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If $X=\mathbb{R}$ and $d\colon\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ given by $d(x,y)={\sqrt{|x-y|}}$, I am able to show that this is indeed a metric.

But what I don't understand is how this metric doesn't come from a norm, like how do I even define a norm on this metric? Thanks

egreg
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1 Answers1

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We have that

$$d(x,0)=\sqrt {|x|}.$$ Thus

$$d(\lambda x,0)=\sqrt{|\lambda|}\sqrt {|x|}\ne |\lambda| d(x,0)$$ if $|\lambda|\ne \pm 1$ and $x\ne 0.$ That is, $d$ is a metric but not comes from a norm.

mfl
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    In other words: this distance function does not obey the scalar multiplication rule that $|\lambda x| = |\lambda||x|$ for scalars $\lambda$ and vectors $x\in X$. (Only vector spaces may have norms.) – kimchi lover Jan 27 '19 at 16:27
  • thanks so the norm of x is always d(x,0)? –  Jan 27 '19 at 16:33
  • also you mean x is not the zero vector right in the last line? –  Jan 27 '19 at 16:36
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    Any norm $|\cdot |$ defines a metric $d(\cdot , \cdot)$ and it is $|x|=d(x,0).$ And yes, I mean $x$ is not the zero vector. – mfl Jan 27 '19 at 16:56