Example of a nonprincipal incomplete ultrafilter I am trying to build my intuition on ultrafilters by finding examples of non-implications between these properties:

*

*ultra

*nonprincipal

*completeness

Any filter (on a regular cardinal $\kappa$) generated by a singleton will be an ultrafilter that's $\kappa$-complete. And the club filter on $\kappa$ is a nonprincipal $\kappa$-complete filter that's not an ultrafilter. So my question is: are there examples of ultrafilters on some $\kappa$ that are nonprincipal, but not $\kappa$-complete? I understand that if $\kappa$ is not a measurable cardinal, then every nonprincipal ultrafilter on it is not $\kappa$-complete. But I wonder if there are more concrete examples. Thank you!
 A: First, note that an incomplete filter cannot be extended to a complete ultrafilter:
If $F$ is a filter and $A_\alpha$ is a sequence of sets in $F$ for $\alpha<\gamma$, such that the intersection of the $A_\alpha$'s is "practically empty" (read: its complement is in $F$), then any ultrafilter extending $F$ will contain the complement of $\bigcap A_\alpha$, and is therefore incomplete.
Now, pick your favourite partition of $\kappa$ into countably many infinite sets, $X_n$, and let $A_n=\kappa\setminus\bigcup_{k<n}X_k$. Now consider the filter generated by $A_n$.
(Fun exercise: any filter that does not contain finite sets can be extended to a $\sigma$-incomplete ultrafilter.)
A: Given the subjectivity of the term "concrete" it's difficult to give a convincing negative answer, but here's some good evidence: $\mathsf{ZF}$ is consistent with "There is no nonprincipal ultrafilter on any set whatsoever." Certain models of $\mathsf{ZF}$ may have definable nonprincipal ultrafilters (see e.g. here), but even these won't be very concrete since the relevant definitions will be quite complicated.
Note that this applies to complete and incomplete ultrafilters alike. Interestingly, models associated to certain types of ultrafilter can be far more concrete than any specific ultrafilter of that type: e.g. if $U_1,U_2$ are measures on the same cardinal $\kappa$ then $L[U_1]=L[U_2]$, so "$L$ with a single measurable is canonical." (See Theorem $20.10$ in Kanamori. Note that $U$ and $W$ there are assumed to be normal in their respective models. This is part of Kanamori's definition of "$M$-ultrafilter," which differs from Kunen's and threw me off at first since in the absence of a normality assumption part of the conclusion of that theorem is blatantly false.)
