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.


  • $\begingroup$ If there are no measurable cardinals, then all ultrafilters must have completeness $\aleph_0$. So it's kinda hard to give you a straight answer. $\endgroup$ – Asaf Karagila Oct 29 '18 at 11:39
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    $\begingroup$ The possible completeness numbers of non-principal ultrafilters are $\aleph_0$ and the measurable cardinals. I"ll write some details as an answer. $\endgroup$ – Andreas Blass Oct 29 '18 at 13:11

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 $V$.

  • $\begingroup$ And we haven't even used the axiom of choice! $\endgroup$ – Asaf Karagila Oct 29 '18 at 16:08
  • $\begingroup$ Also, arguably $\aleph_0$ is measurable. $\endgroup$ – Asaf Karagila Oct 29 '18 at 16:10
  • $\begingroup$ This is a very nice argument. Another way of thinking about it is that the completeness number of an ultrafilter is exactly the critical point of the associated ultrapower embedding, and the discontinuity is witnessed by the function $f$. $\endgroup$ – Miha Habič Oct 29 '18 at 16:10
  • $\begingroup$ @Miha: But then choice is needed and $\aleph_0$ is definitely not measurable (I mean, choice is needed to prove $\aleph_0$ is measurable as a completeness number, but it is outright disprovable that $\aleph_0$ is a critical point of an elementary embedding, unless you count embedding into non-$\omega$ models. (Although if you count $\aleph_0$ as a measurable, might as well consider embedding into non-$\omega$ models too.)) $\endgroup$ – Asaf Karagila Oct 29 '18 at 16:15
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    $\begingroup$ This is actually useful in at least once choiceless situation. It shows that, if there are no nonprincipal ultrafilters on $\omega$ (for example under AD), then all ultrafilters (on any set) are countably complete. $\endgroup$ – Andreas Blass Oct 29 '18 at 23:19

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