There is a famous quote of Paul Cohen which reads “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 (...)”

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?

  • $\begingroup$ Would you care to explain what's "stupid" about this question? :-) $\endgroup$
    – joriki
    Oct 20, 2011 at 16:29
  • $\begingroup$ I believe some answer lies within real-measurable cardinals which are not measurable themselves. The consistency strength, however, is more than just weakly inaccessible. $\endgroup$
    – Asaf Karagila
    Oct 20, 2011 at 16:47
  • $\begingroup$ @ joriki : it's because I know very little about forcing. If it's not stupid, it might be. $\endgroup$ Oct 20, 2011 at 16:56
  • $\begingroup$ @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. $\endgroup$
    – Asaf Karagila
    Oct 20, 2011 at 18:53
  • $\begingroup$ @Asaf : Glad to hear it's not a stupid question! $\endgroup$ Oct 20, 2011 at 20:25

1 Answer 1


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.

  • $\begingroup$ Welcome to math.SE! $\endgroup$
    – Asaf Karagila
    Oct 27, 2011 at 19:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.