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Metric Space, closed ball is a closed set. prove this

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Let (X, d) be a metric space. The set Y in X , d(x; y) less than equal to r is called a closed set with radius r centred at point X.

    Show that a closed ball is a closed set.

    2. Relevant equations

    In a topological space, a set is closed if and only if it coincides with its closure. Equivalently, a set is closed if and only if it contains all of its limit points.

    3. The attempt at a solution

    I have no idea, please help!
  2. jcsd
  3. Nov 19, 2009 #2
    What have you tried? There are a few ways to show a set is closed. You can try showing that it contains all of its limit points, or try showing that the complement is open.
  4. Nov 19, 2009 #3
    Remember, what is the definition of a limit point? Have you tried using the second definition of a closed set?
  5. Nov 19, 2009 #4
    Well I wrote that the boundary points in a closed ball are contained, but i dont know how to prove it?
  6. Nov 20, 2009 #5


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    The closed about p, of radius r, is defined as [itex]\{ q| d(p, q)\le r\}[/itex]. What are the boundary points of that set?
  7. Oct 1, 2011 #6
    Here is my proof. I would appreciate any comments. Assuming it's correct, I like this proof because it doesn't require knowledge of sequences.

    Let B[x0,ε] be the closed ball.
    We will show that this set contains all of its limit points.
    Case 1: |B[x0,ε]| < ∞
    A finite set contains all of its limit points.
    Case 2: Otherwise
    Let x' be a limit point of B[x0,ε]. W.t.s. x' is in B[x0,ε].
    Suppose x' is not in B[x0,ε]. Then d(x0,x') > ε.
    Let γ = 1/2 * (d(x0,x')-ε). Then γ > 0.
    Consider the open set B(x',γ). Since x' is a limit point of B[x0,ε], (B(x',γ) intersect B[x0,ε]) - {x'} ≠ null set.
    Let a be an element of (B(x',γ) intersect B[x0,ε]) - {x'}.
    By definition, d(x0,x') = ε + 2γ.
    d(x0,a) + d(x',a) < ε + γ < ε + 2γ = d(x0,x').
    By the triangle inequality, d(x0,x') < d(x0,a) + d(x',a).
    This is a contradiction.
    In all cases, x' is an element of B[x0,ε].

    Since B[x0,ε] contains all of its limit points, it is a closed set.
    Last edited: Oct 1, 2011
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