A worldly cardinal permits a countable transitive model of ZFC and a transitive model of ZFC with , , and . First collapse a countable elementary substructure of . Then collapse an -sized Skolem hull containing every countable ordinal and every member of , as well as itself. The transitive collapse fixes transitive subsets lemma preserves and those ordinals.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 121 1 v a Solution Created 2026-10-03 Updated 2026-10-05
We construct the two transitive models together, taking care that the later collapse really retains the earlier model. First, a worldly cardinal is greater than : ZFC in contains the full power set , and the axiom of choice there gives a bijection between it and an ordinal . That power set is externally uncountable by Cantor theorem, so . Infinity rules out the finite cardinals and before this argument is applied.
By the Downward Lowenheim-Skolem theorem, choose a countable elementary substructure . Its inherited membership is a well-founded relation, and elementarity transfers extensionality from . Apply the Mostowski collapse theorem to obtain a countable transitive model of ZFC. Because is a countable transitive set, it is a hereditarily countable set and its rank of a set is below . In particular, .
Now take the Skolem hull in of the seedCall it . The seed has external cardinality and lies inside . The hull is an elementary substructure of with : closure under countably many finitary Skolem functions gives the upper bound, and gives the lower bound. Let be its Mostowski collapse theorem map. Then is a transitive model of ZFC.
The transitive collapse fixes transitive subsets lemma gives for every , since is transitive. Since also , the collapse sends that element to itself. Thus , and transitivity of yields the required inclusion:This is the construction of nested transitive models from a worldly cardinal; the next two parts verify its size and ordinal requirements.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 121 1 v c Solution Created 2026-10-03 Updated 2026-10-05
The collapse is a bijection, so . Every and every element of belong to . The transitive collapse fixes transitive subsets lemma therefore gives , so every ambient countable ordinal belongs to .
Moreover, correctly regards each such as countable. An injection has rank below , so a witness belongs to . Elementarity gives “ is a countable ordinal”. Transporting this statement across the collapse, which fixes , gives the same statement in .
The collapse also fixes the element . No injection exists externally, hence none belongs to ; together with the countability witnesses below it, this shows that correctly identifies as its first uncountable ordinal. Elementarity and the collapse preserve that identification. Thus the construction gives