Let $U$ be a countably complete free ultrafilter over S. Prove that there is an uncountable cardinal $\kappa, \kappa \le |S|$ with a $\kappa$-complete free ultrafilter over $\kappa$.
Is the following attempt correct:
I say that $\kappa$ is the smallest cardinal such that the intersection of $\kappa$ many members of $U$ is not a member of $U$ and assume a sequence {$A_{\alpha} : \alpha \in \kappa$} and $B_{\beta}$= $\bigcap$ {$A_{\alpha} : \alpha \lt \beta$}\ $A_{\beta}$. Now I need to prove that there is an ultrafilter that contains $B$ and I am stuck there.