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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
May 15, 2012 at 12:47 vote accept Ioannis Souldatos
May 15, 2012 at 12:46 vote accept Ioannis Souldatos
May 15, 2012 at 12:46
May 15, 2012 at 12:46 vote accept Ioannis Souldatos
May 15, 2012 at 12:46
May 15, 2012 at 12:45 answer added Ioannis Souldatos timeline score: 4
Nov 21, 2011 at 15:45 answer added mmm timeline score: 0
Aug 8, 2011 at 14:14 comment added Ioannis Souldatos ...The paper SIX CLASSES OF THEORIES is where the question "Is $Ded(\kappa)<Ded(\kappa)^\omega$ consistent?" appears.
Aug 8, 2011 at 14:13 comment added Ioannis Souldatos Good point. $Ded(\kappa)<Ded(\kappa)^\omega$ can be consistent only if $\kappa<\kappa^\omega<Ded(\kappa)<2^\kappa$. This is remarked by Keisler in SIX CLASSES OF THEORIES, J. Austral. Math. Soc. 21 (Series A) (1976), 257-266. He attributes the proof to Kunen, but I didn't find a reference. (To obtain the article follow the link: journals.cambridge.org/… ) This is where the question
Aug 6, 2011 at 1:23 comment added Artem Chernikov Perhaps you could add to the list of known things that if $\kappa=\kappa^{\omega}$, then $Ded(\kappa)=Ded(\kappa)^{\omega}$. This is an observation of Kunen mentioned in the paper of Keisler.
Aug 5, 2011 at 13:32 vote accept Ioannis Souldatos
Aug 5, 2011 at 13:33
Aug 5, 2011 at 6:25 comment added David Milovich My answer to (1) was fatally flawed, so please un-accept my answer, which I think should be deleted. Andy Voellmer's comment about (2) is still correct though: you can force $2^\omega=2^\kappa=\kappa^+$ (with GCH in the ground model) to show that (2) is consistently true.
Aug 4, 2011 at 21:34 vote accept Ioannis Souldatos
Aug 5, 2011 at 13:26
Aug 4, 2011 at 17:12 history asked Ioannis Souldatos CC BY-SA 3.0