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Aug 3, 2013 at 3:31 answer added Mohammad Golshani timeline score: 9
Aug 2, 2013 at 1:08 answer added Andrés E. Caicedo timeline score: 13
Aug 2, 2013 at 0:40 comment added Andrés E. Caicedo (I'm turning my comments into an answer.)
Aug 1, 2013 at 14:33 comment added Ramiro de la Vega @Andres: why a comment and not an answer?
Aug 1, 2013 at 14:21 comment added Andrés E. Caicedo His argument assumes $\mathsf{GCH}$ and that $\kappa$ is what is either called $\kappa+n$-strong or $\kappa^{+n+1}$-strong, that is, there is $j:V\to M$ with $\mathrm{cp}(j)=\kappa$ and $V_{\kappa+n}\subset M$. His assumptions are an overkill, since in the final model he preserves inaccessibility of $\kappa$, and $V_\kappa$ is the model where $\forall \lambda,(2^\lambda=\lambda^{+n})$.
Aug 1, 2013 at 14:15 comment added Andrés E. Caicedo As you say, Hugh's precise result is unpublished (it does not use the supercompact version of Radin's forcing, but what you call the ordinary version). The only published full account of the result that I can think of currently is Carmi Merimovich. A Power Function with a Fixed Finite Gap Everywhere, The Journal of Symbolic Logic, 72 (2), (2007), 361-417. Merimovich uses extender based Radin forcing, his argument can give $\forall\lambda\,(2^\lambda=\lambda^{+n})$ for any fixed $n$, $1<n<\omega$, though he presents the details for $n=3$.
Aug 1, 2013 at 13:07 comment added Péter Komjáth My understanding is that Foreman-Woodin: $\forall\lambda, 2^\lambda\geq\lambda^{++}$, Woodin, later: $\forall\lambda, 2^\lambda=\lambda^{++}$.
Aug 1, 2013 at 13:00 comment added Joel David Hamkins Isn't the result due to Foreman and Woodin? Also, see mathoverflow.net/questions/79920/failure-of-the-gch.
Aug 1, 2013 at 9:30 history asked Mohammad Golshani CC BY-SA 3.0