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Thread: 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 $\displaystyle x \neq y\$ then

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

    but $\displaystyle d$ is a metric on $\displaystyle X$ so:

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

    so:

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

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

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

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

    Suppose $\displaystyle x \neq y\$ then

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

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

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

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

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

    for any $\displaystyle z\ \epsilon\ X$. So

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

    Hence:

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

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

    $\displaystyle D(x,y)\ =\ 1$,

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

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

    for any $\displaystyle z\ \epsilon\ X$.

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

    RonL
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