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Math Help - Help with proving metric space

  1. #1
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    Help with proving metric space

    Hi All, i'm getting lost here

    Let (X,d) be a metric space. Show that (X,c) is a metric space where c(x,y) = d(x,y)/(d(x,y)+1).

    It's just the triangle inequality that I can't seem to nut out!

    Any help would be much appreciated!

    Thanks!
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  2. #2
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    Re: Help with proving metric space

    Quote Originally Posted by metric89 View Post
    Let (X,d) be a metric space. Show that (X,c) is a metric space where c(x,y) = d(x,y)/(d(x,y)+1).
    It's just the triangle inequality that I can't seem to nut out!
    Lemma
    If 0\le a\le 0 then \dfrac{a}{1+a}\le\dfrac{b}{1+b}

    Then let a=d(x,y)~\&~b=d(x,z)+d(z,y).
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  3. #3
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    Re: Help with proving metric space

    Quote Originally Posted by metric89 View Post
    Hi All, i'm getting lost here

    Let (X,d) be a metric space. Show that (X,c) is a metric space where c(x,y) = d(x,y)/(d(x,y)+1).

    It's just the triangle inequality that I can't seem to nut out!

    Any help would be much appreciated!

    Thanks!
    O.K

    We have:

     c(x,y)=\frac{d(x,y)}{1+d(x,y)} \leq\frac{d(x,z)}{1+d(x,y)} + \frac{d(x,z)}{1+d(x,y)}.

    Now if you can prove that: \frac{d(x,z)}{1+d(x,y)}\leq\frac{dx,z)}{1+d(x,z)} =c(x,z)

    AND

    \frac{d(z,y)}{1+d(x,y)}\leq\frac{d(z,y)}{1+d(z,y)}  =c(z,y) ,Then you are done

    Start with : d(x,z)\geq 0 AND  d(z,y)\geq 0
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