Call Us 07766496223

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