Is there any calculus whose algebraization is a sigma-complete Lindenbaum algebra, i.e., a sigma-complete Boolean algebra after identification of equivalent formulas?
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$\begingroup$ Could you clarify a little more exactly what you want? I believe that any Boolean algebra can arise as a Lindenbaum algebra for some suitably chosen first-order theory. And do you really want it to be $\sigma$-complete? In that case, it can't be countably infinite, since there is no countably infinite $\sigma$-complete Boolean algebra. Once the algebra is infinite, you get a countable antichain, and then if it is $\sigma$-complete, you will find continuum many elements arising as joins of subsets of that antichain. $\endgroup$– Joel David HamkinsCommented Mar 31, 2014 at 20:50
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$\begingroup$ Joel, do you have a proof of your first assertion? for propositional theories is obvious, but for a first-order theory I'm not even sure it's true... $\endgroup$– godelianCommented Mar 31, 2014 at 21:26
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$\begingroup$ Let's see, I had in mind to build the theory out of the Boolean algebra itself, but I didn't think it all the way through. But I don't understand your comment, since I thought the problem was: given a Boolean algebra $\mathbb{B}$, find a first-order language and a theory $T$ in that language, such that the collection of formulas, modulo provable equivalence in $T$, is a copy of $\mathbb{B}$. Is that true? $\endgroup$– Joel David HamkinsCommented Mar 31, 2014 at 21:32
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$\begingroup$ Yes, that was the assertion I was referring to. Is that true? Sorry, I cannot see it immediately, perhaps it's obvious to you... $\endgroup$– godelianCommented Mar 31, 2014 at 22:21
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$\begingroup$ I had thought it would be easy, but now I'm not sure whether it is true or not... $\endgroup$– Joel David HamkinsCommented Mar 31, 2014 at 22:33
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There is no countably infinite $\sigma$-complete Boolean algebra. Once the Boolean algebra is infinite, it must have an countably infinite antichain $A\subset\mathbb{B}$, and then by $\sigma$-completeness every subset $X\subset A$ will have a distinct join $\bigvee X$, making the Boolean algebra $\mathbb{B}$ have size at least continuum.
Perhaps you meant something else...?
answered Mar 31, 2014 at 21:21
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$\begingroup$ OP does not say "infinte" anywhere... $\endgroup$ Commented Apr 1, 2014 at 10:59
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$\begingroup$ Yes, I had noticed that, but of course, the finite Boolean algebras are trivially realized as Lindenbaum algebras. For example, in the theory asserting that there are exactly $n$ objects, with $n$ distinct constants $c_i$ and one more constant $d$, which must be one of the $c_i$. The assertions $d=c_i$ are atoms, and every assertion is some Boolean combination of the these atoms. $\endgroup$ Commented Apr 1, 2014 at 11:30
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$\begingroup$ Or: the theory $T$ asserting that one of $n$ distinct finitely-axiomatizable complete theories holds. Each such theory is an atom, and every statement will be a join of atoms. $\endgroup$ Commented Apr 1, 2014 at 12:47