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.

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    $\begingroup$ 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 $a\in A$. $\endgroup$ Jul 25 '21 at 0:40

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