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

The Lévy reflection theorem states that for every finite collection $\Phi$ of formulas and every ordinal $\gamma$, there is an ordinal $\alpha >\gamma$ such that, for every $\phi \in \Phi$ and all parameters $\vec{a}\in V_{\alpha }$,Indeed, the ordinals $\alpha$ reflecting all formulas in $\Phi$ form a closed unbounded class.

Let $\alpha =({\omega }_1{)}^M$ and takeThis formula is upward absolute between transitive models: if the smaller model contains such an $f$, the assumed absoluteness of “function”, domain, range, and $\omega$ shows that the same witness works in the larger model.

The generic union $g=\bigcup G$ is a total map $\omega \to \alpha$, because the conditions deciding each input form a dense set. For every $\beta <\alpha$, the conditions putting $\beta$ somewhere in the range are also dense, so $g$ is surjective. Thus $M[G]\models {\phi }_1(\alpha )$. But $M/\models {\phi }_1(\alpha )$ because $M$ regards $\alpha$ as its first uncountable ordinal. Hence ${\phi }_1$ is not downward absolute between $M$ and $M[G]$.

With the same parameter $\alpha =({\omega }_1{)}^M$, letThis formula is downward absolute: if the larger transitive model has no such function, then neither can the smaller model, since any witness in the smaller model would remain a witness in the larger one. The model $M$ satisfies ${\phi }_2(\alpha )$, while the generic surjection $\bigcup G$ makes it false in $M[G]$. Therefore it is not upward absolute.

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 ${\Delta }_1^{\mathrm{ZF}}$.