Why is the metric distance defined as maximum of the set of metric distances on the finite product of metric spaces? 
Why is the metric distance defined as maximum of the set of metric distances on the finite product of metric spaces ?
I think the triangular inequality will not be satisfied but l cannot understand the bigger picture. Will appreciate a good example. 
 A: As Kavi Rama Murthy said, in terms of topology, the $max$-metric is only one of many choices that induce the same topology. Other typical choices include 
$$d_{A \times B}((x,y),(x',y')) := d_A(x,x') + d_B(y,y')$$
and 
$$d_{A \times B}((x,y),(x',y')) := \sqrt{d_A(x,x')^2 + d_B(y,y')^2}$$
(At least) if the metric is induced by a norm, then 
$$d_{A \times B}((x,y),(x',y')) := (d_A(x,x')^p + d_B(y,y')^p)^{1/p}$$
for $p \in [1, \infty)$ is another choice. However, the $\max$-metric has a universal property: 
Consider the category where objects are metric spaces and morphisms are non-expansive maps i.e. maps $f: A \rightarrow B$ such that $d_B(f(x),f(y)) \leq d_A(x,y)$. Let $(A,d_A)$ and $(B,d_B)$ be two metric spaces in this category, then $(A \times B, d_{A \times B})$ with $d_{A \times B} ((x,y), (x',y')):= \max\{d_A(x,x'), d_B(y,y')\}$ is the categorical product of the two spaces. 
Essentially, this comes down to checking the definition and the fact that $\max\{d_A(x,x'), d_B(y,y')\}$ being the least upper bound of the two values. That is, if for some other space $(Z,d_Z)$ there exists a maps $f_A: Z \rightarrow A$, $f_B: Z \rightarrow B$ with $\forall z,z' \in Z: d_A(f_A(z),f_A(z')) \leq d_Z(z,z')$ and $d_B(f_B(z),f_B(z')) \leq d_Z(z,z')$, then $d_Z(z,z')$ also dominates their supremum, which is the metric on $A \times B$. 
