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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

$1)$ $x_n$ 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 $n_k > x_*+1$.



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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.