Fraïssé Limit and Ultraproduct

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 $$M^*$$?

• Could you ask a more specific question?
– YCor
Jul 29 '19 at 16:24

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.

• Could you explain why $M^*$, in the second example, has endpoints? Jul 29 '19 at 16:40
• @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. Jul 29 '19 at 16:41
• @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! Jul 29 '19 at 16:49
• @MostafaMirabi I think is is right for arbitrary ultraproducts. But I don’t think an ultraproduct of finite structures can be countably infinite. Jul 29 '19 at 17:05
• @GabeConant Yes, you're right. Jul 29 '19 at 17:15