Closure of extender ultrapowers

Question

Some elementary embeddings $j : V \to M$ can be defined as ultrapower embeddings by extenders. Extenders are defined using finite indices, and as I've noted in a previous question, that makes it not obvious that $M$ is closed under countable subsets, much less subsets of size at most the critical point of $j$. This question proved that the ultrapower by a $(\kappa, \lambda)$-extender is not countably closed, if $cf (\lambda) = \omega$. Are extender ultrapowers $\lt cf (\lambda)$-closed? Edit: are they $\lt min (\kappa^+, cf (\lambda)$-closed? Edit 3: I have moved my remaining question to a separate question

Answer 1

For the edited version look at the end of the answer.

---

This is not true in general. For example, let $\kappa$ be a measurable cardinal and let $U$ be a normal measure witnessing that. And let $j_U:V\rightarrow M=\operatorname{Ult}(V; U)$ be the corresponding ultrapower mapping. Note that by basic theorems for normal measures(e.g. proposition 5.7(d) in Kanamori's "The Higher Infinite"), we have that ${^{\kappa^+}M}\not\subset M$. So let $E$ be the $(\kappa, \kappa^{++})$-extender derived from $j_U$. Then if $j_E:V\rightarrow \operatorname{Ult}(V; E)$ is the corresponding ultrapower mapping, by applying the factor lemma twice, we can see that $\operatorname{Ult}(V; E)=\operatorname{Ult}(V; U)$, as the two factor maps are inverse to eachother. So this gives that the ultrapower by $E$ is not $\kappa^+$-closed, which amounts to saying that it is also not $<\operatorname{cof}(\kappa^{++})$-closed.

---

For the edited version, it is still not true in general. For example consider the situation in the first link you mention: $\kappa \lt \theta$, $\theta$ is a strong limit of cofinality $\omega$. Let $E$ be a $(\kappa, \theta)$-extender and $j_E:V\rightarrow M_E$ be the corresponding map such that $^\omega M_E \not\subset M_E$. Now let $\lambda = \theta^+$. Then if $F$ is the $(\kappa, \lambda)$-extender derived from $j_E$, you can again by an argument similar to the above see that $M_F=M_E$ and so $^\omega M_F \not\subset M_F$, but $\omega < \min\{\kappa^+, \operatorname{cof}(\lambda)\}$.