# Is it consistent that the gaps between cardinals $\kappa$ and $2^\kappa$ “get larger and larger”?

Is the following statement consistent in $\mathsf{ZFC}$?

For every ordinal $\beta$ there is an ordinal $\lambda_0$ such that for all ordinals $\lambda\geq\lambda_0$ we have $2^{\aleph_{\lambda}}\geq \aleph_{\lambda+\beta}.$

• Easton's theorem will give you the answer, methinks – David Roberts May 10 '16 at 8:33
• @DavidRoberts Will it? The question includes those $\lambda$ for which $\aleph_\lambda$ is singular, and unlike regular cardinals, those are not an easy matter to control. I seem to remember a past question on the consistency of $(\forall\lambda)\,2^{\aleph_\lambda}=\aleph_{\lambda+2}$, and while I can't remember the conclusion, I think it wasn't at all trivial. – Gro-Tsen May 10 '16 at 9:04
• @Gro-Tsen: Of course it will give the answer. Even if you only apply Easton's theorem to the class of successors of limits of cofinality $\omega$, it will still suffice to ensure the gaps are arbitrarily large. – Asaf Karagila May 10 '16 at 10:39
• @AsafKaragila If you notice how Dominic had asked his question, he inquires not just that there are arbitrarily large gaps, but that the gaps become eventually as large as desired, and so this would include singular cardinals. So I think we really need the sophisticated tools of the Foreman-Woodin model. – Joel David Hamkins May 10 '16 at 11:40

Yes, in the Foreman-Woodin model for the global failure of $GCH$ your statement is true.
• Thanks for this link! So if I understand it correctly, as a special case, given an ordinal $\beta$, it is consistent that $2^{\aleph_0} \geq \aleph_\beta$? Is it also possible that $2^{\aleph_0} = \aleph_{2^{\aleph_0}}$? (The answer to the latter is perhaps "No", for some simple reason that escapes me at the moment.) – Dominic van der Zypen May 10 '16 at 11:47
• @DominicvanderZypen Now this, unlike your original question, is reducible to Easton's theorem, which is a much easier result than Foreman-Woodin. Basically, you can make $2^{\aleph_0}$ equal to whatever $\kappa$ you like provided only that it has uncountable cofinality. (In particular, of $\kappa$ is the smallest fixed point of uncountable cofinality of $\alpha\mapsto\omega_\alpha$, you can make $2^{\aleph_0}=\kappa$.) – Gro-Tsen May 10 '16 at 12:00