# Proof metric space with distance function

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. $d((a_n),(c_n) \leq d((a_n),(b_n))+d((b_n),(c_n)) \ Triangle \ inequality$

• 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. – Lost May 6 '14 at 21:43
• And $t\mapsto \frac{t}{1+t}$ is increasing on $[0,\infty)$. – Daniel Fischer May 6 '14 at 21:46
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.