# Upwards reflection

Classical reflection principles have the general form $\phi \rightarrow \exists \alpha \ V_\alpha \vDash \phi$; their strength, as is well known, depends on what language one allows for $\phi$. The first-order case is provable, the second is Bernays-Lévy reflection, and it has a minimal model in $V_{\delta^2_0}$, where $\delta^2_0$ is the least second-order indescribable (or second-order indescribable-without-parameters, if $\phi$ is restricted to sentences) ordinal. Third-order parameters lead to inconsistency.

But can we find a reasonable upward reflection principle to complement this downward one? A first thought is $\forall \alpha \ (V_\alpha \vDash \phi \rightarrow \exists \beta > \alpha \ V_\beta \vDash \phi)$, but this has obvious counterexamples (consider, e.g., $\alpha$ finite). So what about

(URP) $\exists \lambda \ \forall \alpha > \lambda \ (V_\alpha \vDash \phi \rightarrow \exists \beta > \alpha \ V_\beta \vDash \phi)$?

This is modelled on the idea behind an extendible cardinal, but it will obviously be much weaker, since we aren't requiring an elementary embedding.

How strong is (URP) for various choices for the language of $\phi$?

What parameters, if any, are allowed in $\phi$?
Case 1: $\phi$ is intended to be a sentence. Then (URP) is true: For each sentence $\phi$, if there exists an upper bound for the ordinals $\alpha$ such that $V_\alpha\models\phi$, then let $\beta_\phi$ be such a bound. (For $\phi$ such that no such bound exists, leave $\beta_\phi$ undefined.) Let $\lambda$ be any ordinal greater than all the $\beta_\phi$'s that are defined.
Case 2: $\phi$ can have arbitrary elements of $V_\alpha$ as parameters. Then (URP) is false. Given any $\lambda$, take $\alpha$ to be a successor ordinal $>\lambda$, say $\alpha=\beta+1$. Take as a parameter some set $p$ of rank $\beta$, so $p$ is an element of $V_\alpha$ but not an element of an element of $V_\alpha$. Let $\phi$ be the formula $(\forall x)\,p\notin x$.