero sets. Then $$\omega(X;\mathscr{L})\cong\beta(X)$$, as you have recognised.
1. $$X$$ is rim-compact $$T_2$$ and $$\mathscr{L}$$ is the set of all finite intersections of regularly closed sets with compact boundaries. Then $$\omega(X;\mathscr{L})=\mathfrak{f}(X)$$ is the Freudenthal compactification of $$X$$.