Club set 2026-10-05
For a limit ordinal , a subset is club if it is unbounded and contains each limit below of an increasing sequence from . At an uncountable regular cardinal , intersections of fewer than club sets are club: above a starting point cycle through the sets and take suprema, using regularity to stay below ; closure puts the resulting limit in all of them. Club sets provide the reflecting ranks in worldly cardinals below an inaccessible cardinal.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 121 1 vi Solution Created 2026-10-03 Updated 2026-10-05
Use the standard convention that a strongly inaccessible cardinal is uncountable, regular and strong limit. The printed explanatory definition omits uncountability: literally it also admits , for which no finite cardinal is a worldly cardinal. Thus that omission makes the requested conclusion false under the literal abbreviated definition. The proof below applies to the usual, intended strongly inaccessible cardinal convention .
First for every . At successor stages this follows from the strong limit cardinal property, and at limits from the regular cardinal property. Consequently . In particular, for Axiom schema of replacement, a domain has size less than , so its functional image has fewer than elements, so the supremum of the rank of a set over its elements is below by regular cardinal structure. All other axioms, including axiom of choice, have their witnesses at bounded ranks below ; uncountability supplies axiom of infinity.
Enumerate the first-order formulas as . For each finite collection, reflection inside the set structure gives a closed unbounded subset of of agreeing ranks. To see why the witness bounds stay below , there are fewer than parameter tuples at each , and the supremum of their least witness ranks remains below by regular cardinal structure. Closing under these bounds and taking increasing countable limits gives the usual reflection theorem for definable hierarchies argument. The countable intersection of these closed unbounded subsets is still closed unbounded because is regular and uncountable. At each resulting ,The infinite cardinal numbers below also form a closed unbounded subset: they are unbounded because for infinite , and a supremum of increasing cardinal numbers is a cardinal number. Intersect the two closed unbounded subsets. Every member of the intersection is a worldly cardinal, and an unbounded subset of a regular cardinal has size . HenceIn fact this proves the stronger worldly cardinals below an inaccessible cardinal result that these worldly cardinals contain a closed unbounded subset of .