# Proving that the infinity metric gives a metric space on $\mathbb{R}^n$

I'm trying to prove that $$\left(\mathbb{R}^n, d_{\infty}\right)$$, where $$d_{\infty} (x,y) = \max\limits_{1 \leq i \leq n} |x_i - y_i|$$, is a metric space. Here is my attempt.

Let $$x = (x_1, \ldots, x_n), y = (y_1, \ldots, y_n), z = (z_1, \ldots, z_n) \in \mathbb{R}$$. We have \begin{align*} d_{\infty} (x,x) = \max\limits_{1 \leq i \leq n} \{|x_i - x_i|\} = \max \{0\} = 0. \end{align*} Furthermore, if $$x \neq y$$, then there exists $$j \in \{1, \ldots, n\}$$ such that $$x_j \neq y_j$$, so $$|x_j - y_j| > 0$$, so $$d_{\infty} (x,y) = \max\limits_{1 \leq i \leq n} |x_i - y_i| \geq |x_j - y_j| > 0$$. Second, we have \begin{align*} d_{\infty} (x,y) = \max\limits_{1 \leq i \leq n} \{|x_i - y_i|\} = \max\limits_{1 \leq i \leq n} \{|y_i - x_i|\} = d_{\infty} (y,x). \end{align*} since $$|x_i - y_i| = |y_i - x_i|$$ for all $$i \in \{1, \ldots, n\}$$. Finally, for any $$i \in \{1, \ldots, n\}$$, we have \begin{align*} |x_i - y_i| \leq |x_i - z_i| + |z_i - y_i| \leq \max\lim\limits_{1 \leq i \leq n} |x_i - z_i| + \max\limits_{1 \leq i \leq n} |z_i - y_i| = d_{\infty} (x,z) + d_{\infty} (z,y). \end{align*} As this holds for each $$i$$, it holds for the maximal $$i$$, so we have \begin{align*} d_{\infty} (x,y) = \max\limits_{1 \leq i \leq n} |x_i - y_i| \leq d_{\infty} (x,z) + d_{\infty} (z,y), \end{align*} as desired. Therefore, $$\left(\mathbb{R}^n, d_{\infty}\right)$$ is a metric space.

The questions I have, besides whether this is accurate/complete, is mainly on notation. In my first line, for example, is tere a difference way to denote the maximum of the set $$\{|x_i - x_i|\}$$? Is it even valid to write $$\max\limits_{1 \leq i \leq n} 0$$? The argument, of course, doesn't depend on $$n$$. This is the only reason I put the "set" notation back, though I defined the distance metric without it.

When proving $$d(x,y)\leq d(x,z)+d(z,y)$$ something weird is happening. The argument goes like this $$d(x,y)=\max |x_i-y_i|=$$$$\max |x_i-z_i+z_i-y_i|\leq$$$$\max (|x_i-z_i|+|z_i-y_i|)\leq$$$$\max |x_i-z_i|+\max |z_i-y_i|)=$$$$d(x,z)+d(z,y).$$ Everything here follows almost immediately except for maybe the last inequality. It is true however that $$\max (a_i+b_i)\leq \max a_i+\max b_i$$. We can see this if we let $$a$$ be the greatest element of $$a_i$$ and $$b$$ the greatest of $$b_i$$. We have $$\max (a_i+b_i)=a_j+b_j$$ for some $$j$$. Surely $$a_j\leq a$$ and $$b_j\leq b$$ and so $$a_j+b_j\leq a+b$$.