Inner models and strongly compact cardinals

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The following question is motivated by a result of Magidor that it is consistent that the least strongly compact cardinal is the least measurable cardinal.

Question. Assume $\kappa$ is a strongly compact cardinal. Is there an inner model $M$ of the universe such that $M \models$``$\kappa$ is strongly compact and the set of measurable cardinals below $\kappa$ is unbounded in $\kappa$'' $?$

asked May 18, 2016 at 6:07

Mohammad Golshani

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$\begingroup$ I don't know enough about inner model theory, but are there any ad hoc inner models for very large cardinals? I mean, are there even results of this type known? $\endgroup$
– Asaf Karagila ♦
Commented May 18, 2016 at 8:28
1

$\begingroup$ @AsafKaragila Yes, there are some arguments producing inner models of very large cardinals. See for example my paper with Apter and Gitman: jdh.hamkins.org/innermodelswithlargecardinals. For example, if there is a supercompact cardinal, then there is an inner model with an indestructible supercompact cardinal, and another inner model with a supercompact cardinal at which the GCH fails. Our methods don't seem, however, to answer the current question. $\endgroup$
– Joel David Hamkins
Commented May 18, 2016 at 11:26
1

$\begingroup$ Mohammad, it is a nice question. I don't think we know even how to get at least one measurable cardinal below the strongly compact cardinal in an inner model, let alone unboundedly many. $\endgroup$
– Joel David Hamkins
Commented May 18, 2016 at 11:44
1

$\begingroup$ @AsafKaragila Let me add more such results other those mentioned by Prof. Hamkins. One is the work of Foreman on inner models for Huge cardinals. See Smoke and mirrors: combinatorial properties of small cardinals equiconsistent with huge cardinals $\endgroup$
– Mohammad Golshani
Commented May 19, 2016 at 3:59
1

$\begingroup$ - and Chang's conjecture, generic elementary embeddings and inner models for huge cardinals. I may also mention the worn of Neeman and Steel ``Equiconsistencies at subcompact cardinals''. $\endgroup$
– Mohammad Golshani
Commented May 19, 2016 at 4:01

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