It is known that if cardinal $\kappa$ is such that there is a cardinal $\lambda$ with $\lambda < \kappa \le 2^\lambda$ then any $\kappa$-complete ultrafilter on $\kappa$ is principal.

Do such $\kappa$ have a special name and is this property equivalent to some other well known properties? Is this equivalent to being measurable?

It seems pretty special to me and relates to GCH and critical points of elementary embeddings.

$\mathfrak{c}$ has the property but $\omega$ doesn't.

A consequence is that any ultrafilter on $\mathbb{R}$ closed under countable intersections is principle (assuming CH).

  • 1
    $\begingroup$ You mean it's not a strong limit? Strong limit cardinals exist just fine, $\beth_\delta$ for any limit ordinal $\delta$ is a strong limit cardinal pretty much by definition. So in a sense "almost all cardinals are strong limit cardinals". If you also require regularity you get strongly inaccessible cardinals, which you need to assume more to actually have, but it's still pretty far from measurable cardinals. $\endgroup$ – Asaf Karagila Aug 25 '18 at 5:06
  • $\begingroup$ Thanks @AsafKaragila you have actually answered my question by saying it is not a strong limit. The other info is also helpful. I probably should have figured it out myself. Regards. $\endgroup$ – Mark Kortink Aug 26 '18 at 1:06
  • $\begingroup$ The least k such that there is a countably closed free (non-principal) ultra-filter on k is also the least measurable cardinal, which is necessarily a regular strong limit. So your last sentence holds regardless of CH, $\endgroup$ – DanielWainfleet Aug 27 '18 at 0:03

As Asaf points out the name of such $\kappa$s is not a strong limit

$\kappa\text{ is not a strong limit}$





When $\kappa$ is not a strong limit then any $\kappa$-complete ultrafilter $U$ on $\kappa$ is principal (ie.has the form $\left\{ A\subseteq\kappa\mid a\in A\right\} $). This is proved here.

Therefore in particular no infinite cardinal of the form $2^{\lambda}$ can have a $2^{\lambda}$-complete free ultrafilter, and that includes $\mathbb{R}$ which has cardinality $2^{\omega}$.

Note that if $U$ is an ultrafilter on $\mathbb{R}$ and CH is false then there is a $\omega<\kappa<2^{\omega}$ so the above theorem would enable us to conclude that if $U$ is $\kappa$-complete then it is principal.

A measurable cardinal is an uncountable cardinal $\kappa$ with a non-principle $\kappa$-complete ultrafilter and so is a strong limit. If the GCH is true then $\lambda^{+}=2^{\lambda}$ for every infinite cardinal so if measurable cardinals exist they are not successor cardinals.

  • 1
    $\begingroup$ I don't understand your last sentence; measurables are not successors, regardless of GCH. $\endgroup$ – Noah Schweber Aug 30 '18 at 23:14
  • $\begingroup$ I am sure that is true, all i point out is it's really obvious with GCH. $\endgroup$ – Mark Kortink Sep 2 '18 at 23:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.