- #1
anhedonia
- 2
- 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?
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]
Definition of a metric space (Defn. 1, p. 37).
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.
Homework Statement
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]
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: