Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple Metric Question

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Let X be a metric space with metric d. Show that the space X, where distance is measured by d' = d/(1+d), is also a metric space.

    2. Relevant equations

    Three requirements of a distance function for X to be a metric space:
    1. d(x,y) = 0 <=> x=y
    2. d(y,x) = d(x,y)
    3. d(x,z) <= d(x,y) + d(y,z)

    3. The attempt at a solution

    It's demonstrating the triangle inequality (3. above) that has me stumped.

    My starting point is to try and use (1) d(x,z) <= d(x,y) + d(y,z) to demonstrate (2) d'(x,z) <= d'(x,y) + d'(y,z).

    I tried manipulating the RHS of (2) to get a common denominator and then tried to 'bash' it out, without any success. I then tried to manipulate the inequalities, but I can't get (2) whilst maintaining <= as opposed to simply <.

    The problem wasn't set as a difficult one, so I presume there is something simple that I can't seem to see.

    Any help would be greatly appreciated!

  2. jcsd
  3. Jan 17, 2010 #2
    Hmmm your first approach should work (I remember it worked for me). Define [itex]f(t) = \frac{1}{1+t}.[/itex] You basically need to show [itex]f(a) + f(b) \geq f(c)[/itex] given [itex]a + b \geq c \geq 0.[/itex] Multiplying out the first inequality gives
    [tex]a + ac + ab + abc + b + ba + bc + abc \geq c + ca + cb + abc,[/tex]
    and yeah this should be true unless I made a mistake somewhere.
  4. Jan 17, 2010 #3
    Thanks for the help, I think that sorts it.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook