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Show a closed subset of a compact set is also compact

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that if E is a closed subset of a compact set F, then E is also compact.

    2. Relevant equations
    I'm pretty sure you refer back to the Heine-Borel theorem to do this.

    "A subset of E of Rk is compact iff it is closed and bounded"

    3. The attempt at a solution
    We are deal with metric spaces here. It should seem that I need to prove the same thing as in the second half of the Heine-Borel theorem. My textbook is proving Heine-Borel in a confusing way without clear statement/reason steps that I can apply to my problem.

    -From the given information, E is in F
    -From the given information, every open cover of F has a finite subcover of F
    -From the given information, S \ E is open

    ...not sure where this leave me
  2. jcsd
  3. Oct 28, 2009 #2
    The Heini-Borel theorem only applies to euclidean spaces. Use the definition of compact.
  4. Oct 28, 2009 #3


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    You don't need Heine-Borel for this. Take an open cover of E. If you add the open set S/E to that (S is the whole space, right?), then you have an open cover of F. Since F is compact... Can you continue?
  5. Oct 29, 2009 #4
    I'm with you so far.....you declare O and open cover of E. Then you take the union of O and S \ E and get an open set. This unified open set has to cover all E and all not-E so it has to be an open cover for all of F.

    So how does the compactness of F flow back down to E? (Clearly, I have no idea what I'm doing).
  6. Oct 29, 2009 #5


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    Since F is compact, there is a finite subcover. So that subcover also covers E. Now you don't need the S\E set to cover E. What's left is a finite subcover of the original cover that covers E.
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