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The following is known (see e.g., [1]): Modulo large cardinals it is consistent that $2^\lambda>\lambda^+$ for all $\lambda$. (Furthermore, it seems that the gap can be always infinite?)

Question: Is it consistent that $2^\lambda=2^{\lambda^+}$ for all $\lambda$? (Which obviously implies infinite gap.)

Remark: For just the regulars this can be done for free by Easton, e.g. setting $2^{\aleph_\alpha}=\aleph_{\alpha+\omega+1}$.

[1] Foreman, Matthew; Woodin, W. Hugh, The generalized continuum hypothesis can fail everywhere, Ann. Math. (2) 133, No. 1, 1-35 (1991). ZBL0718.03040.

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2 Answers 2

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This also holds in Foreman-Woodin model. In this model, for every infinite cardinal $\lambda, 2^\lambda > \lambda$ is weakly inaccessible and for all $\lambda < \mu < 2^\lambda$ we have $2^\mu=2^\lambda.$

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Oh sorry, I just saw that e.g. the model in [2] has this property. (Maybe the one in [1] does as well; I do not understand it enough to see it. In [2] it follows explicitly from item 1 of "the star property".)

(Not sure whether the question should be removed...)

[2] Cummings, James; Shelah, Saharon, A model in which every Boolean algebra has many subalgebras, J. Symb. Log. 60, No. 3, 992-1004 (1995). ZBL0838.03038.

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    $\begingroup$ The question need not necessarily be removed, as it may be of interest to others. (I'm not in the target audience, so can't say myself, but you might as well leave it up to see how it's received; you can always close it later.) If this answers your question, then don't forget to accept your own answer. $\endgroup$
    – LSpice
    Commented Aug 12 at 21:56
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    $\begingroup$ Great question, and I'm glad you answered. No need to delete. It would be great, however, if you could add more explanation, to make the answer somewhat self-contained. Links to articles are great and valuable, but one must recognize that most readers will not follow the links. Often one can just explain the idea quickly, and that is the best. $\endgroup$
    – Joel David Hamkins
    Commented Aug 12 at 22:36

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