About $\lambda$-descending completeness

Question

I am trying to show that

An ultrafilter $\mathcal{U}$, over some cardinal set $I$, is $\lambda$-descendingly complete iff it is $cf(\lambda)$-descendingly complete.

Now, for $\lambda$ regular is trivial. And, for $\lambda$ singular, one side is trivial too ($\lambda$-descending completenes implies $\beta$-desceding completeness, for any $\beta<\lambda$). The other implication is the one i am struggling with.

Answer 1

If $\lambda$ is singular, and $\mu=\operatorname{cf}(\lambda)$, take a $\lambda$-descending sequence, $A_\alpha$ for $\alpha<\lambda$ and a cofinal sequence $F\colon\mu\to\lambda$. What can you say about $\bigcap\{A_\alpha\mid\alpha<\lambda\}$ and $\bigcap\{A_{F(\alpha)}\mid\alpha<\mu\}$?