What is the counterexample for "The sequence {$d(x_n,y_n)$} converges as $n\rightarrow\infty$ only if $X$ is complete."? Let ($X,d$) be a metric space and let  $&lt;x_n&gt;$ and $&lt;y_n&gt;$ be  arbitrary Cauchy sequences in $X$.Which of the following statement is true?

a.The sequence {$d(x_n,y_n)$} converges as $n\rightarrow\infty$.

b.The sequence {$d(x_n,y_n)$} converges as $n\rightarrow\infty$ only if $X$ is complete.

Solution:
the proof for part (a) is same as amsmath's answer

What is the counterexample for (b) part?
Get $\mathbb{Q}$ with the Euclidean distance. Then if $(x_n)_{n\in\mathbb{N}}$ and $(y_n)_{n\in\mathbb{N}}$ are arbitrary Cauchy sequences in $\mathbb{Q}$, they are also Cauchy sequences in $\mathbb{R}$, so by the completeness of $\mathbb{R}$ they have limit in $\mathbb{R}$, say to $x\in\mathbb{R}$ and to $y\in\mathbb{R}$ respectively. So:
$$d(x_n,y_n)=|x_n-y_n|\rightarrow|x-y|, n\rightarrow\infty.$$
On the other hand, $\mathbb{Q}$ with the Euclidean distance is not a complete metric space, because for example the truncated decimal expansion of $\sqrt2$ is a Cauchy sequence in $\mathbb{Q}$ that doesn't admit limit in $\mathbb{Q}$.
Take

$$X=\Bbb Q$$
$$d=| \;\; |$$

$$x_n=1+\frac 11 +\frac{1}{2!}+...+\frac{1}{n!}$$

$$y_n=2x_n$$

$$\lim_{n\to+\infty}|y_n-x_n|=e\notin \Bbb Q$$

$(x_n)$ and $(y_n)$ are Cauchy in $\Bbb Q$ but the differene $(y_n-x_n)$ is not convergent in $\Bbb Q$.