# Showing boundedness of metric space

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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You mean $\mathrm{diam}(A) := \sup\{d(x,y) : x,y \in A\}$. What's the biggest $\frac{|x_n - y_n|}{1 + |x_n - y_n|}$ can be? – Matt Feb 19 '12 at 1:59
Note, by using a comparison test, $d(x,y)\le\sum_{n=1}^\infty 2^{-n}=1$ – David Mitra Feb 19 '12 at 2:04
@Matt -- Thanks, I was reading the definition of the diameter from a pdf file and I must have accidently made a typo. Would the biggest of $\frac{|x_n - y_n|}{1 + |x_n - y_n|}$ be 1? David -- Thanks, but how would $\sum_{n=1}^\infty 2^{-n}=1$? – Roberto Feb 19 '12 at 2:14
It's a geometric series: ${1\over2}+{1\over4}+{1\over8}+\cdots={1/2\over 1-(1/2)}=1$. – David Mitra Feb 19 '12 at 2:24

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 $$\sup_{x,y} d(x,y) \le 1.$$

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