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Given structures $A_i$ each of cardinality $<\kappa$ where $\kappa$ is a measurable cardinal, the cardinalities of the $A_i$ are not uniformly bounded by a cardinal $\lambda <\kappa$, and $\mathcal{U}$ a $\kappa$-complete ultrafilter over $\kappa$, what is the cardinality of the ultraproduct $\prod A_i/\mathcal{U}$?

Edit: The ultrafilter $\mathcal{U}$ is also assumed to be normal.

For all the other missing data, which I am not aware of, please let me know of the case in which the ultraproduct is only of cardinality $\kappa$ (and as a by product is not isomorphic to any of the $A_i$).

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    $\begingroup$ You have not given us enough information. Is $\mathcal U$ an ultrafilter over $\kappa$? If so, and each $A_i$ has cardinality $\aleph_i$, the structure has cardinality $\kappa$. If all the $A_i$ are countable, so is the ultraproduct. Etc. $\endgroup$
    – Andrés E. Caicedo
    Commented Jan 15, 2015 at 22:06
  • $\begingroup$ Thank you. This is still not enough information, I'm afraid. For instance, I misspoke in my previous comment and assumed the ultrafilter was normal. Are you assuming this? $\endgroup$
    – Andrés E. Caicedo
    Commented Jan 15, 2015 at 22:17
  • $\begingroup$ In the OP, the cardinality of the ultraproduct is independent of the similarity type of the structures $A_{i}$, all of which have the same vocabulary. So it is enough to consider ultraproducts of cardinals. $\endgroup$
    – Avshalom
    Commented Jan 16, 2015 at 12:55

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It depends on the function $f : i \mapsto |A_i|$. The ultraproduct $\prod A_i/ U$ has the same cardinality as $\prod f(i)/U$. $f$ represents an ordinal $\alpha <j(\kappa)$ in the ultrapower $M$ of the universe $V$, where $j : V \to M$ is the canonical embedding. It's a standard fact that $2^\kappa < j(\kappa) < (2^\kappa)^+$. So the cardinality of $\prod f(i)/U$ is $|\alpha|$, which can be anything between $\kappa$ and $2^\kappa$, because we can select an $f$ corresponding to each such $\alpha$. For example, if $f$ is the identity function, then the ultraproduct has size $\kappa$. If $f(i) = i^+$ for $U$-almost all $i$, then $\prod f(i) / U = \kappa^+$.

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  • $\begingroup$ For the standard fact, see Proposition 5.7 and section 5 generally in The Higher Infinite (A. Kanamori). $\endgroup$
    – Avshalom
    Commented Jan 16, 2015 at 12:03

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