Theorem
Letbe a metric. Ifandare closed subsets ofwith then open setsandexist such thatand
Proof
For eachdefineand for each
Defineand
Both setsandare open because each is the union of a family of open sets. Since includes eachSimilarly B subset B_1 .
Supposethen for someand for some
Supposethen
withbut
This is a contradiction hence