I have the show that the distance function $d_0(x,y)=\max_{1\leq k\leq n}\ |x_k-y_k|$ is a metric.
I showed the triangle inequality as follows:
$d_0(x,z)=\max_{1\leq k\leq n}\ |x_k-z_k| \leq \max_{1\leq k\leq n} |x_k - y_k + y_k - z_k| \leq \max_{1\leq k\leq n} (|x_k - y_k| + |y_k - z_k|) \leq \max_{1\leq k\leq n} |x_k - y_k| + \max_{1\leq k\leq n} |y_k - z_k| = d_0(x,y) + d_0(y,z)$
I showed symmetry as follows:
$d_0(x,y) =\max_{1\leq k\leq n}\ |x_k-y_k|=\max_{1\leq k\leq n}\ |y_k-x_k|=d_0(y,x)$
I'm unsure how to show positivity. I have the following:
If $x=y$, then $x_k=y_k$ for $1 \leq k \leq n$. Hence $\max_{1\leq k\leq n}\ |x_k-y_k|=0$. If $\max_{1\leq k\leq n}\ |x_k-y_k|=0$, then each term must be $0$ since the maximum is $0$ and each term cannot be negative.
Is this a valid proof? If not, how can I fix it. Thank you.
You have shown every part correctly except:
What you called positivity was actually $$d(x,y)=0 \iff x=y $$ Which is good.
You still need to show that $$ d(x,y) \ge 0 $$ for all $x$,$y$.
Complete the proof by showing $$ d(x,y) \ge 0 $$ for all $x$,$y$.