proving a metric space is complete.

Question

Consider the metric space $(\mathbb{R}, d)$, where $d(x, y) = | \arctan x − \arctan y|, x, y \in \mathbb{R} $. Is this space complete?

My idea of proof is to introduce a sequence in $\mathbb{R}$ to be cauchy and try to show that it is convergent or not. But I don't know how to use the definition of $d(x,y)$ which is given above.

Answer 1

We assume that $\arctan : \mathbb R\to(-\pi/2,\pi/2)$.

Take $x_n=\tan(\pi/2-1/n)$. Then $$ d(x_m,x_n)=\lvert \arctan x_m-\arctan x_n\rvert= \left\lvert\frac 1m -\frac 1n\right\rvert \to 0, $$ hence $\{x_n\}$ is $d-$Cauchy. But it does not converge. For if $d(x_n,x)\to 0$, for some $x\in\mathbb R$, then $$ \lvert \arctan x_n-\arctan x\rvert=\left| \frac{\pi}{2}-\frac{1}{n}-\arctan x\,\right|\to 0, $$ and thus $\,\,\dfrac{\pi}{2}=\arctan x$. But $\,\,\arctan x<\pi/2$.

Answer 2

Using $a_n = n$, we see that $(a_n)$ is Cauchy as

$$|\arctan m - \arctan n | \leq | \arctan m - \pi/2| + | \arctan n - \pi/2 |$$ can be made arbitrarily small by making $m, n$ sufficiently large.

However $(a_n)$ has no limit in $(\mathbb R, d)$.