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..
 A: We have
$d(x_n,y_n) \le d(x_n,x_m)+d(x_m,y_n) \le d(x_n,x_m)+d(x_m,y_m)+d(y_m,y_n)$
hence
$d(x_n,y_n)-d(x_m,y_m) \le d(x_n,x_m)+d(y_m,y_n)$.
Can you now complete the proof , that $(d(x_n,y_n))$ is Cauchy ?
A: 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 $n,m>N$.
A: Suppose $(x_n), (y_n)$ are Cauchy. We will indeed show that the sequence $(r_n)=(d(x_n,y_n))$ is Cauchy, so let $\varepsilon>0$. Then there exists $N\in\mathbb N$ such that whenever $n,m>N$ we have $d(x_n,x_m), d(y_n,y_m) < \frac\varepsilon 2$. Then if $n,m>N$, we have
$$\vert r_n-r_m \vert = \vert d(x_n,y_n)-d(x_m,y_m)\vert$$
$$=\vert d(x_n, y_n) - d(x_m, y_n)+d(x_m,y_n) -d(x_m, y_m)\vert$$
$$\leq \vert d(x_n, y_n) - d(x_m, y_n)\vert+\vert d(x_m,y_n) -d(x_m, y_m)\vert$$
$$\leq d(x_n,x_m)+d(y_n,y_m) < \varepsilon$$
