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Math Help - Proof: Metric Space

  1. #1
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    Proof: Metric Space

    Let (X,d) be a metric space and let D:X*X-->R be defined by:

    D(x,y) = (0 if x=y), (max(1,d(x,y)) otherwise)

    Prove that (X,D) is a metric space.


    I know that I have to show (X,D) is positive definite, symmetric and satisfies the triangle inequality.

    The first I am comfortable with. However, I am not sure of how to go about showing symmetry for the x/=y case...same for the triangle inequality. Anybody have any hints or suggestions?
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  2. #2
    hpe
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    Quote Originally Posted by TexasGirl
    Let (X,d) be a metric space and let D:X*X-->R be defined by:

    D(x,y) = (0 if x=y), (max(1,d(x,y)) otherwise)

    Prove that (X,D) is a metric space.
    Do you mean

    D(x,y) = (0 if x=y), (min(1,d(x,y)) otherwise)

    ? What you have written down is essentially the discrete metric.
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  3. #3
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    Quote Originally Posted by TexasGirl
    Let (X,d) be a metric space and let D:X*X-->R be defined by:

    D(x,y) = (0 if x=y), (max(1,d(x,y)) otherwise)

    Prove that (X,D) is a metric space.


    I know that I have to show (X,D) is positive definite, symmetric and satisfies the triangle inequality.

    The first I am comfortable with. However, I am not sure of how to go about showing symmetry for the x/=y case...same for the triangle inequality. Anybody have any hints or suggestions?
    SYMMETRY

    suppose x \neq y\ then

     D(x,y)\ =\ max(1,\ d(x,y)),

    but d is a metric on X so:

    d(x,y)\ =\ d(y,x),

    so:

     D(x,y)\ =\ max(1,\ d(x,y))\ =\ max(1,d(y,x))\ =\ D(y,x).

    If x = y\ then D(x,y) = 0 = D(y,x)

    Hence D is symmetric in its arguments (and this works whether max or
    min is used in the definition of D).

    RonL
    Last edited by CaptainBlack; December 7th 2005 at 08:11 AM.
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  4. #4
    Grand Panjandrum
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    TRIANGLE INEQUALITY

    Suppose x \neq y\ then

     D(x,y)\ =\ max(1,\ d(x,y)).

    Suppose: d(x,y)\  \geq\ 1, then

    D(x,y)\ =\ d(x,y)

    but d(x,y) is a metric on X, so:

    d(x,y)\ \leq \ d(x,z)\ +\ d(z,y),

    for any z\ \epsilon\ X. So

    d(x,y)\ \leq \ max(1,d(x,z))\ +\ max(1,d(z,y)).

    Hence:

    D(x,y)\ \leq \ D(x,z)\ +\ D(z,y).

    Now suppose: d(x,y)\  <\ 1, then

    D(x,y)\ =\ 1,

    but D(u,v)\ \geq \ 1 for all u, v\ \epsilon \ X, so:

    D(x,y)\ \leq \ D(x,z)\ +\ D(z,y).

    for any z\ \epsilon\ X.

    When x\ =\ y,\ D(x,y)=0 so the triangle inequality
    holds trivialy.

    RonL
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