I'm trying to prove the statement:
"Consider the metric on $\mathbb{R}$ given by $d(x,y)=|\tan^{-1}x-\tan^{-1}y|$. Show that $d$ is equivalent to the usual metric on $\mathbb{R}$ but that $(\mathbb{R},d)$ is not complete."
I believe I have properly shown that $(\mathbb{R},d)$ is not complete, but I am unsure as to how to show the two metrics are equivalent.
This is my proof so far:
"To show that $(\mathbb{R},d)$ is not complete, consider the sequence $x_n=n$.
Pick $\epsilon>0$ and let $n \in \mathbb{N}$ such that for all $n \geq N$, $|\tan^{-1}{n}-\frac{\pi}{2}|<\frac{\epsilon}{2}$.
For all $n,m \geq N$, we have that $d(x_n,x_m)=|\tan^{-1}{n}-\tan^{-1}{m}|\leq|\tan^{-1}{n}-\frac{\pi}{2}|+|\tan^{-1}{m}-\frac{\pi}{2}|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.
In other words, $\{x_n\}$ is Cauchy.
However, $\{x_n\}$ is not convergent.
For a contradiction, suppose that there exists an $x \in \mathbb{R}$ such that $d(x_n,x) \to 0$ as $n \to \infty$.
We have
$d(x_n,x)=|\tan^{-1}{n}-\tan^{-1}{x}|$, and as $n \to \infty$, $\tan^{-1}{n} \to \frac{\pi}{2}$.
Thus, if $d(x_n,x) \to 0$ as $n \to \infty$, then $\tan^{-1}{x}=\frac{\pi}{2}$.
However, there does not exist an $x \in \mathbb{R}$ for which this is true.
Therefore, $(\mathbb{R},d)$ is not complete, as was to be shown."
Any suggestions?
Your proof that $(\mathbb{R}, d)$ is complete looks fine to me (see Real numbers equipped with the metric $ d (x,y) = | \arctan(x) - \arctan(y)| $ is an incomplete metric space).
We can show that $d$ and the usual metric on $\mathbb{R}$ are equivalent by demonstrating that $\arctan: \mathbb{R} \to ((-\pi/2, \pi/2), d)$ is an isometry.
Any real numbers $x_1$ and $x_2$ can be written as $\tan(y_1)$ and $\tan(y_2)$ where $y_1, y_2 \in (-\pi/2, \pi/2)$, and so $\tan$ is surjective. Additionally $\tan$ strictly increases over $(-\pi/2, \pi/2)$, and so it is injective (why?). So $\tan$ (and $\arctan$) are bijective. This means that if $x = \tan(y)$, then $y$ is the only number in $(-\pi/2, \pi/2)$ that maps to $x$, and if $y = \arctan(x)$ then $x$ is the only number in $\mathbb{R}$ that maps to $y$.
If $|x_1 - x_2| < \epsilon$ then $|\tan(y_1) - \tan(y_2)| < \epsilon$ and so $|\arctan(x_1) - \arctan(x_2)| < \epsilon$ and $d(x_1, x_2) < \epsilon$.