David Marker, Model Theory: An Introduction, Exercise 2.5.10

A compactness argument relating universal consequences of a theory to extensions of structures.

Exercise 2.5.10. Let \(T\) be an \(\mathcal{L}\)-theory and \(T_\forall\) be the set of all universal sentences \(\phi\) such that \(T \vDash \phi\). Show that \(\mathcal{A} \vDash T_\forall\) if and only if there is \(\mathcal{M} \vDash T\) with \(\mathcal{A} \subseteq \mathcal{M}\).

Proof. \((\Rightarrow)\) Consider the theory \(T' := T \cup \diag \mathcal{A}\). Any model of \(T'\) is an extension of \(\mathcal{A}\) which is also a model of \(T\).

For a contradiction, assume that \(T'\) is unsatisfiable. By compactness, there exists a quantifier-free \(\mathcal{L}\)-formula \(\phi(\vec{a})\) with \(\vec{a} \in A\) such that \(T \cup \{\phi(\vec{a})\} \vDash \bot\), implying \(T \vDash \forall \vec{x} \, \neg \phi(\vec{x})\). Therefore, \(\forall \vec{x} \, \neg \phi(\vec{x}) \in T_\forall\), and hence \(\mathcal{A} \vDash \forall \vec{x} \, \neg \phi(\vec{x})\). In particular, \(\mathcal{A} \vDash \neg \phi(\vec{a})\), which is a contradiction.

\((\Leftarrow)\) Suppose that \(\mathcal{A} \subseteq \mathcal{M}\) for some \(\mathcal{M} \vDash T\). Every sentence in \(T_\forall\) is true in \(\mathcal{M}\), and universal sentences are preserved under substructures. Thus \(\mathcal{A} \vDash T_\forall\). \(\square\)