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Math Help - Diameter of an Open Ball

  1. #1
    FLT
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    Diameter of an Open Ball

    I want to prove that for any ball B_r(a), one has \text{diam} B_r(a) \leq 2r

    So far I know the following:

    \text{diam} B_r(a) = \sup B_r(a). My problem is, what on earth does \sup B_r(a) = \sup\{\rho(x,a) < r | x,a \in X\} actually mean; how do I go about proving that the supremum is indeed less than or equal to 2r?

    The statement seems obvious, but the proof for me is a little tricky because of this part. Thanks.
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  2. #2
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    Quote Originally Posted by FLT View Post
    I want to prove that for any ball B_r(a), one has \text{diam} B_r(a) \leq 2r

    So far I know the following:

    \text{diam} B_r(a) = \sup B_r(a). My problem is, what on earth does \sup B_r(a) = \sup\{\rho(x,a) < r | x,a \in X\} actually mean; how do I go about proving that the supremum is indeed less than or equal to 2r?

    The statement seems obvious, but the proof for me is a little tricky because of this part. Thanks.

    I think it must be Diam\, B_r(a)=\sup\{\rho(x,y)\;;\;x,y\in B_r(a)\}, but then: \forall\,x,y\in B_r(a) :

    \rho(x,y)\le \rho(x,a)+\rho(a,y)=2r\Longrightarrow this is also true for the supremum and we're done

    Tonio
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  3. #3
    FLT
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    Thanks, I think it was a case of mis-applying the definition then!
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