The question is in the title. Most of this post is a contextual preamble and some of my thoughts on the matter, I have not made much solid progress (and don't even know if this is true…)
$\newcommand{\U}{\mathscr{U}}\newcommand{\top}{\mathsf{Top}}\newcommand{\F}{\mathcal{F}}\newcommand{\N}{\mathcal{N}}$There is an ultrafilter functor $\U:\top\to\top$. To every space $X$, $\U(X)$ is defined to be the space with underlying set the set of all ultrafilters on $X$ (in the power set of $X$) and with the following topology: for $G\subseteq X$ we define $G^\ast$ to be the set of all ultrafilters on $X$ which contain $G$; $(G^\ast)_{G\text{ open}}$ defines a basis for the topology on $\U(X)$. $\U$ becomes a functor by sending continuous $f:X\to Y$ to the map $\U(X)\to\U(Y)$ which assigns to every ultrafilter $U$ the ultrafilter $\U(f)(U):=\{A\subseteq Y:f^{-1}(A)\in U\}$ – $\U(f)$ is clearly continuous and the assignment $f\mapsto\U(f)$ is clearly functorial.
I was interested in $\U$ a while ago because I was investigating the following remarkable facts:
- $\U$ is a compactification functor and has the same universal property as the Stone-Cech compactification $\beta$
- If $X$ is a discrete space, $\U(X)\cong\beta(X)$ canonically (this is the fundamental reason for "compact Hausdorff spaces are algebras for the ultrafilter monad", which is the (co)restriction of $\U$ to the category of sets)
I was recently reviewing my notes on this, and read:
Due to $(1)$, it is clear $\U(X)\cong\beta(X)$ iff. $\U(X)$ is Hausdorff iff. $X$ is discrete
It's pretty easy to prove $\U(X)$ is Hausdorff if $X$ is discrete, but I'm really struggling to prove:
If $\U(X)$ is Hausdorff, then $X$ is discrete
The proof I had in my notes was nonsensical, completely invalid. I don't even know if this claim is true – I have found nothing on the subject.
Fix a space $X$ with more than one point with $\U(X)$ being Hausdorff. We seek to prove or disprove that $X$ is discrete.
Some notes:
- It's possible to prove that for any open $G$ of $X$, $G^\ast$ is compact
- It then follows every $G^\ast$ is in fact also closed, so we have a local base of compact, clopen neighbourhoods
- Since it suffices to prove every $p\in X$ is an open point, there may be merit in considering the principal ultrafilter $\F_p$ at the point $p$
- By the ultrafilter principle, any nonempty proper filter on $X$ extends to (several) elements of $\U(X)$
In fact, my old nonsensical proof tried to rely on the "principal ultrafilters" associated to an arbitrary set $A$, but these do not in fact exist (unless $A$ is a singleton). If I want to show $A$ is open (or closed) I could discuss that $\F_A,\F_{X\setminus A}$ are distinct proper filters on $X$ which lift to (obviously distinct) ultrafilters $U_1$ and $U_2$ on $X$. So, these are separated: there are open $G_1,G_2$ of $X$ with $G_1\in U_1$, $G_2\in U_2$ but $G_1\notin U_2$, $G_2\notin U_1$.
Great… it's not been possible for me to go further, since "$G_1\in U_1$" doesn't conclude that $G_1$ contains $A$ since $U_1$ strictly extends $\F_A$ (most of the time).
The only thing that looks like 'progress' on this problem for is the observation that, if $\N(p)$ denotes the neighbourhood filter at $p$, it's clear that there are ultrafilters $U$ extending $\N(p)$, for example $\F_p$, and that the Hausdorff property implies each such $U$ must in fact equal $\F_p$. It feels like it may be possible to conclude $X$ is a discrete space, given:
The only ultrafilter extending $\N(p)$ is the principal ultrafilter $\F_p$ for any $p\in X$
I'm struggling to find a way to turn that into "$p$ is open". If $p$ is not open, then every neighbourhood of $p$ contains some other elements, so every neighbourhood of $\F_p$ contains some other elements, and so what? Or perhaps if $p$ is not itself open it's possible, by some set-theoretic shenanigans, to prove the existence of non-principal extensions of $\N(p)$ which would form a contradiction – but I have absolutely no idea how to do that, and indeed don't even know if it's true.
Or maybe there is a clever way to exploit the universal property of $\U$. Indeed, if $\U(X)$ is Hausdorff I know $\beta(X)\cong\U(X)$ in a 'nice' way. Constructions of $\beta$ are typically unpleasant, though; I can't see a way to continue that thought.
Is the claim true? If so, does anyone have a reference for the proof or a proof of their own?
Best Answer
Yes, this is true.
$\mathcal U(X)$ as a set is just $\beta(X^\delta)$, where $X^\delta$ is the discretisation of $X$.
The only difference is that you put on it a coarser topology. Since $\beta(X^\delta)$ is compact Hausdorff, a strictly coarser topology cannot be Hausdorff.
Thus, if $\mathcal U(X)$ is Hausdorff, then in fact it is equal (as a topological space) to $\beta(X^\delta)$. In particular, if you identify $X$ with the set of principal ultrafilters, then for every $x\in X$, $\{x\}$ is an open subset of $\mathcal U(X)$, so it must also be open as a subset of $X$.