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?



Can an ultraproduct be infinite countable?
Yes, but the existence of such ultrafilters is a large cardinal hypothesis; it is equivalent to the existence of a [measurable cardinal](http://en.wikipedia.org/wiki/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 ${\cal U}$. 


