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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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    $\begingroup$ 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. $\endgroup$
    – Chris Eagle
    May 31, 2012 at 22:54
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    $\begingroup$ Try the sequence $x_n = n$. Draw a picture. $\endgroup$
    – copper.hat
    May 31, 2012 at 23:00
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    $\begingroup$ 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. $\endgroup$
    – copper.hat
    May 31, 2012 at 23:10
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    $\begingroup$ $d(n,y) = | \arctan(n) - \arctan(y)|$. So $d(n,y) \to (\frac{\pi}{2}-y)$. $y$ is a fixed number, so $\arctan(y) < \frac{\pi}{2}$. $\endgroup$
    – copper.hat
    May 31, 2012 at 23:26
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    $\begingroup$ 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. $\endgroup$
    – copper.hat
    May 31, 2012 at 23:37

3 Answers 3

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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 $(-\pi/2, \pi/2)$.

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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 $\arctan x = \pi/2$.

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    $\begingroup$ I think you meant : Observe that as $n \to \infty$, $\arctan(x_n) \to \pi/2$. $\endgroup$
    – Jonathan
    Sep 29, 2017 at 17:11
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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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  • $\begingroup$ thanks i took sequence $x_n = n$ as suggested by copper.hat $\endgroup$
    – Srijan
    May 31, 2012 at 23:33

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