Let $\displaystyle (X,\| \cdot \|)$ be a normed space.
Show the function $\displaystyle d(x,y)= \| x - y \|$ is a metric.
Hi biddoggy!
a metric satisfies:
(1) d(x,x) = 0
and $\displaystyle d(x,y) \not= 0$, if $\displaystyle x \not= y$
So: ||x-x|| = ||0|| = 0
OKAY
||x-y|| > 0, because$\displaystyle x \not= y$, thus $\displaystyle x-y \not= 0$
OKAY
(2) d(x,y) = d(y,x)
||x-y|| =||y-x||
||x-y|| = ||-(x-y)||
||x-y|| = |-1|*||x-y|
||x-y|| = ||x-y||
OKAY
(3) $\displaystyle d(x+z,y+z) \le d(x,y) + d(y,z)$
||x+z - (y+z)|| \le ||x+z|| + ||-(y+z)|| = ||x+z|| + ||y+z||
OKAY
Are there any questions on that?
Yours
Rapha