# Real numbers equipped with the metric $d (x,y) = | \arctan(x) - \arctan(y)|$ is an incomplete metric space

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.

• 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. May 31, 2012 at 22:54
• Try the sequence $x_n = n$. Draw a picture. May 31, 2012 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. May 31, 2012 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 $\arctan(y) < \frac{\pi}{2}$. May 31, 2012 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. May 31, 2012 at 23:37

## 3 Answers

Sorry for reviving such an old problem...

Anyways, what is important here is that $$\text{arctan}$$ is a bijection from $$\mathbb{R}$$ to $$( -\pi/2, \pi/2 )$$, and it is an isometry if we give $$\left(-\pi/2, \pi/2\right)$$ the metric it carries as a subspace of $$\mathbb{R}$$ with the usual metric. If $$f: X \to Y$$ is a surjective isometry then the Cauchy sequences in $$Y$$ are the images of Cauchy sequences under $$f$$, and the convergent sequences in $$Y$$ are the images of convergent sequences under $$f$$, so $$Y$$ is complete if and only if $$X$$ is. Thus $$(\mathbb{R}, d)$$ is complete if and only if $$(-\pi/2, \pi/2)$$ with the metric $$\text{dist}(x, y) = |x - y|$$ is complete. But $$(-\pi/2, \pi/2)$$ is not complete since $$\{\pi/2 - 1/n\}_{n=1}^\infty$$ is Cauchy and does not converge in $$(-\pi/2, \pi/2)$$.

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 $$n \to \infty$$, $$\arctan x_n \to \pi/2$$. But $$(x_n)$$ does not converge to any element in $$\mathbb R$$ since there is no $$x \in \mathbb R$$ such that $$\arctan x = \pi/2$$.

• I think you meant : Observe that as $n \to \infty$, $\arctan(x_n) \to \pi/2$. Sep 29, 2017 at 17:11

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.

• thanks i took sequence $x_n = n$ as suggested by copper.hat May 31, 2012 at 23:33