Is there a theory which is not the theory of finitely many of its models? Let $L$ be a signature, in the sense of model theory. Does there exist an $L$-theory $T$ such that for no finite set of models $\{M_1,...,M_n\}$ of $T$ is it the case that $Th(\{M_1,...,M_n\})=T$? In other words, is there a signature $L$ and a theory $T$ such that $T$ is not the theory of finitely many of its models?
 A: Recall that for a class $\mathbb{K}$ of structures, $Th(\mathbb{K})$ is the common theory of the elements of $\mathbb{K}$, that is, $Th(\mathbb{K})=\bigcap_{\mathcal{M}\in\mathbb{K}}Th(\mathcal{M})$.

I claim that the theories of the form $Th(\mathbb{K})$ for some finite set of structures $\mathbb{K}$ are exactly those which have finitely many completions - that is, whose model classes are finite up to elementarily equivalence.

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*First, suppose $T$ has only finitely many completions. Let $M_1,...,M_n$ be models of those completions. Then we clearly have $Th(\{M_1,...,M_n\})\supseteq T$. Conversely, if $\varphi\in Th(\{M_1,...,M_n\})$, then every completion of $T$ entails $\varphi$ which means that $T\models\varphi$ (think about $T\cup\{\neg\varphi\}$).


*In the other direction, suppose that $T=Th(\{M_1,...,M_n\})$. Then I claim that $T$ is (up to model class) just the set $$X:=\{\mu_1\vee...\vee\mu_n: M_1\models \mu_1,...,M_n\models\mu_n\}.$$ Clearly we have $Mod(T)\supseteq Mod(X)$; conversely, if $N\not\models T$ then there must be sentences $\nu_i\in Th(M_i)$ ($1\le i\le n$) such that $N\models\neg\nu_i$, but then $N\not\models X$.
As an example, suppose $T$ is computably axiomatizable but essentially undecidable (= has no computable completion). Then no completion of $T$ is finitely axiomatizable over $T$; consequently, $T$ has continuum-many completions (combinatorially: every dead-end-free binary tree with no isolated paths has continuum-many paths), and so cannot be the common theory of a finite set of structures.
