# A pseudofinite structure that does not have the finite model property

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 $$M$$.

• 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 $M$. Commented Mar 9, 2021 at 2:10

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 $$2^m$$.