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I'm not talking about Stone spaces! A Stonean space is a compact Hausdorff extremally disconnected space i.e. a compact Hausdorff space such that for all open $U$, its closure $\overline{U}$ is also open.

I've been thinking about applications of regular open sets in topology, and I came up with the following:

If $X$ is a topological space then the set $\text{RO}(X)$ of regular open sets of $X$ forms a complete Boolean algebra. This Boolean algebra induces a Stonean space $S(X)$ by Stone duality. Thus to any topological space we can assign a Stonean space $S(X)$.

If $X$ is a Stonean space, then $\text{RO}(X)$ is a Boolean algebra of clopen sets of $X$, so that Stone duality tells us $S(X)$ gives us $X$ back.

If $f:X\to Y$ is a continuous map, then $f^{-1}(\text{RO}(Y))\subseteq \text{RO}(X)$ is not true in general (see comments below). Thus I want to restrict to continuous open maps. In this case, if $U\in\text{RO}(Y)$ then $$\text{int }\overline{f^{-1}(U)} = \text{int }f^{-1}(\overline{U}) = f^{-1}(\text{int }\overline{U})$$ see also this post by Paul Frost. So $f^{-1}$ induces a map from $\text{RO}(Y)$ to $\text{RO}(X)$. Then the Stone duality induces a map $S(f):S(X)\to S(Y)$.

Now, using this, does that mean that Stonean spaces are a reflective subcategory of topological spaces with continuous open maps?

Jakobian
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    Do you mean for the morphisms of Stonean spaces to be all continuous functions? Those would correspond to arbitrary Boolean homomorphisms, not complete homomorphisms. – Eric Wofsey Feb 29 '24 at 20:16
  • @EricWofsey yes, because I want the subcategory to be full – Jakobian Feb 29 '24 at 20:37
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    @Jakobian: If $X,Y$ are topological spaces and $f\colon X\to Y$ continuous and $U$ is a regular open subset of $Y$, $f^{-1}(U)$ does not have to be regular open in $X$. For example $f(x)=x^2$ on $\Bbb{R}$. Then $f^{-1}((0,+\infty))=(-\infty,0)\sqcup(0,\infty)$ which is not regular open. The map $f^{-1}$ does not even preserve the Heyting-algebra structure on the open sets of $Y$. – Chad K Feb 29 '24 at 20:55
  • @ChadK are you saying that we should use continuous open maps on the category of Stonean spaces in order to establish meaningful results? – Jakobian Feb 29 '24 at 20:59
  • @Jakobian: I don't know what the functor is. You're implying there's a functor, but you only said what the functor does to objects, not to maps. – Chad K Feb 29 '24 at 21:01
  • @ChadK I've realized that I do want them to be continuous and open, thank you. – Jakobian Feb 29 '24 at 22:53
  • @Jakobian: See Andrew M. Gleason "Projective Topological Spaces", May 1958, Theorem 3.2. – Chad K Mar 02 '24 at 18:21

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No, Stonean spaces are not reflective in all topological spaces with continuous open maps. Consider an arbitrary Stone space $X$. Note that every continuous map $X\to\{0,1\}$ is automatically open. So, if a reflection $i:X\to Y$ of $X$ into Stonean spaces existed, every continuous map $X\to\{0,1\}$ would factor through $i$ and so $i$ would be injective since these maps separate points of $X$. But since $i$ is open and $X$ is compact, this would mean that $i$ is an embedding of $X$ as a clopen subspace of $Y$, so $X$ itself must be Stonean. So any non-Stonean Stone space does not have a Stonean reflection.

(The problem with the construction you propose is that there is not actually a canonical continuous open map $X\to S(X)$. Very concretely, a point of $S(X)$ is an ultrafilter on $RO(X)$. But a point of $X$ doesn't naturally give an ultrafilter on $RO(X)$, since if $U\in RO(X)$, then $U$ and its complement in $RO(X)$ may not actually cover all of $X$, so the filter on $RO(X)$ given by which regular open sets contain a given point of $X$ may not be an ultrafilter.)

Eric Wofsey
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