Timeline for Removing large cardinals from an uncountable transitive model

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Jun 15, 2020 at 7:27	history	edited	CommunityBot		Commonmark migration
Jan 28, 2016 at 22:46	comment	added	Rachid Atmai		This essentially follows from a theorem of Mitchell/Jensen (Mitchell covering lemma article th. 1.16): assume no inner model with a Woodin, then the comparison of $K^{M}$ with $K$ ends with a comon final model $\mathcal{P}$ (actually $K$ never moves), so there are stationarily many Ramseys in $K$, but no regular Jonsson in $V$, contradiction, so Mitchell/Jensen theorem applies and we get $\mathcal{M}$ with a Woodin cardinal.
Jan 28, 2016 at 22:45	comment	added	Rachid Atmai		To extend your answer Grigor (it seems like you originally wanted a Woodin instead of a proper class of measurables): Assume $\kappa$ is inaccessible. If there is an inner model $M$ such that $o(M)$ is $\kappa$, stationarily many $\alpha$'s below $\kappa$ have their successor computed correctly in $M$, stationarily many $\alpha$'s below $\kappa$ are Ramseys in $M$ and there are no regular Jonssons below $\kappa$, then there is an $\mathcal{M}$ with a Woodin.
Dec 14, 2015 at 6:09	comment	added	Mohammad Golshani		For my argument, a measurable cardinal $\kappa$ with $o(\kappa)=\kappa^+$ is sufficient. But in fact we can reduce the large cardinal assumption more: It suffices to have an inaccessible $\kappa$ so that for any $\delta<\kappa, \{ \alpha<\kappa: o(\alpha) \geq \delta \}$ is stationary
Dec 13, 2015 at 18:32	history	edited	Grigor	CC BY-SA 3.0	deleted 151 characters in body
Dec 13, 2015 at 18:29	comment	added	Grigor		ok, the problem is in my use of covering, i forgot that it doesn't quite hold when there are measurables around...so the argument from my note would only give you a proper class of measurables...this matches well with your example, i guess, i wonder if there is exact equiconsistency.
Dec 13, 2015 at 17:29	comment	added	Grigor		You have K^M and K that coiterate, if K^M only has measurable with a few measures then any non-measurable inaccessible of K^M is inaccessible in K. So K has a stationary set of inacessibles. Then because K has covering, you get that this inaccessibles are inacessibles in V. Of course one has to check the details and i haven't done it, but on the face of it, you def need at least a strong...a woodin may be a stretch, as i know one has to be careful...
Dec 13, 2015 at 17:29	comment	added	Grigor		are you sure about the argument? i know i did the K stuff fast and there are details there to check, but kappa is kappa+2 strong sounds something that shouldn't be enough.
Dec 13, 2015 at 11:58	comment	added	Mohammad Golshani		In the final extension, there are no inaccessibles below $\kappa,$ but the elements of $C$ (except its first element that we can assume is below the least inaccessible) were inaccessible in $V$, hence in $M$>
Dec 13, 2015 at 11:58	comment	added	Mohammad Golshani		Can't we use Radin forcing to produce such a model M. Assume $GCH+\kappa$ is $(\kappa+2)-$strong. Let $M=V_\kappa$. Force with Radin forcing, using a measure sequence of length $\kappa^+,$ to add a club of former regulars into $\kappa,$ and preserve $\kappa$ inaccessible. All limit points in the club are singular in the extension. Let $C$ be the set of limit points of the club, and force with Easton forcing to make $2^{\alpha^+}=\alpha_^+,$ where $\alpha < \alpha_$ are successive points in $C$..
Dec 9, 2015 at 17:34	history	edited	Grigor	CC BY-SA 3.0	deleted 7 characters in body
Dec 9, 2015 at 15:26	history	edited	Grigor	CC BY-SA 3.0	edited body
Dec 9, 2015 at 12:49	history	edited	Grigor	CC BY-SA 3.0	deleted 20 characters in body
Dec 9, 2015 at 12:40	history	edited	Grigor	CC BY-SA 3.0	added 569 characters in body
Dec 9, 2015 at 12:35	history	answered	Grigor	CC BY-SA 3.0