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](https://math.stackexchange.com/questions/592357/can-the-ultrafilters-in-the-poset-of-open-subsets-be-made-into-a-topological-spa/4361689#4361689), where I prove that it is the “Gleason space” of $X$.



Ultrafilters of closed sets
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.



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> 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 $\omega(X)\cong\beta(X)$.
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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.



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> 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 $X$.
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> 2. $X$ 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.
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> 3. $X$ 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 $X$.
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