In a metric space $(X,d)$ if two sequences {$x_n$} and {$y_n$} are Cauchy then d($x_n$,$y_n$) is convergent Let (X,d) be a metric space and{ ${x_n}$} , {$y_n$} be two arbitrary Cauchy sequences in X then   {$d(x_n,y_n)$} is convergent.
I think it is sufficient to show that {$d(x_n,y_n)$} is Cauchy in $\mathbb R$. But I can't use the triangle inequality properly so that I can't bring the result.
Need someone's help please..

Let $\epsilon>0$ be given. As $\{x_n\}$ is Cauchy, there exists $N_1$ such that $d(x_n,x_m)<\frac\epsilon2$ for all $n,m>N_1$. Similarly, there exists $N_2$ such that $d(y_n,y_m)<\frac\epsilon2$ for all $n,m>N_1$.
Therefore, with $N:=\max\{N_1,N_2\}$, we have 
$$d(x_n,y_n)\le d(x_n,x_m)+d(x_m,y_m)+d(y_m,y_n)< d(x_m,y_m)+\epsilon $$
and the same with $n\leftrightarrow m$, hence 
$$|d(x_n,y_n)-d(x_m,y_m)|<\epsilon $$
for all $n,m>N$.
