Completeness number of ultrafilters In what I write below, by "ultrafilter" I mean a non-principal ultrafilter.
Given an ultrafilter $U$ on some set $S$, let $\mu$ be the least cardinal such that $U$ is $\mu$-complete but not $\mu^+$-complete. Call this number the completeness number of $U$. It is easy to check that this must always be an infinite regular cardinal.
The question is which cardinals can appear as completeness numbers for ultrafilters. Stated otherwise: for which cardinals $\mu$ we can find examples of ultrafilters that have completeness number $\mu$.
For instance, if an ultrafilter is not countably complete, then its completeness number is $\aleph_0$ (and if there are no measurable cardinals then in fact every ultrafilter has completeness number $\aleph_0$). This means that $\aleph_0$ can be completeness number.
What about $\aleph_1$? or $\aleph_n$? or any other regular or even large cardinal $\mu$?
Perhaps this is well-known but I cannot find a reference.
Thanks.
 A: Any countably incomplete ultrafilter has completeness number $\aleph_0$, and if $\kappa$ is measurable then any $\kappa$-complete non-principal ultrafilter on $\kappa$ has completeness number $\kappa$. I claim that these are the only possible completeness numbers.
To prove it, suppose $U$ is a non-principal ultrafilter on some set $A$ and that its completeness number is an uncountable cardinal $\kappa$. I'll prove that $\kappa$ is measurable by producing a $\kappa$-complete non-principal ultrafilter on $\kappa$. 
Since $U$ is not $\kappa^+$-complete, we can fix $\kappa$ sets $X_\alpha\in U$ (where $\alpha$ ranges over ordinals $<\kappa$) such that then intersection $\bigcap_{\alpha<\kappa}X_\alpha\notin U$.  By subtracting this intersection from each $X_\alpha$, we can assume, without loss of generality, that $\bigcap_{\alpha<\kappa}X_\alpha=\varnothing$. So we can define a function $f:A\to\kappa$ by letting $f(p)$ (for any $p\in A$) be the smallest $\alpha$ such that $p\notin X_\alpha$.  Then let $V$ be the image of $U$ under this map $f$; that is, $V=\{Y\subseteq\kappa:f^{-1}[Y]\in U\}$. Since $f^{-1}$ preserves all Boolean operations, including infinitary ones, it is immediate that $V$ is, like $U$, a $\kappa$-complete ultrafilter.
It remains to check that $V$ is non-principal, but this is easy. For any $\alpha<\kappa$, the definition of $f$ implies that $f^{-1}[\{\alpha\}]$ is disjoint from $X_\alpha$ and is therefore not in $U$.  So $\{\alpha\}$ is not in $V$.
