# Help with proving metric space

• Mar 3rd 2012, 07:56 PM
metric89
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!
• Mar 4th 2012, 03:06 AM
Plato
Re: Help with proving metric space
Quote:

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 $\displaystyle 0\le a\le 0$ then $\displaystyle \dfrac{a}{1+a}\le\dfrac{b}{1+b}$

Then let $\displaystyle a=d(x,y)~\&~b=d(x,z)+d(z,y).$
• Mar 4th 2012, 03:53 AM
psolaki
Re: Help with proving metric space
Quote:

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:

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

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

AND

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

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