# Compact subsets of compact sets

1. Oct 22, 2008

### variety

This may be a stupid question, but I just confused myself on compactness. For some reason I can't convince myself that ANY subset of a compact set isn't compact in general; just closed subsets. Suppose K is a compact set and $$F \subset K$$. Then if $$(V_{\alpha})$$ is an open cover of K, $$K \subset \cup_1^n V_{\alpha}$$ for some n. But since $$F \subset K$$, doesn't that mean that $$F \subset \cup_1^n V_{\alpha}$$, which means that F...oh I just answered my own question. Ha. I guess it helps to write things out.

But just to make sure I understand what went wrong in the argument: The argument doesn't prove anything because it doesn't show that EVERY open cover of F contains a finite subcover. It just shows that every open cover of F that is also an open cover of K has a finite subcover, which is obvious.

Last edited: Oct 22, 2008
2. Oct 22, 2008

### HallsofIvy

Staff Emeritus
Exactly. [0,1] is compact, (0,1) is not. That is because it is possible to find open covers of (0,1) that are not covers of [0,1]. For example, the collection of open sets {(1/n, 1-1/n)} for n a positive integer is an open cover of (0,1). It does not have any finite subcover because any finite collection of those must have a largest n: call it N. Then the collection does not include any x< 1/N or larger than 1-1/N. Of course, none of those sets contains 0 or 1 so it is not an open cover of [0,1]. Well, done!