Metric space

**Connected** · March 29th 2011, 01:42 PM

Let $(X,d)$ be a metric space, prove that $\phi(x,y)=\dfrac{d(x,y)}{1+d(x,y)}$ is a metric over $X.$ Moreover $X$ is bounded with the metric $\phi$ and $\text{diam}\, X<1.$

First part is easy, now I wanted to see how to prove that $X$ is bounded with the metric $\phi,$ is it because $\phi(x,y)<1$ ? Now since $\text{diam}\,X=\underset{x,y\in X}{\mathop{\sup }}\,\phi(x,y)$ so since $\phi(x,y)<1$ then $\underset{x,y\in X}{\mathop{\sup }}\,\phi(x,y)<1,$ then $\text{diam}\,X<1,$ is this correct?

**Drexel28** · March 29th 2011, 01:49 PM

Originally Posted by Connected

Let $(X,d)$ be a metric space, prove that $\phi(x,y)=\dfrac{d(x,y)}{1+d(x,y)}$ is a metric over $X.$ Moreover $X$ is bounded with the metric $\phi$ and $\text{diam}\, X<1.$

First part is easy, now I wanted to see how to prove that $X$ is bounded with the metric $\phi,$ is it because $\phi(x,y)<1$ ? Now since $\text{diam}\,X=\underset{x,y\in X}{\mathop{\sup }}\,\phi(x,y)$ so since $\phi(x,y)<1$ then $\underset{x,y\in X}{\mathop{\sup }}\,\phi(x,y)<1,$ then $\text{diam}\,X<1,$ is this correct?

Yes except $\text{diam}X\leqslant 1$

**Connected** · March 29th 2011, 02:11 PM

I actually edited that before, probably you didn't see it, so it was okay anyway!

**bkarpuz** · March 29th 2011, 08:18 PM

In general, you can prove this:
Let $(X,d)$ be a metric space, and $\varphi:[0,\infty)\to[0,\infty)$ be a increasing positive concave function with $\varphi(0)=0$ , roughly,