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An ultrafilter $U$ is $(\mu,\kappa)$-regular if there is a sequence $\langle X_\alpha : \alpha < \kappa \rangle \subseteq U$ such that for all $y \in [\kappa]^\mu$, $\bigcap_{\alpha \in y} X_\alpha = \emptyset$. Countable incompleteness is equivalent to $(\omega,\omega)$-regularity, and for every $\kappa$, there is always an $(\omega,\kappa)$-regular ultrafilter. The failure of $(\omega,\omega_1)$-regularity for an ultrafilter on $\omega_1$ is consistent relative to large cardinals, and carries some large cardinal strength.

In the 1970s, Kanamori proved that if $\lambda$ is singular, then every ultrafilter on $\lambda^+$ is $(\lambda,\lambda^+)$-regular. Question: Is it known whether Kanamori's result can be improved to, "If $\lambda$ is singular, then every ultrafilter on $\lambda^+$ is $(\kappa,\lambda^+)$-regular for some $\kappa < \lambda$"? What about in the case $\lambda=\aleph_\omega$?

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    $\begingroup$ Hi Monroe. You should take a look at Lipparini's work, particularly the nice More on regular and decomposable ultrafilters in ZFC. $\endgroup$ – Andrés E. Caicedo Apr 3 '14 at 21:24
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Assuming large cardinals, the answer is no even for $\aleph_\omega$.

Given a supercompact cardinal, Ben-David and Magidor constructed a model in which $\aleph_{\omega+1}$ carries a uniform indecomposable ultrafilter $U$, which means that $U$ is $\theta$-indecomposable for $\aleph_0<\theta<\aleph_\omega$.

For regular $\theta$, $\theta$-indecomposability means that the ultrafilter is closed closed under intersections of decreasing sequences of length $\theta$, and for ultrafilters this is equivalent to the failure of $(\theta,\theta)$-regularity. [due to Kanamori or Ketonen --- I'll need to check.]

Said another way, this means that any point-$<\theta$ family of sets from the ultrafilter has size $<\theta$, and hence certainly $(\theta,\aleph_{\omega+1})$-regularity fails.

The ultrafilter in the Ben-David/Magidor paper is $\aleph_n$-indecomposable whenever $0<n<\omega$, hence it cannot be $(\kappa,\aleph_{\omega+1})$ regular for any $\kappa<\aleph_\omega$.

[Note: In the other direction, the answer is surely yes if $V=L$ and the proof ought to go by showing that a counterexample implies some variety of stationary reflection, but I haven't had time to investigate.]

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    $\begingroup$ I think Prikry proved that if $V=L$ then every uniform ultrafilter on any infinite $\kappa$ is $(\omega,\kappa)$-regular. $\endgroup$ – Andreas Blass Apr 4 '14 at 16:02
  • $\begingroup$ Thanks, Andreas! I'll fill in the references as I find them! $\endgroup$ – Todd Eisworth Apr 4 '14 at 16:07
  • $\begingroup$ Oops. I found the Prikry reference, "On a problem of Gillman and Keisler" [Ann. Math. Logic 2 (1970) 179-187], but according to the review on MathSciNet, it only proves the case $\kappa=\aleph_1$. $\endgroup$ – Andreas Blass Apr 4 '14 at 17:13
  • $\begingroup$ In the Kanamori-Magidor "Evolution of Large Cardinal Axioms" paper, the question is mentioned as still open. $\endgroup$ – Todd Eisworth Apr 4 '14 at 17:16
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    $\begingroup$ According to the MathSciNet review of Dieter Donder's paper "Regularity of ultrafilters and the core model" [Israel J. Math. 63 (1988) 289-322] all uniform ultrafilters on $\kappa$ are $(\omega,\kappa)$-regular provided there is no inner model with a measurable cardinal and the core model computes $\kappa^+$ correctly. (Presumably this refers to the original Jensen-Dodd core model.) So it seems we're OK with room to spare if $V=L$. $\endgroup$ – Andreas Blass Apr 4 '14 at 17:22

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