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Whenever $M$ is some fine-structural $L$-like model we can prove the implication $V=M\Rightarrow\textsf{GCH}$. For $L$ this is due to Gödel, and for the modern extender models it follows simply by construction. The most recent direction in the inner model programme searches for fine-structural models of the form $\textsf{HOD}^N$ with $N$ some "nice" model (most recently this includes $N=M_n(x,g)$ for a Turing cone of reals $x$ and $g$ generic for making its least $N$-inaccessible countable). My question is then whether $V=\textsf{HOD}\Rightarrow\textsf{GCH}$ has been shown to be unprovable in general (modulo large cardinals of course)? Is there some pathological $\textsf{HOD}$ which "obviously" doesn't satisfy $\textsf{GCH}$? Also, I'm not even sure if $\textsf{HOD}^{\textsf{HOD}^N}=\textsf{HOD}^N$ holds for these models, as this (apparently) doesn't hold in general.

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Yes, one can produce a model of $ZFC+V=HOD$ in which the $CH$ fails. This should follow from Consistency results about ordinal definability.

In fact, I think the arguments of my paper HOD, V and the GCH can be used to produce a model of $ZFC+V=HOD$ in which the $GCH$ fails everywhere.

Also by a result of Apter (see Large cardinals need not be large in HOD for a proof), if there are class many Laver indestructible supercompact cardinals, then $V=HOD$ holds. Note that in such a model $GCH$ must fail at class many regular cardinals.

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  • $\begingroup$ That's great, thanks for the swift answer Mohammad! I also tried digging through the citations in the papers you mentioned, and found where this was perhaps first shown (in the weakest form): in McAloon ('70) he shows that $V=\textsf{HOD}+\lnot\textsf{CH}$ is consistent. $\endgroup$
    – Dan Saattrup Nielsen
    Commented Jan 29, 2018 at 10:33
  • $\begingroup$ Possibly a silly question - why must GCH fail frequently in a model with class many Laver indestructible supercompacts? $\endgroup$
    – Noah Schweber
    Commented Jan 29, 2018 at 15:21
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    $\begingroup$ @Noah For instance, $\mathsf{GCH}$ cannot first fail above a supercompact $\kappa$, but you can violate it with $\kappa$-directed closed forcing. So, if $\kappa$ is Laver indestructible, $\mathsf{GCH}$ must fail (frequently) below $\kappa$. $\endgroup$
    – Andrés E. Caicedo
    Commented Jan 29, 2018 at 16:11
  • $\begingroup$ @AndrésE.Caicedo Ah, yes, silly of me. $\endgroup$
    – Noah Schweber
    Commented Jan 29, 2018 at 16:16

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