Let E be a (Mitchell-Steel) extender over some M. Recall that the natural length of E, $\nu\_E$, is the strict sup of the generators of E and $\kappa^{+M}$. It is claimed in both "Fine structure and iteration trees" and Steel's handbook article that if the natural length is a limit ordinal then it's also a cardinal in both M and in the ultrapower of M by E.


I've been trying to prove this, but the argument eludes me. I suppose you want to use the surjection f witnessing the non-cardinalness of the natural length in M to define some element $[a,f]\_E$ of the ultrapower, where $a\subset\nu\_E$. But then I get stuck. Steel and Mitchell claim that it's an easy argument, so I hope I'm only missing a minor detail.



Natural length is a cardinal if it's a limit ordinal
it should follow from the fact that generators are critical points. if $ \xi<\nu\_E$ is a generator of E then $\xi$ is a critical point of the canonical factor map


$ k: Ult(M, E|\xi)\rightarrow Ult(M, E)$


Let $\alpha$ be the index of $E$, we claim that if $\nu\_E$ is a limit ordinal then $J\_\alpha^M\models ``\nu$ is a cardinal".


Suppose not and let $f:\eta \rightarrow \nu\_E$ be onto and $f\in J\_\alpha^M$. It follows that $f\in Ult(J\_\alpha^M, E)$ and hence, $f=\pi\_E(g)(s)$ for some $s\in \nu\_E^{<\omega}$. Let $\xi\in (max(s), \nu\_E)$ be a generator. Then


(key fact) $f\in rng(k)$


where $k$ is defined as above. But then $cp(k)=xi$ is not a cardinal in $Ult(J\_\alpha^M, E|\xi)$ contradicting the fact that $\xi$ is a generator. 


