$\forall \exists$-sentences and existentially closed model I'm working my way through the lecture notes on Model Theory by Henson and van den Dries. Proposition $20.10$ on p. $130$:

Let $\Sigma$ be a set of $L$-sentences and let $\mathcal{M}$ be an $L$ structure. Then $\mathcal{M} \models \Sigma_{\forall \exists}$ if and only if there is a model $\mathcal{N}$ of $\Sigma$ such that $\mathcal{M} \subset \mathcal{N}$ and $\mathcal{M}$ is existentially closed in $\mathcal{N}$.

Per definition just before it

We denote by $\Sigma_{\forall \exists}$ the set of all $\forall \exists L$-sentences $\sigma$ such that $\Sigma \models \sigma$.

Yet $20.10$ is only proved in the “$\Rightarrow$” direction.
I tried to prove it in the other direction: suppose there is a model $\mathcal{N}$ of $\Sigma$ such that $\mathcal{M} \subset \mathcal{N}$ and $\mathcal{M}$ is existentially closed in $\mathcal{N}.$ Then, for every existential $L$-formula $\varphi(x)$ and each $a \in M^{n}$, if $\mathcal{N} \models \varphi(x)[a]$, then $\mathcal{M} \models \varphi(x)[a]$.
I do not know how to proceed further. I would appreciate any help.
 A: Suppose $\mathcal{N}$ is an extension of $\mathcal{M}$ such that $\mathcal{N}\models\Sigma$ and $\mathcal{M}$ is existentially closed in $\mathcal{N}$. To show that $\mathcal{M}\models\Sigma_{\forall\exists}$, we will pick a $\varphi\in\Sigma_{\forall\exists}$ and try to show that $\mathcal{M}\models\varphi$.
Now since $\varphi\in\Sigma_{\forall\exists}$ we know two things:

*

*$\mathcal{N}\models\varphi$.


*$\varphi$ has the form $\forall x\exists y\theta(x,y)$ for $\theta$ quantifier-free. (I'm writing in terms of individual variables, rather than tuples of variables, for simplicity; this doesn't change the idea of the argument at all.)
Now to show $\mathcal{M}\models\varphi$, we pick some arbitrary $a\in\mathcal{M}$ and will try to show that $\mathcal{M}\models\exists y\theta(a,y)$. Since $\mathcal{N}\models\varphi$ we have $\mathcal{N}\models\exists y\theta(a,y)$, that is, there is some $b\in\mathcal{N}$ such that $\mathcal{N}\models\theta(a,b)$. But we have an assumption which implies that such a $b$ must in fact lie in $\mathcal{M}$:

 $\mathcal{M}$ is existentially closed in $\mathcal{N}$ (note that $a\in\mathcal{M}$).

So in fact $\mathcal{M}\models\varphi$.
