# Closure of extender ultrapowers

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

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 $$\omega < \min\{\kappa^+, \operatorname{cof}(\lambda)\}$$.
• Yes, I know that $^{2^\kappa} M \subset M$ is stronger than superstrongness, but $^\kappa M \subset M$ is possible if $\kappa$ is merely measurable, which is at the bottom on the strongnes hierarchy. Sep 17 at 10:15