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Denote $X$, the space of all sequences $\in$ $\mathbb R$. I have a metric $d(x,y):=\sum_{n=1}^\infty 2^{-n}\frac{| x_n-y_n|}{1+| x_n-y_n|}$

If $(X,d)$ is a metric space and if $A$ is a subset of $X$, the diameter of $A$ is defined as $\operatorname {diam} \left({A}\right) := \sup \left\{{d \left({x, y}\right): x, y \in A}\right\}$. I know that the set $A$ is bounded if it has finite diameter. What Im stuck at is how would I show that $\operatorname {diam} \left({A}\right) \le 1$?

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    It's a geometric series: ${1\over2}+{1\over4}+{1\over8}+\cdots={1/2\over 1-(1/2)}=1$.2012-02-19

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For any $x$ and $y$, $d(x,y) = \sum_{n=1}^\infty {1\over 2^N} {|x_n - y_n|\over1 + |x_n-y_n|} < \sum_{n=1}^\infty {1\over 2^n} = 1. $ Hence $\sup_{x,y} d(x,y) \le 1.$