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Compactness of subsets

  1. Aug 24, 2012 #1
    I'm trying to show that "closed subsets of compact sets are compact". I think I proved (or didn't) that every subset of a compact set is compact, which may be wrong. Here is what I've done so far, please correct me.

    q in A, q not in B, p in B implies p in A. Let {V_a} an open cover of A where V_a = N_r (p) if r = max(p,q). By compactness {V_a} has finite subcover {V_a_i}. Due to our choice of V_a, B can also covered by that finite subcover of {V_a}. Hence B is compact.
     
  2. jcsd
  3. Aug 24, 2012 #2

    jgens

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    This is pretty much nonsense.
     
  4. Aug 24, 2012 #3
    OK. Are you working in topological spaces, metric spaces, [itex]\mathbb{R}^n[/itex], [itex]\mathbb{R}[/itex]?
     
  5. Aug 24, 2012 #4
    I'm working in some metric space. It seems like I picked a really bad way to show that. Although I've seen rudin's solution and it does make sense to me, I just wanted to know if my proof could be corrected somehow
     
  6. Aug 24, 2012 #5
    OK, here is a criticism of the proof:

    What is A? What is B?

    What is [itex]N_r(p)[/itex]? Why can we choose [itex]V_a=N_r(p)[/itex]?

    p and q are not real numbers, how can you take the max?

    Where did you that the subset of the compact set is closed?
     
  7. Aug 24, 2012 #6
    Oops I forgot to mention a lot of things... But I still can't figure out how to use that the subset is closed. I'll be happy with rudin's proof. Thanks
     
  8. Aug 24, 2012 #7

    Bacle2

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    Just to note an implicit point I think Micromass brought up. Your result is true in

    general for closed subsets of compact Hausdorff spaces; metric spaces are Hausdorff,

    so Hausdorff condition is not explicitly used.
     
  9. Aug 24, 2012 #8
    The result is actually always true, even without Hausdorff condition. The "converse" that every compact set is closed uses Hausdorff though...
     
  10. Aug 24, 2012 #9

    Bacle2

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    Ah, yea, I misread it, my bad....
     
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