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Woodin's HOD Dichotomy Theorem says that if an extendible cardinal exists, then either $V$ and $HOD$ are rather close or rather far apart. My question is whether the "far" case can be strengthened in analogy to Jensen's Covering Lemma to say something more about large singulars. Suppose $\delta$ is extendible and every regular $\kappa \geq \delta$ is measurable in $HOD$. Does there exist a singular $\lambda > \delta$ which is singular in $HOD$?

EDIT: The comment by Gabe Goldberg gives an easy "yes" answer. Here's a harder version of the question. Suppose $\delta$ is extendible and $\lambda$ is the least cardinal above $\delta$ such that $V_\lambda \models ZFC$. Are all singular cardinals in the interval $(\delta,\lambda]$ singular in $HOD$?

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    $\begingroup$ A typical example is $\lambda = \delta^{+\omega}$, which has cofinality $\omega$ in $\text{HOD}$ since $\langle\delta^{+n} : n < \omega\rangle$ is ordinal definable. $\endgroup$
    – Gabe Goldberg
    Commented Jul 9, 2019 at 11:41
  • $\begingroup$ @GabeGoldberg Oh I see, thanks. So any non-$\aleph$-fixed point is also singular in HOD as well. Also the least $\aleph$-fixed point above $\alpha$, the $\omega^{th}$, etc. What is the most general thing we can say here? $\endgroup$
    – Monroe Eskew
    Commented Jul 9, 2019 at 12:06
  • $\begingroup$ I'm not sure. Since every regular cardinal $\kappa \geq\delta$ is $\omega$-strongly measurable in $\text{HOD}$, there is an $\omega$-club below $\kappa$ of cardinals of cofinality $\omega$ which are regular in $\text{HOD}$. Is it interesting to look at whether the ordinals that are singular in $\text{HOD}$ are nonstationary in $\kappa$? $\endgroup$
    – Gabe Goldberg
    Commented Jul 9, 2019 at 12:31

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This is independent relative to the failure of the HOD hypothesis in the presence of large cardinals.

We first give a positive answer under GCH. (Note that if it is consistent for the HOD Hypothesis to fail in the presence of an extendible, then this is consistent with GCH.) More generally, we show that under GCH, for any singular cardinal $\lambda$ that is regular in $\text{HOD}$, $V_\lambda\vDash \text{ZFC}$. Suppose towards a contradiction that $V_\lambda$ does not satisfy ZFC. Then there is a singularization of $\lambda$ definable over $V_\lambda$ from a parameter $a\in V_\lambda$. We may code $a$ by a set of ordinals $A\subseteq \alpha$ for some $\alpha < \lambda$. Thus $\lambda$ is singular in $\text{HOD}_A$. But $\text{HOD}_A$ is a generic extension of $\text{HOD}$ for a forcing of size less than $\lambda$ in $\text{HOD}$. (This bound falls out of Vopenka's theorem given GCH: recall that the Vopenka algebra is in bijection with the OD powerset of $P(\alpha)$.) One cannot destroy the $\text{HOD}$-inaccessible cardinal $\lambda$ by small forcing over $\text{HOD}$, so we have reached a contradiction.

On the other hand, the GCH assumption is necessary, which is proved easily by forcing. Start with a model $M$ of the failure of the HOD Hypothesis and an extendible cardinal $\delta$. Let $\lambda$ be the least singular cardinal of $M$ above $\delta$ that is regular in $\text{HOD}$. Force preserving cardinals, preserving the extendibility of $\delta$, and without increasing HOD to blow up $2^\delta$ above $\lambda$. This gives us a generic extension $N$. In $N$, the least $\lambda' > \delta$ such that $V_{\lambda'}$ is a model of ZFC is such that $\lambda' > 2^\delta > \lambda$. But $\lambda$ is still a singular of $N$ that is regular in $\text{HOD}$.

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    $\begingroup$ Can you do that thing with the forcing? I always had the impression that extendible cardinals are very destructible under "large forcing" (that was my first question to my advisor when I was thinking about the Bristol model in light of the AC/HOD Conjectures). You make this off-hand remark, but this appears as an open question in Usuba's preprint on extendible cardinals and the mantle (arxiv.org/abs/1803.03944). What is this forcing that you want to use, and why does it preserve extendibility? $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 10, 2019 at 23:11
  • $\begingroup$ @AsafKaragila I think it works if you iterate following the rule, “At stage $\alpha$, if the iteration so far has size $\leq \alpha$, find the least $\lambda_\alpha>\alpha$ such that $\lambda_\alpha$ is singular but regular in $HOD^V$, and add $\lambda_\alpha$ Cohen subsets of $\alpha$.” Then we would want to take a $j : V_\alpha \to V_\beta$ in $V$ with critical point $\delta$ such that $\alpha > $ the least blah as above, but also where $V_\alpha$ and $V_\beta$ reflect locally the true disparity between singulars and HOD-regulars. Now I’m stuck. $\endgroup$
    – Monroe Eskew
    Commented Jul 11, 2019 at 6:50
  • $\begingroup$ @Monroe: If my memory serves me right, adding Cohen subsets above an extendible will definitely violate its extendibility. Again, I am not an expert on the topic, I was just under the impression that this is an open and difficult problem, one that requires a lot more than a "this is so trivial I'm not even gonna give you the details" approach. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 11, 2019 at 7:01
  • $\begingroup$ @AsafKaragila What might happen is that adding some Cohen subsets resurrects extendibility, simply because it’s the tail end of an appropriate iteration. I’m certain that this works if we want to, for example, have an extendible along with $2^\kappa=\kappa^{++}$ for all regular $\kappa$. $\endgroup$
    – Monroe Eskew
    Commented Jul 11, 2019 at 7:35
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    $\begingroup$ Monroe's forcing is the one I had in mind. Asaf is right, of course, that I should have been more careful. The preservation of extendibility seems to require that we started with a HOD-extendible, i.e., that for all $\alpha$, there is a $j : V_{\alpha+1}\to V_{j(\alpha)+1}$ such that $j(\text{HOD}\cap V_\alpha) = \text{HOD}\cap V_{j(\alpha)}$. (This follows from somewhat stronger large cardinals, for example Vopenka's Principle.) Given this I think the lifting argument is just like the $2^\kappa = \kappa^{++}$ case. $\endgroup$
    – Gabe Goldberg
    Commented Jul 11, 2019 at 14:08

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