Take and the proper filter on a set . In the first-order language with unary predicates , let both factors be one-element first-order structures. Set true and false in the first factor, and reverse these truth values in the second.
The reduced product also has one element . Neither nor holds there: their truth sets are and , neither belonging to . Thus
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.

Articles by others on the same topic (0)

There are currently no matching articles.