The character of an ultrafilter $U$, denoted $\chi(U)$, is the minimal size of an $A \subseteq U$ such that $(\forall x \in U ) (\exists y \in A) y \subseteq x$. This cardinal characteristic has been studied for ultrafilters on $\omega$. For all nonprincipal $U$, $\omega_1 \leq \chi(U)$, and it is known that, consistently, there exists nonprincipal $U$ with $\chi(U) < 2^\omega$. My question is, for normal ultrafilters $U$ on a measurable cardinal $\kappa$, is it possible that $\chi(U) < 2^\kappa$?

Here are some relevant papers about ultrafilters on $\omega$, by Shelah, Brendle, and Hart:

MR1686797, MR0987317, MR2365799, MR2847327

(available through http://www.ams.org/mathscinet/)

  • $\begingroup$ Could you summarize or link to the method of showing $\chi(U)\lt 2^\omega$ for an ultrafilter on $\omega$? $\endgroup$ – Joel David Hamkins Sep 13 '13 at 20:29
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    $\begingroup$ @JoelDavidHamkins I think this result for $\omega$ is due to Kunen. His construction is an $\omega_{\omega_1}$-long, finite support iteration of a Mathias-style forcing. The Mathias reals he adds form an almost-decreasing sequence of infinite subsets of $\omega$, and because of genericity they generate an ultrafilter in the final extension. In that almost-decreasing $\omega_{\omega_1}$-sequence, any cofinal subsequence of length $\omega_1$ is a base for this ultrafilter. $\endgroup$ – Andreas Blass Sep 13 '13 at 21:34
  • $\begingroup$ Thanks, @Andreas. Perhaps this suggests a $\lt\kappa$-support $\kappa^{+\kappa^+}$-iteration of long Prikry forcing (or some other $\kappa$ analogue of Mathias forcing). If it is $\lt\kappa$-directed closed, one could do it after the Laver preparation and maintain the measurability of $\kappa$... $\endgroup$ – Joel David Hamkins Sep 13 '13 at 21:53
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    $\begingroup$ Perhaps the right analogue has conditions of the form $(s,C)$ where $s$ is a bounded subset of $\kappa$ and $C$ is a club of possible extensions. This seems Mathias-like and is $\lt\kappa$-directed closed. $\endgroup$ – Joel David Hamkins Sep 13 '13 at 22:08

To get the ball rolling...

One can show easily that $\chi(U)$ must be at least $\kappa^+$, since otherwise one can take the diagonal intersection of a $\kappa$-sized family and find a single set that supposedly generates $U$, which is impossible.

Thus, your situation would require that $\kappa$ is measurable and $2^\kappa\gt\kappa^+$, a situation already whose consistency strength strictly exceeds that of a measurable cardinal. Silver showed that this is consistent relative to a $\kappa^+$-supercompact cardinal, but it is now known to be equiconsistent with a cardinal $\kappa$ that is $\kappa^{++}$-tall. But it follows from this that one cannot prove that your situation is consistent if one starts only from the assumption that measurable cardinals are consistent (unless that assumption is inconsistent). We need to use a stronger hypothesis.

But it is conceivable that we might hope to mimic the methods on $\omega$ higher up, if we start with a supercompactness assumption on $\kappa$....


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