Let $T$ be a complete theory, $L$ a countable first order language. We call a model $A$ of $T$:

  • Weakly Saturated if it realises every $n$-type ($\forall n \in \mathbb{N}$).
  • Saturated if in every expansion $A'$ of $A$ by finitely many constant symbols, every 1-type in $\mathrm{Th}(A')$ is realised in $A'$.

I am hoping for an example of countable $L$, complete $T$ and countable $A$ which is weakly saturated but not saturated.

Note: with the above assumptions on $L$ and $T$,

  • a countable $A$ is saturated iff it's $\omega$-universal,

  • and such an $A$ exists iff $T$ is small.

  • 2
    $\begingroup$ Is your definition of "$w$-universal" that every countable model embeds elementarily in $A$? If so, that should be an omega not a w, and it's not true that saturated is equivalent to $\omega$-universal. What's true is saturated iff ($\omega$-universal and $\omega$-homogeneous) implies $\omega$-universal implies weakly saturated. But the last two implications are not equivalences. $\endgroup$ Commented May 5, 2023 at 12:26
  • 1
    $\begingroup$ Alex is right, what you write about $\omega$-universality is not true. Still, the intuition is partially correct: $\omega$-universality is saturation-like property. It is equivalent to requiring that all parameter-free types with $\omega$ free variables are realized. $\endgroup$ Commented May 5, 2023 at 14:41
  • $\begingroup$ In our lectures, we defined a model B to be $w$- universal if every countable model elementarily embeds into it $\endgroup$ Commented May 5, 2023 at 16:54
  • $\begingroup$ Is it true T has a countable saturated model iff it's a small theory? The proof in our lecture notes goes via $\omega$-universal on the forward implication. $\endgroup$ Commented May 5, 2023 at 17:00
  • 1
    $\begingroup$ @user1044791 Yes, that's true. Every countable saturated model is $\omega$-universal, so your proof of the forward implication is fine. $\endgroup$ Commented May 8, 2023 at 19:58

1 Answer 1


The language contains $<$ and infinitely many constants $c_i$ for $i\in\omega$. Let $M$ be the model with domain $\mathbb Q∪\{\sqrt 2\}$. The interpretation of $c_i$ is a sequence that converges to $\sqrt 2$ from below.

The type $\{c_i<x<\sqrt 2 : i\in\omega\}$ is finitely consistent in $M$ but is is not realized. Hence $M$ is not saturated.

I claim that $M$ is weakly saturated.

Let $p(x)$ be a complete parameter free type. There are two cases:

  1. it contains the formula $x\le c_i$ for some $i$. Then it is realized in $M$ (essentially, because countable DLOs are saturated).

  2. it contains $x>c_i$ for every $i$. Then any $a\ge\sqrt2$ realizes $p(x)$ because all these elements of $M$ have the same type over $\varnothing$.

For types of the form $p(x_1,\dots,x_n)$ the argument is similar.

  • 2
    $\begingroup$ Nice. This example also shows that $\omega$-universality is not equivalent to saturation for countable models. Another way to argue for weak saturation and $\omega$-universality is to note that the saturated model of this theory, namely $\mathbb{Q}$, is an elementary substructure of $M$, and weak saturation and $\omega$-universality are obviously preserved in moving to elementary extensions. $\endgroup$ Commented May 5, 2023 at 12:41
  • $\begingroup$ Thank you so much $\endgroup$ Commented May 5, 2023 at 16:56

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