# What is the topology associated with the algebras for the ultrafilter monad?

It is easy to find references stating that the category of compact Hausdorff spaces $$\mathbf{CompHaus}$$ is equivalent to the category of algebras for the ultrafilter monad, $$\mathbf{\beta Alg}$$. After doing some digging, the $$\mathbf{CompHaus}\to \mathbf{\beta Alg}$$ half of the equivalence is simple enough, but I haven't been able to find a description of the $$\mathbf{\beta Alg}\to \mathbf{CompHaus}$$ half of this equivalence. I have tried to work it out, but I have little experience with topological spaces and am not sure what the associated topology ought to look like.

I'm wondering if anyone has a good reference that describes the $$\mathbf{\beta Alg}\to \mathbf{CompHaus}$$ half of the equivalence, or can describe it here.

• A while ago I found this, which is quite detailed. – Arnaud D. Apr 19 at 9:47

The other half of the equivalence is described on the nLab page on ultrafilters.

Given an algebra structure $$\xi\colon \beta X \to X$$, we define the topology on $$X$$ by declaring that a subset $$U\subseteq X$$ is open if and only if for every point $$x\in U$$ and every ultrafilter $$F\in \beta X$$ such that $$\xi(F) = x$$, we have $$U\in F$$.

• Of course I looked at precisely the wrong nLab pages to answer my question :P Thanks! – Malice Vidrine Apr 18 at 20:12

The idea is simple: in a topological space $$X$$, a subset $$A\subseteq X$$ is closed iff it is closed under limits of ultrafilters. That is, $$A$$ is closed iff for any ultrafilter $$F$$ supported on $$A$$ (i.e., $$A\in F$$), all limits of $$F$$ are in $$A$$.

Now suppose we are given a $$\beta$$-algebra $$L:\beta X\to X$$. The idea is that $$L$$ takes each ultrafilter to its limit. So, a subset $$A\subseteq X$$ should be closed iff it is closed under limits of ultrafilters according to $$L$$: that iff $$L(F)\in A$$ for all $$F\in\beta X$$ such that $$A\in F$$. Or, restating this contrapositively in terms of open sets, a set $$U$$ is open iff for all $$F\in\beta X$$, $$L(F)\in U$$ implies $$U\in F$$.

It is easy to verify that this does define a topology on $$X$$, and that every ultrafilter $$F$$ converges to $$L(F)$$ with respect to this topology (it is essentially by definition the finest topology such that this is true). Verifying that $$L(F)$$ is the unique limit of $$F$$ (and therefore the topology is compact Hausdorff since each ultrafilter has a unique limit) is trickier and requires you to use the associativity and unit properties of $$L$$ as a $$\beta$$-algebra.

In detail, first let $$A\subseteq X$$ and define $$C(A)=\{L(F):F\in\beta X,A\in F\}.$$ I claim that $$C(A)$$ is closed. Suppose $$F\in\beta X$$ and $$C(A)\in F$$. Let $$\mathcal{F}$$ be the filter on $$\beta X$$ generated by the sets $$L^{-1}(B)$$ for $$B\in F$$ together with $$\{G\in\beta X:A\in G\}$$; these have the finite intersection property since every $$B\in F$$ has nonempty intersection with $$C(A)$$. Let $$\mathcal{G}$$ be an ultrafilter on $$\beta X$$ extending $$\mathcal{F}$$ (so $$\mathcal{G}\in\beta\beta X$$). We now use the associativity property of $$L$$, which says that $$L(\lim \mathcal{G})=L(\beta L(\mathcal{G}))$$ where $$\lim:\beta\beta X\to \beta X$$ is the structure map of the monad $$\beta$$ at $$X$$. Explicitly, $$\lim\mathcal{G}$$ is defined as the set of $$B\subseteq X$$ such that $$\{G\in\beta X:B\in G\}\in \mathcal{G}$$. In particular, by our choice of $$\mathcal{G}$$, we have $$A\in\lim\mathcal{G}$$. On the other hand, $$\beta L(\mathcal{G})$$ is by definition the set of $$B\subseteq X$$ such that $$L^{-1}(B)\in\mathcal{G}$$, and so $$F\subseteq \beta L(\mathcal{G})$$. Since $$F$$ is an ultrafilter, this means $$\beta L(\mathcal{G})=F$$. We thus conclude that $$L(\lim\mathcal{G})=L(F)$$ where $$A\in\lim\mathcal{G}$$, and thus $$L(F)\in C(A)$$.

So, for any $$A\subseteq X$$, $$C(A)$$ is closed. Note also that $$A\subseteq C(A)$$ by the unit property of $$\beta$$: $$L$$ sends each principal ultrafilter to the corresponding point. (It follows easily that in fact $$C(A)$$ is the closure of $$A$$, though we will not use this.)

Now let $$F\in\beta X$$ and $$x\in X$$ and suppose $$F$$ converges to $$x$$ in our topology; we will show that $$x=L(F)$$. Since $$F$$ converges to $$x$$, every closed set in $$F$$ contains $$x$$. Thus for all $$A\in F$$, $$x\in C(A)$$, since $$C(A)$$ is a closed set and $$C(A)\in F$$ since $$A\subseteq C(A)$$. Let $$\mathcal{F}$$ be the filter on $$\beta X$$ generated by the sets $$T_A=\{G\in\beta X:L(G)=x,A\in G\}$$ for $$A\in F$$; they have the finite intersection property since $$T_A\cap T_B=T_{A\cap B}$$ and $$x\in C(A)$$ for all $$A\in F$$. Extend $$\mathcal{F}$$ to an ultrafilter $$\mathcal{G}$$ on $$\beta X$$. Similar to the argument above, we then have $$F=\lim\mathcal{G}$$ and $$\{x\}\in \beta L(\mathcal{G})$$ so $$\beta L(\mathcal{G})$$ is the principal ultrafilter at $$x$$. By the associativity and unit properties of $$L$$, we thus have $$L(F)=L(\lim\mathcal{G})=L(\beta L(\mathcal{G}))=x.$$

Morally, what is going on in these arguments is that the associativity property of $$L$$ says that $$L$$ is "continuous" from the standard topology on $$\beta X$$ to $$X$$ (see this answer of mine for some elaboration on that idea). So, to prove that $$C(A)$$ is closed, if we have a net in $$C(A)$$ converging to a point $$x$$, we can choose ultrafilters supported on $$A$$ which $$L$$ maps to each point of this net. Then if we take an appropriate accumulation point of these ultrafilters as points of $$\beta X$$, $$L$$ will map this accumulation point to $$x$$ by continuity. This will then give an ultrafilter which is supported on $$A$$ (because it is a limit of ultrafilters supported on $$A$$) which $$L$$ maps to $$x$$, to prove that $$x\in C(A)$$.

Then, if $$F$$ converges to $$x$$, we have $$x\in C(A)$$ for all $$A\in F$$, so we can pick ultrafilters $$G$$ with $$L(G)=x$$ which are supported on arbitrarily small elements of $$F$$. These ultrafilters then converge in $$\beta X$$ too $$F$$, and so continuity of $$L$$ says $$L(F)=x$$ as well.