## Intersection of filters

Let and be filters on a same set . Then is also a filter on . Proof: belongs to both and , hence it belongs to , and . Since is not an element of , it is not an element of . Let be an element of and a superset of . Then and…

A category is by definition concrete if there exists a faithful functor from to the category of sets . (A functor is faithful if it is one-to-one (injective) regarding morphisms.) It happens to be the case that in the category , monomorphisms and one-to-one mappings are the same thing; similarly for epimorphisms and surjections. We…