1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Metric Space, closed ball is a closed set. prove this
Loading...