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I have to show that the real numbers equipped with the metric $ d (x,y) = | \arctan(x) - \arctan(y)| $ is an incomplete metric space.

Certainly, I have to search for a cauchy sequence of real numbers with respect to given metric that must not be convergent. But I am unable to figure out that. Can anybody help me with this.

Thanks for helping me.

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The simplest way to do this is to notice that $x \mapsto \arctan(x)$ is an isometry for your space to the open interval $(-\pi/2,\pi/2)$ with its usual metric. –  Chris Eagle May 31 '12 at 22:54
Try the sequence $x_n = n$. Draw a picture. –  copper.hat May 31 '12 at 23:00
You can show easily that it does not converge to any number, ie, for any $y$, show $d(n,y)$ does not converge to zero. –  copper.hat May 31 '12 at 23:10
$d(n,y) = | \arctan(n) - \arctan(y)|$. So $d(n,y) \to (\frac{\pi}{2}-y)$. $y$ is a fixed number, so $\arctan(y) < \frac{\pi}{2}$. –  copper.hat May 31 '12 at 23:26
I am showing that $x_n$ does not converge to any fixed y. $\arctan y$ is a fixed number, it doesn't tend towards anything but itself. –  copper.hat May 31 '12 at 23:37

2 Answers 2

Consider a sequence that grows without bound. Such a sequence isn't Cauchy in the usual metric on $\mathbb{R}$, but will be under this metric.

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thanks i took sequence $x_n = n$ as suggested by copper.hat –  srijan May 31 '12 at 23:33

Consider $x_n = n$. Let $\varepsilon > 0$ and choose $\displaystyle N > \tan\bigg(\frac\pi2 - \varepsilon\bigg)$. If $m, n > N$ then $\displaystyle \{\arctan m, \arctan m\} \subseteq \bigg(\arctan N, \frac\pi2\bigg)$. Thus $$d(x_m, x_n) = \vert \arctan m - \arctan n \vert \leq \bigg \vert \frac\pi2 - \arctan N \bigg\vert < \bigg \vert \frac\pi2 - \frac\pi2 + \varepsilon \bigg\vert= \varepsilon$$ Thus $(x_n)$ is a Cauchy sequence. Observe that as $n \to \infty$, $x_n \to \pi/2$. But $(x_n)$ does not converge to any element in $\mathbb R$ since there is no $x \in \mathbb R$ such that $\arctan x = \pi/2$.

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