Let $(M,d)$ be a metric space. Define the following function $d'(x,y):=\dfrac{d(x,y)}{d(x,y)+1}$ for every $x,y$ in $M$. Is $d'$ a distance? Does it define the same topology on $M$ as $d$? I didn't think $d'$ was a distance as $d'(x,y)$ could be less than $0$, of because the $x $ and $y$ are in $M$ does that mean that they have to be positive? Also not sure how to prove that the topology is the same? Thanks
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1How could $d'(x,y)$ be less than $0$ unless $d(x,y)$ committed the same faux pas? – Andreas Blass Apr 17 '13 at 23:08
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1$\rm{Duplicate}^7$. But where...? – Julien Apr 18 '13 at 00:01
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@julian Duplicates. Duplicates everywhere. – Pedro Apr 18 '13 at 00:16
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1@skullpatrol What...? – Pedro Apr 18 '13 at 00:19
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@PeterTamaroff I'll show this to my nephew. That's his favorite movie. – Julien Apr 18 '13 at 00:30
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I will let you prove by yourself that $$d'(x,y)\geq 0$$ $$d'(x,y)=0\iff x=y$$ $$d'(x,y)=d'(y,x)$$ Remember that $$d(x,y)\geq 0$$ $$d(x,y)=0\iff x=y$$ $$d(x,y)=d(y,x)$$
See what this implies for $d'=\frac{d}{1+d}$. Observe, for example, that $$\displaylines{ 0 \leqslant d(x,y) \cr 1 \leqslant d(x,y) + 1 \cr 1 \geqslant \frac{1}{{d(x,y) + 1}} \geqslant 0 \cr d(x,y) \geqslant \frac{{d(x,y)}}{{d(x,y) + 1}} \geqslant 0 \cr} $$
The only bothersome property is the triangle inequality. To sort this out, we look at the function $f(x)=\dfrac x{1+x}$. It is increasing for positive values, so $x\leq y+z\implies f(x)\leq f(y+z)$. You should now prove that $$f(z+w)\leq f(z)+f(w)$$ and see that we obtain $d'$ by applying $f$ to $d$. You will have to show that for positive values, $$\frac{{z + w}}{{1 + z + w}} \leqslant \frac{z}{{1 + z}} + \frac{w}{{1 + w}}$$ is true. To prove it generates the same topology, refer to this question.
