Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 121 1 i a Solution 2026-10-03
U+D. By absoluteness of infinitude between transitive models, two transitive models of ZFC agree on the natural numbers and therefore on whether a shared set is a finite set or an infinite set.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 121 1 i b Solution 2026-10-03
D. Downward absoluteness of cardinalhood holds because, if the larger transitive model sees no bijection with a smaller ordinal, neither can the smaller model, whose functions form only a subset of those in the larger model. Cardinalhood is not described by an upward absolute formula, because the larger model can contain a new bijection collapsing an ordinal that the smaller model regards as a cardinal number.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 121 1 i c Solution 2026-10-03
U+D. The assertion that a given relation is a partial order quantifies only over its underlying set and checks that it is a reflexive relation, an antisymmetric relation and a transitive relation. It is therefore a bounded formula in set theory and is absolute between transitive models.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 121 1 i d Solution 2026-10-03
U+D. The natural numbers are absolute between transitive models of ZFC, and the statement is the bounded assertion . Hence the property of being a subset of is absolute.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 121 1 iii Solution 2026-10-03
Let be the given transitive model and suppose that its ordinal height of a model of set theory were a countable set. For every , the internal Axiom of choice gives a bijection from to an ordinal of ; transitivity makes this an actual bijection, and the ordinal is externally countable. Thus every element of is externally countable.
For each , the internal rank belongs to and is therefore countable. Every lies in one of these ranks, so is a countable union of countable sets and is itself countable, contrary to the hypothesis. By uncountable transitive set model has uncountable ordinal height, contains uncountably many ordinals.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 121 2 ii Solution 2026-10-03
Fix . Inside the ambient transitive model , the Axiom of power set makes the collection of constructible subsets of a set. For each such subset , choose the least stage of the constructible hierarchy at which appears. The Axiom schema of replacement and the supremum of a set of ordinals give an ordinal bounding all these stages; enlarge so that .
Nowis definable over with parameter . It therefore belongs to the definable power set . This set contains exactly the subsets of that belong to the constructible universe, so it witnesses the Axiom of power set in . Therefore Power Set.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 121 2 iv Solution 2026-10-03
Take a transitive model . In , choose a bijection and encode its graph by a set , using a fixed bijection between and . The relative constructible universe can decode , and therefore contains every real number of ; being an inner model of , it has no additional reals.
Models with the same reals have the same first uncountable ordinal, because their reals code exactly the same countable well-orders. If satisfied the Continuum hypothesis, its bijection between and the reals would also belong to , contradicting . This is the construction in relative constructible universe can violate the continuum hypothesis, and it gives
Assume a transitive model satisfies . In , encode a bijection by one set , using a fixed pairing of with . Then decodes and contains every real of . The two models consequently have the same , and any bijection between and the real numbers in would also be one in . Thus .