Interpretable structure saturated Suppose that the $\mathcal{L}_0$-structure $\mathcal N$ is interpretable in the $\mathcal L$-structure $\mathcal M$. Is $\mathcal N$ saturated if $\mathcal M$ is saturated?
That $\mathcal N$ is interpretable in $\mathcal M$ means that (Definition 1.3.9. in Marker's book) there is

*

*a $\mathcal L$-definable set $X\subseteq M^n$ and

*a $\mathcal L$-definable equivalence relation $E$ on $X$ and

*for each symbol of $\mathcal{L}_0$ a $\mathcal L$-definable $E$-invariant set on $X$ such that $X/E$ with the induced structure is isomorphic to $\mathcal N$.

Let $p$ be a complete type in $S^{\mathcal N}(A)$ for some small $A\subset N$.
The interpretation of $\mathcal N$ in $\mathcal M$ translates the type $p$ to a set of $\mathcal L$-formulas over a small subset of $\mathcal M$. If this set of formulas is consistent, then the saturation of $\mathcal M$ would imply that this partial type is realized in $\mathcal M$. This realization translates back to a realization of $p$. However, I do not know why the obtained set of $\mathcal L$-formulas is consistent in $\mathcal M$?
 A: Your argument is correct, so I'll just answer your last question.

However, I do not know why the obtained set of $\mathcal{L}$-formulas is consistent in $\mathcal{M}$?

Let's recall two basic facts:

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*A set of formulas is consistent if and only if every finite subset is consistent. (Compactness)

*A finite set of formulas with parameters from a model $M$ is consistent if and only if it is realized in $M$. (If $\{\varphi_1(x),\dots,\varphi_n(x)\}$ is consistent, then the sentence $\exists x\bigwedge_{i=1}^n\varphi_i(x)$ is true in some elementary extension of $M$, so it is true in $M$).

Ok, so we have a complete type $p\in S^{\mathcal{N}}(A)$ which pulls back along the interpretation to a set of formulas $q$ over a small subset $B\subseteq \mathcal{M}$. Let $q'$ be a finite subset of $q$. The formulas in $q'$ correspond to a finite subset $p'\subseteq p$. Since $p$ is consistent, $p'$ is consistent and realized in $N$. Pulling back the witness along the interpretation, $q'$ is realized in $M$. We've shown every finite subset of $q$ is consistent, so $q$ is consistent.
