# Why in countably saturated models, types that are consistent with $TH(\mathcal{M_a})$ are finitely realizable?

I'm learning about countably saturated ($\alpha$-saturated) models. There is a hidden presupposition everywhere used:

The type $\Gamma(x)$ is consistent with $TH(\mathcal{M_a})$ iff $\Gamma(x)$ is finitely realizable in $\mathcal{M_a}$,

where $\mathcal{M_a}$ is the extended model of the model $\mathcal{M}$ over anextended language $\mathcal{L_a}=\mathcal L \cup \{c_a : a\in A\}$ that $A \subseteq M$.

Why is that? All I know (if right!) is that: from consistency of $\Gamma(x) \cup TH(\mathcal{M_a})$ and the model existence lemma, this union is satisfiable. So there is a model of $TH(\mathcal{M_a})$ that also satisfies $\Gamma(x)$. Then by compactness, $\Gamma(x)$ is finitely satisfiable. Now what?! We know that satisfiablity and realization are not the same.

I'm confused now! Please tell me what's happening here.

• The theory of $M_a$ is a complete theory. So if an element of $\phi(x)\in\Gamma(x)$ is not witnessed by some elements of $M_a$, you would have $\forall x\lnot\phi(x)\in Th(M_a)$. This would imply that the type is not finitely consistent with $Th(M_a)$ a contradiction. At least I think so. – Apostolos Nov 20 '15 at 0:29

## 1 Answer

Satisfiability and realization in a given model $M$ are the same for formulas, relative to a complete theory $T$.

That is, suppose $T$ is a complete theory and $M\models T$. Let $\varphi(x)$ be a formula. If $\varphi$ is satisfiable relative to $T$, then there is some other model $M'\models T$ containing a realization of $\varphi$. Then $M'\models \exists x\, \varphi(x)$. But $T$ is complete, so $\exists x\, \varphi(x)\in T$. Hence $M\models \exists x\, \varphi(x)$, so $M$ contains a realization of $\varphi$.