# Compact sets definition

by Useful nucleus
Tags: compact, definition, sets
 PF Patron P: 220 A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover. Does not this imply that every open set is compact. Because let F is open, then F= F $\bigcup$ ∅. Since F and ∅ are open , we obtained a finite subcover of F. Am I missing something here?
P: 1,667
 Quote by Useful nucleus A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover. Does not this imply that every open set is compact. Because let F is open, then F= F $\bigcup$ ∅. Since F and ∅ are open , we obtained a finite subcover of F. Am I missing something here?
Every cover must have a finite subcover not just one or some of them.
 PF Patron P: 220 But F$\bigcup$∅ is contained in every open cover for F?
P: 791

## Compact sets definition

 Quote by Useful nucleus But F$\bigcup$∅ is contained in every open cover?
As a subset, yes. But F and ∅ might not be elements of the open cover.

Compactness says that if you have an open cover -- a collection of open sets whose union covers F -- then some finite subcollection will also cover F. In other words you have to be able to pick out a finite collection of sets from your open cover, and show that the finite collection is also an open cover.

As an example, take F = (0,1), the open unit interval. Let A_n = (1/n, 1). Then the collection {A_n} for n = 1, 2, 3, ... is an open cover of F; but no finite subcollection of the A_n's covers (0,1). Working through this example will be helpful.
 PF Patron P: 220 Then, I think the definition of the open cover of a set F has to emphasize that F has to be a proper subset of the cover. Otherwise the argument in my first post will be correct.
P: 1,667
 Quote by Useful nucleus But F$\bigcup$∅ is contained in every open cover for F?
As SteveL7 said, F and ∅ may not be elements of the open cover. An open cover is a collection of open sets. F may not be in the collection.
P: 791
 Quote by Useful nucleus Then, I think the definition of the open cover of a set F has to emphasize that F has to be a proper subset of the cover. Otherwise the argument in my first post will be correct.
Well if F is an open set, then {F} is indeed an open cover of F consisting of one set. So thats an example of one open cover of F that happens to have a finite subcover.

But compactness requires that EVERY open cover has a finite subcover.

I think you would find it helpful to work through the example of F = (0,1) that I gave earlier.

FWIW pretty much everyone has trouble learning about compactness. It's very unintuitive. Not like connectedness, say, whose technical definition matches our intuition about what a connected set should be. With compactness, it's like, "How did they ever think of that?"
 PF Patron P: 220 SteveL27 and lavinia thank you very much for your help! I have a better understanding now of the definition. The open unit interval example was particularly helpful.
 Sci Advisor P: 905 the usual example "in the other direction" (unbounded sets, rather than open sets) is R: certainly {R} is an open cover of R. yet it hardly seems "right" to call R "compact", given how large it is. and, of course, it is not: {(n,n+2): n in Z} is an open cover of R, but we need "all |Z| of them" (and |Z| is not finite) to cover R, no finite subcover will do.