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

Question

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.

Answer 1

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 $U\in F$.

Answer 2

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 $$L(F)=L(\lim\mathcal{G})=L(\beta L(\mathcal{G}))=x.$$

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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.