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If I equip $\mathbb{R}$ with the metric

$$ \rho(x,y) := \left|\arctan(x) - \arctan(y)\right| $$

then sequences like for example $x_n = n$ are Cauchy sequences, so it is clear that $\mathbb{R}$ is not complete with respect to this metric.

I need to give the completion, and it seems to be rather immediate that it can only be the extended real line. (where else could a non-oscillating non convergent sequence tend to after all ?)

I am a little unhappy with my reasoning, how would I rigorously construct the completion? I would be very grateful for some hints.

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Depends on what tools you have available. One approach would be to notice that the arctan function itself is an isometry from $\mathbb R$ with the $\rho$ metric, to $(-\pi,\pi)$ with the ordinary distance, so you can complete the space in the latter representation instead. – Henning Makholm Oct 3 '12 at 17:41
@HenningMakholm I think the course my be leading to the kind of tools you mention eventually, but I haven't met them yet. Thanks for your comment though ! would you say then, that for the moment an argument like mine is perhaps quite acceptable ? (in the sense that it s not really wrong) – Beltrame Oct 3 '12 at 17:46
Prove (1) any bounded Cauchy sequence has a limit already in $\Bbb R$; (2) each Cauchy sequence tending to $+\infty$ (respectively $-\infty$) tend to the same thing in the completion; and (3) any sequence that is unbounded in both directions is not Cauchy. Hence the completion is $\Bbb R\cup\{\pm\infty\}$ with the values $\arctan(\pm\infty)=\pm\pi/2$ (respectively), where $+\infty$ and $-\infty$ stand for the obvious corresponding equivalence classes of Cauchy sequences (this is one way to construct completions of course). @HenningMakholm I believe you mean $(-\pi/2,+\pi/2)$. – anon Oct 3 '12 at 17:55
@anon: Indeed I do. Alas, too late to edit now. – Henning Makholm Oct 3 '12 at 18:00
up vote 7 down vote accepted

The map $\tan : ((-\frac \pi 2, + \frac \pi 2),d) \to (\mathbb R,\rho)$, where $d$ is the usual metric $d(x,y) = |x-y|$, is an isomorphism of metric spaces : it is bijective and $\rho(\tan x, \tan y) = d(x,y)$. So your question is equivalent to : "What is the completion of $(-\frac \pi 2, + \frac \pi 2)$ equipped with the usual distance ?"

Well, since $(-\frac \pi 2, + \frac \pi 2)$ is a subset (can I say a sub-metric space ?) of $\mathbb R$, which we know is complete, its completion is its closure in $\mathbb R$, which is $[-\frac \pi 2, + \frac \pi 2]$ : We only have to add two new points, corresponding to sequences converging to $\pm \frac \pi 2$, or in the original context, sequences diverging to $\pm \infty$.

Of course you can try to show this directly, but you will only hide all of this and make it harder. Here is what you would have to prove :

  • If a sequence $(y_n)$ diverges to $+ \infty$, then it is equivalent to your sequence $(x_n = n)$ : $\rho(x_n,y_n) = \arctan(n)- \arctan(y_n)$ and since both sequence diverge to $+ \infty$, $\arctan(n)$ and $\arctan(y_n)$ both converge to $\pi/2$, thus $\rho(x_n,y_n)$ converges to $0$.
  • If a sequence $(y_n)$ diverges to $- \infty$, then it is equivalent to the sequence $(x_n = -n)$
  • If a sequence $(y_n)$ converges to $a \in \mathbb R$, then it is equivalent to the constant sequence $(x_n = a)$. This results from the continuity of $\arctan$
  • If a sequence is in none of the above three cases, it is not a Cauchy sequence. This is a bit painful and needs to use the fact that $\mathbb R$ is complete
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