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Strong and supercompact cardinals are $\varSigma_2$-reflecting; extendible cardinals are $\varSigma_3$-reflecting. It is of course possible to build larger degrees of correctness into the definitions of large cardinals; if I understand correctly, this is essentially what Joan Bagaria does in his paper "$C^{(n)}$-Cardinals" (Arch. Math. Logic 51 (2012): 213–40; doi: 10.1007/s00153-011-0261-8).

But is there any 'standard' large cardinal notion—basically, one not specifically formulated using the notion of $\varSigma_n$-correctness or something equivalent—such that cardinals of that type are $\varSigma_n$-reflecting for $n > 3$?

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    $\begingroup$ I see, belatedly, that this issue has been discussed at length elsewhere: mathoverflow.net/questions/71524/how-elementary-can-we-go?rq=1 I'm not sure whether I should delete this as duplicative, but I'll leave it up in case the link is useful to anyone. $\endgroup$
    – Beau Madison Mount
    Jan 3, 2018 at 9:35
  • $\begingroup$ Generally, no. Most large cardinal properties are $\Sigma_2$-definable, if not $\Sigma_3$-definable. If a large cardinal property implied being $\Sigma_3$-reflecting then it wouldn't be $\Sigma_3$-definable. Note that $\Sigma_3$ is a class containing all statements of the form "There is a __" for __ a $\Pi_2$-formula, and most cardinal properties of the form "for each ordinal $\alpha$, $\kappa$ is $\alpha$-____" are $\Pi_2$ of themselves (for example supercompact cardinals). $\endgroup$
    – Keith Millar
    Oct 15, 2018 at 18:25

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There is actually a type of cardinal that satisfies this, called the "stationarily superhuge" cardinals. A cardinal $\kappa$ is called stationarily superhuge if the $\{\lambda|\kappa\text{ is huge with target }\lambda\}$ is stationary. If $\kappa$ is stationarily superhuge, $V_\kappa\prec V$, and moreover $L_\kappa\prec L$. The reason for this is that $\{\lambda|V_\lambda\prec V\}$ is club, assuming the existence of a stationarily superhuge cardinal. Then $V_\kappa\vDash\phi$ if and only if $M\vDash(V_\lambda\vDash\phi)$ if and only if $V_\lambda\vDash\phi$. Setting $\lambda$ is correct, we have $V_\lambda$ reflects $\phi$.

A similar argument works for $L_\kappa$, and for $H_\kappa$ (Or more simply, as $\kappa$ is superhuge and so inaccessible, $H_\kappa=V_\kappa$). Moreover, the set of such cardinals form a normal measure beneath $\kappa$. Let $D=\{X\subseteq\kappa|\kappa\in j(X)\}$. Then $M\vDash(j(\kappa)\text{ is reflecting})$, and so $U\in D$, where $U=\{\lambda<\kappa|\lambda\text{ is reflecting}\}$.

Reference:

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    $\begingroup$ I don't understand your statement that $\{\lambda\text{ is inaccessible}|V_\lambda\prec V\}$ is club. The $\omega$-th member of a club has cofinality $\omega$ and is therefore not inaccessible. (I also don't understand "and moreover" in the first paragraph. If $\kappa$ is inaccessible and $V_\kappa\prec V$ then automatically $L_\kappa\prec L$.) $\endgroup$
    – Andreas Blass
    Jun 3, 2019 at 4:09
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    $\begingroup$ You are right. I reworked the argument to adjust that. I added the "and moreover" to simply put emphasis on it, because it was relevant for the second part. I realize now it is redundant. $\endgroup$
    – Master
    Jun 3, 2019 at 4:18

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