The converse of Vaught's Test

Question

I work in first order logic.

I noticed that a complete theory that is satisfied by finite models can only be satisfied by models of a given fixed finite cardinality. It made me think about the converse of Vaught's test given all assumptions but $\kappa$-categoricity. Namely,

is there a consistent complete theory without finite models in a countable language such that for every infinite cardinal $\kappa$ the theory is not $\kappa$-categorical?

Given an example of one such theory I would also find interesting knowing how has completeness been proved.

Answer 1

Russoo gave (in a comment, unfortunately), the example of $(\mathbb{N},+,\cdot,0,1)$. While that works, there are tamer examples. Let $L$ be the language with a unary predicate $P_n$ for each $n \in \mathbb{N}$, and let $T$ assert these predicates are disjoint, and that each one has infinitely many realizations. $T$ is complete, and is not categorical for any $\kappa$: for any $\kappa$, there are models of size $\kappa$ containing a point not in any $P_n$, and there are models with no such point. $T$ is tame in that it is stable, if you know what that means.

For this $T$, and for similar theories, there is another test that works if you're trying to show completeness:

Let $T$ be any theory. If, for every $\aleph_0$-saturated $M,N$ modeling $T$, the family of isomorphisms between finitely generated substructures of $M$ and $N$ has the back-and-forth property, then $T$ eliminates quantifiers. If furthermore, for all such models, that family is non-empty, $T$ is complete. (A family $\mathcal{F}$ of partial functions $M \to N$ has the back-and-forth property if, for any $f \in \mathcal{F}, a \in M, b \in N$, there are $f_1,f_2 \in \mathcal{F}$ extending $f$ such that $a$ is in the domain of $f_1$ and $b$ is in the range of $f_2$.)

Answer 2

A mathematically natural example is the theory of real closed fields. Reals which are not algebraic can be included or omitted in arbitrary-cardinality models as you desire, via compactness and omitting types; and a model of RCF, though it may have nontrivial automorphisms, will embed the algebraic reals in a unique way, so which rational cuts are realized is an isomorphism invariant.

Answer 3

I might have found another example here.