Countable Stability implies Stability for All Infinite Cardinalities/Lemma

== Lemma for Countable Stability implies Stability for All Infinite Cardinalities ==
Let $T$ be a complete $\LL$-theory whose language $\LL$ is countable.
Let $\kappa$ be an infinite cardinal.
Let 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 $\sqbrk \phi$ be the set of complete $n$-types over $A$ which contain $\phi$, such that:
:$\sqbrk \phi = \set {p \in \map { {S_n}^\MM} A: \phi \in p}$
Suppose:
:$\card {\sqbrk \phi} > \kappa$.
Then there exists some $\LL_A$-formula $\psi$ such that both:
:$\card {\sqbrk {\phi \land \psi} } > \kappa$
and:
:$\card {\sqbrk {\phi \land \neg \psi} } > \kappa$
== Proof ==
The argument is a Proof by Contradiction.
 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:
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:
:$\card {\sqbrk \phi} \le \kappa$
However, this contradicts our supposition.
The lemma follows by Proof by Contradiction.
Category:Countable Stability implies Stability for All Infinite Cardinalities