# Thread: Is it a metric space?

1. ## 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?

2. 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?

3. Haven't found one yet.

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

4. 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.

5. Okay, thanks!

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