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Is there a pseudofinite structure $M$ that does not have the finite model property? Pseudofinite means that any sentence $M$ satisfies is satisfied in a finite structure, and finite model property means, in addition, that we can take the finite structure to be a substructure of $M$.

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    $\begingroup$ A note on terminology: Usually, "pseudofinite" and "finite model property" are taken to be synonyms, and the term "finite submodel property" is used for the property that every sentence true in $M$ is true in a finite substructure of $M$. $\endgroup$ Commented Mar 9, 2021 at 2:10

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Sure. Consider any pseudofinite field of characteristic zero (e.g. take a nonprincipal ultrapower of a sequence of finite fields of increasing characteristic). The field axioms, conjoined into a single sentence, are satisfied in such a structure but no finite field embeds into any field of characteristic zero.

A simpler, if less mathematically interesting, example is the successor graph $S$ on $\mathbb{Z}$. The sentence "Every vertex has degree $2$" has as its finite models only (unions of) cycles, but $S$ is acyclic. To see that $S$ is pseudofinite, you can show that for every $m$, Duplicator has a winning strategy in the Ehrenfeucht-Fraisse game between $S$ and the graph consisting of a single cycle of length $2^m$.

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