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.

  • 2
    $\begingroup$ 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. $\endgroup$ – Apostolos Nov 20 '15 at 0:29

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 $\varphi$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.