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Define the Baire algebra $Ba(X)$ of a Boolean space $X$ as the $\sigma$-algebra generated by the class of clopen subsets of $X$.

Clearly, every clopen set is a Baire set. An example of open Baire set is the countable union of a family of clopen subsets of $X$, and, as it happens, these are the only one open Baire sets. Dually, a closed Baire set is a countable intersection of a family of clopen subsets of $X$.

Are there other Baire sets in $X$?

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  • $\begingroup$ Are you asking whether every Baire set is open or closed? Just take the union of an open set and a closed set - this will usually be neither open nor closed. $\endgroup$ – Alex Kruckman Aug 14 at 14:05
  • $\begingroup$ Thanks @AlexKruckman for your comment. The definition is from Halmos's Lectures on Boolean algebras. $\endgroup$ – puzzled Aug 14 at 14:45
  • $\begingroup$ So, basically, there exist neither open nor closed Baire sets. $\endgroup$ – puzzled Aug 14 at 14:51
  • $\begingroup$ Oh, I retract my comment about the Baire algebra being the same as the Borel algebra. This is only true when the space is second countable: in general it's not true that every open set is a countable union of clopen sets. Also, I removed the (the-baire-space) tag. The Baire space is a particular topological space, it has nothing to do with the Baire algebra in a Boolean space. $\endgroup$ – Alex Kruckman Aug 14 at 14:54
  • $\begingroup$ Baire sets in this form are defined that way in Boolean spaces (zero-dimensional compact Hausdorff spaces). In locally compact Hausdorff spaces (in the context of measure spaces) they are defined as the $\sigma$-algebra generated by the compact $G_\delta$ sets instead (such spaces could well be connected and have no non-trivial clopen sets); this generalises the idea. It's still confusing (IMHO) to call such sets Baire (overload of the term), why not "strongly Borel" or some such notion? $\endgroup$ – Henno Brandsma Aug 23 at 22:21
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Suppose $X$ is a Boolean space (also known as a Stone space). Let $C\subseteq X$ be a clopen set, let $U\subseteq C$ be a Baire set which is open but not closed, and let $V\subseteq (X\setminus C)$ be a Baire set which is closed but not open. Then $W = U\cup V$ is a Baire set which is neither open nor closed.

Indeed, if $W$ is closed, then $W\cap C = U$ is closed, contradicting our choice of $U$. And if $W$ is open, then $W\cap (X\setminus C) = V$ is open, contradicting our choice of $V$.

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  • $\begingroup$ That's great! Thanks $\endgroup$ – puzzled Aug 14 at 14:54
  • $\begingroup$ To the proposer: An example: The Cantor Set is a Boolean space with a countable base (basis) of clopen sets so the Baire algebra is the Borel algebra on this space. $\endgroup$ – DanielWainfleet Aug 14 at 15:19
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Of course there are other Baire sets: the standard Cantor set $X$ has a base of clopen sets and is second countable, so every open set is Baire too and so $\text{Ba}(X)=\text{Bor}(X)$, the Borel sets of $X$. And the rationals in $X$ are countable (so Borel hence Baire) and dense (so not closed) and not open, e.g.

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  • $\begingroup$ Thanks @HennoBrandsma for your reply. I have a doubt: Is a non-clopen Baire set necessarily infinite? $\endgroup$ – puzzled Aug 16 at 9:23
  • $\begingroup$ @puzzled in the above example no, any finite set is closed, not clopen and Borel and Baire. $\endgroup$ – Henno Brandsma Aug 16 at 9:27
  • $\begingroup$ But Halmos says that every open Baire set in a Boolean space is the union of a countable class of clopen sets. Since the union of a countable class of clopen sets is of course an open Baire set, it seems that a Baire set is open if and only if it is the union of a countable class of open sets. Hence, a Baire set is closed if and only if its the intersection of a countable class of clopen sets. So how can a finite closed set be Baire in a Boolean space? $\endgroup$ – puzzled Aug 16 at 10:01
  • $\begingroup$ @puzzled A singleton in the Cantor set is clearly a countable intersection of clopen sets, and of course so are finite sets. I see no problem. $\endgroup$ – Henno Brandsma Aug 16 at 10:56

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