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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.

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  • $\begingroup$ So does one allow $\infty$ as a possible value of $d(x,y)$ when the sup is not finite? $\endgroup$
    – coffeemath
    Commented Mar 24, 2017 at 15:06
  • $\begingroup$ In your example of $x,y$ each one has only $n$ coordinates, making it seem they lie in a finite product, not an infinite product over $i \in \mathbb{N}.$ $\endgroup$
    – coffeemath
    Commented Mar 24, 2017 at 15:45
  • $\begingroup$ @coffeemath I realized there might be a problem here. I asked a new one:math.stackexchange.com/questions/2201440/…. $\endgroup$
    – user369792
    Commented Mar 24, 2017 at 15:58

1 Answer 1

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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.

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  • $\begingroup$ Thanks so much for your answer. I just realized there might be a problem here because $sup$ cannot be infinite. So I asked a new question here:math.stackexchange.com/questions/2201440/… $\endgroup$
    – user369792
    Commented Mar 24, 2017 at 15:59

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