A cardinal $\kappa$ is worldly if $V_\kappa$ is a model of ZFC.

1. How many $\beth$-fixed points are there smaller than the smallest worldly cardinal?

2. How many worldly cardinals are there smaller than the smallest worldly cardinal of cofinality $\omega_1$?

• See here. – Andrés E. Caicedo Mar 24 '14 at 14:30
• I like the terminology "worldly cardinals"! – Roland Bacher Mar 24 '14 at 20:40
• I introduced the "worldly cardinal" terminology at our seminars here in New York, since the concept often comes up and it is very useful to be able to refer to it directly. I'm very glad that the name has stuck! – Joel David Hamkins Mar 24 '14 at 20:53
• @JoelDavidHamkins That's something that may be good to add at Cantor's attic: References for where the term is first introduced in the literature. – Andrés E. Caicedo Mar 24 '14 at 21:11
• OK, I'll do that. As for the literature, I'm not sure what the first reference would be, but I did use the terminology in my recent paper on the multiverse perspective on V=L. I'm not sure if there is an earlier use... – Joel David Hamkins Mar 24 '14 at 22:23

Since $V_\kappa\models{\rm ZFC}$, it satisfies that there are ${\rm ORD}$-many $\beth$-fixed points. But each of those must really be a $\beth$-fixed point because every $V_\alpha$ of $V_\kappa$ is a true $V_\alpha$ of $V$. Thus, there are $\kappa$-many $\beth$-fixed points below $\kappa$.
If $\kappa$ has cofinality $\omega_1$, then the worldly cardinals are unbounded in $\kappa$. Suppose $\xi<\kappa$. Let $V_{\alpha_0}\prec_{\Sigma_1} V_\kappa$ such that $\alpha_0>\xi$. Now inductively, choose $V_{\alpha_{n+1}}\prec_{\Sigma_{n+1}}V_\kappa$. Let $V_\alpha$ be the union of the $V_{\alpha_n}$. It follows that $V_\alpha\models{\rm ZFC}$ and since the cofinality of $\kappa$ is $\omega_1$ it follows that $\alpha<\kappa$.
• And one does not even need partial elementarity, or the mention of $\xi$: It suffices that $\alpha_{n+1}>\alpha_n$ and that each $V_{\alpha_{n}}$ satisfies the restriction of $\mathsf{ZFC}$ to $\Sigma_{n}$-replacement. For this, the reflection theorem and some coding suffice, we do not need to formalize satisfiability or any other tool that would be required to produce partially elementary substructures. – Andrés E. Caicedo Mar 24 '14 at 14:57
• Don't you need $\xi$ to formally verify unboundedness? – Victoria Gitman Mar 24 '14 at 16:02