# $\mathbb{R}$ under the metric $|\tan^{-1}x - \tan^{-1}y|$

I'm trying to show that the real numbers with the metric described above is an incomplete metric space. To do this, I want to come up with an example of a sequence that is Cauchy but does not have a limit in $\mathbb{R}$ under this metric. First of all, would this be enough?

• What is the simplest sequence you can think of? Apr 27, 2017 at 16:54
• Yes, it would be enough.
– user228113
Apr 27, 2017 at 16:58

Proof: Consider the sequence $x_n=n$. Let $\epsilon > 0$ and then there exists some $N \in \mathbb{N}$ such that $\forall n \geq N$ $|arctan(n)-\frac{\pi}{2}|<\frac{\epsilon}{2}$. So for all $n,m > N$ $d(x_n,x_m)=|arctan(n)-arctan(m)| \leq |arctan(n)-\frac{\pi}{2}|+|arctan(m)-\frac{\pi}{2}| < \epsilon$, i.e this sequence is Cauchy. Note however that $\{x_n\}$ is not convergent, for if it was, there would exist an $x \in \mathbb{R}$ such that $d(x_n,x) \to 0$ as $n \to \infty$. However, $d(x_n,x)=|arctan(n)-arctan(x)|$, and we know as $n \to \infty$ that $arctan(n) \to \frac{\pi}{2}$. Therefore, if $d(x_n,x) \to 0$ as $n \to \infty$ $arctan(x)=\frac{\pi}{2}$. Contradiction, there does not exist an $x \in \mathbb{R}$ such this is true. In other words, we know that for any $x \in \mathbb{R}$ $arctan(x) < \frac{\pi}{2}$. Therefore, $\mathbb{R}$ with this metric is incomplete.
Hint : $$\lim_{x \to \infty} \arctan (x) =\frac\pi 2$$ So you can consider a function which tends to $\infty$ and use the fact that this limit is not real.
Recall how the arctangent function behaves: $\tan^{-1} x$ approaches a finite limit as $x\to\infty$. Therefore the distance between $0$ and $n\in\mathbb N$ approches a finite limit as $n\to\infty$. So there you have a Cauchy sequence with no limit within the space.
Hint Pulling back by the homeomorphism $\tan: (-\frac{\pi}{2}, \frac{\pi}{2}) \to \Bbb R$ shows that $(\Bbb R, d)$ is isometric to $((-\frac{\pi}{2}, \frac{\pi}{2}), \bar d)$, where $\bar d(x, y) = ||x - y||$ is the restriction of the standard metric on $\Bbb R$. If you can find a sequence $(x_n)$ in the latter that is not Cauchy, then $(\tan x_n)$ is not Cauchy in the latter.