# Metric space and absolute value of difference.

1. Jul 9, 2012

### anhedonia

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 $(X, \rho)$, prove that
$$a) \ \ | \rho (x, z) - \rho (y,u) | \leq \rho (x, y) + \rho (z, u) \ \ \ \ (x, y, z, u \in X);$$
$$b) \ \ | \rho (x, z) - \rho (y, z) | \leq \rho (x, y) \ \ \ \ (x, y, z \in X).$$

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. Jul 10, 2012

### tiny-tim

welcome to pf!

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

in b), suppose ρ(x,z) > ρ(y,z) … how would you prove it then?

3. Jul 10, 2012

### anhedonia

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 $(x,y,z) \in X$.

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

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