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Compact subset of a locally compact space

  1. Nov 6, 2007 #1
    1. The problem statement, all variables and given/known data
    How would I prove that if X is locally compact and a subset of X, V, is compact, then there is an open set G with [tex]V \subset G[/tex] and closure(G) compact?


    EDIT: X is also Hausdorff (which with local compactness implies that it is regular) if that matters

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Nov 6, 2007
  2. jcsd
  3. Nov 6, 2007 #2

    morphism

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    If X is regular, then given a point x in X and a nbhd* U of x, we can find a nbhd W of x whose closure sits in U. Do this for each x in V. Now use the facts that X is locally compact and that V is compact.

    (*: I'm using nbhd to mean an open set containing x.)
     
    Last edited: Nov 6, 2007
  4. Nov 6, 2007 #3
    So because X is locally compact, we have a nbhd around every point x whose closure is compact, call it N_x. Now, like you said, for all x in V, we take find a nbhd around x, call it M_x whose closure sits in N_x.

    The union of M_x over x in V form an open cover of V, so we have a finite subcover, say {M_x_1 to M_x_n}. Union M_x_i is an open set that contains V. Is its closure compact, though?
     
  5. Nov 6, 2007 #4
    Okay. I see why the closure is compact. But also, how would I prove that the boundary of our open set G is compact?
     
  6. Nov 6, 2007 #5

    morphism

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    It's a closed subset of a compact space, isn't it?
     
  7. Nov 7, 2007 #6
    Got it. Thanks.
     
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