Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 121 1 e Solution Created 2026-10-03 Updated 2026-10-06
It suffices to obtain a model of ZFC without weakly inaccessible cardinals. Starting with any model of ZFC, pass to its constructible universe, which satisfies ZFC and the Generalized continuum hypothesis. If it has no inaccessible cardinal, use that model. Otherwise pass to its rank segment at its least inaccessible cardinal . This segment satisfies ZFC, retains the Generalized continuum hypothesis, and has no inaccessible cardinals. Under the Generalized continuum hypothesis, every weakly inaccessible cardinal is strongly inaccessible: if and is a limit cardinal, then . Thus in either case the resulting model has no weakly inaccessible cardinals.
Work inside . If is a nonzero limit ordinal, let . It is an uncountable limit cardinal. If it were regular, it would be a weakly inaccessible cardinal, which is impossible in . It is therefore singular, and the singular cardinal enumeration is continuous at this index:The cofinality of an increasing ordinal supremum now givesfor every nonzero limit ordinal in . Hence satisfies the negation of the proposed existential assertion. By the soundness theorem for first-order logic, consistency of ZFC prevents ZFC from proving that assertion. The model construction is a relative-consistency argument; it does not assume that consistency alone supplies a countable transitive model. Here, as usual, a limit ordinal excludes zero.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 121 2 a Solution Created 2026-10-03 Updated 2026-10-06
The Freiling axiom of symmetry isThe values of are countable subsets of the real numbers, including finite ones. One may require without changing the axiom: apply the displayed version to .
The Generalized continuum hypothesis statesHere is the successor cardinal; equivalently for every ordinal .
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 121 2 iv Solution Created 2026-10-03 Updated 2026-10-05
We give a relative-consistency argument using a finite-function collapse to countable size. The assumed consistency implies that ZFC is consistent. Passing to the constructible universe gives consistency of , which also satisfies the Generalized continuum hypothesis. Work in this ground theory, let , and force with finite partial functions , ordered by reverse inclusion.
The union of a generic filter is a total surjection : prescribing a new domain coordinate is dense, and putting any specified into the range is dense. Hence becomes countable. Set forcing preserves ordinals, and absoluteness of constructible levels implies that the extension has the same constructible universe as the ground model. Its is therefore still the old , so it satisfies the stated countability of constructible omega-one condition.
We must also verify GCH after the collapse. The order has ground-model size , hence the -chain condition. By cardinal preservation by chain-condition forcing, all old cardinals at least survive. Every old ordinal below has size at most and becomes countable, so the old is exactly the new .
A name for a subset of a fixed ground-model ordinal can be chosen as a set of pairs with and a condition: use an antichain deciding membership at each coordinate. Thus the number of such names in the ground model is at most . For subsets of this is by ground-model GCH. In the extension it gives , and Cantor theorem gives the reverse lower bound. For every old cardinal the bound isin the ground model. Both and remain cardinals, so again the upper bound and Cantor theorem give the extension's equality . These are all its uncountable cardinals. This proves GCH preservation by a finite-function collapse in the required case.
The forcing relative-consistency theorem now yieldsThe construction in fact only needs consistency of ZFC. This argument uses the formal inner-model and forcing consistency theorems; it does not infer the existence of a countable transitive model merely from consistency.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 116 2 f Solution 2026-09-28
Fix and write , , and . Every ordinal below is represented in the ultrapower by a function . Consequently, in ,where the last equality uses the Generalized continuum hypothesis.
By elementarity, regards as measurable and hence as a strong limit cardinal. Moreover , so and have the same subsets of and the same . It follows inside thatThus, in the ambient , the ordinal is strictly larger than but has cardinality at most . It cannot be a cardinal number. Applying this argument to both and proves that neither nor is a cardinal in , exactly as in moved critical point is not an ambient cardinal under GCH.
Singular cardinals hypothesis 2026-10-06
The singular cardinals hypothesis asserts that every infinite singular cardinal satisfiesEquivalently, for every infinite singular cardinal. Here is the gimel function. The Generalized continuum hypothesis implies this principle, but the principle only constrains the indicated singular-cardinal exponentiation.
Weakly inaccessible cardinal 2026-10-06
A weakly inaccessible cardinal is an uncountable regular cardinal that is a limit cardinal. It need not be a strong limit cardinal. Under the Generalized continuum hypothesis, it is a strongly inaccessible cardinal, since every smaller infinite satisfies .