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Regular metric d(x,y)=|x-y|

  1. Jun 22, 2009 #1
    Hi, I read the following theorem in a book:
    If {[tex]A_n[/tex]} is a sequence of nowhere dense sets in a complete metric space X, then there exists a point in X which is not in any of the [tex]A_n[/tex]'s.

    But what if I say X={1,1/2,1/3, ...} [tex]\cup[/tex] {0} with the regular metric d(x,y)=|x-y|, and [tex]A_n[/tex]={1/n,0}.
    Why wouldn't this be a counterexample?
    As I see it, X is a complete metric space, each [tex]A_n[/tex] is nowhere dense, and their union equals X.
     
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  3. Jun 23, 2009 #2

    gel

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    Re: Counterexample?

    An is dense at 1/n
     
  4. Jun 23, 2009 #3
    Re: Counterexample?

    I don't understand what you mean by "An is dense at 1/n ".
    The definition I have is: A subset S of a metric space is said to be nowhere dense if its closure has empty interior.
    The closure of each set in the sequence I propose, {1/n,0}, is the set itself. These sets have empty interiors because 1/n and 0 are not interior points of {1/n,0}.
     
  5. Jun 23, 2009 #4

    Hurkyl

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    Re: Counterexample?

    Why not?
     
  6. Jun 23, 2009 #5
    Re: Counterexample?

    Because no open ball centered at 1/n or 0 belongs to {1/n,0}
     
  7. Jun 23, 2009 #6

    Hurkyl

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    Re: Counterexample?

    Are you sure about that? What about the one with radius 1/(2n)?
     
  8. Jun 23, 2009 #7
    Re: Counterexample?

    A ball of radius 1/2n would contain points that are not in the set. For example, take the set {1/3,0}. A ball of radius 1/6 centered at 1/3 would contain the point 1/4 which is not in {1/3,0}. So the ball does not belong to the set. The same goes for all sets {1/n,0}.
     
  9. Jun 24, 2009 #8

    Hurkyl

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    Re: Counterexample?

    Hrm. Yep, I made an error. But it is obvious how to fix my statement, right?
     
  10. Jun 24, 2009 #9
    Re: Counterexample?

    I think I see now why the points 1/n and 0 are interior points of each set.
    The problem was that all this time I've been under the assumption that a set with only one point could not be an open ball (this happens when the metric space is [tex]\mathbb{R}[/tex], but it does not happen in a metric space like the one I am proposing). So if I center balls at 1/n and 0 with a very small radius, then the only point of the ball that is inside the set will be the center of the each ball. So the ball is inside the set (it does not matter that the ball has only one point).
    Thus 1/n and 0 are interior points.
    So the sequence I proposed is NOT a sequence of nowhere dense sets.

    Am I right?
     
  11. Jun 25, 2009 #10

    Hurkyl

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    Re: Counterexample?

    Correct for 1/n. Incorrect for 0. (Every ball, no matter how small of a radius, does contain another point) While 1/n is, 0 is not an interior point of any of your sets.
     
  12. Jun 25, 2009 #11
    Re: Counterexample?

    That's right Hurkyl.
    Thanks for your help.
     
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