Proving a metric space - $(\mathbb R^n, d_\infty)$

Question

For every $x = (x_1, x_2,..., x_n) \in \mathbb R^n$, define $$||x||_\infty = \max_{1\leq k\leq n} |x_k|$$

For every $x, y \in \mathbb R^n,$ define $$d_\infty (x,y) = ||x-y||_\infty$$

Prove that ($\mathbb R^n, d_\infty$) is a metric space.

(i) $d_\infty (x, y) \geq 0$:

$d_\infty (x, y) = ||x-y||_\infty = \max_{1\leq k\leq n} |x_k - y_k| \geq 0$

$$$$

if $x = y$,

$d_\infty (x, y) = \max_{1\leq k\leq n} |x_k - y_k| = 0 \Rightarrow x_k = y_k$

$$$$

(ii) $d_\infty (x, y) = d_\infty (y, x)$:

$d_\infty (x, y) = \max_{1\leq k\leq n} |x_k - y_k| = \max_{1\leq k\leq n} |y_k - x_k| = d_\infty (y, x)$

$$$$

(iii) $d_\infty (x, z) \leq d_\infty (x, y) + d_\infty (y, z)$:

$d_\infty (x, z) = \max_{1\leq k\leq n} |x_k - z_k|$

$ = \max_{1\leq k\leq n} |x_k - y_k + y_k - z_k| \leq \max_{1\leq k\leq n} |x_k - y_k| + \max_{1\leq k\leq n} |y_k - z_k|$

$ = d_\infty (x, y) + d_\infty (y, z)$

$$$$

Hence $(\mathbb R^n, d_\infty)$ is a metric space.

$$$$

does this proof make sense?

Answer 1

Everything looks fine except (i), the second part of it specifically. To show its a metric space you should show that $d_\infty(x,y) = 0$ implies that $x=y$ , you assume $x=y$ and conclude $x_k= y_k$ which doesn't seem to be saying much.

Generally putting more detail about why $\max_{1\leq k\leq n} |x_k - y_k + y_k - z_k| \leq \max_{1\leq k\leq n} |x_k - y_k| + \max_{1\leq k\leq n} |y_k - z_k|$ would be better!