Is there a first-order theory which is not the theory of countably many of its models?

Question

This is a natural follow-up to my previous question, here: 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. Is there an $L$-theory $T$ such that $T$ is not the theory of countably many of its models?

Answer 1

Sure. We can even do this in propositional logic. Let $(P_i)_{i\in I}$ be an uncountable set of proposition symbols. Let $T$ be the theory asserting that at most one of the $P_i$ is true, i.e., $T$ is axiomatized by $P_i\rightarrow \lnot P_j$ for all $i\neq j$ in $I$. Let $K$ be a countable set of models of $T$. Then the set $J = \{i\in I\mid \exists M\in K\text{ such that }M\models P_i\}$ is at most countable (since each $M\in K$ satisfies at most one of the $P_i$). Since $I$ is uncountable, there exists $i\in I\setminus J$, and $\lnot P_i\in \mathrm{Th}(K)$, while $T\not\models \lnot P_i$ (since $T$ has a model in which $P_i$ is true). So $T$ is not the theory of countably many of its models.

On the other hand, if the language is countable, then the answer is no. Let $T$ be an arbitrary theory which is closed under entailment. If $T$ is inconsistent, it is the theory of the empty set of models. So assume $T$ is consistent, and let $M_*$ be a model of $T$. Now enumerate all $L$-sentences as $(\varphi_n)_{n\in \omega}$. For each $n\in \omega$, if $T\not\models \varphi_n$, pick some model $M_n\models T\cup \{\lnot \varphi_n\}$. Otherwise, define $M_n = M_*$. Let $K = \{M_n\mid n\in \omega\}$. Since all structures in $K$ are models of $T$, $T\subseteq \mathrm{Th}(K)$. To prove the reverse inclusion, suppose $\varphi\in \mathrm{Th}(K)$. Then $\varphi = \varphi_n$ for some $n$. If $T\not\models \varphi_n$, then $M_n\models \lnot \varphi_n$, contradicting that $\varphi_n\in \mathrm{Th}(K)$. Thus $T\models \varphi_n$.

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Here is a more abstract view on the question: A theory $T$ has an associated space $S_T$ of completions (the Stone space of the Lindenbaum-Tarski algebra of $L$-sentences modulo $T$-equivalence). We have $T = \mathrm{Th}(K)$ if and only if $\{\mathrm{Th}(M)\mid M\in K\}$ is dense in $S_T$. So $T$ is the theory of countably many of its models if and only if $S_T$ is separable (has a countable dense subset). If $L$ is countable, then $S_T$ is second-countable and hence separable. On the other hand, there are certainly Stone spaces which are not separable.