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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).

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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. –  Tomek Kania Dec 28 '12 at 21:07
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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. –  Asaf Karagila Dec 28 '12 at 22:09
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Wouldn't you just need $(2^\kappa)^+$-compactness for this? –  François G. Dorais Dec 28 '12 at 22:25
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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. –  Andres Caicedo Dec 28 '12 at 22:43

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up vote 9 down vote accepted

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.

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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. –  Andres Caicedo Dec 28 '12 at 22:48

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