# Weakly Saturated model not Saturated

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.

• 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. Commented May 5, 2023 at 12:26
• 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. Commented May 5, 2023 at 14:41
• In our lectures, we defined a model B to be $w$- universal if every countable model elementarily embeds into it Commented May 5, 2023 at 16:54
• 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. Commented May 5, 2023 at 17:00
• @user1044791 Yes, that's true. Every countable saturated model is $\omega$-universal, so your proof of the forward implication is fine. Commented May 8, 2023 at 19:58

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 $$\{c_i is finitely consistent in $$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.

• 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. Commented May 5, 2023 at 12:41
• Thank you so much Commented May 5, 2023 at 16:56