Can I construct a complete (as a Boolean algebra) saturated elementary extension of a given Boolean algbera?

102 Views Asked by At

I have been interested in the following problem:

Let $B$ be an arbitrary Boolean algebra and $\kappa$ be an arbitrary cardinal. Can one construct a $\kappa$-saturated $B^* \succ B$ that is complete, i.e., all joins and meets exist in $B^*$?

What are relevant sources for this problem? I bet this problem appears in very old literature in model theory. (I'm aware of Tarski's complete up-to-elementary-equivalence classification of Boolean algebras, but I'm not sure if that's very relevant.)

1

There are 1 best solutions below

2
On BEST ANSWER

If $B$ is infinite and $\kappa$ is uncountable, then the answer is no. The reason is that every nontrivial instance of completeness is also an instance of non-saturation. For example, if $B$ is infinite, then it must have an infinite antichain $\langle a_n\mid n\in\omega\rangle$, by a theorem of Tarski. Suppose that $b=\bigvee_n a_n$ in $B^*$. Now consider the type $p(x,b,a_n)_n$ asserting that $x<b$ and that $a_n<x$ for each $n$. This type is consistent with the elementary diagram of $B^*$, but it is not realized in $B^*$, because $b$ is the join of the $a_n$. This type used countably many parameters, so it shows that if $B^*$ is $\sigma$-complete, then it is not $\sigma$-saturated.