Results 1 to 4 of 4
Like Tree3Thanks
  • 1 Post By Deveno
  • 2 Post By Plato

Math Help - I can't prove the triangle inequality holds for this metric space

  1. #1
    Newbie
    Joined
    Jul 2012
    From
    Canada
    Posts
    2

    I can't prove the triangle inequality holds for this metric space

    In this real analysis book I have there is a proposition that if (X,d) is a metric space then so is (X,p) where p=d(x,y)/(1+d(x,y))

    The proof that this is a metric space is left for the reader.

    I have shown that it satisfies all requirements for a metric space apart from the triangle inequality i.e. p(x,y)<=p(x,z)+p(z,y)

    Could someone give assistance please?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,150
    Thanks
    591

    Re: I can't prove the triangle inequality holds for this metric space

    to simplify the algebra, let:

    a = d(x,y)
    b = d(x,z)
    c = d(z,y), and we are given that a ≤ b+c, and also that a,b,c are all non-negative.

    then:

    p(x,y) = a/(1+a)
    p(x,z) = b/(1+b)
    p(z,y) = c/(1+c)

    what we want to prove is:

    \frac{a}{1+a} \leq \frac{b}{1+b} + \frac{c}{1+c},

    that is:

    \frac{a}{1+a} \leq \frac{b+c+2bc}{1+b+c+bc}

    or, equivalently, that:

    a(1 + b + c + bc) \leq (1+a)(b+c+2bc)

    which is to say:

    a + ab + ac + abc \leq b + c + 2bc + ab + ac + 2abc

    so:

    a \leq b + c + 2bc + abc

    since

    a \leq b+c, it follows that:

    a \leq b+c = b+c+0 \leq b + c + 2bc + abc, from which the desired result follows.
    Thanks from Hefotos2004
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,386
    Thanks
    1476
    Awards
    1

    Re: I can't prove the triangle inequality holds for this metric space

    Quote Originally Posted by Hefotos2004 View Post
    In this real analysis book I have there is a proposition that if (X,d) is a metric space then so is (X,p) where p=d(x,y)/(1+d(x,y))
    Here is a second way.
    Lemma: If 0\le a\le b
     \begin{align*}a+ab &\le b+ab \\ a(1+b)&\le b(1+a) \\ \frac{a}{1+a}&\le\frac{b}{1 +b}\end{align*}

    We know that d(x,y)\le d(x,z)+d(z,y).
    So
     \frac{d(x,y)}{1+ d(x,y)} &\le \frac{d(x,z)+d(z,y)}{1+ d(x,z)+d(z,y)} \le \frac{d(x,z)}{1+d(x,z)}+\frac{d(z,y)}{1+d(z,y)}
    Thanks from Deveno and Hefotos2004
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2012
    From
    Canada
    Posts
    2

    Re: I can't prove the triangle inequality holds for this metric space

    Thank you both very much. Very helpful.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof of Levy's metric, triangle inequality.
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: October 10th 2011, 10:23 AM
  2. Prove that the inequality holds
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 4th 2011, 11:29 AM
  3. metric space triangle inequality
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: September 28th 2010, 07:07 PM
  4. Prove a metric space
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 27th 2009, 09:53 AM
  5. [SOLVED] Metric Space and Inequality
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: July 31st 2008, 04:08 PM

Search Tags


/mathhelpforum @mathhelpforum