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In [KwoM] it is proved that the the core model K exists under the assumption that there is no inner model having a Woodin cardinal and satisfying ZFC. Furthermore, they also mention that the result is the "best possible", in the sense that we cannot weaken this anti-large-cardinal hypothesis without strengthening the remainder of the hypothesis.

From what I understand, using the core model induction (which I must admit I don't know the details of yet), K can be constructed below any finite number of Woodin cardinals. So my question is then whether or not K has been constructed below $\omega$ many Woodins and if so, what the extra hypothesis is? If this is indeed the case, I would greatly appreciate a reference to the given paper.

[KwoM] Jensen and Steel: "K without the measurable"

Edit: In the introduction to Schindler and Steel's "The Core Model Induction", they write that if we assume that there is no proper class model with $n+1$ Woodins and over every set there is a proper class model with $n$ Woodins, then $K$ exists (in the sense of the above [KwoM]).

Can this be extended to: if there is no proper class model with $\omega$ Woodins and over every set there is a proper class model with $n$ Woodins, for every $n<\omega$, then $K$ exists?

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    $\begingroup$ There is a construction of inner model for assumptions up to Woodin cardinal which is a limit of strong cardinals (due to Neeman). The issue is that once there is a Woodin cardinal, those inner models can be changed by forcing (they are no longer generic absolute). $\endgroup$
    – Yair Hayut
    Commented Oct 9, 2016 at 7:33
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    $\begingroup$ Thanks for the comment Yair. It seems that we can get around this absoluteness issue by considering relative premice, from what I can tell by Andrés' answer to the following MO question: mathoverflow.net/questions/153745/…. He mentions that one of these extra assumptions we could use is that V is closed under sharps. I'm just interested in how exactly the theorem is stated, if other assumptions is needed as well. $\endgroup$
    – Dan Saattrup Nielsen
    Commented Oct 9, 2016 at 13:10
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    $\begingroup$ In the introduction to Schindler and Steel's "The Core Model Induction", they write that if we assume that there is no proper class model with $n+1$ Woodins and over every set there is a proper class model with $n$ Woodins, then $K$ exists (in the sense of the above [KwoM]). Can this be extended to: if there is no proper class model with $\omega$ Woodins and over every set there is a proper class model with $n$ Woodins, for every $n<\omega$, then $K$ exists? $\endgroup$
    – Dan Saattrup Nielsen
    Commented Nov 9, 2016 at 17:59

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Dan, the answer is "no." You might be in a situation where Kc is a mouse (i.e., its ctble. substructures are ω1+1 iterable), but Kc is not fully iterable. E.g. it could be that Kc has M2F,#, where F is the operator simultaneously closing under all Mn#, n<ω, but for some θ, V = LF(Vθ) (same F, V could even be an extender model): the Q-structures needed to fully iterate M2F,# would be given by the M1F,# operator which is not available thruout the universe. If Kc is not fully iterable, one can't use it to isolate K, and in such a situation usually there is no inner model which has enough of the relevant properties that it might be called "the core model." See e.g. the first two pages of https://ivv5hpp.uni-muenster.de/u/rds/K.pdf for a list of such properties. (There are exceptions to this, though, see e.g. https://ivv5hpp.uni-muenster.de/u/rds/CMPW.pdf and also https://ivv5hpp.uni-muenster.de/u/rds/RalfGrigor_version_Sept11_2017.pdf .)

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  • $\begingroup$ Thanks Ralf! Just to see if I get the argument: if we consider the operator $G:=M_1^{F,\sharp}$ and assume $K^c$ reaches $M_1^{G,\sharp}$ then to show that $M_1^{G,\sharp}$ is fully iterable we need to be able to build $L^G(\mathcal M(\mathcal T))$ in $V$ to be able to form the necessary $\mathcal Q$-structures, but we can arrange that $V$ isn't closed under $G$. So this shows that we can't even get $K$ if both $V$ is closed under $F$ and $M_2^{F,\sharp}$ exists, which is possible even with the requirement that there be no inner model with $\omega$ Woodins. Is this correctly understood? $\endgroup$
    – Dan Saattrup Nielsen
    Commented Sep 18, 2017 at 9:52
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    $\begingroup$ Yep, that's the thing :) $\endgroup$
    – Ralf Schindler
    Commented Sep 18, 2017 at 10:41

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