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Thread: Proving the 'Closed Ball' is closed

  1. #1
    Super Member Deadstar's Avatar
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    Proving the 'Closed Ball' is closed

    Let (X, d) be a metric space and, for $\displaystyle a \in X$ and r > 0, define the closed ball with centre a and radius r to be set B*(a, r) given by B*(a, r) = $\displaystyle (x \in X : d(x, a) \leq r)$.

    Prove that B*(a,r) is closed.

    Here's my thoughts... (not full proofs obviously, just rough ideas)

    I thought about using the idea of sequences in closed sets to show that for every sequence $\displaystyle x_n$ in B*(a,r) that converge to some limit x, x is contained in B*(a,r).
    And do i do this by saying let x some point contained in X\B*(a,r), then there is a $\displaystyle k \in \mathbb{N}$ such that $\displaystyle x_n$ is in X\B*(a,r) for all $\displaystyle k \geq n$. Hence $\displaystyle x_n \notin $ B*(a,r).
    Is this looking like an actual proof or is it potentially a bit of a ramble..?

    I also thought about just proving that its complement is open. Might be easier?
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  2. #2
    Junior Member
    Joined
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    $\displaystyle A^c=[x \in X: d(x,a)>r]$

    Given $\displaystyle b \in A^c$ then $\displaystyle d(a,b)>r$ so let $\displaystyle e=d(a,b)-r>0$

    Then (considering the ball around b with radius e):

    $\displaystyle d(x,b)<e=d(a,b)-r$
    $\displaystyle \implies d(a,b)-d(x,b)>r$
    $\displaystyle \implies d(x,a) \geq d(a,b)-d(x,b)>r$

    Therefore $\displaystyle B_{e}(b)\subset A^c$

    Complement is therefore Open, hence A is closed.

    [Sorry i renamed your original ball with A]
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