# Can this ultrafilter convergence condition be expressed as a compactness condition?

Suppose that $$X$$ is a topological space. Let us say that an ultrafilter $$\mathcal U$$ on the Boolean algebra $$C_X$$ of clopen subsets of $$X$$ is partition-prime if whenever $$X = \amalg_{i \in I} X_i$$ is a clopen partition of $$X$$ we have $$X_i \in \mathcal U$$ for some $$i \in I$$. The set $$X'$$ of partition-prime ultrafilters on $$X$$ forms a topological space under the usual Stone topology with basis $$[V] = \{ \mathcal U \in X' \mid V \in \mathcal U\}$$ for $$V \in C_X$$, and we have a continuous map $$\eta \colon X \to X'$$ sending $$x$$ to $$\{ V \in C_X \mid x \in V \}$$. This is clearly the reflector onto a certain full subcategory of topological spaces; the game is to figure out which one.

It is easy to see that $$\eta$$ is injective just when $$X$$ is totally disconnected, and that $$\eta$$ is a subspace inclusion just when $$X$$ is zero-dimensional Hausdorff. We would like to know when, moreover, $$\eta$$ is surjective, and hence a homeomorphism. This will happen precisely when each partition-prime ultrafilter on $$C_X$$ is of the form $$\eta(x)$$ for some $$x \in X$$. It is easy to re-express this condition (under the assumption that $$X$$ is zero-dimensional Hausdorff) as saying that $$\eta$$ is surjective precisely when each partition-prime ultrafilter on the full powerset $$\mathcal P X$$ converges.

My question is whether there is a natural way of re-expressing this convergence condition as a compactness condition. The obvious guess is that this is the same as ultraparacompactness: every open cover of $$X$$ can be refined to a disjoint clopen cover.

It is not so difficult to show that our convergence condition is implied by ultraparacompactness. Indeed, if $$X$$ is ultraparacompact, and $$\mathcal U$$ is a completely partition-prime ultrafilter on $$\mathcal P X$$ with no convergent point, then the open sets not in $$\mathcal U$$ cover $$X$$; we can refine this cover to a disjoint cover $$X = \amalg_i X_i$$; and so $$X_i \in \mathcal U$$ for some $$i$$, contradicting the fact that $$X_i \subseteq U$$ for some open $$U \notin \mathcal U$$.

The other direction is much less clear. Suppose $$X$$ is zero-dimensional Hausdorff and ultraparacompact, and suppose towards a contradiction that $$X = \bigcup_i U_i$$ were an open cover with no disjoint refinement. Clearly it does no harm to assume that this is a cover by clopens, and the assumption of no disjoint refinement implies in particular that there is no finite subcover (since a finite cover by clopens can always be refined to a disjoint one). So the sets $$V_i = U_i^c$$ generate a proper filter $$\mathcal F$$ with no adherent point. Moreover, any clopen partition $$X = \amalg_j X_j$$ must fail to refine the cover $$(U_i)_{i \in I}$$, and so there must exist some part $$X_j$$ of the partition which meets each set in $$\mathcal F$$. Clearly, this is a necessary condition for $$\mathcal F$$ to be extendable to a partition-prime ultrafilter. If it were also sufficient we would be done: for then we could choose a partition-prime $$\mathcal U$$ extending $$\mathcal F$$ which, since $$\mathcal F$$ has no adherent point, would not converge: a contradiction.

So, my claimed characterisation of the property that every partition-prime ultrafilter converges as ultraparacompactness would follow if we could prove the following ultrafilter lemma:

If $$\mathcal F$$ is a proper filter such that for any clopen partition $$X = \amalg_j X_j$$ there is some $$X_j$$ which meets each set in $$\mathcal F$$, then there is a partition-prime ultrafilter extending $$\mathcal F$$.

An obvious attempt at a transfinite argument works well at successor stages, but falls over at limit ordinals. So there remain three possibilities: this is true; this is false; this is independent of the axioms of set theory, and I don't know which of these it is.

Another way of looking at this ultrafilter lemma is via locale theory. Given any locale $$L$$, we can obtain a new locale $$L'$$ by taking sheaves on the Boolean algebra of complemented elements of $$L$$, for the topology whose covers are all clopen partitions. Now $$L \to L'$$ is a reflector into zero-dimensional ultraparacompact (aka "strongly zero dimensional") locales. Clearly the passage $$X \mapsto X'$$ described above sends $$X$$ to the space of points of the strongly zero-dimensional reflection $$\mathcal O(X)'$$: which would thus be ultraparacompact so long as $$\mathcal O(X)'$$ were spatial.

Now clearly not every strongly zero-dimensional locale is spatial (take any atomless complete Boolean algebra). But it's possible that the strongly zero-dimensional reflection of any spatial locale is spatial, and this is in fact exactly equivalent to the ultrafilter lemma I quote above.

Any ideas, either about the ultrafilter lemma, or about a different characterisation of the spaces of the form $$X'$$, would be gratefully received!

• Thanks, yes, I meant clopen subsets. Edited to correct – Richard Garner Aug 4 '20 at 20:48

Your condition amounts to saying that the uniformity $$\mathcal{C}$$ generated by all clopen partitions is complete: 'partition-prime' is the same as '$$\mathcal{C}$$-Cauchy'. It is also implied by the condition that the space is $$\mathbb{N}$$-compact, which means that every clopen ultrafilter with the countable intersection property is fixed: if $$\mathcal{U}$$ is $$\mathcal{C}$$-Cauchy then it has the countable intersectionproperty; if $$\{U_n:n\in\omega\}$$ is a decreasing sequence in $$\mathcal{U}$$ with empty intersection (and $$U_0=X$$) then $$\{U_n\setminus U_{n+1}:n\in\omega\}$$ is a clopen partition with no member in $$\mathcal{U}$$.
In New properties of Mrowka's space $$\nu\mu_0$$ Kulesza showed the space $$\nu\mu_0$$ from the title is $$\mathbb{N}$$-compact; it is not strongly zero-dimensional, hance not ultraparacompact.