Prove that it is a metric Hello I have problems with this exercises:
Denote by ${\mathbb{R}}^\infty$ the set of all sequences in $\mathbb{R}$. For $x,y$ define $d$
$d(x,y)= \displaystyle\sum_{i=1}^\infty{} \displaystyle\frac{1}{n!} \displaystyle\frac{|x_n-y_n|}{1+|x_n-y_n|}$
Prove that $d$ is a metric
I have to prove the following axioms:
i) $ d(x,x)=0 $
ii) If $ x \neq y $, then $ d(x,y) > 0 $
iii) $ d(x,y)=d(y,x) $
iv) $ d(x,z)\leqslant d(x,y)+d(y,z) $
$d(x,y)= \displaystyle\sum_{i=1}^\infty{} \displaystyle\frac{1}{n!} \displaystyle\frac{|x_n-y_n|}{1+|x_n-y_n|} \rightarrow{0}$ as $n\rightarrow{\infty}$
This means that, for $\epsilon > 0$ there exists an integer $ n_0(\epsilon$ such that:
$\displaystyle\sum_{i=1}^\infty{} \displaystyle\frac{1}{n!} \displaystyle\frac{|x_n-y_n|}{1+|x_n-y_n|} < \epsilon $
and so
$\displaystyle\frac{1}{n!} \displaystyle\frac{|x_n-y_n|}{1+|x_n-y_n|} < \epsilon$
I don't know how to continue
Thanks
 A: The proofs of i), ii) and iii) are trivial (very easy), so we are going to prove iv).
Since $\;\left|x_n-z_n\right|\leqslant\left|x_n-y_n\right|+\left|y_n-z_n\right|\;,\;$ it results that
\begin{align}\dfrac{\left|x_n-z_n\right|}{1+\left|x_n-z_n\right|}&=1-\dfrac1{1+\left|x_n-z_n\right|}\leqslant\\&\leqslant1-\dfrac1{1+\left|x_n-y_n\right|+\left|y_n-z_n\right|}=\\&=\dfrac{\left|x_n+y_n\right|+\left|y_n-z_n\right|}{1+\left|x_n+y_n\right|+\left|y_n-z_n\right|}=\\&=\dfrac{\left|x_n+y_n\right|}{1+\left|x_n+y_n\right|+\left|y_n-z_n\right|}+\dfrac{\left|y_n-z_n\right|}{1+\left|x_n+y_n\right|+\left|y_n-z_n\right|}\leqslant\\&\leqslant\dfrac{\left|x_n+y_n\right|}{1+\left|x_n+y_n\right|}+\dfrac{\left|y_n-z_n\right|}{1+\left|y_n-z_n\right|}\;.\end{align}
Consequently,
\begin{align}\sum\limits_{i=1}^\infty\dfrac1{n!}\dfrac{|x_n-z_n|}{1+|x_n-z_n|}&\leqslant\sum\limits_{i=1}^\infty\dfrac1{n!}\dfrac{|x_n-y_n|}{1+|x_n-y_n|}+\sum\limits_{i=1}^\infty\dfrac1{n!}\dfrac{|y_n-z_n|}{1+|y_n-z_n|}\end{align}
that is
$d(x,z)\leqslant d(x,y)+d(y,z)\;.$
