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Homework Help: Metric space and absolute value of difference.

  1. Jul 9, 2012 #1
    I'm beginning self-study of real analysis based on 'Introductory Real Analysis' by Kolmogorov and Fomin. This is from section 5.2: 'Continuous mappings and homeomorphisms. Isometric Spaces', on page 45, Problem 1. This is my first post to these forums, but I'll try to get the latex right. Incidentally, I didn't find this particular question by searching; can I use latex in searches?

    1. The problem statement, all variables and given/known data
    Given a metric space [itex](X, \rho)[/itex], prove that
    [tex] a) \ \ | \rho (x, z) - \rho (y,u) | \leq \rho (x, y) + \rho (z, u) \ \ \ \ (x, y, z, u \in X);[/tex]
    [tex]b) \ \ | \rho (x, z) - \rho (y, z) | \leq \rho (x, y) \ \ \ \ (x, y, z \in X).[/tex]

    2. Relevant equations
    Definition of a metric space (Defn. 1, p. 37).

    3. The attempt at a solution
    Things that come to mind:
    - absolute value is equivalent to taking square and root
    - the signs change on (a): on the left is absolute value of a difference, on the right, the regular sum, but
    - group of terms changes: (x,z) - (y,u) -> (x,y) + (z,u)

    So I gather that the absolute value is significant here, but I don't see the steps to make the connection (square both sides, say). The definition of triangle inequality uses the same elements (x,y,z), but I take that to be coincidental and that I should not assume the (x,y,z,u) in this problem have that same relationship.
     
    Last edited: Jul 9, 2012
  2. jcsd
  3. Jul 10, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    welcome to pf!

    hi anhedonia! welcome to pf! :smile:
    i have no idea what you mean by this :confused:

    in b), suppose ρ(x,z) > ρ(y,z) … how would you prove it then? :wink:
     
  4. Jul 10, 2012 #3
    Re: welcome to pf!

    Just that I shouldn't assume the relation of (x,y,z) given in the definition of the triangle inequality applies to the specific (x,y,z) given in this problem statement -- these are different instances of some [itex](x,y,z) \in X[/itex].

    I gather you're saying, essentially, refer back to the general definition of 'absolute value', which is piece-wise:
    [tex]|a| =
    \left\{
    \begin{array}{l}
    a,\ \ \text{if}\ a \geq 0 \\
    -a,\ \text{if}\ a < 0
    \end{array}
    \right.
    [/tex]

    Meaning, split the problem in two and handle each case. That makes sense. I'll see where that gets me.
     
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