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Thread: Metric space

  1. #1
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    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?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Connected View Post
    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 $\displaystyle \text{diam}X\leqslant 1$
    Last edited by Plato; Mar 29th 2011 at 02:07 PM. Reason: LaTeX fix
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  3. #3
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    I actually edited that before, probably you didn't see it, so it was okay anyway!
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  4. #4
    Senior Member bkarpuz's Avatar
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    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 $\displaystyle x,y\in X$.
    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 $\displaystyle \mathrm{diam}_{u}(X)=\sup_{0\leq\lambda\leq\mathrm {diam}_{d}(X)}\varphi(\lambda)$.
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