Is it a metric space?

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Apr 3rd 2011, 03:31 PM
Connected

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?
Apr 3rd 2011, 05:18 PM
Drexel28

Quote:

Originally Posted by Connected

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?
Apr 3rd 2011, 05:22 PM
Connected

Haven't found one yet.

But are you implying that it doesn't satisfy the triangle inequality?
Apr 3rd 2011, 05:23 PM
Drexel28

Quote:

Originally Posted by Connected

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.
Apr 3rd 2011, 05:25 PM
Connected

Okay, thanks!
Apr 3rd 2011, 06:44 PM
theodds

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