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It is consistent that the least measurable cardinal can carry exactly one normal measure but in almost all models for this theory there is no supercompact cardinal. It seems existence of a supercompact cadinal forces the least measurable cardinal to have many normal measures. Equivalently this means that the assumption "there exists exactly one normal measure on the least measurable cardinal" is an anti-large cardinal axiom which refutes existence of any large cardinal larger than supercompacts.

Question: Assuming consistency of $\text{ZFC}$ and some suitable large cardinal axiom, is the following consistent?

ZFC + There exists at least one supercompact cardinal +

The least measurable cardinal carries exactly one normal measure

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  • $\begingroup$ You'll need to assume the consistency of much more than $ZFC$ to make this question nontrivial, since $ZFC+$"There exists at least one supercompact cardinal" already has consistency strength well above that of $ZFC$. (That said, I really like the question!) $\endgroup$ Nov 5, 2013 at 6:23
  • $\begingroup$ @NoahS: Thank you for your notification. I fixed it. $\endgroup$
    – user42090
    Nov 5, 2013 at 6:32
  • $\begingroup$ Well, the least measurable has to have Mitchell order of $0$ anyway. So that's a promising start. I can't contribute much more, but as @Noah said, it's a nice question. $\endgroup$
    – Asaf Karagila
    Nov 5, 2013 at 7:31
  • $\begingroup$ The natural idea is to show that if V satisfies GCH and $\kappa$ is the least measurable in it, then there is a set forcing notion which ensures that only one ultrapower embedding can lift. $\endgroup$ Nov 5, 2013 at 7:35
  • $\begingroup$ @Mohammad: Can one do that? If yes, then one does not destroy the supercompactness of $\delta$, and the answer is positive. $\endgroup$
    – Asaf Karagila
    Nov 5, 2013 at 7:38

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