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

Question

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?

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.