# Large cardinals and reflection properties

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 $n > 3$?

• 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. Jan 3, 2018 at 9:35
• 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). Oct 15, 2018 at 18:25

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 $$U=\{\lambda<\kappa|\lambda\text{ is reflecting}\}$$.
• 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 $L_\kappa\prec L$.) Jun 3, 2019 at 4:09