Prove that $\mathbb{R}$ with metric $\rho(x,y)=|\arctan x-\arctan y|$ is not a complete metric space. Determine its completion.
Here is the link to prove the given metric is not complete.
Here is the definition of the completion of a metric space:
A completion of a metric space $(X, d)$ is a pair consisting of a complete metric space $(X^*,d^*)$ and an isometry $\varphi: X\to X^*$ such that $\varphi [X]$ is dense in $X^*.$
I don't know how to find its completion. Can someone help me?
As mentioned in your link, $\arctan$ is a bijection from $\mathbb{R}$ onto $( -\pi/2, \pi/2 )$, and it is an isometry if $\left(-\pi/2, \pi/2\right)$ carries its usual metric and $\mathbb{R}$ carries the metric $\rho(x,y)=|\arctan x-\arctan y|$. As we know, the completion of $\left(-\pi/2, \pi/2\right)$ is $\left[-\pi/2, \pi/2\right]$, thus the completion of $(\mathbb{R}, \rho)$ is a space that is isometric to $[-\pi/2, \pi/2]$ with the usual metric. The isometry here is essentially an extension of $\arctan$ to the boundaries at $\pm \infty$ of $\mathbb{R}$.
We define $(\hat{\mathbb{R}},\hat\rho)$, the completion of $(\mathbb{R},\rho)$, to be $\{-\infty\} \cup \mathbb{R}\cup \{+\infty\}$, where $\{-\infty\}$ and $\{+\infty\}$ correspond to $-\pi/2$ and $\pi/2$. We define $\hat\rho(x,y)=|\arctan^* x-\arctan^* y|$, where $\arctan^*$ is an extension of $\arctan$ with the additional rule that $\arctan^*(-\infty)=-\pi/2$ and $\text{arctan*}(+\infty)=\pi/2$. It is easy to verify that $\text{arctan*}:(\hat{\mathbb{R}},\hat\rho)\to [-\pi/2, \pi/2]$ is a bijective isometry. Therefore, the completion of $(\mathbb{R}, \rho)$ is isometric to the closed interval $[-\pi/2, \pi/2]$ with the usual metric.
In other words, we can think of the completion $(\hat{\mathbb{R}},\hat\rho)$ as $\mathbb{R}$ together with two ideal points at infinity, with the understanding that these points correspond to the $\arctan$ function approaching its asymptotes. And of course there are other completions, but all of them are isometric.