# $\sigma$-space to complete space

The terminology is from Halmos's Lectures on Boolean Algebras.

Let $$X$$ be a Boolean space. A subset of $$X$$ is a Baire set if it an element of the $$\sigma$$-field generated by the clopen subsets of $$X$$. A Boolean space is a $$\sigma$$-space if the closure of every open Baire set is open; and a complete space if the closure of every open set is open.

With these definitions in mind, Halmos proves (Theorem 13, p.102) that if $$X$$ a Boolean $$\sigma$$-space, then the dual algebra $$A$$ of $$X$$ (the field of clopen subsets of $$X$$) is isomorphic to the quotient $$B/M$$, where $$B$$ is the $$\sigma$$-field of Baire subsets of $$X$$ and $$M$$ is the $$\sigma$$-ideal of meager Baire sets.

Since, every complete space is a $$\sigma$$-space, can I safely replace "$$\sigma$$-space" by "complete space" in the above theorem, or is there any pitfall I should pay attention to?

I don’t see a problem, a complete ( extremally disconnected is the topological term, and its clopen algebra is complete, hence Halmos’ term) space is indeed is a $$\sigma$$-space (aka basically disconnected spaces, the clopen algebra is then $$\sigma$$-complete) so the theorem applies (complete implies $$\sigma$$-complete and open Baire implies open, so in both the BA view as the topological view the implication is trivial).
• Thank you for confirming. In passing, I think I can also safely replace "Baire set" by "Borel set" in the original theorem if $X$ is metrisable. And, since a subspace of a metrisable space is metrisable, I should be also entitled to replace Baire by Borel in case $X$ is complete or extremally disconnected.. – puzzled Oct 11 '18 at 9:57
• @puzzled There are no non-trivial metrisable $\sigma$-spaces, only finite discrete ones. – Henno Brandsma Oct 11 '18 at 22:30