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Metric space

  1. Dec 9, 2007 #1
    [SOLVED] metric space

    1. The problem statement, all variables and given/known data

    If x and y are two points in a metric space and d(x,y) = 1, is it always true that the closure of B(x,1/2) does not contain y?

    In general, is [tex] closure( B(x,r)) = \{z | r \geq d(x,z)\}[/tex]

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Dec 9, 2007
  2. jcsd
  3. Dec 9, 2007 #2
    Never mind. The answer is of course because closed balls are closed in a metric space. Why are closed balls in a metric space closed. Because their complement is open. Why is their complement open? Because, for any point x in the complement, let d be the distance from x to the closed ball. Then B(x,d/2) is a nbhd of x that lies in the complement. d is always nonzero and well-defined because otherwise x would be a limit point of the closed ball.
     
  4. Dec 10, 2007 #3

    morphism

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    Hold on - this is not true. In the reals with the discrete metric, closure(B(x,1))={x}, while {z : 1>=d(x,z)}=R.

    But what is true is that the closure of B(x,r) will always sit inside {z : r>=d(x,z)}.
     
  5. Dec 10, 2007 #4
    I see. Thanks.
     
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