Suppose $\kappa$ is a measurable cardinal and let $\mathcal{F}\subset\wp(\kappa)$ be a $\kappa$-complete non-principal filter. Can we extend $\mathcal{F}$ to a $\kappa$-complete ultrafilter?

My motivation comes from a problem I have just encountered. I need a $\kappa$-complete ultrafilter whereas the best I can do is to construct only some filter.

Of course, the standard Zorn argument does not work here by a simple meta-argument. Take the filter on $\omega_1$ consisting of sets with countable complement. It is $\omega_1$-complete but there is no $\omega_1$-complete ultrafilter on $\omega_1$, since this cardinal is not measurable (in ZFC).

  • 2
    $\begingroup$ en.wikipedia.org/wiki/Strongly_compact_cardinal $\endgroup$ – Emil Jeřábek Dec 28 '12 at 21:02
  • 1
    $\begingroup$ Actually, this counts as an answer since the filter I am dealing with is rather mysterious and, consequently, I have no chance for a "painless" extension, by the above. $\endgroup$ – Tomek Kania Dec 28 '12 at 21:07
  • 2
    $\begingroup$ If the theory "ZFC+There is a strongly compact cardinal" is consistent then it is impossible that it would be a theorem of ZFC. But depending on your background assumptions (which are generally thought of as plain ZFC, I suppose) we cannot prove nor disprove the existence of large cardinals. If your theory is, for example, ZFC+"There is a supercompact cardinal" then we can prove that there is a model in which there is a strongly compact cardinal. So we cannot disprove it from ZFC. $\endgroup$ – Asaf Karagila Dec 28 '12 at 22:09
  • 1
    $\begingroup$ Wouldn't you just need $(2^\kappa)^+$-compactness for this? $\endgroup$ – François G. Dorais Dec 28 '12 at 22:25
  • 1
    $\begingroup$ Asaf, what you said is not quite correct. Any supercompact is strongly compact, but the general expectation is that the theories ZFC+"There is a supercompact cardinal" and ZFC+"There is a strongly compact cardinal" should be equiconsistent, so neither should be able to prove the existence of (set) models of the other. $\endgroup$ – Andrés E. Caicedo Dec 28 '12 at 22:43

If your filter is generated by $\kappa$ many sets, then indeed the conclusion you seek can be made, by a direct argument that does not go through strong compactness.

Theorem. The following are equivalent, for any uncountable regular cardinal $\kappa$.

  1. $\kappa$ is a measurable cardinal.
  2. Every $\kappa$ complete filter $F$, generated by at most $\kappa$-many sets, extends to a $\kappa$-complete ultrafilter.

Proof: It is easy to see that $2$ implies $1$, since the filter of co-bounded sets in $\kappa$ is $\kappa$-complete and generated by the tails, so there is a $\kappa$-complete non-principal ultrafilter on $\kappa$.

For the main direction, assume $\kappa$ is measurable and $F$ is a $\kappa$-complete filter on a set $D$ with $F$ generated by at most $\kappa$ many sets $X_\alpha$, for $\alpha\lt\kappa$. Let $j:V\to M$ be an elementary embedding with critical point $\kappa$. By applying $j$ to $\vec X=\langle X_\alpha\lt\kappa\rangle$ and restricting to $\kappa$, we see that $\langle j(X_\alpha)\mid\alpha\lt\kappa\rangle$ is in $M$. And since this is fewer than $j(\kappa)$ many elements of $j(F)$, which is $j(\kappa)$-complete in $M$, it follows that $\bigcap_{\alpha\lt\kappa}j(X_\alpha)\in j(F)$, and in particular, there is some $a\in \bigcap_\alpha j(X_\alpha)$. Define $U=\{X\subset D\mid a\in j(X)\}$. It is easy to verify that $U$ is a $\kappa$-complete ultrafilter on $D$ and $F\subset U$, as desired. QED

For $\theta$-generated filters, one generally needs $\theta$-strong compactness, as mentioned in the comments, and this is in fact equivalent to $\theta$-strong compactness. The essence of the argument above, then, is that a cardinal $\kappa$ is measurable if and only if it is $\kappa$-strongly compact.

That said, if you want this filter extension property, then I encourage you to go ahead and make the strong compactness assumption. There are many beautiful theorems using strongly compact cardinals.

  • 6
    $\begingroup$ To complement Joel's nice answer, since you asked about consistency strength in the comments: The consistency of ZFC+"there exists a cardinal $\kappa$ that is $\kappa^+$-strongly compact" is already significantly beyond all large cardinal assumptions that we can currently capture through inner model theory. It implies determinacy in $L(\mathbb R)$, the existence of proper class inner models with a proper class of Woodin cardinals, and much more. $\endgroup$ – Andrés E. Caicedo Dec 28 '12 at 22:48

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.