Let $(X_i , d_i), i ∈ \Bbb N$, be a collection of metric spaces. Here $x = (x_1, . . . , x_n)$ and $y = (y_1, . . . , y_n)$ are elements of $\prod_{i \in \Bbb N} X_i.$
Define the metric $d(x,y)=\sup \lbrace d_i(x_i,y_i) \rbrace$ on the infinite product $\prod_{i \in \Bbb N} X_i.$
My question is how to prove that the metric $d(x,y)=\sup \lbrace d_i(x_i,y_i) \rbrace$ satisfy the triangle inequality? Can I ask for someone's help? Thanks so much.
For every $i$ we have $d_i(x_i,x_i)+d_i(y_i,y_i) \ge d_i(x_i+y_i,x_i+y_i)$ because metric space $X_i,d_i$ satisfies triangle inequality.
Suppose that : $\sup \lbrace d_i(x_i,x_i) \rbrace +\sup \lbrace d_i(y_i,y_i) \rbrace \lt \sup \lbrace d_i(x_i+y_i,x_i+y_i) \rbrace $. That would mean that for some $i$ : $ d_i(x_i,x_i) + d_i(y_i,y_i) \lt d_i(x_i+y_i,x_i+y_i) $. This contradicts with the fact that the triangle inequality is valid in all spaces $X_i$ individually.