Thats the first time i have to do such an proof but don't know how, never seen or done this before. Especially (iii).

Let $X$ be the Set of all complex sequences. $$ d((a_n),(b_n)) := \sum^\infty_{i=0} \frac{1}{2^{i+1}}\frac{\left | a_i-b_i \right |}{1+\left | a_i-b_i \right|}, ((a_n),(b_n) \in X) $$ Proof that $(X,d)$ is an metric space.

Definition of metric space says:

  1. $d((a_n),(b_n)) \geq 0 $ and $d((a_n),(b_n))=0 \Leftrightarrow (a_n)=(b_n)$
  2. $d((a_n),(b_n)) = d((b_n),(a_n))$
  3. $d((a_n),(c_n) \leq d((a_n),(b_n))+d((b_n),(c_n)) \ Triangle \ inequality $

Can someone help me please


  • 1
    $\begingroup$ Just verify the axioms. $1$ and $2$ should be clear if you look at the definition of $d$. It's $3$ that requires work, but note that you have absolute values in the expression for $d$ and you know that the triangle inequality is valid for them. $\endgroup$ – Lost May 6 '14 at 21:43
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    $\begingroup$ And $t\mapsto \frac{t}{1+t}$ is increasing on $[0,\infty)$. $\endgroup$ – Daniel Fischer May 6 '14 at 21:46

Add on Daniel Fisher's comment

When $f(t)= \frac{t}{1+t}>0$ is concave for $t>0$, then $$ f(|a-b|) +f(|b-c|)\geq f(|a-b| + |b-c|)\geq f(|a-c|)$$ where $a,\ b,\ c$ are points in a metric space $(X,d=|\ |)$. Hence $(X,\frac{d}{1+d})$ is a metric space.

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