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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.

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  • $\begingroup$ i know that $\lim_{x\rightarrow +\infty} \arctan(x)=\frac{\pi}{2}$ $\endgroup$ – Vrouvrou Jan 3 '15 at 10:20
  • $\begingroup$ Can you find any $\ell\in\mathbb R$ s.t. indeed $\arctan(\ell)=\frac{\pi}{2}$?? If not then there you have the contradiction. $\endgroup$ – drhab Jan 3 '15 at 11:40
  • $\begingroup$ yes i understand $tag(\pi/2)$ do not exist $\endgroup$ – Vrouvrou Jan 3 '15 at 12:04
  • $\begingroup$ The existence of a convergent subsequence implies the existence of an $\ell$ with $\arctan(\ell)=\frac{\pi}{2}$. Backwards the non-existence of this $\ell$ implies that no such convergent subsequence exists. This is exactly what had to be proved. So you are ready. $\endgroup$ – drhab Jan 3 '15 at 12:09
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    $\begingroup$ That's something else. Btw, I did not downvote. Don't worry about that. You have your answers, haven't you? That is more important. Cheers. $\endgroup$ – drhab Jan 3 '15 at 12:13
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$1)$ $x_n$ 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.

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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?

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  • $\begingroup$ i want to prove that $x_n=n$ is bounded and have no convergent subsequence not $1/n$ $\endgroup$ – Vrouvrou Jan 3 '15 at 10:31
  • $\begingroup$ Right. But my point is that is that if $\phi : (X,d) \to (X',d')$ is an isometric isomorphism from one metric space to another, then a sequence $x_n$ in $X$ is bounded and non convergent if and only if the sequence $\phi(x_n)$ in $X'$ is bounded and non convergent. Here "isometric isomorphism" is a synonym for "distance preserving bijection". $\endgroup$ – Mike F Jan 3 '15 at 10:33
  • $\begingroup$ don't understand $\endgroup$ – Vrouvrou Jan 3 '15 at 10:34
  • $\begingroup$ but why $\frac1n$ ? see my edit $\endgroup$ – Vrouvrou Jan 3 '15 at 10:42
  • $\begingroup$ Sorry it was a typo. I'll fix it. Regarding your edit, yes your observation is enough since then $d(n,m) \leq d(n,0) + d(m,0) \leq 2 \frac{\pi}{2}$, so the distance between any two terms is at most $\pi$. I think it sort of misses the point, though, that the whole metric $d$ is actually bounded. Since $\arctan(x)$ and $\arctan(y)$ are both numbers in $(-\pi/2,\pi/2)$, the distance $d(x,y) = |\arctan(x)-\arctan(y)|$ is less than $\pi$ for any $x,y \in \mathbb{R}$. $\endgroup$ – Mike F Jan 3 '15 at 10:51
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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.

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  • $\begingroup$ Can you tel me where is the contradiction with the fact that $\frac{\pi}{2}-\arctan{\ell} =0$ $\endgroup$ – Vrouvrou Jan 3 '15 at 11:30
  • $\begingroup$ You should use what you know of what a limit really is. If there is a limit, then for ALL $\epsilon>0$ you must be able to find a $N$ s.t. bla bla bla. You don't do any of that. You just have to work a little harder to find the contradiction. But basically my answer here is in the same vein as your attempt, but you just did not quite finish it. $\endgroup$ – user2520938 Jan 3 '15 at 11:36

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