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If $F$ is a $\kappa$-complete filter on some set $S$, and $F$ is generated by a basis of size $\lambda$, then $F$ extends to a $\kappa$-complete ultrafilter on $S$ when we assume that $\kappa$ is $\lambda$-compact (Thm 22.17 in Kanamori's book).

What about normal filters? If $F$ is a $\kappa$-complete normal filter on $S$, it can be extended to a $\kappa$-complete ultrafilter $U$, but can we assume that $U$ is normal as well?

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In general, no, because $\kappa$ might not be $\lambda$-supercompact, even if it is $\lambda$-strongly compact. The two large cardinal notions are not provably equivalent (although it is an open question whether they are equiconsistent).

Suppose $\kappa$ is $\lambda$-strongly compact but not $\lambda$-supercompact, and consider the club filter $F$ on $S=P_\kappa\lambda$, consisting of the subsets $X\subseteq P_\kappa\lambda$ that contain a club set $C\subseteq X$. This is a normal filter, and fine, and so if $F$ could be extended to a normal ultrafilter $\mu$, it would mean $\kappa$ should be $\lambda$-supercompact, which it is not.

So it is consistent with ZFC relative to large cardinals that your desired property fails.

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  • $\begingroup$ Thank you! Does this work in the other direction, i.e., if we assume $\kappa$ is $\lambda$-supercompact, then does my desired property hold? $\endgroup$
    – Tristan vd Vlugt
    Commented Jun 27 at 11:51
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    $\begingroup$ I think even with $\lambda$-supercompactness, however, that the property fails. I think there will always be normal filters not contained in any normal ultrafilter. But I don't have an argument at hand... I recall once thinking about this many years ago... $\endgroup$
    – Joel David Hamkins
    Commented Jun 27 at 11:51
  • $\begingroup$ You must be reading my mind by responding to my question within 6 seconds... $\endgroup$
    – Tristan vd Vlugt
    Commented Jun 27 at 11:52
  • $\begingroup$ Ha! I was writing the comment already before you had posted. $\endgroup$
    – Joel David Hamkins
    Commented Jun 27 at 11:52
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As Joel suggested in the comments, there are always normal filters that do not extend to normal ultrafilters. For example, let $F$ be the filter on $\kappa$ of all sets containing the set $S$ of singular cardinals below $\kappa$. Then $F$ is $\kappa$-complete and normal, but doesn't extend to a normal ultrafilter on $\kappa$. If it did, $\kappa$ would have to be measurable, but at the same time, the corresponding ultrapower would think $\kappa$ is singular, which is a contradiction.

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