We call a topological space $X$ extremally disconnected if the closures of its open sets remain open. Obviously, Hausdorff extremally disconnected spaces are totally disconnected in the sense that their connected components are singletons.

Suppose $\mathbb{Z}_p = \varprojlim_{n \ge 1} \mathbb{Z}/p^n\mathbb{Z}$ is the ring of $p$-adic integers, viewed as a closed subspace of the compact Hausdorff space $\prod_{n \ge 1} \mathbb{Z}/p^n\mathbb{Z}$, where the latter is equipped with the product topology and each factor $\mathbb{Z}/p^n\mathbb{Z}$ is discret. Then it is well-known that $\mathbb{Z}_p$ is totally disconnected. But I can't answer the seemingly naive question: is $\mathbb{Z}_p$ extremally disconnected?

Following the traditions of operator algebraists, we call a compact Hausdorff extremally disconnected space a stonean space. And we call a stonean space hyperstonean if it admits enough normal measures---a technical condition which characterises hyperstonean spaces exactly as the spectrum of abelian von Neumann algebras. A series of possibly less naive questions are: if $\mathbb{Z}_p$ is stonean (which is the same as its extremal disconnectedness), is it hyperstonean? And if it is hyperstonean, can we associate some natural Hilbert space (like $L^2(\mathbb{Z}_p, \mu)$, where $\mu$ is the normalized Haar measure on $\mathbb{Z}_p$ viewed as a compact abelian topological group), and an abelian von Neumann algebra (hopefully the algebra $C(\mathbb{Z}_p)$ of complex-valued continuous functions on $\mathbb{Z}_p$) acting on this Hilbert space, such that $\mathbb{Z}_p$ is exactly the spectrum of this latter von Neumann algebra?

Finally, a perhaps stupid and somewhat less well-formulated question: do we have some "nice" examples of stonean, even hyperstonean spaces, that are infinite, and are not constructed as the spectrums of their associated operator algebras (of course they do appear as such spectrums, I mean one describe the relevant topologies without referring first to the algebra of functions on it)?

I ask these questions because as far as I know, unlike the widely spread notion of totally disconnected spaces (or $0$-dimensional spaces), (hyper)stonean spaces seems to be only of a minor interest to operator algebraists as spectrums of abelian von Neumann algebras (except Gleason's theorem characterising them as the projective object in the category of compact Hausdorff spaces). And the only way of producing nontrivial stonean spaces that I am aware of is taking the spectrum of some algebras. If $\mathbb{Z}_p$ does turn out to be stonean, it would be a cute concrete example in my humble opinion.

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    $\begingroup$ It's not "extremely" but "extremally". Clearly the Cantor space is not extremally disconnected and $\mathbf{Z}_p$ is a Cantor space. $\endgroup$
    – YCor
    Apr 2, 2019 at 13:56
  • $\begingroup$ proofwiki.org/wiki/Cantor_Space_is_not_Extremally_Disconnected $\endgroup$
    – R.P.
    Apr 2, 2019 at 14:01
  • $\begingroup$ As for the question "is there an example of a "nice" infinite stonean space", I could formulate the more precise: is there an infinite stonean space of cardinal $<2^c$; I'm not aware of any. (Caution, stonean space does not mean Stone space, bad terminology!) $\endgroup$
    – YCor
    Apr 2, 2019 at 14:05
  • $\begingroup$ @YCor, nice answer. I know cantor space is not extremally disconnected. Clearly I should be more familiar with Brower's theorem characterizing cantor space as compact Hausdorff space without isolated points and admits a countable clopen basis. $\endgroup$ Apr 2, 2019 at 14:10
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    $\begingroup$ Actually it's easy to check that every extremally disconnected, metrizable space is discrete. Indeed otherwise, choose an accumulation point $x$, choose an injective sequence $(x_n)$ converging to $x$ and choose an open subset $U$ (a suitable union of balls) containing each $x_{2n}$ but whose closure contains none of the $x_{2n+1}$. Then the closure of $U$ is not open. $\endgroup$
    – YCor
    Apr 2, 2019 at 14:18

2 Answers 2


Since questions recirculate to the front page forever if left unanswered, I will amalgamate the comments into an answer, which I've made community wiki.

Firstly, by the characterization of the Cantor space as the only metrizable Stone space with no isolated points (up to homeomorphism), $\newcommand{\Z}{\mathbb{Z}}\Z_p$ is homeomorphic to the Cantor space. A proof that the Cantor space is not extremally disconnected can be found here: https://proofwiki.org/wiki/Cantor_Space_is_not_Extremally_Disconnected

In fact, any convergent sequence in a Stonean space $X$ is eventually constant. If we have a sequence that is not eventually constant, we can pass to a subsequence $\newcommand{\N}{\mathbb{N}}(x_i)_{i \in \N}$ such that the mapping $\N \rightarrow X$ defined by $i \mapsto x_i$ is injective. Using the Hausdorffness of $X$, we can build up a disjoint sequence $(U_i)_{i \in \N}$ of open sets such that $x_i \in U_i$ but not in any other. Then $U = \bigcup_{i=0}^\infty U_{2i}$ and $V = \bigcup_{i=0}^\infty U_{2i+1}$ are disjoint open sets, but $x \in \overline{U} \cap \overline{V}$, so they do not have disjoint closures, which contradicts $X$ being Stonean. It follows that any metrizable stonean space is discrete.

To answer the question in your second-to-last paragraph, proving the existence of infinite Stonean spaces requires the axiom of choice, so there are no "explicit" examples. What I mean by this is that ZF + DC (dependent choice) + "all Stonean spaces are finite" is relatively consistent to ZF. So, unless you make a set-theoretic assumption, you can't find a simpler description of an infinite Stonean space than "the Stone space of the complete Boolean algebra $A$" or "the Gelfand spectrum of the commutative AW*-algebra $B$" or "the Yosida spectrum of the Dedekind-complete Riesz space $C$".

To prove the relative consistency of ZF + DC + "all Stonean spaces are finite", let $X$ be an infinite Stonean space, and $A$ be the set of clopens of $X$, as a complete Boolean algebra (the join of a set of clopens is the closure of their union, this requires no AC). If $A$ were finite, then the intersection of all clopens containing a point would be clopen, so $X$ would be discrete, which contradicts it being compact and infinite. So $A$ is an infinite complete Boolean algebra.

Now, under DC, every infinite complete Boolean algebra $A$ contains a countable disjoint sequence, by the following argument. If $A$ contains infinitely many atoms, we use DC to get a sequence $(a_i)_{i \in \N}$ of disjoint atoms of $A$ (apply DC to the poset of finite disjoint sequences of atoms ordered by extension). If it does not, let $b$ be the join of the atoms, and so the complete Boolean algebra $B = \downarrow(\lnot b)$ is atomless and infinite. If we apply DC to the poset of non-zero elements of $B$, ordered by $<$ (not by $\leq$), then we get a strictly descending sequence $(b_i)_{i \in \N}$ of elements of $B$, which are also elements of $A$. So $a_i = b_i \land \lnot\bigvee_{j = i+1}^\infty b_j$ defines a disjoint family $(a_i)_{i \in \N}$ in $A$.

By the previous paragraph, there must be a disjoint family $(C_i)_{i \in \N}$ of clopens in $X$. If $U = \bigcup_{i \in \N}C_i$ were closed, it would be compact, and therefore the union of some finite subfamily, which it can't be by disjointness. So there exists $x \in \overline{U} \setminus U$. Define $u \subseteq \mathcal{P}(\N)$ by $$ u = \left\{ S \subseteq \N \,\left\lvert\, x \not\in \overline{\bigcup_{i \in S}C_i} \right.\right\} $$ It is easy to verify from properties of the closure (monotonicity and preservation of finite unions) that $u$ is an ultrafilter and all finite subsets of $\N$ are in $u$, so it is non-principal.

Now, by constructing a model, Solovay and Pincus proved that ZF + DC (dependent choice) + "there exist no non-principal ultrafilters on any set" is relatively consistent to ZF in Theorem 2 of that article (apparently this was previously proved by Andreas Blass, but I can't find the article anywhere). So by the previous paragraphs, in Solovay and Pincus's model all Stonean spaces are finite.


Here is a short explicit proof of the fact that $\mathbb{Z}_p$ is not extremally disconnected:

Let $n \in \mathbb{Z}$, $n \geq 0$ and let $U_n = p^n + p^{n+1} \mathbb{Z}_p$ be the open ball of radius $\frac{1}{p^{n+1}}$ and center $p^n$. Consider the open set $U= \bigcup_{n\geq 0} U_n$. The closure of $U$ is $\overline{U} = U \cup \{0\}$. But $\overline{U}$ is not open, because none of the balls $p^n \mathbb{Z}_p$ with center $0$ is fully contained in $\overline{U}$.

  • 2
    $\begingroup$ This example as written needs $p > 2$. If $p = 2$ then $U_n$ is all $p$-adic integers with absolute value $1/p^n$, so $U = \mathbf Z_2 - \{0\}$ and $\overline{U} = \mathbf Z_2$ is open in $\mathbf Z_2$. For the case $p = 2$ you could use $U_n = 2^n + 2^{n+2}\mathbf Z_2$ instead. $\endgroup$
    – KConrad
    Sep 2, 2019 at 18:21

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