Prove that $d_1\left({x,y}\right) = \max\{\left|{x_j-y_j}\right|:j=1,2,...,k\}$ is a metric on $\mathbb{R}^k$ I am trying to prove that $d_1\left({x,y}\right) = \max\{\left|{x_j-y_j}\right|:j=1,2,...,k\}$ is a metric on $\mathbb{R}^k$. So far I know that $\text{dist}\left({x,y}\right) \leq d_1\left({x,y}\right)$ where $\text{dist}\left({x,y}\right)$ is the standard distance function in k-dimensional Euclidian space. I also know that $d_1\left({x,y}\right) \leq \text{dist}\left({x,y}\right) \leq \sqrt{k}\cdot d_1\left({x,y}\right)$. I have already proven that $d_1\left({x,y}\right) \geq 0$, that $d_1\left({x,y}\right) = 0 \iff x=y$, and that $d_1\left({x,y}\right) = d_1\left({x,y}\right)$, but I am stuck on the triangle inequality property of this particular metric. Any hints or suggestions would be greatly appreciated. 
 A: Hint: For any particular $j$, we have $|x_j-z_j|\le |x_j-y_j|+|y_j-z_j|$. 
Note also that for example $|x_j-y_j|\le \max|x_i-y_i|$. 
A: Let $d_1\left({x,y}\right) = \max\{\left|{x_j-y_j}\right|:j=1,2,...,k\}$. Since $d_1\left({x,y}\right)=\left|{x_j-y_j}\right|$ for some $j$ it is obvious that $d_1\left({x,y}\right) \geq 0$, that $d_1\left({x,y}\right) = d_1\left({y,x}\right)$, and that $d_1\left({x,y}\right)=0\iff{x=y}$. Now, by the definition of $d_1\left({x,y}\right)$ we have $\left|{x_j-y_j}\right| \leq d_1\left({x,y}\right)$ for any $j$. It follows that for any $j$ we have $$\left|{x_j-y_j}\right| \leq \left|{x_j-z_j}\right| + \left|{z_j-y_j}\right| \leq \max\{\left|{x_j-z_j}\right|:j=1,2,...,k\}+\max\{\left|{z_j-y_j}\right|:j=1,2,...k\}$$ Thus $$\max\{\left|{x_j-y_j}\right|:j=1,2,...,k\} \leq \max\{\left|{x_j-z_j}\right|:j=1,2,...,k\} + \max\{\left|{z_j-y_j}\right|:j=1,2,...,k\} $$ Or $d_1\left({x,y}\right) \leq d_1\left({x,z}\right) + d_1\left({z,y}\right)$, completing the proof that $d_1\left({x,y}\right)$ is a metric on $\mathbb{R}^k$. 
A: Three properties need to be checked:
*(1) $d(x,x)=0$
*(2) $d(x,y)=d(y,x)$
*(3) $d(x,z)\le d(x,y)+d(y,z)$
Usually the first two are easy to see.
So we check (3) suppose that when choose n,then the left side gets max, 
$$ ∣∣x_n−z_n∣∣ =||(x_n-y_n) + (y_n-z_n)|| \le ||x_n-y_n|| +||y_n-z_n||$$
$$\le max\{∣∣x_j−y_j∣∣:j=1,2,...,k\} + max\{∣∣y_j−z_j∣∣:j=1,2,...,k\} $$
so $d$ is metric on $R^k$
