1
$\begingroup$

I have a proof of the following:

[*] Let $A$ be a countable $\omega$-saturated model of a complete, countable theory $T$ (with infinite models). There is a bijection between the orbits under the action of $\text{Aut}(A)$ on $A^n$ and the types $S_n(T)$.

Here the orbit of $(a_1, \dots, a_n)$ under the action of $\text{Aut}(A)$ on $A^n$ is defined as the set $\{(f(a_1), \dots, f(a_n)) : f \in \text{Aut}(A)\}$.

I want to use this to prove:

If for every countable model $A \models T$ and $n \in \mathbb{N}$, the set of orbits under the action of $\text{Aut}(A)$ on $A^n$ is finite, then $T$ is $\omega$-categorical. (The reverse is not so difficult.)

My attempt so far: it suffices to find a countable $\omega$-saturated model of $T$, because then the result follows from the assumptions, lemma [*], and the theorem $\omega$-categorical $\Leftrightarrow$ every $S_n(T)$ is finite. Such a model exists iff every $S_n(T)$ is countable, so it would be enough to prove that $T$ has only countably many countable models. I thought this would be easy enough, but then I found out that there are theories with a countable $\omega$-saturated model that have continuum-many countable models. So things are a bit more complicated...

$\endgroup$
1
$\begingroup$

If you already know that $\omega$-categoricity is equivalent to the condition that every $S_n(T)$ is finite, then there's an easy argument that doesn't use [*].

Suppose for contradiction that there is some $n\in \omega$ such that $S_n(T)$ is infinite. Pick some countable set of these and realize them: Then you have a countable model $A\models T$ which realizes infinitely many $n$-types. But $n$-tuples which realize different types cannot be in the same orbit of the action of $\text{Aut}(A)$ on $A^n$, so there are infinitely many orbits, contradiction.

The point is that [*] is really two statements:

(1) If $a$ and $b$ are in the same orbit, they realize the same type.

(2) If $a$ and $b$ realize the same type, they are in the same orbit.

Point (1) is true in any model, while point (2) is only true in certain models, e.g. saturated ones. And the argument above only uses (1), not (2).

$\endgroup$
  • $\begingroup$ Thanks Alex! I did not know you could take a countable set of types in $S_n(T)$ and find a model that realizes all those types. I'm trying to think if I have seen this before. But I only recall the theorem that for complete theories $T$ a model $M \models T$ finitely satisfies a partial type $p$ $\Leftrightarrow$ $p$ is consistent with $T$. How do you show your claim? $\endgroup$ – Joachim Apr 24 at 9:46
  • $\begingroup$ I also know $p$ is finitely satisfiable in $M$ $\Leftrightarrow$ p is realized in some elementary extension of $M$. Wait... maybe I can create an infinite chain of elementary embeddings each realizing one of the countable-many types $(p_n)_{n\in\omega}$. I know that these types are consistent with $T$, so they are finitely realized in every model of $T$. Let $M_0$ realize $p_0$, and let $M_1$ be some elementary extension of $M_0$ where $p_1$ is realized (which exists by the theorem mentioned above). This $M_1$ still realizes $p_0$ because it is an elementary extension of $M_0$. (continued.) $\endgroup$ – Joachim Apr 24 at 10:02
  • $\begingroup$ We can continue to define $M_n$ as a model realizing $p_0, \dots, p_n$. The colimit of all these models realizes all types $p_n$, and is countable if all the $M_n$ are countable. But the $M_n$ can all be taken countable: $M_0$ can be taken countable by Lowenheim-Skolem; $M_{n+1}$ can be taken countable if $M_{n}$ is countable, by inspecting the proof of the theorem I mentioned at the beginning of my previous comment & using Lowenheim-Skolem. Is this how you show your claim? $\endgroup$ – Joachim Apr 24 at 10:11
  • 1
    $\begingroup$ @Joachim yes, that's exactly right! $\endgroup$ – Alex Kruckman Apr 24 at 13:26
  • $\begingroup$ Thanks again for your help. It seems that this should be a standard theorem, but it is not in my lecture notes, even though it is easy to prove. $\endgroup$ – Joachim Apr 24 at 16:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.