Is it a metric space?

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

I think is not, because of the triangle inequality.

What do you guys think?

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Originally Posted by Connected

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

I think is not, because of the triangle inequality.

What do you guys think?

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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?

Haven't found one yet.

But are you implying that it doesn't satisfy the triangle inequality?

Quote:

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Originally Posted by Connected

Haven't found one yet.

But are you implying that it doesn't satisfy the triangle inequality?

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I'm saying it probably doesn't. Just an inkling.

Okay, thanks!

Indeed, the triangle inequality fails. Keep looking for a counterexample; it isn't particularly tricky in this case.