Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Compact subsets of compact sets

  1. Oct 22, 2008 #1
    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 [tex]F \subset K[/tex]. Then if [tex](V_{\alpha})[/tex] is an open cover of K, [tex]K \subset \cup_1^n V_{\alpha}[/tex] for some n. But since [tex]F \subset K[/tex], doesn't that mean that [tex]F \subset \cup_1^n V_{\alpha}[/tex], 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. jcsd
  3. Oct 22, 2008 #2


    User Avatar
    Science Advisor

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook