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Math Help - metric space proof

  1. #1
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    metric space proof

    How will you prove that d(x,y)>=0 in a metric space.

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  2. #2
    Senior Member Tinyboss's Avatar
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    That is part of the definition of a metric space, not something to be proved about metric spaces.

    Unless you've been given a specific set and distance function on that set, and need to prove that the distance function is non-negative, in which case we'd need to see the details before we can help.
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  3. #3
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    I know, but some books say that d(x,y)>=0 can be proved from the other axioms:
    d(x,y)=0 if x = y
    and the triangle inequality.
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  4. #4
    Senior Member Tinyboss's Avatar
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    d(x,y)=x-y satisfies those but isn't non-negative.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    We can define a metric on a set A as a map d:A\times A\rightarrow \mathbb{R} satisfying:


    M.1\quad d(x,y)=0 \Leftrightarrow x=y
    M.2\quad d(x,y)+d(x,z)\geq d(y,z)


    From M.1 and M.2 it is easy to prove:

    (i)\quad d(x,y)=d(y,x)
    (ii)\;\; d(x,y)\geq 0

    For example, to prove (ii) take z=y .


    Fernando Revilla
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