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About closed sets in the plane

  1. Jul 19, 2013 #1
    I'm wondering if the following is true: Every closed subset of ##\mathbb{R}^2## is the boundary of some set of ##\mathbb{R}^2##.

    It seems false to me, does anybody know a good counterexample?
     
  2. jcsd
  3. Jul 19, 2013 #2

    lavinia

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    a closed disk is not a boundary
     
  4. Jul 19, 2013 #3

    micromass

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    Let ##A## be the closed disk, then ##A## is the boundary of ##A\cap(\mathbb{Q}\times \mathbb{Q})##.
     
  5. Jul 19, 2013 #4
    The result in the OP is infact true.
     
  6. Jul 19, 2013 #5

    lavinia

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    No. The question was a boundary of a subset of the plane. The closed disk is not a boundary of a subset of the plane.
     
  7. Jul 19, 2013 #6

    micromass

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    :confused: ##A\cap (\mathbb{Q}\times \mathbb{Q})## is a subset of the plane. ##A## is its boundary.
     
  8. Jul 19, 2013 #7
    Consider a closed subset of R2, A. Let B be a countable dense subset of A. B has an empty interior so A is the boundary of B. There are some details to fill in, but that sketches the idea.
     
  9. Jul 19, 2013 #8

    lavinia

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    right.

    So I guess the Cantor set is the boundary of itself.
     
    Last edited: Jul 19, 2013
  10. Jul 20, 2013 #9

    jbunniii

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    This is correct. The Cantor set ##C## is closed, so it contains its boundary. On the other hand, ##C## contains no intervals, so if ##x \in C##, then any neighborhood of ##x## contains points not in ##C##. Therefore ##x## is a boundary point of ##C##.
     
  11. Jul 20, 2013 #10
    Awesome! Thanks a lot!!
     
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