Past exam of the mathematics course of the University of Cambridge 2009 ii Paper 1 16G Solution Created 2026-09-24 Updated 2026-10-14
Call a set function with ordinal domain good if for every . Any two good functions agree on their common domain: otherwise their first disagreement has identical earlier ranges and hence identical values under . This is transfinite induction, not an assumption of the desired recursion.
Good functions exist with every ordinal domain. At a successor stage extend the unique previous function by its prescribed value under . At a limit stage, Axiom schema of replacement collects the unique good functions on all smaller domains into a set, and their compatible union is a good function on the limit domain. Each earlier range is a set, so is applied only to sets. Transfinite induction now establishes existence at all stages. Define to be the value at of the unique good function on . This is a definable class function because goodness and the uniqueness assertion are formulas in the defining formula for . Compatibility gives
An initial ordinal is an ordinal not bijective with any smaller ordinal. It is the canonical ordinal representative of a well-orderable cardinal, so cardinal arithmetic can be defined by taking the initial ordinal equinumerous with a disjoint union, Cartesian product or function set. In the usual setting with the axiom of choice, every cardinal has such a representative. Hartogs theorem says that for every set there is an ordinal which does not inject into , and there is a least such ordinal. For an initial ordinal it is the next larger initial ordinal: every smaller ordinal injects into , and if were equinumerous with a smaller ordinal it too would inject into .
Use the recursion just proved to set , and for nonzero limit . These instructions use the earlier value set alone: at a successor it has a maximum, and at a limit its supremum is taken. They are therefore implemented by an appropriate definable . The limit supremum is initial: if it were bijective with below it, choose an earlier ; then would inject into , contradicting initiality. The sequence is strictly increasing and continuous. There are no omitted infinite initial ordinals, since each successor chooses the least next one and each limit is the supremum; more formally, a least omitted initial ordinal cannot lie between successive terms, and the least term above it cannot be a limit term. Terms eventually exceed any prescribed ordinal, since a strictly increasing ordinal sequence satisfies .
There is a fixed point of a normal ordinal function. Explicitly let , and . The sequence is strictly increasing, by induction from and strict increase of the enumeration. It is cofinal in the limit , so continuity gives