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In exercise 4 page 456 of Hodges "Model Theory" it is required to show that if an ultrafilter $\mathcal{U}$ is not $\omega_1$-complete, then every ultraproduct $\prod_I A_i/ \mathcal{U}$ has cardinality $< \omega$ or $\geq 2^\omega$.

Does this suggest there can be ultraproducts $\prod_I A_i/ \mathcal{U}$ which have cardinality $\omega$ if we assume that the ultrafilter is (non-principal) $\omega_1$-complete? Is this possible?

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up vote 10 down vote accepted

Yes, but the existence of such ultrafilters is a large cardinal hypothesis; it is equivalent to the existence of a measurable cardinal.

If each $A_i$ is countably infinite and $\cal U$ is countably complete ($\omega_1$-complete), then the ultraproduct $\Pi_i A_i/{\cal U}$ will be countable. To see this, fix enumerations of each $A_i$, and observe that any element $x\in\Pi_iA_i/{\cal U}$ is represented by some function $\vec x=\langle x_i\rangle_i$, where $x_i$ is the $n_i^{th}$ element of $A_i$. But by countable completeness, the value of $n_i$ must be constant on a set in $\cal U$, and so the function $n\mapsto [g_n]_{\cal U}$ where $g_n(i)$ is the $n^{th}$ element of $A_i$ is a bijection of $\omega$ with the ultraproduct. So it is countably infinite, as desired.

Meanwhile, it is important to note that every countably complete nonprincipal ultrafilter $\cal U$ on any set is actually $\kappa$-complete for some measurable cardinal $\kappa$, and so the hypothesis carries large cardinal strength.

A stronger result is possible, once you realize that any countably complete ultrafilter is actually $\kappa$-complete for a much larger cardinal. If $\cal U$ is a $\kappa$-complete ultrafilter on a set $I$ and the structures $A_i$ are uniformly bounded in size below $\kappa$, in a language of size less than $\kappa$, then the ultrapower $\Pi_i A_i/{\cal U}$ is isomorphic to one of the $A_i$. The reason is that we may assume $\kappa$ is a measurable cardinal, and in particular, it is also inaccessible. And so there are fewer than $\kappa$ many structures in that language of size at most $\delta$, for any fixed $\delta<\kappa$. It follows by $\kappa$-completeness that the measure $\cal U$ must concentrate on a set of indices $i$ for which the $A_i$ are all isomorphic to each other. And it follows by the reasoning in the first argument that this common $A_i$ is also isomorphic to the ultrapower $\Pi_i A_i/{\cal U}$.

The result mentioned in the previous paragraph is the size $\kappa$ analogue of the fact for ultrafilters $\cal U$ on $\omega$ that the ultrapower $\Pi_i A_i/{\cal U}$ of structures $A_i$ having uniformly bounded finite size in a finite language is isomorphic to one of them, namely, to the one whose isomorphism type occurs on a set in ${\cal U}$.

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Does your answer mean that if the $A_i$ are countable, then if the ultraproduct has countable cardinality then it must be isomorphic to one of the $A_i$'s? – user38200 Jan 29 '14 at 10:55
Yes. In the case you describe, the ultrafilter must have been countably complete (for otherwise the ultraproduct would have had size continuum), and then the argument at the end of my answer kicks in. – Joel David Hamkins Jan 29 '14 at 11:26
If the structures $A_i$ are not uniformly bounded below $\kappa$, what would be the cardinality of $\prod_I A_i/\mathcal{U}$? (For instance if we take the cardinality of any $A_i$ equal to $\kappa$). – user38200 Jan 29 '14 at 12:12
That is, this holds if one interprets “$A_i$ are countable” as also bounding the cardinality of the language. It is not true in general. For instance, let the $A_i$’s be one-element structures in a language with predicates $P_j$ for $j\in I$, where $P_j$ holds in $A_i$ iff $i=j$. Then no $P_i$ holds in the ultraproduct, so it is not isomorphic to any $A_i$. One can make $A_i$ countably infinite if desired. – Emil Jeřábek Jan 29 '14 at 12:12
@EmilJeřábek: So by your trick we can obtain an ultraproduct of countable cardinality when the structures $A_i$ are each countable. – user38200 Jan 29 '14 at 12:21

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