Bounded sequence has no convergent subsequence How can you prove that in the metric space $(\mathbb{R},d)$ where $d(x,y)=|\arctan{x}-\arctan{y}|$ the sequence $(x_n)=n$ is bounded but it has no convergent subsequence ?
Edit 1. Can I say that $(x_n)$ is bounded because $d(n,0)\leq \frac{\pi}{2}$ ?
Edit 2. to see that $x_n$ has no convergent subsequence, I suppose by contradiction that there exists a subsequence $x_{\varphi(n)}$ which converges to $\ell$ it means that $d(x_{\varphi(n)},\ell)\rightarrow 0$ it means that $|\arctan(x_{\varphi(n)})-\arctan(\ell)|\rightarrow 0,$ so $\frac{\pi}{2}-\arctan{\ell} =0$ but $\frac{\pi}{2}-\arctan{\ell}\neq 0$ contradiction.
Thank you.
 A: $1)$ $x_n$ is bounded as the maximum distance between $x$ and $y$ is $|1-(-1)|=2$ as $-1 \leq |\arctan x|\leq 1$.
$2)$ Suppose that $x_n$ has a convergent subsequence $x_{n_k}$. Suppose further that $\displaystyle \lim_{k \rightarrow \infty}x_{n_k}=x_*$ say.
Then $d(x_{n_k},x_*)=|\arctan n_k-\arctan x_*|> \arctan  (x_*+1)-\arctan x_*$ for $n_k > x_*+1$.
Contradiction.
A: Boundedness is simple, by your own comment.
To show it has no convergent subsequences:
Assume it has a convergent subsequence, $(x_{n_k})$, with limit $K$. Then $n_k\ge n=x_n$. Now take $\epsilon=\frac{\frac{\pi}{2}-\arctan{K}}{2}$. Then take $N$ s.t. $\arctan{N}>\arctan{K}+\epsilon$ (possible again by your own comment out the limit of $\arctan$). Then we easily see that for every $n_k\ge N$, $d(x_{n_k},K)>\epsilon$ (because $\arctan$ is increasing). This is a contradiction, so $(x_n)$ cannot have converging subsequences.
A: Note that $x \mapsto \arctan x : \mathbb{R} \to (-\pi/2,\pi/2)$ is a bijection. The metric on $\mathbb{R}$ which you are using is the pullback of the standard metric on $(-\pi/2,\pi/2)$ through this bijection. In other words, the arctangent function is, by design, an isometric isomorphism from $\mathbb{R}$ in the given metric onto $(-\pi/2,\pi/2)$ in the standard metric.
Thus, it is equivalent to prove that the sequence $y_n = \arctan(n)$ in $(-\pi/2,\pi/2)$ is bounded, but not convergent, for the standard metric. Can you take it from here?
