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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-23/1/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 23 / 1 / b / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-06

Take

I = {0, 1}

and the proper filter on a set

F = {I}

. In the first-order language with unary predicates

P, Q

, let both factors be one-element first-order structures. Set

P

true and

Q

false in the first factor, and reverse these truth values in the second.

The reduced product also has one element

a

. Neither

P (a)

nor

Q (a)

holds there: their truth sets are

{0}

and

{1}

, neither belonging to

F

. Thus

R \neq ⊨ P (a) \lor Q (a), {i : M_{i} ⊨ P (a_{i}) \lor Q (a_{i})} = I \in F .

Logical disjunction can therefore break the equivalence. A union can belong to a filter on a set without either summand belonging to it; the corresponding union property does hold for an ultrafilter.

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