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This question already has an answer here:

I guess it is not difficult, but I spent an hour thinking about this without success.

Elementary embedding implies elementary equivalence, but elementary equivalence between structures does not implies, necessarily, the existence of an elementary embedding between these structures. I`m looking for an example of two structures that are elementary equivalent but there is no elementary embedding between them.

any ideas?

Thanks!

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marked as duplicate by tomasz, Amzoti, rschwieb, Dan Rust, Lord_Farin Jun 14 '13 at 17:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Consider the theory of equivalence relations with exactly two infinite equivalence classes, which is a complete theory, so every pair of models is elementarily equivalent. Let $M$ be a model with classes of size $\aleph_0$ and $\aleph_2$ and let $N$ be a model with classes of size $\aleph_1$ and $\aleph_1$. Then $M$ cannot embed in $N$ because a class of size $\aleph_2$ cannot embed in a class of size $\aleph_1$, and $N$ cannot embed in $M$ for a similar reason. So there is no embedding at all, much less an elementary embedding.

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Consider the following two fields (as structures in the language $(0,1,+,\cdot)$): the real numbers $\Bbb{R}$ on the one hand, and on the other some countable nonarchimidean real-closed field $F$. They're both real-closed, so they're elementarily equivalent. But $\Bbb{R}$ doesn't embed in $F$ (due to cardinality) and $F$ doesn't embed in $\Bbb{R}$ (due to not being archimedean).

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