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

Question

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.

Answer 1

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 $a_j+b_j\leq a+b$.