# A theory whose models are closed under chains has a $\forall\exists$-axiomatization

My question refers to exercise 2.5.15 in Marker's book on Model Theory (I found another thread on this exercise, but I could not find the answers I am looking for there...). Let $\mathcal L$ be any language. One has to show that any first order theory $T$ with the property that if $(\mathcal M_i\vert i\in I)$ is a chain of models of $T$, then $\bigcup_{i\in I}\mathcal M_i\models T$, has a $\forall\exists$-(or more precisely a $\Pi_2$-) axiomatization. The idea is to show that $\Gamma=\{\phi\vert\ T\models\phi\}\cap\Pi_2$ is the desired axiomatization.

The first step is to show that for $\mathcal M\models\Gamma$ there is $\mathcal N$ such that $\mathcal M\subseteq\mathcal N, \mathcal N\models T$ and whenever $\phi$ is $\Sigma_2$ with $\mathcal M\models \phi$ then $\mathcal N\models\phi$. I have managed to do this.

The next step is to show that there is $\mathcal M'$ such that $\mathcal N\subseteq \mathcal M'$ and $\mathcal M\prec\mathcal M'$. I have only managed to show that there is such a model such that there is an elementary embedding $j:\mathcal M\prec \mathcal M'$ which is not necessarily the inclusion map, this is not enough for the next step where one iterates this construction to get $$\mathcal M = \mathcal M_0\subseteq \mathcal N_0\subseteq\mathcal M_1\subseteq\mathcal N_1 \subseteq\dots$$ One then wants to look at $\mathcal M_\ast=\bigcup_{i\in\mathbb N}\mathcal M_i$. By our assumption $\mathcal M_\ast\models T$ since $\mathcal M_\ast=\bigcup_{i\in\mathbb N}\mathcal N_i$. To complete the proof one needs that $\mathcal M\prec \mathcal M_\ast$ which only holds in general if the elementary embeddings $j_i:\mathcal M_i\prec\mathcal M_{i+1}$ are the inclusion maps.

What is the correct way to do step 2?

My proof goes as follows: Let $Diag_{el}(\mathcal M)$ be the elementary diagramm of $\mathcal M$ in the language $\mathcal L_\mathcal M$ and $Diag (\mathcal N)$ be the atomic diagramm in the language $\mathcal L_\mathcal N$. I put an emphasis on the fact that I need $(\mathcal L_\mathcal M\setminus\mathcal L)\cap( \mathcal L_\mathcal N\setminus\mathcal L)=\emptyset$ (that means that if $m\in \mathcal M$ then there are two different new constant symbols for $m$, one in $\mathcal L_\mathcal M$ and one in $\mathcal L_\mathcal N$). I show that the theory $S=Diag_{el}(\mathcal M)\cup Diag(\mathcal N)$ is consistent. If not, there are $\phi_1,\dots \phi_n\in Diag(\mathcal N)$ such that $Diag_{el}(\mathcal M)\cup\{\phi_1,\dots\phi_n\}$ is inconsistent. Write $\psi(\vec c)=\bigwedge_{i=1}^n\phi_i$ where $\vec c$ are the new constant symbols of $\mathcal L_\mathcal N$ that appear in the $\phi_i$. We have $Diag_{el}(\mathcal M)\models \lnot \psi(\vec c)$ and hence $\mathcal M\models \forall\vec v\lnot\psi(\vec v)$. Since $\forall\vec v\lnot\psi(\vec v)$ is $\Pi_1$, we have $\mathcal N\models \forall\vec v\lnot\psi(\vec v)$, contradicting $\mathcal N\models \psi(\vec c)$.

Let $\mathcal M'\models S$ and wlog $\mathcal N\subseteq\mathcal M'$. Then there is an elementary embedding $j:\mathcal M\prec \mathcal M'$, but this need not be the inclusion map because of my emphasis on $(\mathcal L_\mathcal M\setminus\mathcal L)\cap( \mathcal L_\mathcal N\setminus\mathcal L)=\emptyset$.

If one instead tries to repeat the proof with the theory $S$ only in the language $\mathcal L_\mathcal N$, then one cannot expect $\mathcal M\models\forall\vec v\lnot\psi(\vec v)$ to hold since $\mathcal M\models Diag_{el}(\mathcal M)$ only if we interpret the new constants in $\mathcal L_\mathcal N$ which correspond to members of $\mathcal M$ "correctly". What one could do instead is to write $\psi(\vec c,\vec d)=\bigwedge_{i=1}^n\phi_i$ where $\vec c$ are the new constants corresponding to members of $\mathcal N\setminus\mathcal M$ and $\vec d$ only constants corresponding to members of $\mathcal M$. Then $\mathcal M\models \forall \vec v\lnot \psi(\vec v, \vec d)$, but this is not an $\mathcal L$-sentence, so we cannot move over to $\mathcal N$. If one wants an $\mathcal L$-sentence one only has $\mathcal M\models \exists\vec w\forall\vec v\lnot \psi(\vec v, \vec w)$ and hence $\mathcal N\models \exists\vec w\forall\vec v\lnot \psi(\vec v, \vec w)$, but I do not see how this helps.

Any help is greatly appreciated!

• Why do you use different names for the elements of M when the occur in Diag(N)? It is self-harm. – Primo Petri Sep 26 '17 at 3:16
• Because my proof does not work otherwise, I explain it in the last paragraph. – Andreas Lietz Sep 26 '17 at 10:05
• Try to strengthen the requrement in the first step to: $M\preceq_1N$. Where the latter symbol means that elementarity is restrected to universal sentences. – Primo Petri Sep 26 '17 at 12:43
• Thank you, that works! – Andreas Lietz Sep 26 '17 at 14:57