Supercompacts are Woodin limits of Woodins

Question

I was trying to prove a couple of implications between large cardinal properties, but there is one I am not able to do, and I don't even have an idea on how to approach it. The implication is: if $\kappa$ is a supercompact cardinal, then it is a Woodin limit of Woodin cardinals. So, I started by considering a normal ultrafilter $U$ on $\mathcal P_\kappa(\kappa)$ and was thinking about: if I consider this as a subset of $\mathcal P(\kappa)$, this is just a normal ultrafilter on $\kappa$, right? Then I should for example prove that the set $\{\delta<\kappa:\ \delta\ \text{is Woodin}\}$ belongs to the ultrafilter, hence $\kappa$ is a limit of Woodin cardinals. But I have no idea on how to continue from here, and also how to prove $\kappa$ itself is Woodin (I know that being a limit of Woodins is not enough)

Answer 1

Kanamori's Book (The higher infinite) shows in the combination of Propositions 26.11 and 26.12 that if $\kappa$ is $2^{\kappa}$-supercompact, then there is a normal ultrafilter $U$ over $\kappa$ such that $\{\alpha<\kappa\;|\;\alpha\text{ is Woodin}\}\in U$. This normality actually implies that $\kappa$ is Woodin itself: If $f:\kappa\longrightarrow\kappa$ is any function, the set of all closure points of $f$ (ordinals $\alpha$ such that $f[\alpha]\subseteq\alpha$) is a club and thus in $U$. So there exists $\alpha<\kappa$ such that $f[\alpha]\subseteq\alpha$ and $\alpha$ is Woodin. By considering $f\upharpoonright\alpha$, we can see that there is $\delta<\alpha$ such that $(f\upharpoonright\alpha)[\delta]\subseteq\delta$ (which implies $f[\delta]\subseteq\delta$) and there is an elementary embedding $j:V\longrightarrow M$ with critical point $\delta$ and $V_{j(f\upharpoonright\alpha)(\delta)}\subseteq M$. But $j(f\upharpoonright\alpha)=j(f)\upharpoonright j(\alpha)$ and thus $j(f\upharpoonright\alpha)(\delta)=j(f)(\delta)$. Therefore $\delta$ witnesses Woodinness of $\kappa$ with respect to $f$.