5
$\begingroup$

My question is about existing of basis of club filter club($\omega_1$) with cardinality $c$. Does it exist?

$\endgroup$
11
$\begingroup$

That's independent of ZFC.

On the one hand, it's consistent with ZFC that $2^{\aleph_1}=\mathfrak c$, in which case the whole club filter on $\aleph_1$ has cardinality $\mathfrak c$.

On the other hand, the continuum hypothesis is consistent with ZFC and implies that the club filter has no basis of size $\mathfrak c=\aleph_1$. That's because the diagonal intersection of any $\aleph_1$ club sets is again club.

$\endgroup$

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.