0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.