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

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

Our goal will be to find a countable set $B$ in $\mathcal M$ such that $|S_n^\mathcal M (B)| = 2^{\aleph_0} \neq \aleph_0$, 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 $|[\phi]| > \kappa$. We argue that we can select some $\mathcal L_A$-formula $\psi$ such that both $|[\phi \wedge \psi]| > \kappa$ and $|[\phi \wedge \neg\psi]| > \kappa$.

Proof of Lemma
We will argue by contradiction, so assume this is not true.

Let $p$ be the subset $\{\psi : |[\phi \wedge \psi]| > \kappa \}$ where each $\psi$ is an $\mathcal L_A$-formula in $n$ free variables.

We will eventually write $[\phi]$ as a union of $\{p\}$ and other sets which are "too small", so that we contradict the cardinality of $[\phi]$. In order to do this, we first need to show that $p$ is a type.

First, note that if both $|[\phi \wedge \neg\psi]| \leq \kappa$ and $|[\phi \wedge \neg\psi]| \leq \kappa$, then $|[\phi]| \leq \kappa$, which is not the case. Hence, for each $\psi$, either $\psi \in p$ or $\neg\psi \in p$.

Next, note that by assumption, we cannot have both $\psi$ and $\neg\psi$ in $p$.

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

If $\psi_1 \wedge \cdots \wedge \psi_m$ is not in $p$, then by the above comment, $\neg (\psi_1 \wedge \cdots \wedge \psi_m)$ is in $p$. But this means that $|[\phi \wedge \neg (\psi_1 \wedge \cdots \wedge \psi_m)]|$ $=$ $|[(\phi \wedge \neg \psi_1) \vee \cdots \vee (\phi \wedge \neg\psi_m)]|$ $=$ $|[\phi \wedge \neg\psi_1] \cup \cdots \cup [\phi \wedge \neg\psi_m]| > \kappa$.

Thus, by Cardinality of Infinite Union of Infinite Sets, at least one of the $\psi_i$ must satisfy $|[\phi \wedge \neg\psi_i]| > \kappa$, which is impossible since $\psi_i \in p$.

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

By definition of $p$ this means $|[\phi \wedge \psi_1 \wedge \cdots \wedge \psi_m]| > \kappa$, and hence there are types containing $\psi_1 \wedge \cdots  \wedge \psi_m$.

So $\Delta$ is satisfiable.

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

Now, since $p$ is a type, we can write $\displaystyle [\phi] = \{p\}\cup \bigcup_{\psi \notin p}[\phi \wedge \psi]$, since every type besides $p$ which contains $\phi$ must contain some $\psi \notin p$.

Note the cardinalities involved in this union:

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

By definition of $p$ each $[\phi\wedge\psi]$ 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, we should conclude that $|[\phi]| \leq \kappa$, but this contradicts our supposition, completing our proof of the lemma by contradiction.

Now we build the tree. This amounts to recursively defining formulas $\phi_\sigma$ for each finite sequence $\sigma$ over $\{0,1\}$.


 * First, we define the root of the tree which we denote by, $\phi_{}$, where the subscript is the empty sequence.

Since we are assuming $\displaystyle |\bigcup [\phi]| = |S_n^\mathcal M \left({A}\right)| > \kappa$ where the union is taken over all $\mathcal L_A$-formulas $\phi$, but there are only $\kappa$ many such  formulas, we have by Cardinality of Infinite Union of Infinite Sets  that there must be some $\mathcal L_A$-formula $\phi_{}$ such  that the cardinality of $[\phi_{}]$ is strictly larger than  $\kappa$.


 * Suppose $\phi_\sigma$ has been defined and $|[\phi_\sigma]| > \kappa$.

Let $\sigma = (\sigma_0, \dots, \sigma_k)$

By the lemma above, we can choose an $\mathcal L_A$ formula $\psi$ such that both $|[\phi \wedge \psi]| > \kappa$ and $|[\phi \wedge \neg\psi]| > \kappa$.

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

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

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

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

We will define an injection from the set of infinite sequences over $\{0,1\}$ to $S_n^\mathcal M (B)$ using our tree. This will demonstrate that our theory $T$ is not $\omega$-stable.

Since $S_n^\mathcal M \left({A}\right)$ is compact (when viewed as a type space), it satisfies the finite intersection axiom by Equivalent Definitions of Compactness.

Since each $[\phi_\sigma]$ is closed (essentially by definition of the type space topology), and any finite intersection $[\phi_{}] \cap [\phi_{(\sigma_0)}] \cap \cdots \cap [\phi_{(\sigma_0,\dots,\sigma_k)}]$ is equal to $[\phi_{(\sigma_0,\dots,\sigma_k)}]$ by construction and hence is nonempty (by its cardinality), we have by the finite intersection axiom that for each infinite sequence $\Sigma = (\Sigma_0,\Sigma_1,\Sigma_2,\dots)$ over $\{0,1\}$, the intersection $\displaystyle \bigcap_{k \mathop \in \N} [\phi_{(\Sigma_0, \Sigma_1,\dots, \Sigma_k)}]$ is nonempty.

Moreover, if $\Sigma = (\Sigma_0,\Sigma_1,\Sigma_2,\dots)$ and $\Sigma' = (\Sigma'_0,\Sigma'_1,\Sigma'_2,\dots)$ are two distinct infinite sequences over $\{0,1\}$, then there is some $k$ for which $\Sigma_i = \Sigma'_i$ for $i \leq k$ and $\Sigma_{k+1} \neq \Sigma_{k+1}$. But, we defined $\phi_{(\Sigma_1, \dots, \Sigma_k, 0)}$ and $\phi_{(\Sigma_1, \dots, \Sigma_k, 1)}$ to imply $\psi$ and $\neg\psi$ respectively for some $\psi$, so no type can satisfy both of them simultaneously. Thus, $\displaystyle \bigcap_{k \mathop \in \N} [\phi_{(\Sigma_0, \Sigma_1,\dots, \Sigma_k)}]$ and $\displaystyle \bigcap_{k \mathop \in \N} [\phi_{(\Sigma'_0, \Sigma'_1,\dots, \Sigma'_k)}]$ cannot both contain the same type.

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

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

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

The theorem now follows by the Rule of Transposition.