Let us call a theory $T$ good if

  • it is complete;
  • it is formulated in a countable language $L$;
  • it has infinite models.

Suppose $T$ is good and $M \models T$, the exercise is to prove

$M$ is $\omega$-saturated iff it is $\omega$-homogeneous and it realizes all types [in $\cup_n S_n(T)$].

Right to left is immediate. We also know that

[*] $M$ is $\omega$-saturated iff it is $\omega$-homogeneous and $\omega^+$-universal.

So it suffices to show that realizing all types implies $\omega^+$-universality (maybe using $\omega$-homogeneity). My idea is to utilize the technique of the $\Rightarrow$-direction proof of [*], but adept it to use that $M$ realizes $\cup_n S_n(T)$ instead of $\omega$-saturation. But I am stuck.

I don't know any theorems that describe/involve models that realize all types, except that saturated models realize all types (but that is not very useful here). What do we know about such models, and how can I use that?


1 Answer 1


The question boils down to showing that an $\omega$-homogeneous model $M$ that realises all the types in $\bigcup_{n <\omega} S_n(T)$ is $\omega^+$-universal, so let's prove that.

Let $A$ be countable and elementarily equivalent to $M$. Enumerate $A$ as $(a_i)_{i < \omega}$, and denote $A_n = \{a_i : i < n\}$ (so this does not include $a_n$). We will build an increasing chain of maps $f_0 \subseteq f_1 \subseteq \ldots$, such that $f_n: A_n \to M$ is a partial elementary embedding for all $n < \omega$.

  • For $n = 0$ we can just take $f_0$ to be the empty map.
  • Suppose we have constructed $f_n: A_n \to M$. Let $p(x_0, \ldots, x_n) = tp(a_0, \ldots, a_n)$. By assumption, there are $b_0, \ldots, b_n$ in $M$ that realise $p(x_0, \ldots, x_n)$. Restricting types, we find $$ tp(b_0, \ldots, b_{n-1}) = tp(a_0, \ldots, a_{n-1}) = tp(f_n(a_0), \ldots, f_n(a_{n-1})). $$ So the map $g: \{b_0, \ldots, b_{n-1}\} \to M$ given by $g(b_i) = f_n(a_i)$ is partial elementary. By $\omega$-homogeneity we can extend $g$ to a partial elementary map $g': \{b_0, \ldots, b_n\} \to M$. Then we have that $$ tp(b_0, \ldots, b_n) = tp(g'(b_0), \ldots, g'(b_n)) = tp(f_n(a_0), \ldots, f_n(a_{n-1}), g'(b_n)). $$ So now we can extend $f_n: A_n \to M$ to $f_{n+1}: A_{n+1} \to M$ by sending $a_n$ to $g'(b_n)$.

Now we have our chain of partial elementary maps, we can just set $f = \bigcup_{n < \omega} f_n$ to obtain an elementary embedding $f: A \to M$.

  • $\begingroup$ Although you explicitly pointed out your convention that $A_n$ doesn't contain $a_n$, you seem to have forgotten that convention in the second bullet item, the induction step, where $f_n(a_n)$ occurs even though $f_n$ has domain $A_n$. (Just replace $n$ by $n-1$ throughout that second bullet item.) $\endgroup$ Apr 18, 2019 at 21:30
  • $\begingroup$ @AndreasBlass you are completely right, I have corrected it now. Thanks for pointing it out! $\endgroup$ Apr 18, 2019 at 21:54
  • $\begingroup$ Thanks for your help. I had something very similar in mind, using the right to left proof of [*]. But I could not show the required elementary embedding. Doing it stepwise like this solves that issue. $\endgroup$
    – Joachim
    Apr 19, 2019 at 9:52

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