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Let $X$ be a Stone space (i.e. totally disconnected compact Hausdorff). Then the following are equivalent:

  1. $X$ is second countable
  2. $X$ is metrizable
  3. $X$ has countably many clopen subsets
  4. $X$ is an inverse limit of a sequence $\{X_n\}_{n \in \mathbb N}$ of finite (discrete) spaces

In (4), the transition maps can be take to be surjective. Such spaces have received new attention in their role in Clausen and Scholze's light condensed mathematics.

By Stone duality and (3), classifying second-countable Stone spaces $X$ is equivalent to classifying countable Boolean algebras, which sounds pretty hopeless. Nevertheless, the Cantor-Bendixson theorem tells us that $X$ is either countable, or the disjoint union of a countable space and a copy of the Cantor set $\mathcal C$.

In the case where the perfect core is empty, the Sierpinski-Mazurkiewicz theorem tells us that the countable part is homeomorphic to $n \times (\omega^\alpha+1)$ for some finite (discrete) $n$, where $\alpha + 1$ is the Cantor-Bendixson rank of $X$ and $\omega^\alpha + 1$ is the set of ordinals $\leq \omega^\alpha$ in the order topology. ($\alpha$ can be an arbitrary countable ordinal)

Unfortunately, when the perfect core is nonempty, the remaining scattered part is generally not itself compact, so the Sierpinski-Mazurkiewicz theorem doesn't apply to it. This still sounds pretty hopeless, but still I wonder:

Question: What more can be said about the classification of second-countable Stone spaces up to homeomorphism?

Is there any analog of the Sierpinski-Mazurkiewicz theorem which can help us here? For instance, is there a classification of second-countable, countable totally disconnected spaces which are not assumed to be compact, which could be applied to the scattered part of $X$ to at least understand that piece?

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    $\begingroup$ Let $X_0$ be the perfect core, $X_1$, the intersection of $X$ with the closure of the set of isolated points, $X_2$, the intersection of $X$ with the closure of the set of scattered point of CB $\ge 1$ (i.e., non-isolated scattered points). And so on. So the classification is at least as complicated as the data of a Cantor set and a decreasing family, eventually empty, indexed by countable ordinals, of closed subsets. $\endgroup$
    – YCor
    Commented Feb 11 at 23:31
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    $\begingroup$ Something more precise can be found here: R. S. Pierce. Existence and uniqueness theorems for extensions of zero-dimensional compact metric spaces. Trans. Amer. Math. Soc., 148:1–21, 1970. link at AMS site $\endgroup$
    – YCor
    Commented Feb 11 at 23:33
  • $\begingroup$ @YCor Thanks, Pierce's paper seems to be a almost-complete answer to my question. I'd gladly accept it i you post it as an answer! The only gap I see is that he only shows how to glue together the Cantor set with a scattered set along a descending sequence of closed susets when the scattered set has $n=1$. I'm not sure how serious that restriction is... especially now that you've alerted me to the fact that the scattered set need not be compact... $\endgroup$
    – Tim Campion
    Commented Feb 11 at 23:54
  • $\begingroup$ Just in case that somebody else is similarly confused as I was at first when reading the question: After ten minutes of confusion how an infinite metrizable Stone space can even exist, I finally noticed that Stone space $\not=$ Stonian space... $\endgroup$ Commented Feb 12 at 7:39
  • $\begingroup$ The scattered part of $X$ is just a countable ordinal with the order topology. $\endgroup$ Commented Feb 27 at 19:26

1 Answer 1

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The paper linked by YCor in the comments,

  • R. S. Pierce. Existence and uniqueness theorems for extensions of zero-dimensional compact metric spaces, Trans. Amer. Math. Soc., 148:1–21, 1970, doi:10.1090/S0002-9947-1970-0254804-4

gives a complete answer. To see this, let's use the term scattered space for a countable totally disconnected Hausdorff space. In Prop 3.2, Pierce shows that every scattered space $S$ is a topologically-disjoint union $S = T \amalg U$ where $U$ is a compact scattered space and $T$ is of the form $\omega^\alpha$. Now if $X$ is a second-countable Stone space with perfect part $P$ and scattered part $S = T \amalg U$, then note that $P \cup T$ is closed in $X$: since $P$ is closed, we need only note that a sequence in $T$ has a limit in $X$, and that limit can't be in $U$ because it is topologically disjoint from $T$. Therefore $P \cup T$ is compact Hausdorff. There is a continuous bijection $(P \cup T) \amalg U \to X$, which must be a homeomorphism because these are compact Hausdorff spaces.

Next, if $X$ is a second-countable Stone space, then write $X^\alpha$ for the $\alpha$th Cantor-Bendixson derivative of $X$, and $P= X^\infty$ for the perfect core of $X$ and $S = X - P$ for the scattered part. Write $X_\alpha = P \cap \overline{X^\alpha - P} = P \cap \overline{S^\alpha}$ for the limit points in $P$ of the $\alpha$th derivative of $S$. Then $(X_\alpha)_{\alpha < \mu}$ is a decreasing filtration of $P$ by nonempty closed subsets indexed by a countable ordinal.

In Thm 1.2,Pierce shows that if $\mathcal C \supseteq \mathcal C_0 \supseteq \cdots$ is any descending filtration of nonempty closed subspaces $(\mathcal C_\alpha)_{\alpha < \mu}$ of Cantor space indexed by a countable ordinal $\mu$, then there is a second-countable Stone space $X$ with perfect part $P$, and scattered part $\omega^\mu$, and $X_\alpha = \mathcal C_\alpha$. By taking a topologically-disjoint union with a scattered Stone space, we obtain that for any such filtration of Cantor space and any scattered space $S$ which is a topologically-disjoint union of $\omega^\mu$ with a scattered Stone space, there is a second-countable Stone space $X$ with perfect part $\mathcal C$, scattered part $S$, and $X_\alpha = \mathcal C_\alpha$.

Finally, Pierce shows in Thm 1.1 that a second-countable Stone space $X$ is determined up to homeomorphism by its scattered part along with the filtration $(X_\alpha)_{\alpha < \mu}$ of its perfect core. More precisely, Given $X,Y$ second-countable Stone spaces, a homeomorphism $f_\infty : P \to Q$ between perfect cores extends to a homeomorphism $f : X \to Y$ if and only of the scattered parts are homeomorphic and $f_\infty(X_\alpha)= Y_\alpha$. Of course, the perfect core is homeomorphic to $\mathcal C$ iff it is nonempty.

Summary:

To sum up, we can construct an uncountable second countable Stone space from the following data:

  1. A scattered space which is of the form $\omega^\mu \amalg (n \times (\omega^\lambda + 1))$ for countable ordinals $\mu \leq \lambda$ and $n \in \mathbb N$, and

  2. A decreasing filtration of Cantor space indexed by $\mu$ and comprising nonempty closed sets.

Every uncountable second-countable Stone space arises this way up to homeomorphism, and two such are homeomorphic iff there is a homeomorphism between the scattered parts in (1) and a self- homeomorphism of Cantor space relating the filtrations of (2). The invariants (1) and (2) can be read off from a general $X$ as indicated above.

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  • $\begingroup$ Might be useful to add concerning the scattered part that duals of scattered spaces are precisely atomic Boolean algebras (every nonzero element has an atom below it). I believe this also gives that second countable scattered Stone spaces are precisely continuous images of the Stone-Čech compactification of a countable discrete space. But maybe this is directly obvious too... $\endgroup$ Commented Feb 12 at 12:33
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    $\begingroup$ @მამუკაჯიბლაძე The Stone Cech compactification of an infinite set is never second countable. $\endgroup$
    – Tim Campion
    Commented Feb 12 at 15:08
  • $\begingroup$ Ouch. Seems these are very special continuous images. What I was trying to use is this. Let $B$ be an atomic BA, then assigning to an element of $B$ the set of atoms below it is an embedding of $B$ into the powerset of atoms. So if $B$ is also countable it embeds into the powerset of a countable set. Dually we get a continuous onto from the dual of the powerset to the dual of $B$. And dual of the powerset is the Stone-Čech. It seems I am overlooking something very special about that embedding, or dually about that onto map... $\endgroup$ Commented Feb 12 at 16:49
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    $\begingroup$ That is interesting, but one should note that given that the dual of any countable boolean algebra can be realized as a closed subset of the Cantor space, I'm not completely sure this classifies anything: Classyfing Filtration of the cantor space by closed subspaces up to homomorphisms subsumes classifying countable boolean algebra. $\endgroup$ Commented Feb 12 at 18:59
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    $\begingroup$ @SimonHenry I see, sorry. I had in mind my previous comment about atomic countable Boolean algebras. Representing them as a quotient of a free Boolean algebra dually they correspond to a closed subset of the Cantor space; but since they are generated by atoms dually you get scattered closed subsets, i. e. those with the subset of isolated points dense. $\endgroup$ Commented Feb 15 at 19:43

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