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Closed sets in a topological space

  1. May 26, 2009 #1
    If [tex]A\subseteq B[/tex] are both subsets of a topological space [tex](X,\tau)[/tex], is it true that any closed subset of A is also a closed subset of B?
     
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  3. May 26, 2009 #2

    Hurkyl

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    Note that A is a closed subset of A....
     
  4. May 26, 2009 #3
    Hmm, I'm not sure. I never said A was closed. A is in B which is in X. And some other set in A, maybe call it U, is closed. I think the answer would be U is closed in X but I don't quite see why.
     
  5. May 26, 2009 #4

    HallsofIvy

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    Hurkyl's point is that any topological space is both open and closed as a subset of itself. If A is NOT closed as a subset of topological space B, since it IS closed as a subset of itself, the statement "any closed subset of A is also a closed subset of B" is false.
     
  6. May 26, 2009 #5
    Ahh i see. Thanks. But is that the only problem here? If we insist that A is compact in B, then that fixes the problem and the statement is true, right?
     
  7. May 26, 2009 #6

    matt grime

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    Um, just insisting that A is closed is all you need, nothing to do with compactness. In fact compactness won't help you at all - compact does not imply closed (e.g. the Zariski topology on R).
     
  8. May 26, 2009 #7

    HallsofIvy

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    ?? I thought compact did imply closed!

    Oh, I see. I started to give the proof and then realized I was saying "given points p and q construct neighborhoods about p and q that do not intersect". That's not possible in some topological spaces.
     
  9. May 26, 2009 #8

    matt grime

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    If you don't know what the Zariski topology is (and Halls does but forget, temporarily) consider the topology on R given by:

    U is open if and only if U contains the interval (0,1) - the set (0,1) is in this and is certainly compact, but not closed.
     
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