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For $x,y \in \mathbb R$, let $d(x,y) = |x-y|/(1+|x-y|)$.

Determine whether $d$ defines a metric on $\mathbb R$.

If it does, determine whether $\mathbb R$ is compact under $d$.

Since $|x-y|=|y-x|$, $d$ is symmetric.

If $x \ne y$, then $|x-y|>0$, so $d(x,y)>0$.

Since $|x-x|=0$, then $d(x,x)=0$. So, $d(x,y) \ge 0$ with equality if and only if $x=y$.

I am having trouble showing the triangle inequality $$\frac{|x-y|}{1+|x-y|} \le \frac{|x-z|}{1+|x-z|}+\frac{|z-y|}{1+|z-y|}.$$

Trying a few values, I'm guessing triangle inequality holds.

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user5826
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1 Answers1

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In general, if $d'(x,y)$ is a metric,

$$ d(x,y)=\frac{d'(x,y)}{1+d'(x,y)} $$

is again a metric. Apply this for $d'(x,y)=|x-y|$.

References:

Showing $\rho (x,y)=\frac{d(x,y)}{1+d(x,y)}$ is a metric

Showing $d(x,y) = \frac{|x-y|}{1+|x-y|}$ is a distance.

Proving $d_1=\frac{d(x,y)}{1+d(x,y)}$ is a metric equivalent with $d$, the Euclidean metric

Dietrich Burde
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