Countable Stability implies Stability for All Infinite Cardinalities

Theorem
Let $T$ be a complete $\LL$-theory whose language $\LL$ 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 $\MM \models T$ and $A \subseteq \MM$ with $\card A = \kappa$ such that:
 * $\card {\map { {S_n}^\MM} A} > \kappa$

Let $\LL_A$ denote $\LL \cup \set {a: a \in A}$, the language obtained from $\LL$ by adding new constant symbols for each $a \in A$.

For each $\LL_A$-formula $\phi$, let $\sqbrk \phi = \set {p \in \map { {S_n}^\MM} A: \phi \in p}$, the set of complete $n$-types over $A$ which contain $\phi$.

Our goal will be to find a countable set $B$ in $\MM$ such that:
 * $\card {\map { {S_n}^\MM} B} = 2^{\aleph_0} \ne \aleph_0$

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
Now the tree is to be built.

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

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

The assumption is:
 * $\ds \card {\bigcup \sqbrk \phi} = \card {\map { {S_n}^\MM} A} > \kappa$

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

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

Thus by Cardinality of Infinite Union of Infinite Sets there must be some $\LL_A$-formula $\phi_{\paren {} }$ such that the cardinality of $\sqbrk {\phi_{\paren {} } }$ is strictly larger than $\kappa$.

Suppose $\phi_\sigma$ has been defined and that:
 * $\card {\sqbrk {\phi_\sigma} } > \kappa$

Let $\sigma = \paren {\sigma_0, \dots, \sigma_k}$.

By the lemma above, we can choose an $\LL_A$ formula $\psi$ such that both:
 * $\card {\sqbrk {\phi \land \psi} } > \kappa$

and:
 * $\card {\sqbrk {\phi \land \neg \psi} } > \kappa$

Define $\phi_{\paren {\sigma_0, \dots, \sigma_k, 0} }$ to be $\phi_\sigma \land \psi$.

Define $\phi_{\paren {\sigma_0, \dots, \sigma_k, 1} }$ 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 $\set {0, 1}$ to $\map { {S_n}^\MM} B$ using our tree.

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

From Type Space is Compact, $\map { {S_n}^\MM} A$ is compact (when viewed as a type space).

Thus it satisfies the finite intersection axiom by Equivalence of Definitions of Compact Topological Space.

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

Also, any finite intersection $\sqbrk {\phi_{\paren {} } } \cap \sqbrk {\phi_{\paren {\sigma_0} } } \cap \cdots \cap \sqbrk {\phi_{\paren {\sigma_0, \dots, \sigma_k} } }$ is equal to $\sqbrk {\phi_{\paren {\sigma_0, \dots, \sigma_k} } }$ by construction.

Hence it is nonempty (by its cardinality)

Thus, by the finite intersection axiom, for each infinite sequence $\Sigma = \paren {\Sigma_0, \Sigma_1, \Sigma_2, \ldots}$ over $\set {0, 1}$, the intersection $\ds \bigcap_{k \mathop \in \N} \sqbrk {\phi_{\paren {\Sigma_0, \Sigma_1, \ldots, \Sigma_k} } }$ is nonempty.

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

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

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

So no type can satisfy both of them simultaneously.

Thus $\ds \bigcap_{k \mathop \in \N} \sqbrk {\phi_{\paren {\Sigma_0, \Sigma_1, \ldots, \Sigma_k} } }$ and $\ds \bigcap_{k \mathop \in \N} \sqbrk {\phi_{\paren {\Sigma'_0, \Sigma'_1, \ldots, \Sigma'_k} } }$ cannot both contain the same type.

Thus, we can define our injection by sending each infinite sequence $\Sigma$ over $\set {0, 1}$ to a type chosen from $\ds \bigcap_{k \mathop \in \N} \sqbrk {\phi_{\paren {\Sigma_0, \Sigma_1, \ldots, \Sigma_k} } }$.

The existence of this injection implies that the cardinality of $\map { {S_n}^\MM} B$ is at least $2^{\aleph_0}$, as this is the cardinality of the set of infinite sequences over $\set {0, 1}$.

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

The theorem now follows by the Rule of Transposition.