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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?

edited Apr 11 2010 at 0:43

Joel David Hamkins
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asked Apr 10 2010 at 22:59

Simon Thomas
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Maybe the Strong Chang Conjecture? – François G. Dorais♦ Apr 11 2010 at 0:20

Is the proof a relatively direct application of Namba forcing? Or is it more indirect? – Simon Thomas Apr 11 2010 at 0:41

I added the forcing tag. – Joel David Hamkins Apr 11 2010 at 0:43

@Francois: Are you referring to Shelah's result that "Namba forcing is semiproper" implies SCC? – Haim Apr 11 2010 at 0:53

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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. – Simon Thomas Apr 11 2010 at 1:01

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saf · Answer 1 · 2010-04-11 02:23:34Z UTC

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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.

answered Apr 11 2010 at 2:23

saf
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Simon Thomas · Accepted Answer · 2010-04-11 13:44:00Z UTC

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.

What are the Martin’s Maximum consequences of Namba forcing?

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