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In short, what can we say about the consistency strength of "$\kappa$ is a singular worldly and inaccessible in an inner model"?

Clearly, $0^\#$ exists since we have a singular cardinal which is regular in an inner model. And clearly, this is not more than a measurable cardinal, since Prikry forcing gives us exactly this scenario.

One might be tempted to think that this will be outright just $0^\#$, but assuming that $0^\#$ exists, $\aleph_\omega$ is inaccessible in $L$, but in $V$ it is not a worldly cardinal. Indeed, work in $L[0^\#]$ below the least worldly cardinal (if it exists), then any uncountable cardinal is inaccessible in $V$, but none is worldly.

The above, of course, does not preclude any intermediate model between $L$ and $L[0^\#]$, but it does show that the argument isn't going to be straightforward, I don't think.

Can we extract higher sharps from the assumption?

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  • $\begingroup$ What do you mean by "higher sharp"? If $0^\sharp$ exists and there exists a singular cardinal $\eta$ such that $V_\eta\models$ ZF and $\eta$ is inaccessible in $L$, then the same hypothesis is true in $L[0^\sharp]$. (In fact, if $0^\sharp$ exists and $V_\eta\models$ ZF, then $\eta$ is inaccessible in $L$, and if $\eta$ is least such that $V_\eta\models$ ZF, then $\eta$ has cofinality $\omega$.) $\endgroup$
    – Farmer S
    Commented Aug 24 at 10:40
  • $\begingroup$ I am confused by your comment. If $\eta$ is the least worldly cardinal in $L[0^\#]$, working in $V_\eta^{L[0^\#]}$, there is no $\alpha$ for which $V_\alpha\models\sf ZF$, but $0^\#$ still exists. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 24 at 10:44
  • $\begingroup$ I'm not working in $V_\eta^{L[0^\sharp]}$. $\endgroup$
    – Farmer S
    Commented Aug 24 at 10:48
  • $\begingroup$ Even more confused now... $V_\eta^{L[0^\#]}\models V=L[0^\#]$... Right? And so it is a model of $\sf ZFC$ in which there are no worldly cardinals. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 24 at 11:05
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    $\begingroup$ So, "$0^\sharp$ exists, and there is a wordly cardinal". $\endgroup$
    – Andrés E. Caicedo
    Commented Aug 24 at 13:13

1 Answer 1

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To summarize the comments, the hypothesis is equiconsistent with (ZFC +) $0^\sharp$ exists + "there is a worldly cardinal". For as mentioned in the original post, under the hypothesis, $0^\sharp$ exists since $\kappa$ is a singular cardinal which is regular in $L$ (and by Jensen's covering lemma). Conversely, assume ZFC + $0^\sharp$ exists and there is a worldly cardinal. Let $\eta$ be the least worldly cardinal. Then $\mathrm{cof}(\eta)=\omega$ (for letting $\eta_n$ be least such that $V_{\eta_n}\preccurlyeq_n V_\eta$, then $\eta_n<\eta_{n+1}<\eta$, and letting $\eta'=\sup_{n<\omega}\eta_n$, then $V_{\eta'}\preccurlyeq V_\eta$, so $V_{\eta'}\models$ ZFC, so $\eta'=\eta$). And since $0^\sharp$ exists and $0^\sharp\in V_\eta$, $V_\eta\models$"$0^\sharp$ exists" and $(0^\sharp)^{V_\eta}=0^\sharp$, so note that $\eta$ is an $L$-indiscernible, so $\eta$ is inaccessible in $L$.

And the hypothesis is also consistent with "$V=L[0^\sharp]$", since the hypothesis is downward absolute to $L[0^\sharp]$. So it does not prove that $(0^\sharp)^\sharp$ exists.

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  • $\begingroup$ Does ZFC+$0^\sharp$ exists prove the consistency of "there is a worldly cardinal"? $\endgroup$
    – Lucenaposition
    Commented Sep 2 at 3:22
  • $\begingroup$ Yes: if $\kappa$ is a Silver indiscernible then in $L$, $\kappa$ is inaccessible, weakly compact, and much more. $\endgroup$
    – Farmer S
    Commented Sep 2 at 11:28

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