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

True. If $\mathrm{TC}(\{x\})$ is countable, then each of its members is a subset of this transitive set and is therefore countable.

Conversely, if $x$ is reasonable, then $x$ itself and every set occurring at every finite membership depth below it are countable. Starting with the countable set $\{x\}$, each next level is a countable union of countable sets, hence countable. Their countable union is $\mathrm{TC}(\{x\})$, so the transitive closure is countable.

True. By part (i), the reasonable sets are exactly the hereditarily countable sets $H_{{\omega }_1}$. The rank of a countable transitive closure is the supremum of countably many countable ordinals, plus at most a finite amount, and is therefore below ${\omega }_1$. Hence every reasonable set belongs to

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.