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Let $d$ be a metric on $X$, and define $\rho: X^{2} \to \mathbb{R}$ as $$\rho(x,y)=\frac{d(x,y)}{1+d(x,y)}$$ The difficulty is in checking the triangle inequality. So, I can prove this by writing $f(t)=\frac{t}{1+t}$ and showing (by doing some algebra) that $$ f(A)+f(B)\ge f(A+B)$$ From which it follows that $A+B\ge C \implies f(A)+f(B) \ge f(C)$, and this is enough to verify the triangle inequality.

I have two problems with this: one, it's quite messy, and two, it doesn't really give me any intuition for what the metric $\rho$ looks like. It's obviously bounded, and it sort of squishes the whole space $X$ into a unit disc. Is there a good way of thinking about it?

For instance, the claim $f(A)+f(B) \ge f(A+B)$ looks similar to (but not quite the same as) the definition of concavity, and I could check $f$ is concave by looking at its derivative.

preferred_anon
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    It feels great to write this, but check out Batman's answer in the link! http://math.stackexchange.com/questions/686792/showing-rho-x-y-fracdx-y1dx-y-is-a-metric-space –  Jun 03 '16 at 09:30
  • The mean value theorem! Yeah that fits what I wanted, thank you! I'vew left an upvote on his answer. – preferred_anon Jun 03 '16 at 09:39
  • You're welcome. All the best. –  Jun 03 '16 at 14:14

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Note $f:[0,\infty]\to\mathcal{R}$ given by $f(x)=\frac{x}{1+x}$ is increasing as $f'(x)=\frac{1}{(1+x)^2}$, Therefore $$\rho(x,z)=\frac{d(x,z)}{1+d(x,z)}\le \frac{d(x,y)+d(y,z)}{1+d(x,y)+d(y,z)}=\frac{d(x,y)}{1+d(x,y)+d(y,z)}+\frac{d(y,z)}{1+d(x,y)+d(y,z)}$$ $$\rho(x,z)\le \frac{d(x,y)}{1+d(x,y)}+\frac{d(y,z)}{1+d(y,z)}=\rho (x,y)+\rho (y,z)$$

Behrouz Maleki
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