Timeline for Inner model in which every uncountable cardinal is large
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 22, 2014 at 1:52 | comment | added | Andrés E. Caicedo | Philip, do you know what the optimal assumption is? Surely $0^{sword}$ is more than needed, but a sharp for a model with proper class many measurables does not seem strong enough. | |
| Apr 21, 2014 at 14:44 | comment | added | Mohammad Golshani | Thanks a lot. So it seems that the result I have stated in the remark is not surprising. The only advance is that in this case, $V$ is a generic extension of $M$. | |
| Apr 21, 2014 at 14:42 | vote | accept | Mohammad Golshani | ||
| Apr 21, 2014 at 14:05 | comment | added | Philip Welch | @Mohammad Is M included in HOD? Yes. Does M satisfy V=HOD? Yes it does. | |
| Apr 21, 2014 at 7:58 | comment | added | Philip Welch | @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? | |
| Apr 21, 2014 at 7:52 | comment | added | Philip Welch | @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.) | |
| Apr 21, 2014 at 7:42 | comment | added | Mohammad Golshani | 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 $V=HOD$? | |
| Apr 21, 2014 at 7:40 | comment | added | Mohammad Golshani | 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? | |
| Apr 21, 2014 at 7:37 | history | answered | Philip Welch | CC BY-SA 3.0 |