# Ryll-Nardzewski theorem and types with parameters

By Ryll-Nardzewski theorem we know that a theory $$T$$ is $$\omega$$-categorical iff for each $$n$$, every type in $$S_n(T)$$ is isolated.

Question. What can we say about types with parameters. Does $$\omega$$-categoricity of $$T$$ imply that for every countable model $$M\models T$$, every $$A\subseteq M$$, any type in $$S_n(A)$$ is isolated?

And if for every countable model $$M\models T$$, every $$A\subseteq M$$, each type in $$S_n(A)$$ is isolated, then can we conclude that $$T$$ is $$\omega$$-categorical?

The answer to your first question is a strong no whenever $$T$$ has infinite models. Indeed, if $$M$$ is any infinite model of $$T$$, then there is a consistent type $$p(v)\in S_1(M)$$ containing the set of formulas $$\{v\neq m:m\in M\}$$. This type cannot be isolated, since any isolated type is realized, and $$p(v)$$ is not realized in $$M$$. If you prefer an example where the type in question is realized in $$M$$, here's a different one; let $$T=\operatorname{DLO}$$ and let $$M=\mathbb{Q}$$ be the standard model, and take $$A$$ to be the subset $$\{1/n:n\in\mathbb{N}\}$$. Then $$\operatorname{tp}(0/A)$$ is not isolated, since for any finite subset $$A_0\subset A$$ there exists $$c>0$$ with the same type as $$0$$ over $$A_0$$, but there does not exist any $$c>0$$ with the same type as $$0$$ over $$A$$.
However, if $$A$$ is finite, then the answer is yes. Indeed, in that case, enumerate $$A$$ as $$a_1\dots a_m$$, and let $$p(v_1,\dots,v_n)$$ be a type in $$S_n(A)$$. Let $$q(v_1,\dots,v_n,w_1,\dots,w_m)\in S_{m+n}(\varnothing)$$ be the type obtained by replacing every instance of $$a_i$$ in a formula of $$p$$ by the variable $$w_i$$. By Ryll-Nardzewski, $$q$$ is isolated, say by a formula $$\phi(\overline{v},\overline{w})$$. Then $$p$$ is isolated by $$\phi(\overline{v},\overline{a})$$ (why?), as desired.
Finally, the answer to your second question is yes; taking $$A=\varnothing$$ gives precisely the Ryll-Nardzewski criterion that you cite at the beginning of your answer.
• Here's another way of seeing the "strong no": if every type in $S_n(A)$ is isolated, then by compactness $S_n(A)$ is finite. But then necessarily $A$ is finite, since $S_n(A)$ at least contains the types isolated by $x_1=\dots= x_n=a$ for each $a\in A$. Jul 25 '21 at 0:40