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. 

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 $\mathbb{R}^k$. 
