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).

  • $\begingroup$ 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.. $\endgroup$ – puzzled Oct 11 '18 at 9:57
  • 1
    $\begingroup$ @puzzled There are no non-trivial metrisable $\sigma$-spaces, only finite discrete ones. $\endgroup$ – Henno Brandsma Oct 11 '18 at 22:30

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