The following is well-known. If $\kappa$ is measurable, $\theta > \kappa$, and $M \prec V_\theta$ has size $<\kappa$, then there is $N\prec V_\theta$ such that $N \supseteq M$, $M \cap \kappa \not= N \cap \kappa$, but $N \cap \sup(M \cap \kappa) = M \cap \kappa$. Repeating this a number of times allows us to also find such $N$ with the ordertype of $N \cap \kappa$ being any ordinal $\leq \kappa$.

The construction of $N$ can be done by adjoining a single ordinal that is in all measure-one sets in $M$ for some normal ultrafilter, and closing under Skolem functions. If $M$ is $\mu$-closed, then so is $N$. However, repeating this $\omega$-times in an elemenatary chain can kill countable closure.

My question: Is there is some large cardinal $\kappa$ with the following stronger property where we demand closure?

Whenever $\mu<\kappa$ is regular, $\theta > \kappa$, and $M \prec V_\theta$ is $\mu$-closed and of size $<\kappa$, then for every $\alpha <\kappa$, there is $N \prec V_\theta$ such that $N$ is $\mu$-closed, $N \supseteq M$, $M \cap \kappa \not= N \cap \kappa$, $N \cap \sup(M \cap \kappa) = M \cap \kappa$, and the ordertype of $N \cap \kappa$ is $\geq \alpha$.


1 Answer 1


It seems unlikely.

Let me use the following consequence of partial global square (which appears in my joint paper with Garti). This partial sequence is consistent with the existence of very large cardinals.

Lemma: Let us assume that there is a partial square sequence $\langle C_\alpha \mid \alpha \in S \rangle$ where $S^{\kappa}_{\omega_2} \subseteq S \subseteq \kappa$ (so $\mathrm{otp}\ C_\alpha < \alpha$ for all $\alpha \neq \omega_2$, each one of them is closed and unbounded at $\alpha$ and the sequence is coherent).

Then, there is a partition of $S^\kappa_{\omega}$ into $\aleph_2$ sets, $\langle S_i \mid i < \omega_2\rangle$ such that for all $\alpha < \kappa$ with $\mathrm{cf}\ \alpha = \omega_2$, $\forall i < \omega_2,\, S_i \cap \alpha$ is stationary.

Proof: Recursively, we can start with a partition of $\omega_2$ into $\omega_2$ disjoint stationary sets and using the coherence we can "copy" them upwards.

Now, let us assume that there is such a partition for $\kappa$ and assume $\mathrm{CH}$. Then, if $M$ is a $\sigma$-closed model of size $\aleph_1$, and $N$ is a $\sigma$-closed model extending it of order type at least $\omega_2$, then let's look at $\delta$ the supremum of the first $\omega_2$ elements of $N \cap \kappa$ - this is an ordinal of cofinality $\omega_2$. $N \cap \delta$ is $\sigma$-closed set of ordinals and in particular, intersects each $S_i$ for all $i < \omega_2$ which means that $\omega_2 \subseteq N$ but it is not a subset of $M$.

  • $\begingroup$ If I understand the gist of your argument correctly, the partial square rules out the existence of certain Chang-type models, which is stronger than saying not every model can be end-extended as desired. This suggests that maybe for some degree of supercompactness or hugeness for $\kappa$, perhaps some special class of elementary submodels might be end-extendible as desired, since these large cardinals tend to contradict squares. $\endgroup$ Commented Nov 4, 2021 at 15:22
  • $\begingroup$ Yes, if $\mu$ is huge, then this type of partition is not possible. I was under the impression that you want this property to hold for every $\mu < \kappa$. $\endgroup$
    – Yair Hayut
    Commented Nov 4, 2021 at 17:37
  • $\begingroup$ @MonroeEskew Isn’t supercompact good enough? If you are looking for a special class of ESMs, then don’t $\kappa$- Magidor models have that property? Or you are just not interested in such models? $\endgroup$
    – Rahman. M
    Commented Nov 4, 2021 at 20:25
  • $\begingroup$ @YairHayut, Do you know if the desired property is consistent? Is there some forcing extension of a model of ZFC+LCs such that some $\kappa$ has this end-extension property with respect to countably closed models? Or maybe other nice models like guessing models? $\endgroup$ Commented Nov 25, 2021 at 19:11

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