Prove that $d(x,y)=|x-y|$ is a metric on any subset of $R^n$?

Asked Sep 07 '16 at 11:17

Active Dec 07 '21 at 17:12

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to prove that $d(x,y)=|x-y|$ is a metric on any subset of $R^n$.

For this it must satisfy these three conditions,

$1)d(x,y)=d(y,x)$

My response is: $|x-y|=|y-x|$.

$2)d(x,y)\gt 0,d(x,y)=0 $ if $x=y$.

My response is: $|x-y|=\sqrt{(x-y)^2}\gt 0\;\;\sqrt{(x-y)^2}= 0$ if $x=y$ .

$3)d(x,z)\le d(x,y)+d(y,z)$.

My response is: $|x-z|\le|x-y|+|y-z|$.

Is my proof correct?

edited Dec 07 '21 at 17:12

amWhy

asked Sep 07 '16 at 11:17

Bogorovich

If it is true on $\Bbb R^n$ then it is true for any subset because all of those statements are "for all" statements, so if you've already shown them for $\Bbb R^n$, you have shown it for all subsets. – Adam Hughes Sep 07 '16 at 11:20
but how to assure that it is proved for $R^n$ – Bogorovich Sep 07 '16 at 11:24
It follows from the Cauchy-Schwarz inequality. – Matthew Leingang Sep 07 '16 at 11:45
$\sqrt {(x-y)^2}$ doesn't mean anything in more than 1 dimension. – DanielWainfleet Sep 07 '16 at 13:07
Yes but how to prove it for $R^n $ – Bogorovich Sep 07 '16 at 13:21

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