Is it possible, starting from any large cardinal assumption, to find a countably closed forcing $\mathbb{P}$ such that for some inaccessible $\kappa$, $\Vdash_\mathbb{P} "\kappa = \lambda^+$ and $\lambda$ is singular"?
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asked Aug 8, 2014 at 16:03
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1 Answer
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No. The following theorem is from a work in progress by Yair Hayut and myself.
Theorem. If $\Bbb P$ is a proper forcing, and it changes the cofinality of $\kappa$ to $\mu>\omega$, then $\Bbb P$ adds a surjection from $\mu$ onto $\kappa$.
Now suppose that you had such countably closed $\Bbb P$, it is certainly proper. And it changes the cofinality of $(\lambda^+)^V$ to be something which is smaller than $\lambda$, and therefore collapses $(\lambda^+)^V$, and as a consequence it must collapse $\lambda$ as well.
(It should be remarked that a countably closed forcing cannot change the cofinality of something $\omega$ anyway.)
answered Aug 8, 2014 at 17:45
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$\begingroup$ Wow! Let me know when it is available. $\endgroup$ Commented Aug 8, 2014 at 19:51
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$\begingroup$ Will do. This is just the first theorem of the paper, the rest is actually far more interesting. I think we're about halfway through, and hopefully in a month or so we'll have something on arXiv. I'll keep you posted. $\endgroup$– Asaf Karagila ♦Commented Aug 8, 2014 at 20:11
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$\begingroup$ (I also have to admit that I found it quite fun that this theorem popped up like this, something I did not expect!) $\endgroup$– Asaf Karagila ♦Commented Aug 8, 2014 at 20:20
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$\begingroup$ @Mohammad: And that's the not-very-interesting part of the work. :-) $\endgroup$– Asaf Karagila ♦Commented Aug 19, 2014 at 6:58