Question. Suppose $\kappa$ is a supercompact cardinal and $\lambda > \kappa$ is measurable (or even larger large cardinal if necessary). Is there a set generic extension of the universe in which $\kappa$ remains supercompact, $\lambda$ is preserved and $cf(\lambda)=\omega?$

Remark 1. By Gitik-Shelah indestructibility result, if supercompact cardinal is replaced with strong cardinal, then the answer is yes.

Remark 2. If we require that the forcing preserves $\lambda^+,$ then the answer is no, as it is shown by Yair.

Remark 3. In A note on sequences witnessing singularity - following Magidor-Sinapova, Gitik has conjectured the following:

Conjecture. Suppose that

  1. $V ⊆ W$ models of ZFC with same ordinals,

  2. $κ$ is a regular cardinal in $V$,

  3. $cof(κ) = ω$ in $W$,

  4. $\aleph_1^V=\aleph_1^W,$

  5. $V, W$ agree about a final segment of cardinals.

Then there is a subclass $V′$ of $V$ which is a model of $ZFC$, agree with $V$ about a final segment of cardinals, and there is a sequence witnessing singularity of $κ$ (in $W$) which is generic over $V′$ for either Namba, Woodin tower or Prikry type forcing.

Assuming this conjecture, it seems quite plausible that the answer to the question might be no in general.

Edition. I realized that the question has connection with recent work of Woodin:

Theorem. Assume $\kappa$ is an extendible cardinal. If Woodin's $HOD$-conjecture holds, then we can not change the cofinality of some large cardinal $\lambda > \kappa$, preserving the supercompactness of $\kappa,$ by set forcing without collapsing $\lambda.$

  • 1
    $\begingroup$ Supercompact can be made immune for $\kappa$-directed closed forcings, unfortunately those are proper, and a proper forcing cannot change cofinality without collapsing cardinals. $\endgroup$
    – Asaf Karagila
    Mar 17, 2016 at 17:46
  • 2
    $\begingroup$ I think that if you allow class forcings then Woodin's stationary tower (with height $\delta = On$, assuming "$On$ is Woodin") can get you the desired situation. By forcing below $S^\lambda_\omega$, the critical point of the embedding $j\colon V\to M$ is $\lambda$, $\text{cf }\lambda = \omega$, and therefore $\kappa$ remains supercompact in $M$. Using the closure properties of $M$, the same holds in $V[G]$. $\endgroup$
    – Yair Hayut
    Mar 18, 2016 at 7:19
  • $\begingroup$ That's quite a conjecture. $\endgroup$
    – Asaf Karagila
    Mar 26, 2016 at 8:30
  • $\begingroup$ Did you realize this by reading the recent review paper about the HOD conjecture? Because I reviewed it recently and felt something similar. But the question is what happens if the HOD conjecture is in fact false, and it is possible to have a dichotomy. Can you still force something like this? $\endgroup$
    – Asaf Karagila
    May 7, 2016 at 12:04
  • 1
    $\begingroup$ I recently asked @YairHayut the same question in an email. He showed me an argument of Magidor showing that the strong compactness of $\kappa$ must be destroyed. $\endgroup$ Jun 13, 2018 at 16:50

1 Answer 1


There is no such forcing that preserves $\lambda^+$. Since $\lambda$ is measurable, $2^{<\lambda} = \lambda$ and therefore $\square_{\lambda,\lambda}$ holds in $V$. Since $\lambda^{+}$ is preserved, the same sequence will witness that there is still a weak square at $\lambda$ in the generic extension. But weak square fails at singular cardinals of small cofinality above a supercompact cardinal, so $\kappa$ cannot be supercompact in the generic extension.

In fact, a theorem of Dzamonja and Shelah shows that if you change the cofinality of an inaccessible cardinal $\lambda$ while preserving $\lambda^+$, there is even a $\square_{\lambda,\omega}$ sequence in the generic extension.


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