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!

  • $\begingroup$ Thanks, yes, I meant clopen subsets. Edited to correct $\endgroup$ Aug 4, 2020 at 20:48

1 Answer 1


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.


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