• Q0. Does anyone know of a refutation, in ZFC, of the following statement?

  • Q1. If not, does ZFC plus large cardinals prove its consistency?

Statement. There exists a transitive model $M$ of ZFC having height strictly less than $\beth_1$ (equivalently, strictly less than $|\mathbb{R}|$) such that for all $\kappa \in M$, we have that if $M$ believes that $\kappa$ is a cardinal number, then $\kappa$ really is a cardinal number (in the ambient universe).

Motivation. It may be possible to formulate one or more "anti-continuum hypotheses" (ACH) based on the existence of such models. I'd be interested in learning whether any of the usual cardinal characteristics of the continuum can be proved distinct (or equal, as the case may be) using ACH axioms.


Start with two inaccessible cardinals and $V=L$. Then add Cohen reals so the continuum equals to the larger of the two, and consider $L_\kappa$ where $\kappa$ is the smaller inaccessible.

Of course you don't need inaccessible cardinals for that, a Worldly cardinal of uncountable cofinality will suffice, since it is the limit of smaller worldly cardinals.

On the other hand, if $M$ is such model, of height $\kappa$ then $\kappa$ is a cardinal in $V$ and therefore in $L$, it is not hard to show that it has to be a $\beth$ fixed point in $L$, and therefore it has to be worldly in $L$.

  • $\begingroup$ I do not understand why you have your first paragraph (it is distracting. Why two? etc). I think the answer is clearer if you just start with the claim that the statement is equiconsistent with the existence of a wordly cardinal of uncountable cofinality (and then mention adding the Cohen reals, etc). $\endgroup$ – Andrés E. Caicedo Nov 29 '14 at 23:53
  • $\begingroup$ @AndresCaicedo, I think that is my fault; I changed the question somewhat after posting it. Asaf answered the original question. $\endgroup$ – goblin Nov 29 '14 at 23:55
  • $\begingroup$ Ah, OK. Carry on, then. $\endgroup$ – Andrés E. Caicedo Nov 29 '14 at 23:56
  • $\begingroup$ @Andres: And he wrote the answer from his phone with one eye already closed. So there wasn't much to do after posting, but shut the other one too! $\endgroup$ – Asaf Karagila Nov 30 '14 at 4:40
  • $\begingroup$ @goblin: What do you mean? $\endgroup$ – Asaf Karagila Nov 30 '14 at 10:01

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