# 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

Suppose $\card {\sqbrk \phi} > \kappa$.

We argue that we can select some $\LL_A$-formula $\psi$ such that both:

$\card {\sqbrk {\phi \land \psi} } > \kappa$

and:

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

### Proof of Lemma

The argument is a Proof by Contradiction.

Suppose the proposition is not true.

Let $p$ be the subset:

$\set {\psi: \card {\sqbrk {\phi \land \psi} } > \kappa}$

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

We will eventually write $\sqbrk \phi$ as a union of $\set p$ and other sets which are "too small", so that we contradict the cardinality of $\sqbrk \phi$.

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

First, note that if both $\card {\sqbrk {\phi \land \neg \psi} } \le \kappa$ and $\card {\sqbrk {\phi \land \neg \psi} } \le \kappa$, then $\card {\sqbrk \phi} \le \kappa$, 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 $\Delta = \set {\psi_1, \ldots, \psi_m} \cup \Delta'$ be a finite subset of $p \cup \map {\operatorname {Th}_A} \MM$, where $\Delta$ is written so that any sentences from $\map {\operatorname {Th}_A} \MM$ are in $\Delta'$.

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

Then by the above comment, $\neg \paren {\psi_1 \land \cdots \land \psi_m}$ is in $p$.

But this means that:

 $\ds \card {\sqbrk {\phi \land \neg \paren {\psi_1 \land \cdots \land \psi_m} } }$ $=$ $\ds \card {\sqbrk {\paren {\phi \land \neg \psi_1} \lor \cdots \lor \paren {\phi \land \neg \psi_m} } }$ De Morgan's Laws (Logic) $\ds$ $=$ $\ds \card {\sqbrk {\phi \land \neg \psi_1} \cup \cdots \cup \sqbrk {\phi \land \neg \psi_m} }$ $\ds$ $>$ $\ds \kappa$

Thus, by Cardinality of Infinite Union of Infinite Sets, at least one of the $\psi_i$ must satisfy $\card {\sqbrk {\phi \land \neg \psi_i} } > \kappa$.

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:

$\card {\sqbrk {\phi \land \psi_1 \land \cdots \land \psi_m} } > \kappa$

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

So $\Delta$ is satisfiable.

By the Compactness Theorem, this means that $p \cup \map {\operatorname {Th}_A} \MM$ is satisfiable.

Hence:

$p \in \map { {S_n}^\MM} A$

We have that $p$ is a type.

So we can write:

$\sqbrk \phi = \set p \cup \bigcup_{\psi \notin p} \sqbrk {\phi \land \psi}$

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

Note the cardinalities involved in this union:

Clearly, $\set p$ has cardinality $1 < \kappa$.

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

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

Thus, by Cardinality of Infinite Union of Infinite Sets, it is to be concluded that:

$\card {\sqbrk \phi} \le \kappa$

The lemma follows by Proof by Contradiction.

$\Box$

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 Equivalent Definitions of Compactness.

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.

$\blacksquare$