Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 23 1 a Solution Created 2026-10-03 Updated 2026-10-06
Let be the reduced product by a proper filter on a set. Its underlying equivalence relation is when . Operations are interpreted coordinatewise, and a relation holds of the classes exactly when its coordinate truth set belongs to .
Evaluation of a first-order term commutes with passage to the quotient, by mathematical induction on terms. Consequently the desired equivalence holds for every atomic formula, including logical equality. For a formula and representatives , write .
For logical conjunction, . The filter on a set axioms giveThus the induction hypothesis transfers a conjunction in both directions.
For existential quantification, first suppose . Choose a representative of a witness. Induction gives , and this set is contained in . Upward closure therefore gives .
Conversely, suppose . For each , choose a coordinate witness , and choose an arbitrary element of outside . These simultaneous choices use the axiom of choice, as does the usual product construction. Then , so that truth set belongs to . Induction gives , providing the required witness. Thereforefor every primitive positive formula. The exam's tame formulas are exactly this fragment, built using logical conjunction and existential quantification. No ultrafilter dichotomy was used.
Reduced product 2026-10-06
Given nonempty first-order structures and a proper filter on a set , a reduced product identifies product functions when . Operations are interpreted coordinatewise, and relations hold when their coordinate truth sets belong to . For an ultrafilter this is an ultraproduct. Primitive positive formulas transfer through an arbitrary proper filter, whereas logical disjunction and logical negation can fail to transfer.