Show that if $\{x_{n}\}$ and $\{y_{n}\}$ are Cauchy sequences in X then $d(x_{n},y_{n})$ converges in $\mathbb{R}$. Could someone help me through this problem?
Let $X$ be a metric space. Show that if $\{x_{n}\}$ and $\{y_{n}\}$ are Cauchy sequences in $X$ then $d(x_{n},y_{n})$ converges in $\mathbb{R}$.
Does this follow from the fact that every Cauchy sequence in $\mathbb{R}$ is convergent?
 A: This is essentially Davide Giraudo's approach, but somewhat shorter: The triangle inequality gives
$$d(x_m,y_m)\leq d(x_m,x_n)+d(x_n,y_n)+d(y_n,y_m)$$
or
$$d(x_m,y_m)-d(x_n,y_n)\leq d(x_m,x_n)+d(y_n,y_m)\ .$$
As the right side is symmetric in $m$ and $n$ we have in fact
$$\bigl|d(x_m,y_m)-d(x_n,y_n)\bigr|\leq d(x_m,x_n)+d(y_n,y_m)\ .$$
A: Hint: use triangular inequality
$$|d(x_n,y_n)-d(x_m,y_m)|\leq |d(x_n,y_n)-d(x_n,y_m)|+|d(x_n,y_m)-d(x_m,y_m)|,$$
then the reversed triangular inequality $|d(x,y)-d(z,y)|\leq d(x,z)$.
A: Every cauchy sequence in $\mathbb{R}$ is convergent, so exists $x, y\in\mathbb{R}$  such that $x_n\to x$ , $y_n\to y$, then  by continuity of the metric, 
$$d(x_n,y_n)\to d(x,y).$$
A: If $X$ is a complete metric space then there exist $ x,y \in X $ such that $ d(x_n,x)   \rightarrow 0$ and $ d(y_n,y) \rightarrow 0$. Thus
\begin{eqnarray*}
 \vert d(x_n,y_n) - d(x,y) \vert & = & \vert d(x_n,y_n) - d(x,y_n) + d(x,y_n) - d(x,y)\vert \\
                                 & \le & \vert d(x_n,y_n) - d(x,y_n) \vert + \vert  d(x,y_n) - d(x,y)\vert \\
                                 & \le d(x_n,x) + d(y_n,y) \rightarrow 0 \\
 \end{eqnarray*}
If $X$ isn't complete, I dont believ the result is true because the natural candidate $d(x,y)$ maybe don't exist.
