I want to prove the following

We have $\kappa$ an infinite regular cardinal such that every $\kappa$-complete filter on $\kappa$ can be extended to an $\omega_1$-complete ultrafilter on $\kappa$.

Show that there exists a measurable cardinal $\lambda\leq\kappa$.

Moreover, if $\lambda$ is the least measurable cardinal less than or equal to $\kappa$, then in fact every $\kappa$-complete filter on $\kappa$ can be extended to a $\lambda$-complete ultrafilter on $\kappa$.

Attempt For the first part, I need to prove that there exist a $\lambda\leq \kappa$ such that there exist a $\lambda$-complete non-principal ultrafilter on $\lambda$. I'm stomped... I tried to the partition equivalence of a $\kappa$-complete ultrafilter, trying to fit in some Ramsey theory and weakly compact cardinal, but i can't reach any conclusion. I also looked at strongly compact cardinal with no solution. I'm now trying a more constructive approach. Using the property of my cardinal to build a $\lambda$-complete non-principal ultrafilter on $\lambda$.

Maybe there are some equivalent definition of measurable that I could use here? Any help is welcomed, thanks

  • $\begingroup$ (I fixed the title: you cannot prove that there is a measurable strictly below $\kappa$.) $\endgroup$ – Andrés E. Caicedo May 31 at 11:24
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    $\begingroup$ Since you've got an $\omega_1$-complete nontrivial ultrafilter on $\kappa$, why don't you define $\lambda$ to be the smallest cardinal which carries an $\omega_1$-complete nontrivial ultrafilter? Then you've got $\lambda\le\kappa$ automatically, and you've got a nontrivial $\omega_1$-complete ultrafilter on $\lambda$. Maybe there's some way you can use minimality to show that the ultrafilter is $\lambda$-complete? $\endgroup$ – bof May 31 at 11:54
  • $\begingroup$ by nontrivial do you mean non-principal? If so I don't have a non-principal $\omega_1$ ultrafilter on $\kappa$. Is there any reason I can assume one extention will be non-trivial? $\endgroup$ – user678462 May 31 at 12:01
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    $\begingroup$ What if you extend the $\kappa$-complete filter $\{X\subseteq\kappa:|\kappa\setminus X|\lt\kappa$ to an $\omega_1$-complete ultrafilter on $\kappa$? $\endgroup$ – bof May 31 at 12:09
  • $\begingroup$ Indeed! I do know that any ultrafilter that extends the Fréchet filter is non-principal! Thanks, That should be a good point to start $\endgroup$ – user678462 May 31 at 12:19

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