# Metric space

• March 29th 2011, 01:42 PM
Connected
Metric space
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?
• March 29th 2011, 01:49 PM
Drexel28
Quote:

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 $\text{diam}X\leqslant 1$
• March 29th 2011, 02:11 PM
Connected
I actually edited that before, probably you didn't see it, so it was okay anyway!
• March 29th 2011, 08:18 PM
bkarpuz
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,
1. $\varphi(0)=0$
2. $\varphi^{\prime}>0$ on $(0,\infty)$
3. $\varphi^{\prime\prime}<0$ on $(0,\infty)$.

Then $(X,u)$ is a metric space, where $u$ defined by $u(x,y):=\varphi(d(x,y))$ for $x,y\in X$. (Itwasntme)
The first condition is required for M1, and the last two is for M3,
you can see that M2 follows directly without any condition. :)
This was what I have proved when I was 2nd year BC. :)
You can also show what if $\mathrm{diam}_{u}(X)=\sup_{0\leq\lambda\leq\mathrm {diam}_{d}(X)}\varphi(\lambda)$. :)