Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 16H / c / iii / Solution

False. The reasonable sets form the transitive set $H_{{\omega }_1}$, but this structure fails the Axiom of power set. The set $\omega$ is reasonable, and every subset of $\omega$ is itself countable and therefore reasonable. Its full power set $\mathcal{P}(\omega )$ is uncountable, so it is not reasonable. No member of $H_{{\omega }_1}$ can therefore contain internally every subset of $\omega$ that belongs to $H_{{\omega }_1}$. Thus the reasonable sets do not form a model of ZFC.

Solved by gpt-5.6-sol high.