Prikry forcing can be used to produce a model $V$ of $ZFC$ such that fo rsome cardinal $\kappa$ we have:

(1) $\kappa$ is singular in $V$ of cofinality $\omega,$

(2) $\kappa$ is regular (and in fact measurable) in $HOD$.

Now my question is can this happen with $\kappa$ having uncountable cofinality in $V$. So

Question Can we find a model $V$ of $ZFC$ which contains a cardinal $\kappa$ so that

(1) $\kappa$ is singular of uncountable cofinality in $V$,

(2) $\kappa$ is regular in $HOD$.

Let me explain why Magidor or Radin forcing do not work in general. Let's start with core model $K$ in which $\kappa$ is large enough, and let $V$ be the generic extension obtained by Magidor or Radin forcing to change the cofinality of $\kappa$ to, say, $\omega_1.$ Let $C$ be the resulting club. We can assume all elements of $C$ were regular in $K$.

Claim. $Lim(C) \in HOD,$ where $Lim(C)$ is the set of limit points of $C$.

Proof. We have $Lim(C)=\{\alpha \leq \kappa: \alpha$ is singular, but regular in $K \} \in HOD.$

In particular $\kappa$ is singular in $HOD$.

  • 1
    $\begingroup$ Probably unrelated, but this is certainly true in V=L($\mathbb{R}$)+AD. In particular, $\omega_3$ is singular in V but measurable in HOD. $\endgroup$
    – Cody Dance
    Mar 21, 2016 at 17:46

1 Answer 1


The answer to the question is yes. In fact, it is possible to prove something stronger:

Theorem (Omer Ben-Neria, Spencer Unger) Assuming the existence of suitable large cardinals, there exists a model $V$ of $ZFC$ which contains a club of cardinals all of them are regular in $HOD$ of $V$.

Their work is under preparation.

Remark. Their result is optimal in the sense that we can not hope to build a model in which all infinite cardinals are regular in $HOD$, as for example $\aleph_\omega$ is always singular in $HOD$.

Update: The paper by Omer Ben-Neria and Spencer Unger is now available:

Homogeneous changes in cofinalities with applications to HOD

  • 1
    $\begingroup$ @Rahman.M It doesn't matter. You can always manage all point in C are singular in $V$. $\endgroup$ May 3, 2016 at 3:49
  • $\begingroup$ What are the large cardinals they're using there? $\endgroup$
    – Asaf Karagila
    May 3, 2016 at 6:35
  • 1
    $\begingroup$ @AsafKaragila I just heard about their result by email correspondence and have not seen their paper. But I think strong type large cardinals are sufficient for it. $\endgroup$ May 3, 2016 at 7:21

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