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There is a famous quote of Paul Cohen which reads $\lt\lt$ A point of view which the author feels may eventually come to be accepted is that the continuum hypothesis is obviously false ... The continuum $\mathfrak c$ is greater than $\aleph_n,\aleph_\omega,\aleph_{\alpha}$ where $\alpha=\aleph_\omega$ etc. This point of view regards $\mathfrak c$ as an incredibly rich set given to us by one bold new axiom (...) $\gt\gt$

Paul Cohen's opinion implies that there is a weakly inacessible cardinal below $\mathfrak c$. Let us denote WIBC (weakly inaccessible below continuum) this hypothesis (does that hypothesis already have a name in the literature?). Since the existence of a weakly inacessible cardinal cannot be shown to be consistent with $ZFC$, we will never be able prove that WIBC is consistent with ZFC (unless of course ZFC is inconsistent).

But if we assume that a weakly inacessible cardinal exists, can one use a variation of Easton's forcing method to show that WIBC is consistent with ZFC?

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Would you care to explain what's "stupid" about this question? :-) – joriki Oct 20 '11 at 16:29
I believe some answer lies within real-measurable cardinals which are not measurable themselves. The consistency strength, however, is more than just weakly inaccessible. – Asaf Karagila Oct 20 '11 at 16:47
@ joriki : it's because I know very little about forcing. If it's not stupid, it might be. – Ewan Delanoy Oct 20 '11 at 16:56
@Ewan: It is not a stupid question, also you push to expand things, so you push the continuum above the weakly inaccessible. It makes more sense. – Asaf Karagila Oct 20 '11 at 18:53
@Asaf : Glad to hear it's not a stupid question! – Ewan Delanoy Oct 20 '11 at 20:25
up vote 8 down vote accepted

You don't need all of the Easton machinery to do this, as simple Cohen forcing will do the job. For example, one might start with an inaccessible cardinal $\kappa$ in $L$, and forcing with the finite partial functions from $\kappa$ to $\{0,1\}$ (i.e., "adding $\kappa$ Cohen reals") will result in a model where $2^{\aleph_0}=\kappa$. This forcing is mild, in that it preserves cofinalities, so $\kappa$ will still be a regular limit cardinal in the extension.

Similar arguments (assuming the existence of the appropriate cardinals in $L$) that the continuum can be weakly Mahlo, etc., so one can have lots of weakly inaccessible cardinals below $2^{\aleph_0}$ given only mild large cardinal assumptions.

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Welcome to math.SE! – Asaf Karagila Oct 27 '11 at 19:10

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