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