 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-3/16i/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

True. Regard each ordinal

α

as its von Neumann representative, the set of all smaller ordinals. Transfinite induction gives

rank (α) = sup_{β < α} (rank (β) + 1) = sup_{β < α} (β + 1) = α .

Thus the set $α$ itself has rank $α$ .

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