# Is it a metric space?

• Apr 3rd 2011, 04:31 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, 06:18 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, 06:22 PM
Connected
Haven't found one yet.

But are you implying that it doesn't satisfy the triangle inequality?
• Apr 3rd 2011, 06:23 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, 06:25 PM
Connected
Okay, thanks!
• Apr 3rd 2011, 07:44 PM
theodds
Indeed, the triangle inequality fails. Keep looking for a counterexample; it isn't particularly tricky in this case.