Every model of a $\forall \exists$-theory has an existentially closed extension

Let $$T$$ be a $$\forall \exists$$-axiomatizable theory, i.e., there exists a theory $$T'$$ wich contains only formulas of the form $$\forall \vec{x} \exists \vec{y} \varphi(\vec{x}, \vec{y})$$, with $$\varphi$$ quantifier free, such that $$\mathcal{A} \vDash T$$ iff $$\mathcal{A} \vDash T'$$, for all $$\tau$$-structures $$\mathcal{A}$$. Then, if $$\mathcal{A} \vDash T$$, there exists an existentially complete $$\tau$$-structure $$\mathcal{B}$$ such that $$\mathcal{B} \vDash T$$ and $$\mathcal{A} \subseteq \mathcal{B}$$.

Here, $$\mathcal{M}$$ is an existentially closed model of $$T$$ iff for any extension $$\mathcal{N} \supseteq \mathcal{M}$$, existential formula $$\exists \vec{x} \varphi(\vec{x})$$, where $$\varphi$$ is quantifier free, and $$m$$-uple $$\vec{b} \in M^m$$, if $$\mathcal{N} \vDash \exists \vec{x} \varphi(\vec{x}, \vec{b})$$, then $$\mathcal{M} \vDash \exists \vec{x} \varphi(\vec{x}, \vec{b})$$.

I don't even know where to start from here, can anyone give me any tips?

The key idea is that since $$T$$ is $$\forall\exists$$-axiomatizable, the union of any chain of models of $$T$$ is still a model of $$T$$. So to extend $$\mathcal{A}$$ to an existentially closed model of $$T$$, you can just keep extending it over and over by transfinite recursion until it is existentially closed.
Fix an enumeration $$(\varphi_\alpha,\vec{b}_\alpha)$$ of all pairs consisting of a quantifier free formula and a tuple of elements of $$\mathcal{A}$$. Construct a nested sequence of models $$\mathcal{A}_\alpha$$ of $$T$$ by transfinite recursion, starting from $$\mathcal{A}_0=\mathcal{A}$$. At successor steps, take an extension which satisfies $$\mathcal{A}_{\alpha+1}\models \exists\vec{x}\varphi_\alpha(\vec{x},\vec{b}_\alpha)$$, if such an extension exists. At limit steps, take unions.
At the end of this process you'll obtain an extension $$\mathcal{B}_1$$ of $$\mathcal{A}$$ which satisfies the definition of existential closedness except that the tuple $$\vec{b}$$ is restricted to be in $$\mathcal{A}$$. But now you can repeat this process with $$\mathcal{B}_1$$ in place of $$\mathcal{A}$$ to get an extension $$\mathcal{B}_2$$ which is existentially closed for tuples in $$\mathcal{B}_1$$, and then repeat with $$\mathcal{B}_2$$ in place of $$\mathcal{A}$$ to get $$\mathcal{B}_3$$, and so on. After iterating this $$\omega$$ times, the union $$\mathcal{B}=\bigcup_n\mathcal{B}_n$$ will be existentially closed since every tuple from it is contained in some $$\mathcal{B}_n$$.
• Also, do I need to assume that the extension in successor steps is also a model of $T$? Dec 4, 2019 at 21:09
• All the structures here are models of $T$. If no such extension exists in a successor step, you can just do nothing (i.e., take $\mathcal{A}_{\alpha+1}=\mathcal{A}_\alpha$). As I said, this is only a sketch of the argument. There are some details to be filled in to check that $\mathcal{B}_1$ has the property I claimed, for instance. Dec 4, 2019 at 22:50