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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

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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 $\omega < \min\{\kappa^+, \operatorname{cof}(\lambda)\}$.

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  • $\begingroup$ Thank you for pointing out an important detail that I forgot to account for. $\endgroup$ Jun 17 at 21:09
  • $\begingroup$ You're welcome. I have added an answer for the edited part. The main point of these examples is that closure properties have high large cardinal strength and you can't get closure, unless you assume large cardinal hypotheses. $\endgroup$ Jun 17 at 21:36
  • $\begingroup$ 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. $\endgroup$ Sep 17 at 10:15

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