# Example of Baire sets

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 $$X$$?

• 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. – Alex Kruckman Aug 14 at 14:05
• Thanks @AlexKruckman for your comment. The definition is from Halmos's Lectures on Boolean algebras. – puzzled Aug 14 at 14:45
• So, basically, there exist neither open nor closed Baire sets. – puzzled Aug 14 at 14:51
• 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. – Alex Kruckman Aug 14 at 14:54
• 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? – Henno Brandsma Aug 23 at 22:21

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 $$V$$.

• That's great! Thanks – puzzled Aug 14 at 14:54
• 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. – DanielWainfleet Aug 14 at 15:19

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.

• Thanks @HennoBrandsma for your reply. I have a doubt: Is a non-clopen Baire set necessarily infinite? – puzzled Aug 16 at 9:23
• @puzzled in the above example no, any finite set is closed, not clopen and Borel and Baire. – Henno Brandsma Aug 16 at 9:27
• 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? – puzzled Aug 16 at 10:01
• @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. – Henno Brandsma Aug 16 at 10:56