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.
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 $\varphi$.