Results 1 to 4 of 4

Math Help - Metric space

  1. #1
    Junior Member
    Joined
    Mar 2011
    Posts
    49

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Connected View Post
    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
    Last edited by Plato; March 29th 2011 at 03:07 PM. Reason: LaTeX fix
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2011
    Posts
    49
    I actually edited that before, probably you didn't see it, so it was okay anyway!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2
    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.
    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).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 8th 2011, 03:16 PM
  2. Is it a metric space?
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: April 3rd 2011, 07:44 PM
  3. Limit of function from one metric space to another metric space
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 17th 2010, 03:04 PM
  4. Sets > Metric Space > Euclidean Space
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 25th 2010, 11:17 PM
  5. Metric Space
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 11th 2009, 04:47 AM

Search Tags


/mathhelpforum @mathhelpforum