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Math Help - Metric Space

  1. #1
    Super Member Showcase_22's Avatar
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    Metric Space

    For 0<p<1 show that d_p(x,y)=(|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}} is not a metric on \mathbb{R}^2.
    I think i've done this correctly, but i'm new to this module so I just want some confirmation that i've done it right (it's one of those modules where questions are unassessed so I could just be doing all the questions wrong!)

    There are two axioms that need to be satisfied:

    d(x,y)=0 iff x=y
    d(x,z) \leq d(x,y)+d(y,z)

    It satisfies the first axiom since d(x,z)=0 gives |x_1-y_1|^p+|x_2-y_2|^p=0 and this is only possible if x_1=y_1 and x_2=y_2.

    I'm aiming to show that the second inequality does not hold.

    I've decided to go with a proof by contradiction and try to contradict the first axiom.

    So suppose d(x,z) \leq d(x,y)+d(y,z).

    Then d(x,z) \leq (|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}}+(|y_1-z_1|^p+|y_2-z_2|^p)^{\frac{1}{p}}

    So when z=x we have d(x,z) \leq 2(|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}} so provided that x_1 \neq y_1 and x_2 \neq y_2, we have that d(x,z) is not necessarily 0. This contradicts the first axiom.

    Is this correct? After writing it down my proof by contradiction seems a bit odd =S
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    Quote Originally Posted by Showcase_22 View Post
    I think i've done this correctly, but i'm new to this module so I just want some confirmation that i've done it right (it's one of those modules where questions are unassessed so I could just be doing all the questions wrong!)

    There are two axioms that need to be satisfied:

    d(x,y)=0 iff x=y
    d(x,z) \leq d(x,y)+d(y,z)

    It satisfies the first axiom since d(x,z)=0 gives |x_1-y_1|^p+|x_2-y_2|^p=0 and this is only possible if x_1=y_1 and x_2=y_2.

    I'm aiming to show that the second inequality does not hold.

    I've decided to go with a proof by contradiction and try to contradict the first axiom.

    So suppose d(x,z) \leq d(x,y)+d(y,z).

    Then d(x,z) \leq (|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}}+(|y_1-z_1|^p+|y_2-z_2|^p)^{\frac{1}{p}}

    So when z=x we have d(x,z) \leq 2(|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}} so provided that x_1 \neq y_1 and x_2 \neq y_2, we have that d(x,z) is not necessarily 0. This contradicts the first axiom.

    Is this correct? After writing it down my proof by contradiction seems a bit odd =S
    But it can be 0, you didn't prove it can't be 0. So this doesn't contradict the first axiom :O

    I don't have time to find how to do the question though, sorry
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    Quote Originally Posted by Showcase_22 View Post
    I think i've done this correctly, but i'm new to this module so I just want some confirmation that i've done it right (it's one of those modules where questions are unassessed so I could just be doing all the questions wrong!)

    There are two axioms that need to be satisfied:

    d(x,y)=0 iff x=y
    d(x,z) \leq d(x,y)+d(y,z)

    It satisfies the first axiom since d(x,z)=0 gives |x_1-y_1|^p+|x_2-y_2|^p=0 and this is only possible if x_1=y_1 and x_2=y_2.

    I'm aiming to show that the second inequality does not hold.

    I've decided to go with a proof by contradiction and try to contradict the first axiom.

    So suppose d(x,z) \leq d(x,y)+d(y,z).

    Then d(x,z) \leq (|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}}+(|y_1-z_1|^p+|y_2-z_2|^p)^{\frac{1}{p}}

    So when z=x we have d(x,z) \leq 2(|x_1-y_1|^p+|x_2-y_2|^p)^{\frac{1}{p}} so provided that x_1 \neq y_1 and x_2 \neq y_2, we have that d(x,z) is not necessarily 0. This contradicts the first axiom.

    Is this correct? After writing it down my proof by contradiction seems a bit odd =S

    Why do you think this question belongs in abstract/linear algebra and not, say in Analysis or at least Calculus? Anyway:

    Take x=(0,0)\,,\,y=(1,1)\,,\,z=(1,0) and now check directly that d_p(x,y) \nleq d_p(x,z)+d_p(z,y)\,\,\,when\,\,\,0<p<1

    Tonio
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