# Show if theory T is ∀∃-axiomatizable, then T has an existentially closed model.

Setting

Definition $\mathcal{M} \models T$ is existentially closed if whenever $\mathcal{N} \models T$, $\mathcal{N} \supseteq \mathcal{M}$, and $\mathcal{N}\models \exists \bar{v} \phi(\bar{v},\bar{a})$, where $\bar{a} \in \mathbb{M}$ and $\phi$ is quantifier free, then $\mathcal{M} \models \exists \bar{v} \phi(\bar{v},\bar{a})$.

1. I would like to show that if T is $\forall\exists$-axiomatizable,, then T has an existentially closed model.

2. From now on we assume that if $\mathcal M \models T$, then there is $\mathcal N \supseteq \mathcal M$ existentially closed with $|\mathbb{N}| = |\mathbb{M}| + |\mathcal L| + \aleph_o$. Suppose that T has an infinite nonexistentially closed model, then I want to prove T has a nonexistentially closed model of cardinality $\kappa$ for any infinite cardinal $\kappa \ge |\mathcal{L}|$.

3. Finally, I would like to show that if T is $\kappa$-catagorical for some infinite $\kappa \ge |\mathcal{L}|$ and axiomatized by $\forall\exists$-sentences, then all models of T are existentially closed.

• The key lemma for this is one you mentioned in another question: if $T$ has an AE axiomatization then the class of models of $T$ is closed under taking unions of chains of models of $T$. So the next step is to verify a lemma: if you have just one formula $(\exists \bar v)\phi(\bar v, \bar a)$ which is consistent with $\text{Diag}(M)$ then there is an $M'$ satisfying that formula with $M \subseteq M'$. Then apply this lemma repeatedly to get a chain of nested models, and take the limit. – Carl Mummert Feb 19 '15 at 14:03
• @CarlMummert could you elaborate on what "taking the limit" means? – chibro2 Feb 19 '15 at 14:07
• Taking the union of the chain of models. This is one example of the concept of 'direct limit' in category theory (it's a somewhat trivial example, but it is a direct limit). – Carl Mummert Feb 19 '15 at 14:14
• @Carl are you assuming there is such a formula (∃v)ϕ(v,a) so that $Diag(M) \models (∃v)ϕ(v,a)$? Because if there is, wouldn't any extension of $M$ (you called it M') also satisfy the formula since the witness to the $\exists$ is contained in $M'$? Or am I showing that there is this formula (∃v)ϕ(v,a)? – chibro2 Feb 19 '15 at 18:27
• @CarlMummert adding to what I previously said, isn't it true that $\mathcal M \models \pmb{T} \Rightarrow \mathcal M \models \forall \bar{w} \exists \bar{v} \phi(\bar{v},\bar{w}) \Rightarrow \mathcal M \models \exists \bar{v} \phi(\bar{v},\bar{a}) \text{ for every$\bar{a} \in \mathbb{M}$}$? And just to summarize you're saying we want a chain $M_1,M_2,\ldots$ where $M_{i+1} \models T_{i} \cup diag_{\exists}(M_i)$ correct? – chibro2 Feb 19 '15 at 19:05