[Math] triangle inequality for a metric space

general-topologymetric-spacesreal-analysis

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.

Best Answer

In general it holds:

$| 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)

Now take the maximum on both sides.

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