Cute Proofs: Continuous Image of Compact Set is Compact.
October 3, 2010
While not an especially difficult proof, this kind of this is standard when we’re using compact sets. What we’re going to do is take an open cover, pull it back, use compactness, then push it forward. Let’s get down to it!
Theorem: If is continuous and is a compact space, then is also compact.
It suffices to show that for every open cover of there exists a finite subcover. Let’s take is an arbitrary cover of . Note that and each is open since is continuous. Since is compact, we have that (with potentially reordering or renaming the indices) is a finite cover of our set . Therefore, we have that
(with equality holding because is surjective onto its image) is a cover for and is therefore a finite subcover of our original cover. This proves that is compact.
This is a nice theorem, because we can push things into spaces that are strange and preserve compactness knowing only that our map is continuous and our original set is compact.