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During my studies, I came across several different Stone spaces, e.g.:

(i) The Cantor cube $X=\{0,1\}^{\omega_1}$, which is the Stone space of the free Boolean algebra on $\omega_1$ free generators;

(ii) The absolute $EX$ of $X$ (i.e. the Gleason cover of $X$ since $X$ is compact), which is the Stone space of the regular closed algebra of $X$;

(iii) The minimal basically-disconnected cover $\Lambda X$ of $X$, which is the Stone space of the $\sigma$-completion of the clopen algebra of $X$, etc…

One Stone space which still eludes me, however, is the Stone space $Y$ of the free $\sigma$-algebra on $\omega_1$ free generators. As a Stone space, it is certainly totally-disconnected compact Hausdorff, and, in fact, it can be shown that it is even basically-disconnected (such spaces are sometimes called quasi-Stonean).

(1) Is there a more precise characterisation of $Y$?

Also, it is known that the free $\sigma$-algebra on $\omega_1$ free generators is isomorphic to the Baire $\sigma$-algebra of $X$. But, as previously noted, $X$ is the Stone of the free Boolean algebra on $\omega_1$ free generators, not the Stone space $Y$ of the free $\sigma$-algebra on $\omega_1$ generators.

(2) What are the pros and cons of working with $X$ instead $Y$?

(3) How workable is $Y$, especially from a measure theory and integration point of view?

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  • $\begingroup$ There are quite flexible characterizations of the space $X$. Keyword: "dyadic space". Information can be found in papers by S. O. Sirota (1968), L.B. Shapiro ( 1976), Yu. V. Tsybenko (1985/86). $\endgroup$
    – YCor
    Commented Jan 22, 2019 at 18:14
  • $\begingroup$ More precisely Tsybenko (link.springer.com/content/pdf/10.1007%2FBF01060948.pdf, sorry for the paywall) says that Sirota proved that if we have an inverse limit, indexed by $\omega_1$, of metrizable Stone spaces, whose projections are open and "have complete preimages of points", are homeomorphic to $2^{\aleph_1}$. I'm not sure what "have complete preimages of points" is supposed to mean, maybe it just means surjective. Possibly somebody reading Russian and checking Tsybenko's Russian 1985 original might help. $\endgroup$
    – YCor
    Commented Jan 22, 2019 at 21:29
  • $\begingroup$ (Well, it can't be just "surjective" because the existence of an isolated point or clopen Cantor set is an obstruction.) $\endgroup$
    – YCor
    Commented Jan 22, 2019 at 23:13
  • $\begingroup$ @YCor I don't know what "have complete preimages of points" means either, but for the result to be true it is enough to say that each point splits cofinally often. Or equivalently, that each point of $X$ has character $\omega_1$. But I think the OP wants to characterize $Y$ not $X$. $\endgroup$
    – Ramiro de la Vega
    Commented Jan 23, 2019 at 16:45

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