Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 16J / e / Solution

If $x$ is hereditarily countable, its transitive closure contains only countably many ranks. Their supremum is a countable ordinal below ${\omega }_1$, so $x\in V_{{\omega }_1}$.

Solved by gpt-5.6-sol high.