# Existence of $\kappa$-saturated models of cardinality $\lambda$

I am having some trouble proving Theorem 1.7(2) from Shelah's Classification Theory, (the proof of which he omits, saying that it is similar to the proof of 1.7(1).) With only a minor simplification, the theorem states:

Suppose $M$ is an $\mathcal{L}$-structure, and $||M||+||\mathcal{L}||\leq\lambda=\lambda^{<\kappa}$. Suppose also that for all $m<\omega$, there are no more than $\lambda$ complete types in $m$ free variables without parameters in $M$. Then $M$ has a $\kappa$-saturated elementary extension of cardinality $\lambda$.

I would like to prove some kind of lemma that generalizes the bound on types without parameters to include types with fewer than $\kappa$ parameters, but I have not yet had any success. Once I have some kind of bound on the number of types, then the construction is clear, and in particular the extension will not exceed size $\lambda$. I have also considered the possibility of inducting on the number of parameters, or perhaps including more and more parameters as part of the recursion, but without success.

Any help is greatly appreciated - thanks!

Here's the bound you're looking for.

Lemma: Suppose that for all finite $m$, there are at most $\lambda$-many $m$-types over $\emptyset$. Then for any nonempty set $A$, there are at most $\lambda^{|A|}$-many $1$-types over $A$.

Proof: First, let's assume $A$ is infinite. Then, $$|S_1(A)| \leq \prod_{B\subseteq_{\text{finite}} A}|S_1(B)|\leq \prod_{B\subseteq_{\text{finite}} A}|S_{|B|+1}(\emptyset)|\leq \lambda^{|A|}$$ First inequality: a $1$-type over $A$ is uniquely determined by its restrictions to all finite subsets of $A$ (since any given formula only mentions finitely many parameters). Second inequality: a $1$-type $p(x,b_1,\dots,b_n)$ over a finite set $B$ is uniquely determined by the $(|B|+1)$-type $p(x,y_1,\dots,b_n)$. Third inequality: Use the assumed bound on the sizes of the type spaces, plus the fact that there are $|A|$-many finite subsets of $A$ (since $A$ is infinite).

In the case that $|A|$ is finite and nonempty, we just get $|S_1(A)| \leq |S_{|A|+1}(\emptyset)| \leq \lambda = \lambda^{|A|}$ directly. $\square$

Note that when $|A|\geq |L|$, it's easy to see that there are at most $2^{|A|}$-many $1$-types over $A$, which is a better bound than the lemma gives us. And when $|A|<|L|$, if $\lambda \geq 2^{|L|}$, then $\lambda^{|A|}\geq 2^{|L|\times |A|} = 2^{|L|}$, and the obvious bound is again better.

So the lemma has limited utility for counting types: it's only interesting when $\aleph_0 \leq |A| < |L|$ and $\lambda<2^{|L|}$.

Proof of the theorem: First, let $M_0$ be any elementary extension of $M$ of size $\lambda$. This is where we use $|M|+|L|\leq \lambda$. Now we start realizing types. To get $\kappa$-saturation, we need to realize all of the $1$-types over $A$, for every set $A\subseteq M_0$ with $|A|<\kappa$. For each such set, the lemma tells us that there are at most $\lambda^{|A|} \leq \lambda^{<\kappa} = \lambda$ types to realize (we already know there are at most $\lambda$ types over $\emptyset$). And since $|M_0| = \lambda$, there are $\lambda^{<\kappa} = \lambda$ such parameter sets. So we have to realize $\lambda$-many types in total: at most $\lambda$ over each of the $\lambda$-many parameter sets.

Great! We can use compactness and Löwenheim-Skolem to realize all of these types in an elementary extension $M_0\preceq M_1$ with $|M_1| = \lambda$. Of course, there are now new parameter sets in $M_1$, so we repeat, building a transfinite elementary chain $M_0\preceq M_1\preceq \dots \preceq M_\alpha \preceq\dots$, where each model $M_{\alpha+1}$ realizes all of the types over sets of size $<\kappa$ contained in $M_\alpha$, and we take unions at limit stages.

How far do we have to go to get $\kappa$-saturation? I claim we can stop at $M_\lambda$. Let $p(x)$ be a type over $A\subseteq M_\lambda$, with $|A| = \mu <\kappa$. If $A$ is contained in $M_\alpha$ for some $\alpha<\lambda$, then $p(x)$ is already realized in $M_{\alpha+1}$. But if not, then new elements of $A$ keep getting added all the way up the $\lambda$-indexed elementary chain, so $\mathrm{cf}(\lambda)\leq \mu$. This is impossible, by König's theorem on cofinalities: $$\lambda < \lambda^{\mathrm{cf}(\lambda)} \leq \lambda^\mu \leq \lambda^{<\kappa} = \lambda.$$

The union of a $\lambda$-length chain of models of size $\lambda$ has size $\lambda$, so $|M_\lambda| = \lambda$, and we're done.

• Thank you for your reply. I understand the proof except for the line where you claim that if $\mu$ is infinite, then there are at most $2^\mu$ types over $A$. How does one justify this? A hint may suffice. – Alex J Jul 26 '17 at 23:27
• Oh, good point - I was implicitly assuming countable language. The actual bound is $2^{\max(\mu+|L|)}$, just by counting the number of formulas with parameters from $A$. I'll have to think about how to patch up that step. – Alex Kruckman Jul 27 '17 at 3:30
• Thanks, I've been going crazy trying to justify this step. Perhaps the theorem is misstated, and we also require $|\mathcal{L}|<\kappa$? I've been unable to find an equivalent statement of such generality elsewhere... – Alex J Jul 27 '17 at 3:59
• @Georg9551 OK, I figured out the missing step. See my edit. – Alex Kruckman Jul 27 '17 at 5:15