Checking whether $X=\mathbb R$ with $d(x,y)=\min\{ \sqrt{|x-y|},|x-y|^2\}$ is a metric space

Asked Sep 17 '18 at 12:46

Active Sep 17 '18 at 13:49

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Examine whether $d$ is a metric on $X=\mathbb{R}$ where $d\left(x,y\right)=\min\{ \sqrt{|x-y|},|x-y|^2\}$ for all $x,y\in \mathbb{R}$

I think it is not. Even though it satisfies all other properties but it doesn't satisfy triangle inequality.

Take $x=2,y=1.5,z=1$ $d\left(x,z\right)=1$

$d\left(x,y\right)=0.25$

$d\left(y,z\right)=0.25$

Can someone confirm and verify. It is correct or not.

edited Sep 17 '18 at 13:49

Martin Sleziak

asked Sep 17 '18 at 12:46

user581912

1

Looks correct. Well done. – Cornman Sep 17 '18 at 12:48
@Cornman Thank you Sir – user581912 Sep 17 '18 at 12:50
This can also be helpful: Explain why $|x - y|^2 $ is not a metric. Found using Approach0. – Martin Sleziak Sep 17 '18 at 13:50
If $0\leq u\leq 1$ then $u^2\leq \sqrt u.$ So if $x,y,z\in [0,1]$ then $d(x,y)+d(y,x)\geq d(x,z)\iff$ $ \iff (x-y)^2+(y-z)^2\geq (x-z)^2 \iff$ $ (y-\frac {x+z}{2})^2\geq (\frac {x-z}{2})^2 \iff$ $ |y-\frac {x+z}{2}|\geq |\frac {x-z}{2}|.$.. So the Triangle Inequality fails when $0\leq x<y<z\leq 1.$ – DanielWainfleet Sep 18 '18 at 05:21
You are right... – DanielWainfleet Sep 18 '18 at 05:23

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