triangle inequality for a metric space

If $d_{\infty}(a,b) =$ max$\{|a_{i} - b_{i}|\}$ for $1 \leq i \leq k$, I want to prove that this is a metric on $\mathbb{R}^k$.

Its pretty clear that $d_{\infty}(a,a) = 0$ and it is also pretty clear that $d_{\infty}(a,b) = d_{\infty}(b,a)$ since absolute value always gives us a positive number. As always with metric spaces, I am having a hard time proving the triangle inequality. Could I do something like, max$\{|a_{i} - b_{i}|\}$

= max$\{a_{i} - c_{i} + c_{i} - b_{i}\}$ and proceed from there? I knew that was a trick one could use to solve the triangle inequality but I didnt know if I could do it here.

• You also need to show that if $d(a,b) = 0$ then $a=b$. – James Sep 8 '14 at 22:03
• For a metric d it have to hold: d(x,y)=0 <=> x=y. You showed only one direction. – Marm Sep 8 '14 at 22:12
• I was under the assumption that $d(x,x) = 0$ was equivalent to solving as you stated. – Nick Freeman Sep 8 '14 at 23:46

$| a_i-b_i | = | a_i-c_i+c_i-b_i | \leq | a_i-c_i | + | c_i-b_i |$ (triangle inequality for the absolute value of real numbers)