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It's known that we can have global failures of GCH---for example, where $\forall \lambda(2^\lambda = \lambda^{++})$---given suitable large cardinal axioms.

My question is whether we can have global failures of GCH where there is a weakly inaccessible cardinal between $\lambda$ and $2^\lambda$ for each $\lambda$. Similarly, whether we can have a cardinal fixed point between $\lambda$ and $2^\lambda$. I'd also be interested in whether $2^\lambda$ can be weakly inaccessible/a cardinal fixed point, for every $\lambda$.

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  • $\begingroup$ Two questions: (1) What is weakly measurable? (2) Remind me what does the original Foreman-Woodin model satisfy? $\endgroup$ Commented Aug 5, 2020 at 19:45
  • $\begingroup$ oh, haha! (I use weaklymeasurable as a handle on instagram.) (1) is fixed. So, I think Foreman-Woodin was just $2^\lambda> \lambda^+$ and then Woodin got $2^\lambda = \lambda^{++}$. $\endgroup$ Commented Aug 5, 2020 at 19:51

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In the Foreman-Woodin model The generalized continuum hypothesis can fail everywhere. for each infinite cardinal $\kappa, 2^\kappa$ is weakly inaccessible.

This answers your last question. The answer to the first two questions can be yes as well. In the case of Foreman-Woodin model, they start with a supercompact $\kappa=\kappa_0$ and infinitely many inaccessibles $\kappa_n, n<\omega,$ above it. They first force to get $2^{\kappa_n}=\kappa_{n+1}$ preserving $\kappa$ supercompact, and this is reflected below for all cardinals. So if for example each $\kappa_n$ is measurable, then what you get in the final model is that for each infinite cardinal $\lambda, 2^\lambda$ has been measurable in $V$, in particular there are both weakly inaccessible and cardinal fixed points between $\lambda$ and $2^\lambda.$

See also the paper A model in which every Boolean algebra has many subalgebras by Cummings and Shelah, where they build a model in which for each infinite cardinal $\kappa, 2^\kappa$ is weakly inaccessible and $Pr(2^\kappa)$ holds. Here $Pr(\lambda)$ is in some sense a large cardinal property (for example it holds if $\lambda$ is a Ramsey cardinal). For its definition see the paper.

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  • $\begingroup$ Thanks, Mohammad! That answers my final question. Does it also answer my first two questions? Sorry if I'm missing something obvious! $\endgroup$ Commented Aug 10, 2020 at 12:18
  • $\begingroup$ @SamRoberts I added details. $\endgroup$ Commented Aug 12, 2020 at 4:39
  • $\begingroup$ Thanks so much, Mohammad! $\endgroup$ Commented Aug 12, 2020 at 8:00

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