Big Implies Saturated

Theorem
Let $\mathcal{M}$ be an $\mathcal{L}$-structure.

Let $\kappa$ be a cardinal.

If $\mathcal{M}$ is $\kappa$-big, then it is $\kappa$-saturated.

Proof
The idea of the proof is to go to some elementary equivalent structure where a type is realized, and interpret a new relation symbol as the singleton containing some realization of the type. This lets us write sentences about the extension saying that there must exist an element satisfying the relation, and every element satisfying the relation satisfies all of the formulas in the type. Then using bigness, these sentences carry over into $\mathcal{M}$.

Let $p$ be a complete $n$-type over $A$, where $A$ is a subset of the universe of $\mathcal{M}$ with cardinality strictly less than $\kappa$.

We must show that $p$ is realized in $\mathcal{M}$.

Let $\mathcal L_A$ be the language obtained from $\mathcal L$ by adding constant symbols for each element of $A$.

Viewing $\mathcal M$ as an $\mathcal L_A$-structure in the natural way, let $\mathrm{Th}_A (\mathcal M)$ be the collection of $\mathcal L_A$-sentences satisfied by $\mathcal M$.

By definition of type, $p\cup \mathrm{Th}_A (\mathcal M)$ is satisfiable by some $\mathcal L_A$-structure $\mathcal{N}$.

Since $\mathcal{N}$ satisfies $p$, some $n$-tuple $\bar b$ in it realizes $p$.

Now, let $\mathcal{L}_A^*$ be the language obtained from $\mathcal{L}_A$ by adding a new relation symbol $R$.

We can extend $\mathcal{N}$ to be an $\mathcal{L}_A^*$-structure by interpreting the symbol $R$ as the singleton $R^\mathcal{N} = \{\bar b\}$.

Since $\mathcal{N}\models \mathrm{Th}_A (\mathcal M)$, we have that $\mathcal{M}$ and $\mathcal{N}$ are elementary equivalent as $\mathcal{L}_A$-structures.

Hence, since $\mathcal{M}$ is $\kappa$-big by assumption, there is some relation $R^\mathcal{M}$ on the universe of $\mathcal{M}$ such that $(\mathcal{M}, R^\mathcal{M})$ is elementary equivalent to $(\mathcal{N},R^\mathcal{N})$ as $\mathcal{L}_A^*$-structures.

But, $(\mathcal{N},R^\mathcal{N})$ satisfies the $\mathcal{L}_A^*$-sentences:
 * $\exists \bar x R(\bar x)$, and
 * $\forall \bar x (R(\bar x) \to \phi(\bar x))$ for each $\phi(\bar x)$ in $p$.

Hence, $(\mathcal{M},R^\mathcal{M})$ also satisfies these sentences, and so there must be some $\bar d \in R^\mathcal{M}$ realizing $p$.

So, $\mathcal{M}$ realizes $p$.