14
$\begingroup$

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 $\lambda.$

$\endgroup$
8
  • 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

11
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.