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What is the (currently known) consistency strength of global failure of the GCH?

I do not have access to the exact statement of the original Foreman-Woodin result. My searches seem to indicate that they used an assumption at the region of a supercompact, although I have seen comments stating that the result has been improved to require something in the region of a hypermeasurable. Is this correct? What this exact upper bound?

Thanks a lot.

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Can't you just use Easton theorem to have $2^\kappa = \kappa^{+++}$ or something like that? – Asaf Karagila Nov 3 '11 at 13:50
Easton does only work for regular cardinals $\kappa$. But the question is about global GCH. – Stefan Hoffelner Nov 3 '11 at 14:33
Closely related. – Andrés E. Caicedo Aug 3 '13 at 3:14
up vote 10 down vote accepted

The following quotations are taken from Matthew Foreman and W. Hugh Woodin, "The generalized continuum hypothesis can fail everywhere," Ann. Math. 133 (1991), 1–35.

THEOREM. Let $\kappa$ be a supercompact cardinal with infinitely many inaccessible cardinals above $\kappa$. Then there is a partial ordering $\mathbf P$ such that in $V^{\mathbf P}$, $V_\kappa \models ZFC + \forall \lambda: 2^\lambda > \lambda^+$.

In fact we can arrange by our choice of partial orderings that $V^{\mathbf P}\models$ $\kappa$ is $\beth_n(\kappa)$-supercompact. Solovay has shown that if $\kappa$ is supercompact then $2^{\beth_\omega(\kappa)} = \beth_\omega(\kappa)^+$; hence this is near best possible. Woodin extended this result to get:

THEOREM (Woodin). If there is a supercompact cardinal then there is a model of ZFC in which $2^\kappa = \kappa^{++}$ for each cardinal $\kappa$.

The last sentence of the paper reads:

The second author has also reduced the consistency strength of "$ZFC + \forall\kappa: 2^\kappa > \kappa^+$" and "$ZFC + \forall\kappa: 2^\kappa = \kappa^{++}$" to that of a ${\mathscr P}^2(\kappa)$-hypermeasurable.

It's not clear to me if the proofs of the two theorems attributed to Woodin have ever been published.

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Timothy, The unpublished proofs are by now standard. You can get $ZFC+\forall\kappa(2^\kappa=\kappa^{+n})$ for any fixed natural $n>0$ from ${\mathcal P}^n(\kappa)$-hypermeasurables. The arguments in, for example, J. Cummings paper (on violating SCH at all limits) or the many papers by Gitik and his students (including his very nice Handbook article) should explain how to fill in any missing details. – Andrés E. Caicedo Nov 3 '11 at 19:12
Thanks a lot for the answers and the references. So, by Andres' comment, we know the exact large cardinal assumption (hypermeasurability) leading to a full control of the degree of failure of GCH. Now, do we also know what kind of large cardinals can survive in these models where GCH fails? Moreover, what if we are generous enough to assume more than what is actually needed for global failure of GCH e.g., say a supercompact. Can we then get larger cardinals together with GCH failure? – eiths Nov 3 '11 at 22:06
SCH must hold above a strongly compact cardinal, so there can be no strong compactness in a model of global failure of GCH. – Monroe Eskew Nov 4 '11 at 20:19

It looks like the other answers only deal with upper bounds, so I thought I'd point out that by a result of Gitik,, $\exists \kappa\; o(\kappa) = \kappa^{++}$ is a lower bound for $\neg$SCH and therefore also for the global failure of GCH. (But we still haven't answered the question of the exact consistency strength of the latter. Is there a better lower bound out there?)

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By work of Gitik-Mitchell a $(\kappa+2)$-strong cardinal $\kappa$ is required, and by work of Merimovich a $(\kappa+3)$-strong cardinal $\kappa$ (in fact a cardinal $\kappa$ with $o(\kappa)=\kappa^{++}+\kappa^{+}$) is enough. Gitik and Merimovich have a project to get the total failure of $GCH$ from optimal hypotheses. It's yet incomplete. If I remember it correctly, it says something like this:

Theorem. The following are equiconsistent:

1-For any $\alpha$, there are stationary many cardinals $\kappa$ with $o(\kappa)=\kappa^{++}+\alpha,$

2-GCH fails everywhere,

3-$\forall \lambda, 2^{\lambda}=\lambda^{++}.$

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