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How badly can the GCH fail globally?

It’s known that we 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 $lambda$.

MathOverflow Asked by Sam Roberts on January 24, 2021

1 Answers

One Answer

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 $kappa_n, n<omega,$ about 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.

Correct answer by Mohammad Golshani on January 24, 2021

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