Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 128 / 4 / c

Assume $\phi (\vec{a})$ is a Delta-one formula in set theory. Thus ZF proves it equivalent to a ${\Sigma }_1$ formula $\sigma$ and to a ${\Pi }_1$ formula $\pi$. Only finitely many axioms of ZF occur in these two formal proofs; collect them, together with the finite fragment needed for bounded-formula absoluteness, into $T$.

Let $M$ be a transitive class containing $\vec{a}$ and satisfying $T$. If $M\models \phi$, then $M\models \sigma$, and upward absoluteness of ${\Sigma }_1$ formulas gives $V\models \sigma$, hence $V\models \phi$. If $V\models \phi$, then $V\models \pi$, and downward absoluteness of ${\Pi }_1$ formulas gives $M\models \pi$, hence $M\models \phi$. Therefore ZF proves that $\phi$ is absolute for every such $M$.

Conversely, suppose a finite $T\subseteq \mathrm{ZF}$ has the stated absoluteness property. LetBecause $T$ is finite, every satisfaction assertion here can be replaced by the corresponding formula relativization to a class. All quantifiers in the matrix are bounded by $N$, so $\sigma$ is ${\Sigma }_1$. Define the ${\Pi }_1$ formula

By the Lévy reflection theorem, ZF proves that for any parameters $\vec{a}$ there is a level $V_{\alpha }$ containing them and satisfying the finite fragment $T$. The assumed absoluteness says that every such transitive set agrees with $V$ about $\phi$. Consequently ZF provesThus $\phi$ is both ${\Sigma }_1^{\mathrm{ZF}}$ and ${\Pi }_1^{\mathrm{ZF}}$, so it is ${\Delta }_1^{\mathrm{ZF}}$.