# The converse of Vaught's Test

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.

• An example is the complete theory of $\mathbb{N}$ in the language $(+,\cdot,0,1)$. Trivially, this theory is complete, consistent and has no finite models. But it is a well known fact that this theory is not categorical for any infinite $\kappa$. Commented Sep 25, 2015 at 12:27
• @russoo The language (<) would do just as well, wouldn't it?
– bof
Commented Sep 25, 2015 at 23:15
• Thanks @russoo. Could you give me any insight on how it is proved that such theory is not categorical for any infinite $\kappa$ ? Commented Sep 25, 2015 at 23:49
• @Anguepa. The only proof I know uses so called $\kappa$-like models of arithmetic. The existence of such models is a somewhat non-trivial fact. Commented Sep 28, 2015 at 10:29
• @bof. Yes, the language $<$ is sufficient. Commented Sep 28, 2015 at 10:30

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 $f_2$.)