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Apologies for asking a basic question on this forum (was once a professional mathematician, now trying to work back towards being 'alive' gain in the Erdos sense).

Stuck on a part of exercise 2.5.15 d) of Marker's Model Theory - An Introduction.

The goal is to prove the result of Chang Los and Suszko that a theory having $\forall\exists$ axiomatization is equivalent to being preserved under direct limits.

So I have my theory $T$ that is preserved under direct limits (i.e. if $(M_{i}, i \in I)$ is a chain of models of $T$ then $\bigcup_{i} M_{i}$ is a model of $T$).

Let $\Gamma = \{\phi:\phi \text{ is a }\forall\exists\text{-sentence and }T \vDash\phi \}$.

I have a model $M \vDash \Gamma$.

... in section b) of the exercise, we're asked to show that there is an $N \vDash T$ s.t. if $\psi$ is an $\exists\forall$-sentence and $M \vDash \psi$, then $N \vDash\psi$. This is fine (proof by contradiction, if there wasn't it would imply that some of $T$'s axioms imply $\neg\psi$, but $\neg\psi$ is a $\forall\exists$-sentence, so then $M\nvDash\psi$.

... then in section c) we're asked to show that there is an $N' \supseteq M$ with $N'\equiv N$. Fine again - just need to show that $Th(N)\cup\text{Diag}(M)$ is satisfialble. Again, proof by contradiction, the only stuff that can contradict stuff in $\text{Diag}(M)$ is universal statements from $Th(N)$ and $M$ has those by construction.

And then in d) they ask us to show that there is an $M' \supseteq N'$ such that $M \prec M'$. And this is where I'm totally stuck. Coz in general, it seems that if you have an embedding of $M$ into $N$, there is no way normally you should be able to get some $M'$ so that $N$ embeds into $M'$ and the composite embedding of $M$ into $M'$ is elementary.

For instances say we have a model of linear orders with one element. Then embed it into an model with two elements. Great. But you can't embed that two element model into another model where the statement 'there is only one element' is true. Oh - and linear orders are universally axiomatized - so if my statement were correct, this would actually be a counterexample to what I'm trying to show.

I'm sure I'm missing something totally obvious (elementary in fact)... again really sorry for bothering people with this. I genuinely have tried for some time on this one.

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    By the same kind of diagram business that you already used above, an $M'\supseteq N'$ such that $M\preceq M'$ exists iff $N'$ satisfies all universal sentences true in $M$. Which it does. – 2017-01-30
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    With regard to your example involving linear orders: If we start with any finite structure $M$ (say of size $n$) and run the proof, the model $N$ will also have size $n$ (since "there are exactly $n$ elements" is an $\exists\forall$ sentence), so we'll have $M = N' = M'$. Then the chain we build will just give us $M$ in the limit. But this is fine: we've shown that $M = N'\models T$. – 2017-01-31
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    @tomasz It's borderline: this looks like a good question that could come from a graduate student who is preparing to do research in model theory, and certainly it would have been accepted with open arms at one point in MO's history. But anyway, I'll migrate it, and I would be grateful, if you think it is too elementary for MO, if you would give an answer there. – 2017-01-31
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    @user43208: I may be guilty of holding mathoverflow to too high a standard, granted, but I think I have seen way more advanced model theory related questions here at math.se. I also recall having solved this exercise during a problem session while I was an undegrad. ;-) – 2017-01-31
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    @tomasz I would have thought either site would be fine. The division of labor between them is not at all strict. – 2017-01-31

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