# Thread: metric space proof

1. ## metric space proof

How will you prove that d(x,y)>=0 in a metric space.

2. That is part of the definition of a metric space, not something to be proved about metric spaces.

Unless you've been given a specific set and distance function on that set, and need to prove that the distance function is non-negative, in which case we'd need to see the details before we can help.

3. I know, but some books say that d(x,y)>=0 can be proved from the other axioms:
d(x,y)=0 if x = y
and the triangle inequality.

4. d(x,y)=x-y satisfies those but isn't non-negative.

5. We can define a metric on a set $\displaystyle A$ as a map $\displaystyle d:A\times A\rightarrow \mathbb{R}$ satisfying:

$\displaystyle M.1\quad d(x,y)=0 \Leftrightarrow x=y$
$\displaystyle M.2\quad d(x,y)+d(x,z)\geq d(y,z)$

From $\displaystyle M.1$ and $\displaystyle M.2$ it is easy to prove:

$\displaystyle (i)\quad d(x,y)=d(y,x)$
$\displaystyle (ii)\;\; d(x,y)\geq 0$

For example, to prove $\displaystyle (ii)$ take $\displaystyle z=y$ .

Fernando Revilla