Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2010 / B / 4 / e / Solution

The countable omitting types theorem says that for a consistent theory in a countable language and countably many nonprincipal types of finite arities, there is a countable model omitting all those types. In particular a nonisolated complete type of $T$ can be omitted in a countable model.

Interpret “definable” here as without parameters. If parameters were allowed freely, every infinite structure would have infinitely many definable singleton sets, and the claimed implication would be false even for a dense linear order without endpoints.

Infinitely many distinct parameter-free definable subsets of $A^n$ imply infinitely many complete $n$-types: if there were only $r$ such types, each formula would select a subset of those $r$ types, giving at most $2^r$ distinct definable subsets. There must also be a nonisolated complete type. To see this, suppose each complete type $p$ had an isolator ${\theta }_p$. The family of their basic open sets covers the type space. Its compactness theorem gives a finite subcover, which would account for only finitely many complete types, a contradiction. Equivalently apply logical compactness theorem to $T\cup \{\neg {\theta }_p(c):p\text{ complete}\}$.

Choose a nonisolated type $p$. The omitting types theorem gives a countable model omitting it, while part (d) gives a countable model realizing it. An isomorphism of structures preserves realization of parameter-free types, so these models are not isomorphic. Thus $\boxed{T\text{ is not }{\aleph }_0\text{-categorical}}$.