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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 $\omega_1$?

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  • $\begingroup$ See here. $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 24, 2014 at 14:30
  • $\begingroup$ I like the terminology "worldly cardinals"! $\endgroup$
    – Roland Bacher
    Commented Mar 24, 2014 at 20:40
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    $\begingroup$ 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! $\endgroup$
    – Joel David Hamkins
    Commented Mar 24, 2014 at 20:53
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    $\begingroup$ @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. $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 24, 2014 at 21:11
  • $\begingroup$ 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... $\endgroup$
    – Joel David Hamkins
    Commented Mar 24, 2014 at 22:23

1 Answer 1

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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 $\alpha<\kappa$.

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  • $\begingroup$ very GOOD answer。 $\endgroup$
    – huhao
    Commented Mar 24, 2014 at 14:34
  • $\begingroup$ 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. $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 24, 2014 at 14:57
  • $\begingroup$ (Though in truth most of this already goes into establishing the relevant instances of reflection.) $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 24, 2014 at 14:58
  • $\begingroup$ Don't you need $\xi$ to formally verify unboundedness? $\endgroup$
    – Victoria Gitman
    Commented Mar 24, 2014 at 16:02
  • $\begingroup$ @VictoriaGitman Ah, yes, unboundedness, I somehow forgot and was shooting for just one. $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 24, 2014 at 17:35

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