# Can forcing push the continuum above a weakly inacessible cardinal?

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?

-
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

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.