Ill-founded iterated ultrapowers I have two related questions.

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*Can I find an ultrafilter on $\kappa$ which is $\theta$-complete for some uncountable $\theta<\kappa$ but not $\kappa$-complete?


*We know that if $\mathcal{U}$ is a $\kappa$-complete ultrafilter on $\kappa$, then the iterated ultrapower of the universe is well-founded. I would like to know a form of converse:
Are there uncountable cardinals $\theta<\kappa$ with a $\theta$-complete but not $\kappa$-complete ultrafilter on $\kappa$ such that there is necessarily $\alpha$ with the $\alpha$th iterate $Ult^\alpha$ ill-founded?
 A: *

*Yes. Here is one silly example: let $\kappa$ be a measurable cardinal and fix a $\kappa$-complete ultrafilter $U$ over $\kappa$. Define $W$ by
$$W = \{X\subseteq \kappa^+ \mid X\cap\kappa\in U\}.$$
Then $W$ is $\kappa$-complete, but not $\kappa^+$-complete.
However, this example is not uniform. (A filter $F$ over $\kappa$ is uniform if $X\in F$ implies $|X|=\kappa$.) Here is another example you want to find which is also uniform:
Let $\kappa_0<\kappa_1$ be measurable cardinals and $U_i$ be $\kappa_i$-complete ultrafilters over $\kappa_i$. Define
$$W = \{X\subseteq \kappa_0\times\kappa_1 \mid \{\alpha<\kappa_0\mid X_\alpha\in U_1\}\in U_0\},$$
where $X_\alpha=\{\eta < \kappa_1\mid (\alpha,\eta)\in X\}$. It is easy to see that $W$ is uniform $\kappa_0$-complete ultrafilter. But if $U_0$ is not $\kappa_0^+$-complete and $\langle Y(\xi)\mid \xi<\kappa_0\rangle$ witnesses it, then $\langle Y(\xi)\times \kappa_1\mid \xi<\kappa_0\rangle$ witnesses $W$ is not $\kappa_0^+$-complete.


*I do not see there is a reason for the usual proof (cf. Theorem 19.7 of Jech) of the well-foundedness of $\mathrm{Ult}^{(\alpha)}$ breaks down when $U$ is not $\kappa$-complete but $\omega_1$-complete. If I checked the proof correctly, there are no such cardinals you referred.
