Marker Exercise 2.5.10: universal part of a theory and supermodel

Question

I'm trying to solve Exercise 2.5.10 in Marker's Model Theory: An Introduction.

It goes:

Let T be an $\mathcal L$-theory and $T_\forall$ be all of the universal sentences $\phi$ such that $T \models \phi$. Show that $\mathcal A \models T_\forall$ if and only if there is $\mathcal M \models T$ with $\mathcal A \subseteq \mathcal M$.

WLOG we may assume that $T$ is consistent and that $T$ is closed under $\models$. $T_\forall$ is then just the set of universal (aka $\forall_1$ or $\Pi_1$) sentences in $T$. Now "only if" part is easy. I try to show the "if" part by resorting to Diagram Lemma: i.e., by showing that the union of the diagram of $\mathcal A$ and $T$ is consistent. Assume not and let a quantifier-free formula $\phi(\vec x)$ be such that $T \models \neg \phi(\vec a)$ and $\mathcal A \models \phi(\vec a)$, where $\vec a \in \mathcal A$. ...

This is where I'm stuck. I'm trying to replace the use of parameter with a quantifier but with no luck. What can I do? Is it possible to provide me with a clue?

Answer 1

Let $a$ be an (infinite) tuple enumerating ${\mathcal A}$. Let $p(x)=\textrm{qf-tp}_{\mathcal A}(a)$, that is, the set of quantifier-free formulas $\varphi(x)$ such that ${\mathcal A}\vDash\varphi(a)$.

I claim that $T\cup p(x)$ is consistent. If not, $T\vdash\neg\exists x\,\varphi(x)$ for some $\varphi(x)\in p$. Hence $\neg\exists x\,\varphi(x)\in T_{\forall}$. This contradicts ${\mathcal A}\vDash\exists x\,\varphi(x)$.

By compactness there is a model ${\mathcal M}\vDash T$ and a tuple $b$ such that ${\mathcal M}\vDash p(b)$. The map $a\mapsto b$ is an immersion of ${\mathcal A}$ in ${\mathcal M}$. So, up to isomorphism we can assume that $\mathcal A \subseteq \mathcal M$

Answer 2

(I'll use Marker's notation)

We want to show that $Diag(\mathcal{A}) \cup T$ is satisfiable. By compactness, it's enough to show that $T \cup \{\psi_1(\bar a), \ldots, \psi_n(\bar a)\}$ is satisfiable for any finite subset $\{\psi_1(\bar a), \ldots, \psi_n(\bar a)\} \subset Diag(\mathcal{A})$. Note that $\bar a$ here is a finite tuple of symbols representing elements of $A$ (in the language $\mathcal{L}_A$).

Suppose not. Then
$T \models \neg \bigwedge_i \psi_i(\bar a)$ in the language $L_A$
Because this is true for any interpretation of the symbols $\bar a$,
$T \models \forall \bar v \:\neg \bigwedge_i \psi_i(\bar v)$

Let $\phi$ be the sentence $\forall \bar v \:\neg \bigwedge_i \psi_i(\bar v)$. It is universal (since the $\psi_i$ don't have quantifiers) and it follows from $T$. Therefore, $\phi \in T_{\forall}$, and so $\mathcal{A} \models \phi$. But $\phi$ is clearly false in $\mathcal{A}$, a contradiction.