Call Us 07766496223

Theorem

A compact set is countably compact.

Proof

Supposeis compact. Let be a subset ofwith no accumulation points in

Each pointbelongs to some open setwhich contains at most one point ofConsider the family of sets

We have

Henceis an open cover ofSinceis compact, a finite subcoverexists with

Since eachcontains at most one point ofis finite. Therefore every infinite subset ofcontains an accumulation point inso thatis countably compact.