It is known that Namba forcing is stationary-preserving and hence can be used in the setting of Martin's Maximum. Does this result in any striking consequences?
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Joel David Hamkins
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asked Apr 10, 2010 at 22:59
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$\begingroup$ Maybe the Strong Chang Conjecture? $\endgroup$– François G. DoraisApr 11, 2010 at 0:20
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$\begingroup$ Is the proof a relatively direct application of Namba forcing? Or is it more indirect? $\endgroup$– Simon ThomasApr 11, 2010 at 0:41
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$\begingroup$ I added the forcing tag. $\endgroup$– Joel David HamkinsApr 11, 2010 at 0:43
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$\begingroup$ @Francois: Are you referring to Shelah's result that "Namba forcing is semiproper" implies SCC? $\endgroup$– HaimApr 11, 2010 at 0:53
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1$\begingroup$ That's what I was wondering. I'd like an application with a proof along the lines of those in Baumgartner's great article on PFA. If the proof of the Strong Chang Conjecture is: MM implies Namba is semi-proper which implies SCC, then I'm not satisfied. $\endgroup$– Simon ThomasApr 11, 2010 at 1:01
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3 Answers
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Magidor's proof that MM implies that there are no good scales of length lambda^+ for lambda>cf(lambda)=w, utilizes a Namba-style forcing.
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I think that I may have found a suitable candidate; namely, the result of Konig and Yoshinobu that $MM$ implies that there are no $\omega_{1}$-regressive $\omega_{2}$-Kurepa trees. The proof seems to have the same relatively direct flavor as those in Baumgartner's $PFA$ article.
answered Apr 11, 2010 at 13:44
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Justin Moore asked a similar question a while ago, and I pointed him to my papers with Claverie and Doebler. Even though it doesn't exactly answer your question: Namba-like forcings (in the sense that they are stationary set preserving and make $\omega_2$ $\omega$-cofinal) have many applications in the presence of ${\sf MM}$. The most recent one is my proof with David Asperó that ${\sf MM}^{++}$ implies Woodin's $(*)$, see https://ivv5hpp.uni-muenster.de/u/rds/MM_implies_star.pdf .
