Claim in Shelah's "A two-cardinal theorem"

Question

Let $T$ be a first-order theory with unary predicates $P$ and $Q$. We say that $M$ is a model of type $(\lambda, \mu)$ if $|P^M| = \lambda$ and $|Q^M| = \mu$. In Shelah's "A two-cardinal theorem", he makes the following claim.

If for every $n$, $T$ has a model $M$, $|P^M| \ge \aleph_0 > |Q^M| \ge n$, then for every $\lambda \ge \mu \ge |T|$, $T$ has a model of type $(\lambda, \mu)$.

Unfortunately, he doesn't give a proof, and I haven't been able to figure out how to prove it. It seems like some kind of compactness argument should do the trick, but it's not clear to me how to apply the compactness theorem while fixing $|P^M| = \lambda$. How can one prove this?

Answer 1

Here's the sketch: By Löwenheim-Skolem, we can find models of $T$ in which $P$ is arbitrarily large and infinite and $Q$ is arbitrarily large and finite. Taking an ultraproduct of such models, we can find a model of $T$ in which the cardinalities of $P$ and $Q$ are arbitrarily far apart. Using the idea of Morley's Omitting Types Theorem (i.e., Erdös-Rado and Skolem hulls), we can obtain models of type $(\lambda,\mu)$ for arbitrary infinite $\lambda\geq \mu\geq |T|$.

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Now in more detail.

Fix an infinite cardinal $\kappa$. For each $n\in \omega$, let $M_n$ be a model of $T$ such that $|P^{M_n}| \geq \aleph_0 > |Q^{M_n}| \geq n$. By Löwenheim-Skolem, find an elementary extension $M_n\preceq N_n$ such that $|P^{N_n}| \geq \kappa$. Since $|Q^{M_n}|$ is finite, $\aleph_0 > |Q^{N_n}| = |Q^{M_n}| \geq n$.

Let $U$ be a non-principal ultrafilter on $\omega$, and consider the ultraproduct $M = \prod_{n\in \omega} N_n/U$. We have $|P^M| \geq \kappa$ and $|Q^M|=2^{\aleph_0}$.

In particular, since $\kappa$ was arbitrary, we can obtain models of $T$ of type $(\kappa,\mu)$, where $\kappa$ and $\mu$ are arbitrarily far apart. In this situation, we can find models of $T$ of type $(\lambda,\mu)$ for arbitrary infinite $\lambda \geq \mu \geq |T|$.

The best reference I have for the last fact is Chang and Keisler. Quoting from p. 518:

The idea behind Theorem 7.2.2 [Morley's Omitting Types Theorem] is used to prove the following theorem of Vaught; the proof is essentially due to Morley.

Theorem 7.2.6. Suppose $L$ is a countable language, $T$ is a theory of $L$, and for each $n\in \omega$, $T$ admits some pair $(\beth_n(\gamma_n),\gamma_n)$ of infinite cardinals. Then $T$ admits every pair of infinite cardinals $(\alpha,\beta)$.

Removing the countable language assumption is Exercise 7.2.11. For uncountable languages, we probably need to strengthen the hypothesis to use cardinals that are father apart than $(\beth_n(\gamma_n),\gamma_n)$ for finite $n$, in order to apply Erdös-Rado with more colors.