Real numbers equipped with the metric $ d (x,y) = | \arctan(x) - \arctan(y)| $ is an incomplete metric space I have to show that the real numbers equipped with the metric
$ d (x,y) = | \arctan(x) - \arctan(y)| $ is an incomplete metric space.
Certainly, I have to search for a Cauchy sequence of real numbers with respect to given metric that must not be convergent. But I am unable to figure out that.
 Can anybody help me with this. 
Thanks for helping me.
 A: Consider a sequence that grows without bound. Such a sequence isn't Cauchy in the usual metric on $\mathbb{R}$, but will be under this metric.
A: Sorry for reviving such an old problem...
Anyways, what is important here is that $\text{arctan}$ is a bijection from $\mathbb{R}$ to $( -\pi/2, \pi/2 )$, and it is an isometry if we give $\left(-\pi/2, \pi/2\right)$ the metric it carries as a subspace of $\mathbb{R}$ with the usual metric. If $f: X \to Y$ is a surjective isometry then the Cauchy sequences in $Y$ are the images of Cauchy sequences under $f$, and the convergent sequences in $Y$ are the images of convergent sequences under $f$, so $Y$ is complete if and only if $X$ is. Thus $(\mathbb{R}, d)$ is complete if and only if $(-\pi/2, \pi/2)$ with the metric $\text{dist}(x, y) = |x - y|$ is complete. But $(-\pi/2, \pi/2)$ is not complete since $\{\pi/2 - 1/n\}_{n=1}^\infty$ is Cauchy and does not converge in $(-\pi/2, \pi/2)$.
A: Consider $x_n =  n$. Let $\varepsilon > 0$ and choose $\displaystyle N > \tan\bigg(\frac\pi2 - \varepsilon\bigg)$. If $m, n > N$ then $\displaystyle \{\arctan m, \arctan m\} \subseteq \bigg(\arctan N, \frac\pi2\bigg)$. Thus $$d(x_m, x_n) = \vert \arctan m - \arctan n \vert \leq \bigg \vert \frac\pi2 - \arctan N \bigg\vert < \bigg \vert \frac\pi2 - \frac\pi2 + \varepsilon \bigg\vert= \varepsilon$$ Thus $(x_n)$ is a Cauchy sequence. Observe that as $n \to \infty$, $\arctan x_n \to \pi/2$. But $(x_n)$ does not converge to any element in $\mathbb R$ since there is no $x \in \mathbb R$ such that $\arctan x = \pi/2$.
