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Show that $(\mathbb{N}, d)$ is a metric space with $$d(x,y) = \frac{|x-y|}{1+|x-y|}$$

My attempt:

let $x,y \in \mathbb{N}$,

1) $d(x,y) = 0 \implies |x-y| = 0 \iff x = y$

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

3) show for $z \in \mathbb{N}$, $d(x,z) \leq d(x,y) + d(y,z)$

Using the triangle ineuqality I managed to get:

$$d(x,z) \leq \frac{|x-y|}{1+|x-y| - |y-z|} + \frac{|y - z|}{1 + |x-y| - |y-z|}$$

but i assume here I must use the fact that $x,y,z \in \mathbb{N}$ but I'm not sure how.

Any hints please:

edit: improvements on solution - Using fact that $|x-y|$ is a metric we have that $$d(x,y) \leq \frac{|x-y|}{1+|x-y| + |y-z|} + \frac{|y - z|}{1 + |x-y| + |y-z|}$$ (Since $|x-z| \leq |x-y| + |y-z|$) Is this correct? From there I can use the fact that $$\frac{1}{1+|x-y|+|y-z|} \leq \frac{1}{1+|x-y|} $$ and $$\frac{1}{1+|x-y|+|y-z|} \leq \frac{1}{1+|y-z|}$$ to conclude?

warez8
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1 Answers1

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That we have $\mathbb N$ here is irrelevant. In fact, if $\delta$ is a metric on a set $X$, then $d(x,y)=f(\delta(x,y))$ is also a metric if $f$ is a function with suitable properties (In your problem: $f(t)=\frac{t}{1+t}$), namely

  • $f(t)\ge t$ for all $t\ge 0$
  • $f(t)=0\iff t=0$
  • $f(t_1+t_2)\le f(t_1)+f(t_2)$

Can you see that a) these properties of $f$ make $d$ a metric? And b) that $f(t)=\frac{t}{1+t}$ has these properties (espeicially, the last one)?

Hagen von Eitzen
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  • I'm not really sure what you're getting at there, so $\delta(x,y) = |x-y|$ by your notative right? I don't see why $f(t)$ is being considered and why it shows that $d$ is a metric. Is there no direct way as I have yet to learn of what you speak of. – warez8 Jan 14 '15 at 22:30
  • @warez8 Hagen von Eitzen's post answers a generalization of your question. – fonini Jan 14 '15 at 22:38
  • @fonini I have tried another approach - is this correct? – warez8 Jan 14 '15 at 22:57