A filter $F$ on $\lambda$ is weakly normal if and only if for all $f : \lambda \rightarrow \lambda$ such that $\{\xi < \lambda : f(\xi) < \xi\} \in F^+$, then there is some $\beta < \lambda$ and $X \in F^+$ such that $f(\xi) < \beta$ for all $\xi \in X$.
Let $\lambda \geq \kappa$ be a regular cardinal.
$S = \{\alpha < \lambda : cf(\alpha) < \kappa\}$
The question is: If $U$ is a $\kappa$-complete uniform weakly normal ultrafilter on $\lambda$, then $S \in U$.
This question comes from a proof of a theorem of Solovay that if $\kappa$ is a supercompact cardinal, then $\lambda^{<\kappa} = \lambda$ for all regular cardinal $\lambda \geq \kappa$. In the proof, if $\kappa$ is $\lambda$-supercompact, then there exists a $\kappa$-complete uniform weakly normal ultrafilter on $\lambda$. The proof claims that $S$ defined above is an element of $U$.
Something that may be relevant is that: Since $U$ is an ultrafilter $U = U^+$. $S$ is stationary in $\lambda$. And any uniform weakly normal filter contains the club subsets of $\lambda$. However this does not imply that $S \in U^+ = U$.
Thanks for any help or clarification on this problem.