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During my beginning PhD-research I have encountered the following question/problem:

  • Let $X$ be a boolean space. Is it possible to reconstruct the space $X$ from its boolean algebra of clopen sets, $Cl(X)$ (viewed as a boolean algebra)?

By a boolean space I mean a (non-empty) zero-dimensional locally compact Hausdorff space, or equivalently, a Hausdorff space with a basis consisting of compact open sets. The particular spaces I am studying are second countable and have no isolated points as well -- in case it makes a difference.

The set of clopen subsets of $X$, $Cl(X)$, forms a boolean algebra under the usual set-operations of union, intersection and complement. It is generally not a complete boolean algebra.

In the case that $X$ is in fact compact (i.e. a Stone space) the answer is of course yes, this being the classical Stone duality. In the locally compact case however, it seems to me like the answer in general is no, based on the references below, as these seem to require the boolean algebra of compact open sets instead. And it is not possible to "detect" compactness of an element $A \in Cl(X)$, as one only has finite unions/joins in this boolean algebra.

  1. Dimov – Some Generalizations of the Stone Duality Theorem (http://rmi.tsu.ge/tolo2/presentations/Dimov.pdf)
  2. Doctor – The Categories of Boolean Lattices, Boolean Rings and Boolean Spaces (https://cms.math.ca/openaccess/cmb/v7/cmb1964v07.0245-0252.pdf)
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Not in general: two non-isomorphic boolean spaces can have isomorphic algebras of clopen sets. Indeed, let $X$ be any boolean space that is not compact, and let $B=Cl(X)$. Then the Stone space $Y$ of $B$ is a boolean space with $Cl(Y)\cong Cl(X)$, but $Y$ is not homeomorphic to $X$ since $Y$ is compact and $X$ is not.

However, if you restrict to second-countable spaces, then $X$ can be reconstructed from $Cl(X)$. Indeed, the compact open sets can be identified as those elements $b\in Cl(X)$ such that $\{c\in Cl(X):c\leq b\}$ is countable. From there, the locally compact version of Stone duality can be used to reconstruct $X$.

To prove this characterization of the compact open sets, note that $\{c\in Cl(X):c\leq b\}$ is just the clopen algebra $Cl(b)$, considering $b$ as a subspace of $X$. So replacing $X$ with $b$, it suffices to show that if $X$ is a second-countable boolean space, then $X$ is compact iff $Cl(X)$ is countable. To prove this, choose a countable basis $(B_n)_{n\in\mathbb{N}}$ for $X$ consisting of compact open sets. If $X$ is compact, every clopen set is a finite union of basis elements (by compactness), and so $Cl(X)$ is countable. If $X$ is not compact, then for each $k\in\mathbb{N}$, $C_k=\bigcup_{n<k}B_n$ is compact and thus a proper subset of $X$. By induction, we can choose a sequence of disjoint nonempty compact open subsets $D_k\subset X\setminus C_k$ for each $k$. Then for every $A\subseteq\mathbb{N}$, $\bigcup_{k\in A} D_k$ is clopen, giving uncountably many distinct elements of $Cl(X)$. (The assumption that $D_k\subset X\setminus C_k$ guarantees that $\bigcup_{k\in A}D_k$ is closed, since its intersection with each $C_k$ is just a finite union and thus closed, and the $C_k$ are an open cover of $X$.)

(Alternatively, using the additional assumption that $X$ has no isolated points, you can just cite the classification in Henno Brandsma's answer, and so you just have to check that $Cl(C)$ is countable and $Cl(C\setminus\{0\})$ is uncountable.)

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  • $\begingroup$ Thank you very much for this wonderful answer!! It is exactly what I was hoping/looking for. What I'm actually trying to do is to reconstruct some action on the space, and so far I've been able to reconstruct the action on the clopens. But with your help I will be able to recover the action on the space itself! $\endgroup$ – Pjolter Jan 13 '18 at 11:06
  • $\begingroup$ Eric Wofsey, I just wanted to tell you that your answer "appears" in my recent preprint (arxiv.org/abs/1806.11087), and you're mentioned in the ackknowledgements. Thanks again! $\endgroup$ – Pjolter Jun 29 '18 at 6:55
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If you know that $X$ is locally compact (but not compact) zero-dimensional Hausdorff and moreover second-countable and crowded (no isolated points) then $X$ is homeomorphic to $C \setminus \{0\}$ (the Cantor set minus a point; which point does not matter). The proof follows easily from the observation that the one-point compactification of such an $X$ is homeomorphic to $C$, by its classical characterisation of $C$ as up to homeomorphism the only crowded compact metrisable zero-dimensional space.

So topologically speaking you're only interested in two spaces: $C$ (compact case) or $C$ minus a point (the non-compact case).

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  • $\begingroup$ And these two spaces can in fact be distinguished by their clopen algebras, since $Cl(C)$ is countable and $Cl(C\setminus\{0\})$ is uncountable. $\endgroup$ – Eric Wofsey Jan 12 '18 at 18:25
  • $\begingroup$ I cannot believe I have not made this observation myself yet. But I'm very glad to have it pointed out! However, I have to accept Eric Wofsey's answer, since what I really was wondering was whether one could "canonically reconstruct" the space from the clopens. I think I was perhaps a bit too vague when I posed the question. The reason is I am really dealing with dynamical problems, so I need more than just the homeomorphism class of the space. $\endgroup$ – Pjolter Jan 13 '18 at 11:13

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