# A sufficient condition for $\omega$-categoricity

I am trying to prove that if a complete theory $$T$$ on a countable language $$L$$ has the following property:

$$\forall n<\omega \exists M\models T\quad\quad\text{s.t.}\ M\ \text{realizes only a finite number of complete n-types}\ S_n(T)$$ then $$T$$ is $$\omega$$-categorical.

What I am trying to show is that, if there was a non isolated type $$p^*(x)$$ for some $$x$$ of some length $$n$$ the for sure: $$p^*(x)\longrightarrow\bigvee_{q_k\in S_n(T),q_k\neq p_1,...,p_n} \ q_k(x)$$ where $$p_1,...,p_n$$ are the types realised, and hence isolated by $$M$$.
Then one can find a formula $$\varphi(x)$$ such that $$p^*(x)\wedge\varphi(x)\longrightarrow\ q_m(x)$$ for some $$m$$ indexing the disjunction, as shown here An excercise about omitting types theorem
Anyway I cannot reach a contradiction.
I aslo tried to reason with orbits but with no sucess.
Any hint or help would be much appreciated.

• are you familiar with saturation at all? Commented Nov 19, 2020 at 5:36
• Yes, I know the basic facts about it, I also know that a theory is countably categorical if it admits a countable model which is both saturated and atomic Commented Nov 19, 2020 at 6:20

I assume you know the Ryll-Nardzewski theorem, at least in the form which states that for a countable and complete theory $$T$$ with infinite models, $$T$$ is $$\aleph_0$$-categorical if and only if $$S_n(T)$$ is finite for all $$n$$.

What we know is that for all $$n$$, there is a model $$M$$ which only realizes finitely many types in $$S_n(T)$$.

Let's fix an arbitrary $$n$$ and show that $$S_n(T)$$ is finite. Let $$M$$ be a model of $$T$$ which only realizes finitely many complete types in $$S_n(T)$$, say $$p_1,\dots,p_k$$. For each $$1\leq i \leq k$$, let $$\varphi_i$$ be a formula in $$p_i$$ such that $$\lnot \varphi_i\in p_j$$ for all $$j\neq i$$. You can find $$\varphi_i$$ by picking, for each $$j\neq i$$, some formula $$\psi_j\in p_i$$ such that $$\lnot \psi_j\in p_j$$, and defining $$\varphi_i = \bigwedge_{j\neq i} \psi_j$$.

Now every $$n$$-tuple from $$M$$ realizes one of the types $$p_i$$, so $$M\models \forall x_1\dots\forall x_n\, \bigvee_{i=1}^k \varphi_i.$$ And for each $$i$$, if an $$n$$-tuple from $$M$$ satisfies $$\varphi_i$$, then the tuple must realize $$p_i$$, so for all $$\theta\in p_i$$, $$M\models \forall x_1\dots\forall x_n\, (\varphi_i\rightarrow \theta).$$

Now $$T$$ is complete, so any sentence satisfied in any model is actually satisfied in every model. So we have $$T\models \forall x_1\dots\forall x_n\, \bigvee_{i=1}^k \varphi_i.$$ And for each $$i$$ and each $$\theta\in p_i$$, $$T\models \forall x_1\dots\forall x_n\, (\varphi_i\rightarrow \theta).$$

This shows that every model of $$T$$ only realizes the types $$p_1,\dots,p_k$$, since any $$n$$-tuple from any model of $$T$$ must satisfy $$\varphi_i$$ for some $$1\leq i \leq k$$, and thus must realize $$p_i$$. So in fact $$S_n(T) = \{p_1,\dots,p_k\}$$ is finite.

Since $$n$$ was arbitrary, $$S_n(T)$$ is finite for all $$n$$, so $$T$$ is $$\aleph_0$$-categorical.