I am a little confused about what I think must be either a standard theorem or a standard counterexample in model theory, and I am hoping that the MathOverflow model theorists can help sort me out about which way it goes.


My situation is that I have a chain of submodels, which is not necessarily an elementary chain,


$$M\_0\subseteq M\_1\subseteq M\_2\subseteq\cdots$$


and I have elementary embeddings $j\_n:M\_n\to M\_n$, which cohere in the sense that $j\_n=j\_{n+1}\upharpoonright M\_n$. So there is a natural limit model $M=\bigcup\_n M\_n$ and limit embedding $j:M\to M$, where $j(x)$ is the eventual common value of $j\_n(x)$. 


**Question.** Is the limit map $j:M\to M$ necessarily elementary? 


A natural generalization would be a coherent system of elementary embeddings $j\_n:M\_n\to N\_n$, with possibly different models on each side. The question is whether the limit embedding $j:M\to N$ is elementary, where $M=\bigcup\_n M\_n$, $N=\bigcup\_n N\_n$ and $j=\bigcup\_n j\_n$. And of course one could generalize to arbitrary chains or indeed, arbitrary directed systems of coherent elementary embeddings, instead of just $\omega$-chains.


I thought either there should be an easy counterexample or an easy proof, perhaps via Ehrenfeucht-Fraïssé games? 



Is the union of a chain of elementary embeddings elementary?
First I'll give a counterexample to the natural generalization, adapted from my earlier comment: 


Let $M\_n = (\mathbb{N},<)$ for all $n$, and let $N\_n = (\mathbb{N}\sqcup \frac{1}{n!}\mathbb{Z},<)$, where all elements of $\frac{1}{n!}\mathbb{Z}$ are greater than all elements of $\mathbb{N}$ in the order. Note that each $N\_n$ is isomorphic to $(\mathbb{N}\sqcup\mathbb{Z},<)$, and there is a natural inclusion $N\_n\subseteq N\_{n+1}$ for all $n$, since $\frac{1}{n!}\mathbb{Z} \subseteq \frac{1}{(n+1)!}\mathbb{Z}$ as subsets of $\mathbb{Q}$. 


Now $M\_n\preceq N\_n$ for all $n$, but $\bigcup\_n M\_n\not\preceq \bigcup\_n N\_n$, since the former is $(\mathbb{N},<)$ and the latter is $(\mathbb{N}\sqcup\mathbb{Q},<)$, which is not discrete. 




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Now I'll explain how to turn any counterexample to the natural generalization into a counterexample to the original question. 


Suppose we have a coherent system of elementary embeddings $j\_n\colon M\_n \to N\_n$ such that the limit map $j\colon M\to N$ is not elementary. Let $L$ be the language of this counterexample, which we assume to be relational, and let $L' = L\cup \{E\}$, where $E$ is a new binary relation. 


For each $n$, we construct a structure $M^\*\_n$ as follows: $E$ is an equivalence relation with countably many classes, which we denote by $(C\_i)\_{i\in \mathbb{Z}}$. We interpret the relations from $L$ on each class $C\_i$ so that $C\_i$ is a copy of $M\_n$ when $i\leq 0$ and a copy of $N\_n$ when $i > 0$. There are no relations between the classes. 


There is a natural inclusion $M\_n^\*\subseteq M\_{n+1}^\*$ for all $n$, in which each class $C\_i$ in $M\_n^\*$ is included in to the class $C\_i$ in $M\_{n+1}^\*$ according to the inclusions $M\_n\subseteq M\_{n+1}$ and $N\_n\subseteq N\_{n+1}$. 


Let $j\_n^\*\colon M^\*\_n\to M^\*\_n$ map $C\_i$ to $C\_{i+1}$ as the identity on $M\_n$ for all $i<0$, as the identity on $N\_n$ for all $i>0$, and as $j\_n \colon M\_n\to N\_n$ for $i = 0$. Then $j\_n^\*$ is an elementary embedding, and $j\_n = j\_{n+1}\restriction M\_n^\*$. 


In the limit, $M^\* = \bigcup\_n M\_n^\*$ has equivalence classes such that $C\_i$ is a copy of $M$ when $i\leq 0$ and a copy of $N$ when $i>0$. And the limit map $j^\*\colon M^\*\to M^\*$ maps $C\_i$ to $C\_{i+1}$ as the identity on $M$ for all $i<0$, as the identity on $N$ for all $i>0$, and as $j\colon M\to N$ for $i=0$. This $j^\*$ is not elementary, since an $L$-formula whose truth is not preserved by $j$ can be relativized to the equivalence class $C\_0$ to give an $L'$-formula whose truth is not preserved by $j^\*$. 


