# Real numbers equipped with the metric d(x,y)=|arctan(x)−arctan(y)| 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.

Anyways, what is important here is that $$arctan\text{arctan}$$ is a bijection from $$R\mathbb{R}$$ to $$(−π/2,π/2)( -\pi/2, \pi/2 )$$, and it is an isometry if we give $$(−π/2,π/2)\left(-\pi/2, \pi/2\right)$$ the metric it carries as a subspace of $$R\mathbb{R}$$ with the usual metric. If $$f:X→Yf: X \to Y$$ is a surjective isometry then the Cauchy sequences in $$YY$$ are the images of Cauchy sequences under $$ff$$, and the convergent sequences in $$YY$$ are the images of convergent sequences under $$ff$$, so $$YY$$ is complete if and only if $$XX$$ is. Thus $$(R,d)(\mathbb{R}, d)$$ is complete if and only if $$(−π/2,π/2)(-\pi/2, \pi/2)$$ with the metric $$dist(x,y)=|x−y|\text{dist}(x, y) = |x - y|$$ is complete. But $$(−π/2,π/2)(-\pi/2, \pi/2)$$ is not complete since $${π/2−1/n}∞n=1\{\pi/2 - 1/n\}_{n=1}^\infty$$ is Cauchy and does not converge in $$(−π/2,π/2)(-\pi/2, \pi/2)$$.