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The following is known:

$(*)$ If $0^\sharp$ exists, then any uncountable cardinal is is an inaccessible cardinal (and even more) in $L$.

My question is that:

Are there any large cardinal property $LP$ and any inner model $M,$ such that if $LP$, then any uncountable cardinal is a measurable cardinal (or even more) in $M$?

Remark. Assuming the existence of a supercompact cardinal and a measurable above it, it is possible to build a model $V$ of $ZFC$ which contains an inner model $M$ such that any uncountable cardinal of $V$ is a measurable cardinal in $M$.

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    $\begingroup$ A weaker result than what you're specifically looking for is that assuming that there are $\omega$ many Woodin cardinals and a measurable above them (i.e $\mathcal{M}_n^{\sharp}$), then every regular cardinal in $L(\mathbb{R})$ is measurable. This is a theorem of Steel. $\endgroup$
    – Rachid Atmai
    Commented Apr 21, 2014 at 20:08
  • $\begingroup$ Sorry, I meant $\mathcal{M}_{\omega}^{\sharp}$ in the previous comment instead of $\mathcal{M}_{n}^{\sharp}$ . $\endgroup$
    – Rachid Atmai
    Commented Apr 21, 2014 at 20:48
  • $\begingroup$ Does $L(\mathbb{R})$ satisfy $AC$ in this case? $\endgroup$
    – Mohammad Golshani
    Commented Apr 22, 2014 at 9:10
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    $\begingroup$ No, $L(\mathbb{R})$ does not satisfy $AC$. It will however satisfy definable uniformization principles and the "Coding Lemma", which is a choice principle. In general $L(A)$ satisfies choice only if $A \subset L$. $\endgroup$
    – Rachid Atmai
    Commented Apr 22, 2014 at 16:15

1 Answer 1

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Yes: an assumption like the one you quote for inaccessibles in L, namely $0^\sharp$. Instead you take a "mouse" i.e. an iterable structure, which has a measure of Mitchell order 1 as the topmost final measure. It is thus a structure N that has two measures with the same critical point, $\kappa$ say. As the second measure gives measure 1 to a set of smaller cardinals below $\kappa$ which are themselves measurable, if one takes a countable such N and iterates the second measure out $On$ many times, then this leaves behind an inner model M, which contains a class of measurable cardinals. Externally this class is cub beneath (and hence contains) all uncountable cardinals.

The mouse N is then a $\sharp$ for the model M; the least such mouse with a measure of order 1 (in the canonical ordering of mice) is indeed countable and is known as ``$0^{sword}$'' in the literature. The assumption of its existence thus ensures the $\sharp$ of a model of the kind you ask for.

(I have just noticed the ``even more'': if one wants higher Mitchell order measures on all uncountable cardinals, then one must assume the existence of mice themselves that have higher Mitchell order and the argument does the same. If one wants a model with uncountable $V-cardinals strong in the model, again this can be effected with the appropriate mouse where the topmost measures concentrates on strong cardinals. And so on and so forth.)

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  • $\begingroup$ Thanks for your interesting answer, there are two more questions: 1) where can I find more information about your answer? 2) Can we weaken the assumption of having a mouse which has a measure of Mitchell order 1? $\endgroup$
    – Mohammad Golshani
    Commented Apr 21, 2014 at 7:40
  • $\begingroup$ Clearly in your case, $V$ is not a class generic extension of $M$. Am I right? I also think your $M$ is included in $HOD$. Is it true? Does $M$ satisfy $V=HOD$? $\endgroup$
    – Mohammad Golshani
    Commented Apr 21, 2014 at 7:42
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    $\begingroup$ @Mohammad 1) The concept of $0^{sword}$ is due to Jensen and inner model it leaves behind that I describe is the core model for "measures of Order Zero" in his manuscripts - see his web page. Much of the first part of Zeman's Book (Inner Models and Large Cardinals) is a construction of a core model under the assumption that $0^{sword}$ does not exist (Zeman call $0^{sword}$ an "s-mouse" {\em cf.} Ch. 6 of the book.) $\endgroup$
    – Philip Welch
    Commented Apr 21, 2014 at 7:52
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    $\begingroup$ @Mohammad 2) The assumption of $0^{sword}$'s existence is indeed the weakest to imply the $\sharp$ of such a model. It would seem hard to think up something that implied the existence of the model without implying its $\sharp$ whilst at the same time being a `natural' statement, I think. About the second Q: you are right V is not a class generic extension of the model (because of the sharp). In $L[0^{sword}]$ one could build a class generic real $r$ that codes up the model M; thus $L[r]$ contains the inner model M and has the same cardinals, but not the $\sharp$. Probably not what you want? $\endgroup$
    – Philip Welch
    Commented Apr 21, 2014 at 7:58
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    $\begingroup$ @Mohammad Is M included in HOD? Yes. Does M satisfy V=HOD? Yes it does. $\endgroup$
    – Philip Welch
    Commented Apr 21, 2014 at 14:05

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