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Is it possible to define a metric on $\mathbb{R}$ such that $\mathbb{R}$ is not complete? I know it for this to happen, we would need to construct a Cauchy sequence that does not converge in $\mathbb{R}$, but I'm having trouble doing this. Would the completion of $\mathbb{R}$, with respect to this metric, be some subset of the complex numbers?

ponchan
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    For the second question: If you change the metric of $\mathbb{R}$ then it can't be a subset of the complex numbers (as the metric of the complex numbers induced to $\mathbb{R}$ is the standard metric). – Yanko Mar 24 '18 at 17:57
  • For the first: https://math.stackexchange.com/questions/1079920/a-not-complete-metric-space – spaceisdarkgreen Mar 24 '18 at 18:00

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Let $f(x)=\arctan x$, define $d(x,y)=|\arctan x - \arctan y|$ then $d$ is a metric since it is trivially positive definite, symmetric and sub-additive. Now can you show it is not complete?

Hint: consider the sequence $1,2,3,4,\cdots$ under this metric.

Vim
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  • So the sequence $(x_n)=n$ diverges. To show that it is cauchy we need for every $\epsilon >0$ there to be an $N\in\mathbb{N}$ such that whenever $n,m>N$ we have $|\arctan(x_n)-\arctan(x_m)|<\epsilon$. – ponchan Mar 24 '18 at 18:08
  • Which is true since the terms become arbitrarily close? I see that the terms get closer together, but not sure why we're guaranteed that they get arbitrarily close. – ponchan Mar 24 '18 at 18:11
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    @ponchan you can calculate it explicitly, or note that $|\arctan x-\pi/2|\to 0$ where you can replace $x$ by either $m$ or $n$. – Vim Mar 24 '18 at 18:14
  • Does it make sense to say: Let $N<n<m$. Then $d(x_n,x_m)=|\arctan(n)-\arctan(m)|<|\arctan(n)-\arctan(N)|\rightarrow 0$ as $N\rightarrow\infty$? – ponchan Mar 24 '18 at 19:01
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    @ponchan no your last comment is incorrect since you fixed $n$. Actually what I meant was that $|\arctan n - \arctan m| \le |\arctan n - \pi/2|+|\arctan m - \pi/2|$ and the two terms on the far right side are both smaller than a prescribed $\epsilon$ when both $n,m$ exceed some corresponding threshold $N_\epsilon$ (recall the $\epsilon-N$ definition of $\lim_{n\to \infty}\arctan n = \pi/2$). – Vim Mar 25 '18 at 06:08
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    @ponchan and to show that the sequence ${n}$ diverges, simply use proof by contradiction: suppose there exists some $a\in\Bbb R$ that it converges to, then we must have $|\arctan n - \arctan a|\to 0$ which makes $\arctan a = \pi/2$ which is impossible. – Vim Mar 25 '18 at 06:13