Metric space and absolute value of difference.

anhedonia
Messages
2
Reaction score
0
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?

Homework Statement


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).

Homework Equations


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

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:
Physics news on Phys.org
welcome to pf!

hi anhedonia! welcome to pf! :smile:
anhedonia said:
b) \ \ | \rho (x, z) - \rho (y, z) | \leq \rho (x, y) \ \ \ \ (x, y, z \in X).

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.

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:
 


tiny-tim said:
i have no idea what you mean by this
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.

tiny-tim said:
...in b), suppose ρ(x,z) > ρ(y,z) … how would you prove it then? :wink:

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

Meaning, split the problem in two and handle each case. That makes sense. I'll see where that gets me.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top