# Thread: Help with proving metric space

1. ## Help with proving metric space

Hi All, i'm getting lost here

Let (X,d) be a metric space. Show that (X,c) is a metric space where c(x,y) = d(x,y)/(d(x,y)+1).

It's just the triangle inequality that I can't seem to nut out!

Any help would be much appreciated!

Thanks!

2. ## Re: Help with proving metric space

Originally Posted by metric89
Let (X,d) be a metric space. Show that (X,c) is a metric space where c(x,y) = d(x,y)/(d(x,y)+1).
It's just the triangle inequality that I can't seem to nut out!
Lemma
If $0\le a\le 0$ then $\dfrac{a}{1+a}\le\dfrac{b}{1+b}$

Then let $a=d(x,y)~\&~b=d(x,z)+d(z,y).$

3. ## Re: Help with proving metric space

Originally Posted by metric89
Hi All, i'm getting lost here

Let (X,d) be a metric space. Show that (X,c) is a metric space where c(x,y) = d(x,y)/(d(x,y)+1).

It's just the triangle inequality that I can't seem to nut out!

Any help would be much appreciated!

Thanks!
O.K

We have:

$c(x,y)=\frac{d(x,y)}{1+d(x,y)}$ $\leq\frac{d(x,z)}{1+d(x,y)} +$ $\frac{d(x,z)}{1+d(x,y)}$.

Now if you can prove that: $\frac{d(x,z)}{1+d(x,y)}\leq\frac{dx,z)}{1+d(x,z)} =c(x,z)$

AND

$\frac{d(z,y)}{1+d(x,y)}\leq\frac{d(z,y)}{1+d(z,y)} =c(z,y)$ ,Then you are done

Start with : $d(x,z)\geq 0$ AND $d(z,y)\geq 0$