Countable Stability implies Stability for All Infinite Cardinalities

Theorem
Let $T$ be a complete $\mathcal L$-theory whose language $\mathcal L$ is countable.

If $T$ is $\omega$-stable, then $T$ is $\kappa$-stable for all infinite $\kappa$.

Proof
We prove the contrapositive.

Let $\kappa$ be an infinite cardinal.

Suppose that $T$ is not $\kappa$-stable.

Then there exists some $\mathcal M \models T$ and $A \subseteq \mathcal M$ with $\left\vert{A}\right\vert = \kappa$ such that:
 * $\left\vert{ {S_n}^\mathcal M \left({A}\right)}\right\vert > \kappa$

Let $\mathcal L_A$ denote $\mathcal L \cup \left\{{a: a \in A}\right\}$, the language obtained from $\mathcal L$ by adding new constant symbols for each $a \in A$.

For each $\mathcal L_A$-formula $\phi$, let $\left[{\phi}\right] = \left\{{p \in {S_n}^\mathcal M \left({A}\right): \phi \in p}\right\}$, the set of complete $n$-types over $A$ which contain $\phi$.

Our goal will be to find a countable set $B$ in $\mathcal M$ such that:
 * $\left\vert{ {S_n}^\mathcal M \left({B}\right)}\right\vert = 2^{\aleph_0} \ne \aleph_0$

which will demonstrate non-$\omega$-stability of $T$.

We will do this by constructing a countable binary tree of formulas such that each of the $2^{\aleph_0}$ distinct simple paths from the root of the tree out to infinity correspond to distinct types.

Before we can build the tree, we need the following lemma.

Lemma
Suppose $\left\vert{ \left[{\phi}\right] }\right\vert > \kappa$.

We argue that we can select some $\mathcal L_A$-formula $\psi$ such that both:
 * $\left\vert{\left[{\phi \land \psi}\right]}\right\vert > \kappa$

and:
 * $\left\vert{\left[{\phi \land \neg \psi}\right]}\right\vert > \kappa$

Proof of Lemma
The argument is a Proof by Contradiction.

Suppose the proposition is not true.

Let $p$ be the subset:
 * $\left\{ {\psi: \left\vert{\left[{\phi \land \psi}\right]}\right\vert > \kappa }\right\}$

where each $\psi$ is an $\mathcal L_A$-formula in $n$ free variables.

We will eventually write $\left[{\phi}\right]$ as a union of $\left\{{p}\right\}$ and other sets which are "too small", so that we contradict the cardinality of $\left[{\phi}\right]$.

In order to do this, we first need to show that $p$ is a type.

First, note that if both $\left\vert{\left[{\phi \land \neg \psi}\right]}\right\vert \le \kappa$ and $\left\vert{\left[{\phi \land \neg \psi}\right]}\right\vert \le \kappa$, then $\left\vert{\left[{\phi}\right]}\right\vert \le \kappa$, which is not the case.

Hence, for each $\psi$, either $\psi \in p$ or $\neg \psi \in p$.

Next, note that by assumption, it cannot be the case that both $\psi$ and $\neg \psi$ in $p$.

Let $\Delta = \left\{{\psi_1, \ldots, \psi_m}\right\} \cup \Delta'$ be a finite subset of $p \cup \operatorname{Th}_A \left({\mathcal M}\right)$, where $\Delta$ is written so that any sentences from $\operatorname{Th}_A \left({\mathcal M}\right)$ are in $\Delta'$.

Suppose $\psi_1 \land \cdots \land \psi_m$ is not in $p$.

Then by the above comment, $\neg \left({\psi_1 \land \cdots \land \psi_m}\right)$ is in $p$.

But this means that:

Thus, by Cardinality of Infinite Union of Infinite Sets, at least one of the $\psi_i$ must satisfy $\left\vert{\left[{\phi \land \neg \psi_i}\right]}\right\vert > \kappa$.

This is impossible, since $\psi_i \in p$.

So, $\psi_1 \land \cdots \land \psi_m$ is in $p$.

By definition of $p$ this means:
 * $\left\vert{\left[{\phi \land \psi_1 \land \cdots \land \psi_m}\right]}\right\vert > \kappa$

Hence there are types containing $\psi_1 \land \cdots \land \psi_m$.

So $\Delta$ is satisfiable.

By the Compactness Theorem, this means that $p \cup \operatorname{Th}_A \left({\mathcal M}\right)$ is satisfiable.

Hence:
 * $p \in S_n^\mathcal M \left({A}\right)$

We have that $p$ is a type.

So we can write:
 * $\displaystyle \left[{\phi}\right] = \left\{{p}\right\} \cup \bigcup_{\psi \notin p} \left[{\phi \land \psi}\right]$

since every type besides $p$ which contains $\phi$ must contain some $\psi \notin p$.

Note the cardinalities involved in this union:

Clearly, $\left\{{p}\right\}$ has cardinality $1 < \kappa$.

By definition of $p$ each $\left[{\phi \land \psi}\right]$ for $\psi \notin p$ has cardinality at most $\kappa$.

We have noted earlier in the main proof that there are only $\kappa$-many $\mathcal L_A$-formulas.

Thus, by Cardinality of Infinite Union of Infinite Sets, it is to be concluded that:
 * $\left\vert{\left[{\phi}\right]}\right\vert \le \kappa$

However, this contradicts our supposition.

The lemma follows by Proof by Contradiction.

Now the tree is to be built.

This amounts to recursively defining formulas $\phi_\sigma$ for each finite sequence $\sigma$ over $\left\{{0, 1}\right\}$.

First, the root of the tree $\phi_{}$ is defined where the subscript is the empty sequence.

The assumption is:
 * $\displaystyle \left\vert{\bigcup \left[{\phi}\right]}\right\vert = \left\vert{S_n^\mathcal M \left({A}\right)}\right\vert > \kappa$

where the union is taken over all $\mathcal L_A$-formulas $\phi$

But there are only $\kappa$ many such formulas.

Thus by Cardinality of Infinite Union of Infinite Sets there must be some $\mathcal L_A$-formula $\phi_{}$ such that the cardinality of $\left[{\phi_{} }\right]$ is strictly larger than $\kappa$.

Suppose $\phi_\sigma$ has been defined and that:
 * $\left\vert{\left[{\phi_\sigma}\right]}\right\vert > \kappa$

Let $\sigma = \left({\sigma_0, \dots, \sigma_k}\right)$.

By the lemma above, we can choose an $\mathcal L_A$ formula $\psi$ such that both:
 * $\left\vert{\left[{\phi \land \psi}\right]}\right\vert > \kappa$

and:
 * $\left\vert{\left[{\phi \land \neg \psi}\right]}\right\vert > \kappa$

Define $\phi_{\left({\sigma_0, \dots, \sigma_k, 0}\right)}$ to be $\phi_\sigma \land \psi$.

Define $\phi_{\left({\sigma_0, \dots, \sigma_k, 1}\right)}$ to be $\phi_\sigma \land \neg \psi$.

Now, let $B$ be the set of elements of $A$ which occur as constant symbols in any of the $\phi_\sigma$.

Since only countably many $\phi_\sigma$ have been defined, $B$ is countable.

We will define an injection from the set of infinite sequences over $\left\{{0, 1}\right\}$ to $S_n^\mathcal M \left({B}\right)$ using our tree.

This will demonstrate that our theory $T$ is not $\omega$-stable.

From Type Space is Compact, $S_n^\mathcal M \left({A}\right)$ is compact (when viewed as a type space).

Thus it satisfies the finite intersection axiom by Equivalent Definitions of Compactness.

We have that each $\left[{\phi_\sigma}\right]$ is closed, essentially by definition of the type space topology.

Also, any finite intersection $\left[{\phi_{} }\right] \cap \left[{\phi_{(\sigma_0)} }\right] \cap \cdots \cap \left[{\phi_{\left({\sigma_0, \dots, \sigma_k}\right)} }\right]$ is equal to $\left[{\phi_{\left({\sigma_0, \dots, \sigma_k}\right)} }\right]$ by construction.

Hence it is nonempty (by its cardinality)

Thus, by the finite intersection axiom, for each infinite sequence $\Sigma = \left({\Sigma_0, \Sigma_1, \Sigma_2, \ldots}\right)$ over $\left\{{0, 1}\right\}$, the intersection $\displaystyle \bigcap_{k \mathop \in \N} \left[{\phi_{\left({\Sigma_0, \Sigma_1, \ldots, \Sigma_k}\right)} }\right]$ is nonempty.

Moreover, let $\Sigma = \left({\Sigma_0, \Sigma_1, \Sigma_2, \ldots}\right)$ and $\Sigma' = \left({\Sigma'_0, \Sigma'_1, \Sigma'_2, \ldots}\right)$ be two distinct infinite sequences over $\left\{{0, 1}\right\}$.

Then there is some $k$ for which $\Sigma_i = \Sigma'_i$ for $i \le k$ and $\Sigma_{k+1} \ne \Sigma_{k+1}$.

But $\phi_{\left({\Sigma_1, \ldots, \Sigma_k, 0}\right)}$ and $\phi_{\left({\Sigma_1, \ldots, \Sigma_k, 1}\right)}$ were defined to imply $\psi$ and $\neg \psi$ respectively for some $\psi$.

So no type can satisfy both of them simultaneously.

Thus $\displaystyle \bigcap_{k \mathop \in \N} \left[{\phi_{\left({\Sigma_0, \Sigma_1, \ldots, \Sigma_k}\right)} }\right]$ and $\displaystyle \bigcap_{k \mathop \in \N} \left[{\phi_{\left({\Sigma'_0, \Sigma'_1, \ldots, \Sigma'_k}\right)} }\right]$ cannot both contain the same type.

Thus, we can define our injection by sending each infinite sequence $\Sigma$ over $\left\{{0, 1}\right\}$ to a type chosen from $\displaystyle \bigcap_{k \mathop \in \N} \left[{\phi_{\left({\Sigma_0, \Sigma_1, \ldots, \Sigma_k}\right)} }\right]$.

The existence of this injection implies that the cardinality of $S_n^\mathcal M \left({B}\right)$ is at least $2^{\aleph_0}$, as this is the cardinality of the set of infinite sequences over $\left\{{0, 1}\right\}$.

Hence, $T$ is not $\omega$-stable.

The theorem now follows by the Rule of Transposition.