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.

  • 1
    $\begingroup$ A while ago I found this, which is quite detailed. $\endgroup$ – 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 $U\in F$.

  • $\begingroup$ Of course I looked at precisely the wrong nLab pages to answer my question :P Thanks! $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.