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Math Help - Is it a metric space?

  1. #1
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    Is it a metric space?

    On X=\mathbb R^2 consider d(x,y)={{\left( {{\left| {{x}_{1}}-{{y}_{1}} \right|}^{\frac{1}{2}}}+{{\left| {{x}_{2}}-{{y}_{2}} \right|}^{\frac{1}{2}}} \right)}^{2}}, where x=(x_1,x_2) and y=(y_1,y_2). Is it (X,d) a metric space?

    I think is not, because of the triangle inequality.

    What do you guys think?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Connected View Post
    On X=\mathbb R^2 consider d(x,y)={{\left( {{\left| {{x}_{1}}-{{y}_{1}} \right|}^{\frac{1}{2}}}+{{\left| {{x}_{2}}-{{y}_{2}} \right|}^{\frac{1}{2}}} \right)}^{2}}, where x=(x_1,x_2) and y=(y_1,y_2). Is it (X,d) a metric space?

    I think is not, because of the triangle inequality.

    What do you guys think?
    I don't know for sure, but the likliehood that this is a metric is a very small...that square makes me doubt, as you suggest, that the triangle inequality doesn't work. Have you tried looking for a counterexample?
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  3. #3
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    Haven't found one yet.

    But are you implying that it doesn't satisfy the triangle inequality?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Connected View Post
    Haven't found one yet.

    But are you implying that it doesn't satisfy the triangle inequality?
    I'm saying it probably doesn't. Just an inkling.
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  5. #5
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    Okay, thanks!
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  6. #6
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    Indeed, the triangle inequality fails. Keep looking for a counterexample; it isn't particularly tricky in this case.
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