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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?

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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.

A sketch of one way to carry this out is hidden below.

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 $\mathcal{B}_n$.

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  • $\begingroup$ Do I have to prove anything about the existence of these extensions with the successor steps in transfinite recursion? Or it is enought to argue that, if some exists, I can just set it as part of this chain of structures? $\endgroup$
    – user480840
    Dec 4, 2019 at 20:25
  • $\begingroup$ Also, do I need to assume that the extension in successor steps is also a model of $T$? $\endgroup$
    – user480840
    Dec 4, 2019 at 21:09
  • $\begingroup$ 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. $\endgroup$ Dec 4, 2019 at 22:50

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