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Sep 23, 2021 at 20:47 comment added Arvid Samuelsson Actually totally otherworldly cardinals are even $\Sigma_3$-correct. Suppose $\psi = \exists x \forall y \phi(x, y)$ is a $\Sigma_3$ formula (where $\phi$ is $\Sigma_1$). If $\kappa$ is totally otherworldly, there is $\lambda$ such that $V_\kappa \prec V_\lambda$ and there is a witness $x \in V_\lambda$ of $\psi$. Then $\phi(x, y)$ is absolute between $V_\kappa$, $V_\lambda$, and $V$, as noted in this blog post, so $\forall y \phi(x, y)$ is downward absolute and $\psi$ is absolute between $V_\kappa$ and $V_\lambda$ by elementarity.
Sep 11, 2020 at 18:24 history answered Jason Zesheng Chen CC BY-SA 4.0