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Let $\mathcal{C}=\{M_i\}$ be a Fraïssé class of finite $\mathcal{L}$-structures with the generic model $M$. Also, let $M^*=\prod_U M_i$ the be ultraproduct of members of $\mathcal{C}$ where $U$ is a non-principal ultrafilter.

Question. What is the relation between $M$ and $M^*$?

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  • $\begingroup$ Could you ask a more specific question? $\endgroup$
    – YCor
    Jul 29 '19 at 16:24
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First note that $M$ is countable while $M^*$ has size continuum. So the two structures are never isomorphic. But you may instead ask when/if they are elementarily equivalent; and the answer is “sometimes”.

Example 1: Let $\mathcal{C}$ be the class of finite graphs. Then $M$ is the countable random graph, and one can have that $M^*$ is elementarily equivalent to $M$ (for example by taking an ultraproduct of Paley graphs).

Example 2: Let $\mathcal{C}$ be the class of finite linear orders. Then $M$ is isomorphic to $(\mathbb{Q},<)$, while $M^*$ is a ($\aleph_1$-saturated) discrete linear order with endpoints.

EDIT: In the question “an ultraproduct of members of $\mathcal{C}$” is ambiguous, since perhaps you allow one to take an ultraproduct of a subclass. This doesn’t make much difference for Example 2, but in Example 1 one could potentially get different answers for $M^*$ by varying the subclass (or even the ultrafilter). For example, an ultraproduct of the class of triangle-free graphs will not be a model of the theory of the random graph.

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  • $\begingroup$ Could you explain why $M^*$, in the second example, has endpoints? $\endgroup$
    – Lajos
    Jul 29 '19 at 16:40
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    $\begingroup$ @Lajos Because having endpoints can be expressed in a single first order sentence, which is satisfied by any finite linear order. So it’s true in $M^*$ by Los’s Theorem. $\endgroup$ Jul 29 '19 at 16:41
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    $\begingroup$ @GabeConant I think under some set theoretic assumption $M^*$ can be countable. It depends on the ultrafilter $U$. The existence of such ultrafilters is equivalent to the existence of a measurable cardinal, I guess! $\endgroup$ Jul 29 '19 at 16:49
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    $\begingroup$ @MostafaMirabi I think is is right for arbitrary ultraproducts. But I don’t think an ultraproduct of finite structures can be countably infinite. $\endgroup$ Jul 29 '19 at 17:05
  • $\begingroup$ @GabeConant Yes, you're right. $\endgroup$ Jul 29 '19 at 17:15

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