Let $\kappa$ be an infinite cardinal. A Boolean algebra $\mathbb{B}$ is said to be $\kappa$-saturated if there is no partition (i.e., collection of elements of $\mathbb{B}$ whose pairwise meet is $0$ and least upper bound is $1$) of $\mathbb{B}$, say $W$, of size $\kappa$. Is there any relationship between this and the model theoretic meaning of $\kappa$-saturated (namely that all types over sets of parameters of size $<\kappa$ are realized)?

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    $\begingroup$ I thought it was kind of funny that you gave it a name, $W$, only to say that no such thing exists :) $\endgroup$ – Amit Kumar Gupta Jun 28 '12 at 5:41
  • $\begingroup$ In their book "Models and Ultraproducts", Bell & Slomson discuss boolean algebras in Chapter 1, and saturated products in Chapter 11. The last section of 11 has historical and bibliographical remarks. I think if there were a relationship between the two uses of "saturated" they would mention it somewhere, but they don't seem to. $\endgroup$ – MikeC Jun 28 '12 at 16:46

As far as I know, there is no connection; it's just an unfortunate clash of terminology. It's especially unfortunate because the model-theoretic notion of saturation comes up in the theory of Boolean algebras. For example, the Boolean algebra of subsets of the natural numbers modulo finite sets is $\aleph_1$-saturated but (by a definitely non-trivial result of Hausdorff) not $\aleph_2$-saturated, in the model-theoretic sense, even if the cardinality of the continuum is large.

When (complete) Boolean algebras are used in connection with forcing, it is customary to say "$\kappa$-chain condition" instead of "$\kappa$-saturated" in the antichain sense.


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