Asaf Karagila
I don't have much choice...

Preserving Properness

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I just posted another problem in the problems page. The prize, by the way, is a bottle of port wine, or equivalent. And I truly hope to make good on that prize.

In another problem there, coming from a work with David Asperó, we asked if an \(\omega_2\)-closed forcing must preserve the property of being proper. Yasou Yoshinobu provided us with a negative answer based on Shelah's "Proper and Improper Forcing" XVII Observation 2.12 (p.826). Take \(\kappa\) to be uncountable, by forcing with \(\Add(\omega,1)\ast\Col(\omega_1,2^\kappa)\) and appealing to the gap lemma, \((2^{<\kappa})\) is a tree with only \(\aleph_1\) branches. It can therefore be specialized by a ccc forcing in that model. The iteration of these three forcing (Cohen real, collapse, specialize) is clearly proper. But now by forcing with \(\Add(\kappa,1)\) we must in fact violate the properness of this forcing, which was defined in the ground model, since the new branch is also generic for the tree and will therefore collapse \(\omega_1\).

Similarly, Shelah gave a solution where a \(\sigma\)-closed forcing is not proper after adding just a Cohen real. You can find the solution, provided kindly by Martin Goldstern, in this MathOverflow answer.

And this leads us to the point where we realize that provably in \(\ZFC\) that for any \(\kappa\) there is always a forcing \(\QQ\) which is proper in \(V\), but not proper after forcing with \(\Add(\kappa,1)\). And that leads us to the following question.

Question. If \(\PP\) is any nontrivial forcing, is there a forcing \(\QQ\) which is proper in \(V\), but not proper after forcing with \(\PP\)?

Well, what are you still doing here? Go! Solve! Earn that bottle of port!

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