# Metric space

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• Mar 29th 2011, 01:42 PM
Connected
Metric space
Let $\displaystyle (X,d)$ be a metric space, prove that $\displaystyle \phi(x,y)=\dfrac{d(x,y)}{1+d(x,y)}$ is a metric over $\displaystyle X.$ Moreover $\displaystyle X$ is bounded with the metric $\displaystyle \phi$ and $\displaystyle \text{diam}\, X<1.$

First part is easy, now I wanted to see how to prove that $\displaystyle X$ is bounded with the metric $\displaystyle \phi,$ is it because $\displaystyle \phi(x,y)<1$ ? Now since $\displaystyle \text{diam}\,X=\underset{x,y\in X}{\mathop{\sup }}\,\phi(x,y)$ so since $\displaystyle \phi(x,y)<1$ then $\displaystyle \underset{x,y\in X}{\mathop{\sup }}\,\phi(x,y)<1,$ then $\displaystyle \text{diam}\,X<1,$ is this correct?
• Mar 29th 2011, 01:49 PM
Drexel28
Quote:

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

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

Yes except $\displaystyle \text{diam}X\leqslant 1$
• Mar 29th 2011, 02:11 PM
Connected
I actually edited that before, probably you didn't see it, so it was okay anyway!
• Mar 29th 2011, 08:18 PM
bkarpuz
In general, you can prove this:
Let $\displaystyle (X,d)$ be a metric space, and $\displaystyle \varphi:[0,\infty)\to[0,\infty)$ be a increasing positive concave function with $\displaystyle \varphi(0)=0$, roughly,
1. $\displaystyle \varphi(0)=0$
2. $\displaystyle \varphi^{\prime}>0$ on $\displaystyle (0,\infty)$
3. $\displaystyle \varphi^{\prime\prime}<0$ on $\displaystyle (0,\infty)$.

Then $\displaystyle (X,u)$ is a metric space, where $\displaystyle u$ defined by $\displaystyle u(x,y):=\varphi(d(x,y))$ for $\displaystyle x,y\in X$. (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 $\displaystyle \mathrm{diam}_{u}(X)=\sup_{0\leq\lambda\leq\mathrm {diam}_{d}(X)}\varphi(\lambda)$. :)