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Let $(\Bbb{R},d)$ be a metric space where $d(x,y)=\left\vert\arctan x−\arctan y\right\vert$. Is the sequence $x_n=\frac{1}{n}$ a Cauchy sequence with this metric?

The definition of Cauchy sequence is that, for any given $\varepsilon>0$ there exists $n_0$ such that $|x_m-x_n|<\varepsilon$ whenever $m>n\ge n-0$.

For this sequence we get $$x_m-x_n = \arctan \frac1m - \arctan \frac1n.$$ But how can the above expression be simplified or estimated?

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closed as off-topic by user26857, user228113, Martin Sleziak, user1551, Antonios-Alexandros Robotis Mar 17 '16 at 0:03

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Since $\arctan(x)-\arctan(y) =\arctan(\frac{x-y}{1+xy}) $, $\arctan(\frac1{n})-\arctan(\frac1{m}) =\arctan(\frac{\frac1{n}-\frac1{m}}{1+\frac1{n}\frac1{m}}) =\arctan(\frac{m-n}{mn+1}) $.

Therefore, if $N < n < m$, since $\frac{m-n}{mn+1} <\frac{m}{mN} =\frac1{N} $, $\arctan(\frac1{n})-\arctan(\frac1{m}) <\arctan(\frac1{N}) <\frac1{N} $ so the sequence is Cauchy.

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Hint: Consider a bijection $(-\pi/2,\ \pi/2)$ to $\Bbb R$ which preserves the metric and check whether the image of this sequence is Cauchy or not.

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By Lagrange theorem, if $x\ne y$ $$\arctan x-\arctan y=\frac{1}{1+\xi^2}(x-y)$$ for some $\xi\in (x,y)\cup(y,x)$.

Therefore, $d(x,y)\le \lvert x-y\rvert$. Which means that \begin{align}id_{\Bbb R}: (\Bbb R,\lvert\bullet\rvert)&\to (\Bbb R,d)\\x&\mapsto x\end{align} is a Lipschitz-continuous map. Since Lipschitz-continuous maps are uniformly continuous, and uniformly continuous maps send Cauchy sequences to Cauchy sequences, $\left(\frac1n\right)_{n\ge 1}$ must be Cauchy in $(\Bbb R,d)$.

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