# Completion of a metric space $\mathbb{R}$ with a special metric

Let $(\mathbb{R},d)$ be a metric space. Define $d(x,y) = |\tan^{-1} x - \tan^{-1} y|$. I want to prove that $(\mathbb{R},d)$ is isometric to $(\{\tan^{-1} x, x \in \mathbb{R}\}, \rho)$ where $\rho$ is a Euclidean distance. Define $f: (\mathbb{R},d) \to (\{\tan^{-1} x, x \in \mathbb{R}\}, \rho), f(x) = \tan^{-1} x$. Obviously, the function is a bijection, and it preserves metric: $d(x,y) = |\tan^{-1} x - \tan^{-1} y| = \rho(f(x),f(y))$. So the metric spaces are isometric.

We know that $(\{\tan^{-1} x, x \in \mathbb{R}\}, \rho)$ is incomplete, since the sequence $x_n = \tan^{-1} n$ is Cauchy, but it converges to $\frac{\pi}{2}$ in $(\mathbb{R},\rho)$, since $\frac{\pi}{2} \notin \{\tan^{-1} x, x \in \mathbb{R}\}$, it is incomplete. So since spaces are isometric, $(\mathbb{R},d)$ is also incomplete.

I assume the completion of $(\{\tan^{-1} x, x \in \mathbb{R}\}, \rho)$ will be $(\{\tan^{-1} x, x \in \mathbb{R}\} \cup \{-\frac{\pi}{2}, \frac{\pi}{2}\}, \rho)$. What will be completion of $(\mathbb{R},d)$ then?

And, please, tell me if any one my statements are wrong.

The completion of $\langle\Bbb R,d\rangle$ is $\left\langle\Bbb R\cup\{-\infty,+\infty\},\overline d\right\rangle,$ where $\overline d$ is the natural (continuous) extension of $d$.