# What are the algebras for the ultrafilter monad on topological spaces?

Motivation: Let $$(X,\tau)$$ be a topological space. Then the set $$\beta X$$ of ultrafilters on $$X$$ admits a natural topology (cf. Example 5.14 in Adámek and Sousa - D-ultrafilters and their monads), giving rise to a functor $$\beta: \operatorname{Top} \to \operatorname{Top}$$ which admits the structure of a monad. It turns out that the algebras for this monad, which I'll call "$$\beta$$-spaces", admit the following description (which one can alternatively take as a definition).

Definition: A $$\beta$$-space consists of a topological space $$(X,\tau)$$ equipped with an additional topology $$\tau^\xi$$ on $$X$$ such that

1. $$(X, \tau^\xi)$$ is compact Hausdorff;
2. The topology $$\tau^\xi$$ refines the topology $$\tau$$; and
3. For every $$x \in X$$ and every $$\tau$$-open neighborhood $$U$$ of $$x$$, there exists a $$\tau$$-open neighborhood $$V$$ of $$x$$ such that the $$\tau^\xi$$-closure of $$V$$ is contained in $$U$$.

Notes:

• From (1) and (2) it follows that $$(X,\tau)$$ is compact.
• So if $$(X,\tau)$$ is additionally Hausdorff, then it admits a unique $$\beta$$-space structure, namely the one with $$\tau^\xi = \tau$$ (since continuous bijections of compact Hausdorff spaces are homeomorphisms).

• $$(X,\tau)$$ need not be Hausdorff—e.g., if $$\tau$$ is the indiscrete topology, then the topology $$\tau^\xi$$ can be an arbitrary compact Hausdorff topology.

• The compact Hausdorff topology $$\tau^\xi$$ traces back to Manes' theorem, which says that the algebras for the ultrafilter monad on $$\operatorname{Set}$$ rather than $$\operatorname{Top}$$ are precisely the compact Hausdorff spaces.

Questions:

1. Are there additional restrictions on the topology $$(X,\tau)$$ such that it admits a refinement $$\tau^\xi$$ satisfying (1), (2), (3) (i.e. constituting a $$\beta$$-space), beyond the fact, as noted, that $$X$$ must be compact?

2. Do $$\beta$$-spaces already have some other name? Or at least, is condition (3) above, relating a topology $$\tau$$ to a refinement $$\tau^\xi$$, something which has a name?

• This is so cool. I missed good point topology problems :) – Andrea Marino Dec 18 '19 at 1:12
• @AndreaMarino I agree! I think it's worthwhile to revisit some of these classic things from time to time. Manes' theorem, in particular, is a gem which deserves to be more widely known. The proof -- once you know what the Stone-Cech compactification of a discrete space is -- is an easy, fun application of the Beck Monadicity Theorem. And there's a whole cottage industry of extensions of these ideas, starting with a description of an arbitrary topological space as a kind of "lax algebra" for the ultrafilter monad. – Tim Campion Dec 18 '19 at 1:42
• Condition (3) looks like some sort of regularity of $\tau$ relative to $\tau^\xi$, for if $\tau = \tau^\xi$ it's just ordinary regularity, isn't it? – Andrej Bauer Dec 18 '19 at 8:19
• Well I found a pretty indirect characterization. There exist an explicitly constructible refinement $\tau'$, and $\tau$ defines a $\beta$ space iff $\tau'$ is compact and locally compact. If you want I can post it, but I think we can do better. A necessary condition though is that $X$ must be also locally compact . – Andrea Marino Dec 18 '19 at 12:23
• @LSpice Eh, you can view it as just being like a prime. It started because I was thinking of a $\beta$ structure in terms of the structure map $\xi: \beta X \to X$, but then the formulation I arrived at didn't mention this map at all. So it's kind of a relic. – Tim Campion Dec 18 '19 at 18:07

$$\DeclareMathOperator\cp{cp}$$We will derive some additional necessary conditions from the following

Observation: Let $$\tau$$ be a topology on $$X$$ and $$\tau'$$ a topology refining $$\tau$$. Suppose that $$(X,\tau')$$ is compact. Then any $$\tau'$$-closed set is $$\tau$$-compact.

Indeed, it is compact in $$\tau'$$ because it is closed in a compact, and so it is compact also in $$\tau$$ because the identity $$\tau' \to \tau$$ is continuous.

Consequences: Let $$(X,\tau)$$ be a topological space admitting a $$\beta$$-structure $$\tau^\xi$$. Then:

1. $$(X,\tau)$$ is compact (as noted in the question).

2. $$(X,\tau)$$ is locally compact (in the sense that for every $$x \in X$$ there is a local base of compact neighborhoods). This follows from condition (3) on a $$\beta$$-space and the Observation.

3. $$(X,\tau)$$ is "c-separated": For every disjoint $$C,D \subseteq X$$ which are either closed or singletons, there exist compact $$K,L \subseteq X$$ such that $$C \cap K = \emptyset$$, $$D \cap L = \emptyset$$, and $$K \cup L = X$$. This follows from the fact that $$(X,\tau^\xi)$$ is Hausdorff, regular, and normal and the Observation.

4. $$(X,\tau)$$ is "c-completely separated": Let $$C,D \subseteq X$$ be disjoint and either closed or singletons. Then there exists a (not necessarily continuous) function $$f: X \to [0,1]$$ such that $$f^{-1}(0) = C$$, $$f^{-1}(1) = D$$, and $$f^{-1}([a,b])$$ is compact for every $$a \leq b$$. This follows from the fact that $$(X,\tau^\xi)$$ has the corresponding separation property and the Observation.

Note also that if the collection of sets with compact complement forms a topology, this this topology is the unique $$\beta$$-structure on $$(X,\tau)$$. But this is not necessarily the case.

• Thanks! Remark 2 seems especially insightful to me. But I'm not sure that $cp(\tau)$ is a topology -- it's closed under finite intersections but is it closed under even finite unions? Also, in the absence of Hausdorffness, one needs to be careful about the definition of "locally compact". I see how condition (3) and remark (2) imply that for every $x$ and every open neighborhood $U$ of $x$, there exists an open neighborhood $V$ of $x$ such that $V \subseteq K \subseteq U$ for some compact $K$ -- is this equivalent to a standard definition? – Tim Campion Dec 18 '19 at 16:57
• Note also that if $cp(\tau)$ is a compact Hausdorff topology, then for any $\beta$-structure $\tau^\xi$, the map $id: (X, cp(\tau)) \to (X,\tau^\xi)$ is a continuous bijection of compact Hausdorff spaces and therefore a homeomorphism. So whenever $cp(\tau)$ is a compact Hausdorff topology, it is the unique $\beta$ structure on $(X,\tau)$. So I think $cp(\tau)$ must not be a topology when $X$ is an infinite indiscrete space. – Tim Campion Dec 18 '19 at 17:02
• Fiuu! You were right! I hope I fixed this on the edge :) however, the infinite indiscrete topology is not compact, so I don't see any contraddiction! – Andrea Marino Dec 18 '19 at 17:40
• By locally compact I mean that - at each point - you have a local basis of compact neighbourhoods. I am quoting for Wikipedia: this is a possible definition of the local compactness, which is equivalent to the classical one if the space is Hausdorff. The definition that I just quoted is indeed equivalent to: for any point x and open $V$ containing x, there exist an open $U$ containing x and a compact $K$ such that $U \subset K \subset V$. – Andrea Marino Dec 18 '19 at 17:59
• Indeed, let $(X,\tau)$ be an infinite indiscrete space. This is compact -- every open cover has a singleton subcover! It's also locally compact, and compactly-separated. Then $K \subseteq X$ is $\tau$-compact for every subset $K$, so $cp(\tau)$ is the discrete topology, which is not compact. – Tim Campion Dec 18 '19 at 18:04