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 $U$.


  • 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.


  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?

  • $\begingroup$ This is so cool. I missed good point topology problems :) $\endgroup$ Dec 18, 2019 at 1:12
  • $\begingroup$ @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. $\endgroup$
    – Tim Campion
    Dec 18, 2019 at 1:42
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    $\begingroup$ 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? $\endgroup$ Dec 18, 2019 at 8:19
  • $\begingroup$ 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 . $\endgroup$ Dec 18, 2019 at 12:23
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    $\begingroup$ @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. $\endgroup$
    – Tim Campion
    Dec 18, 2019 at 18:07

1 Answer 1


$\DeclareMathOperator\cp{cp}$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.

  • $\begingroup$ 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? $\endgroup$
    – Tim Campion
    Dec 18, 2019 at 16:57
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    $\begingroup$ 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. $\endgroup$
    – Tim Campion
    Dec 18, 2019 at 17:02
  • $\begingroup$ 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! $\endgroup$ Dec 18, 2019 at 17:40
  • $\begingroup$ 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 $U \subset K \subset V$. $\endgroup$ Dec 18, 2019 at 17:59
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    $\begingroup$ 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. $\endgroup$
    – Tim Campion
    Dec 18, 2019 at 18:04

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