# If every $\kappa$-complete filter can be extended to a $\aleph_1$-complete ultrafilter, then some $\lambda\le\kappa$ is measurable

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

• (I fixed the title: you cannot prove that there is a measurable strictly below $\kappa$.) – Andrés E. Caicedo May 31 at 11:24
• 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? – bof May 31 at 11:54
• 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? – user678462 May 31 at 12:01
• What if you extend the $\kappa$-complete filter $\{X\subseteq\kappa:|\kappa\setminus X|\lt\kappa$ to an $\omega_1$-complete ultrafilter on $\kappa$? – bof May 31 at 12:09
• Indeed! I do know that any ultrafilter that extends the Fréchet filter is non-principal! Thanks, That should be a good point to start – user678462 May 31 at 12:19