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The free Boolean algebra on countably many generators is closely related to the classical (two-valued) propositional calculus (after identification of logically equivalent formulas).

By the Stone representation theorem, it is isomorphic to the field of open-closed subsets of the Cantor space $2^\omega$. Suppose now that I consider the $\sigma$-field generated by these open-closed sets. It is the free Boolean $\sigma$-algebra on countably many generators.

To what logic (if any) does the free Boolean $\sigma$-algebra on countably many generators correspond to? Could it be the classical (two-valued) propositional calculus with some kind of axiom of monotone continuity (like e.g. the one needed in probability theory to go from a Boolean algebra of events to the $\sigma$-algebra generated by these events)?

Thank you in advance.

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  • $\begingroup$ Thank you Andreas for your comment. I am a bit confused, though, for Halmos, in his "Lectures on Boolean Algebras", Theorem 14, p. 102 states: " For every set $I$, there exists a free $\sigma$-algebra generated by $I$, and, in fact, that algebra is isomorphic to the $\sigma$-field of all Baire sets in the Cantor space $2^I$". (By a Baire set, Halmos means "a set belonging to the $\sigma$-field generated by class of all clopen sets" (p. 97 of the same book).) What do I misunderstand? $\endgroup$
    – Beginner
    Commented Aug 2, 2018 at 7:26
  • $\begingroup$ @AndreasBlass I think the issue is that the term on the right hand side of the $\omega$-$2$-distributive law involves a continuum-sized disjunction. So it's not a term in the language of $\sigma$-algebras, and hence it isn't necessarily preserved by homomorphisms. $\endgroup$
    – Alex Kruckman
    Commented Aug 2, 2018 at 13:54
  • $\begingroup$ @AlexKruckman I think you're right. I'll delete my previous comment after I add a new comment so that your comment and the preceding one make sense without mine. $\endgroup$
    – Andreas Blass
    Commented Aug 2, 2018 at 14:04
  • $\begingroup$ The comments from @puzzled and Alex Kruckman refer to an earlier comment of mine, which I've deleted because it was, as they pointed out, incorrect. $\endgroup$
    – Andreas Blass
    Commented Aug 2, 2018 at 14:06

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The correspondence between Boolean algebras and propositional logic extends to a correspondence between $\sigma$-algebras (the term Boolean $\sigma$-algebra is redundant, since all $\sigma$-algebras are Boolean algebras) and the infinitary propositional logic which allows conjunctions and disjunctions of any countable set of formulas.

This is the propositional fragment of the infinitary predicate logic $L_{\omega_1,\omega}$ described here or here.

You could also take a look at Chapters 4 and 5 of Carol Karp's classic book Languages with Expressions of Infinite Length, which cover infinitary propositional logics and proof systems for these logics, respectively.

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    $\begingroup$ Halmos uses "Boolean $\sigma$-algebra" to mean a $\sigma$-complete (abstract) Boolean algebra. This is not redundant, because a $\sigma$-algebra then has its usual definition in measure theory (a field of sets closed under complement and countable union). However, the word Boolean is acting in a peculiar way, so I avoid this terminology in my own writing. $\endgroup$
    – Robert Furber
    Commented Sep 14, 2018 at 10:17

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