Prove that the real line equipped with the distance $p(x,y)= |\arctan (x)-\arctan (y)|$ is an incomplete metric space.

My attempt

Duplicate but I want to ask about my specific proof and also how to prove that it is a metric space

A metric space is called complete if every cauchy sequence of points in $M$ converges in $M$.

A sequence $x=(x_1,x_2,...x_n)$ is cauchy if for all $\epsilon>0$ there exists an $N \in \mathbb{N}$ such that $m,n \geq N$ implies $d(x_n,x_m)< \epsilon$.

The $\arctan$ function is well-defined from $-\pi/2$ to $\pi/2$.

Fix $\epsilon >0$. Let $X_n=n$. $$d(n,\pi/2)<\epsilon$$

$$|\arctan(n)-\arctan(\pi/2)| < \epsilon$$

Not sure where to go from here. I think I also need to prove that this is a metric

  • 3
    $\begingroup$ Is the sequence $n \mapsto n$ Cauchy? Does it converge? $\endgroup$ – copper.hat May 1 '17 at 1:55
  • $\begingroup$ The domain of $\arctan$ is all of $\mathbb{R}$. Its range is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. $\endgroup$ – kccu May 1 '17 at 1:55
  • $\begingroup$ Note that $\arctan(x)-\arctan(y) = \arctan{\frac{x-y}{1+xy}}$. $\endgroup$ – marty cohen May 1 '17 at 4:49
  • $\begingroup$ Ah, yes, this problem comes from the most recent Putnam exam, doesn't it? $\endgroup$ – Feryll May 1 '17 at 8:03
  • $\begingroup$ $\arctan(x)$ is well-defined on all of $\mathbb{R}$, with values in $(-\frac{\pi}{2}, \frac{\pi}{2})$. The precise function is irrelevant for proving it to be a metric, it only needs to 1-1. $\endgroup$ – Henno Brandsma May 1 '17 at 8:50

(I). If $f:\mathbb R\to \mathbb R$ is injective then $d(x,y)=|f(x)-f(y)|$ is a metric on the set $\mathbb R:$

(i). $x\ne y\implies d(x,y)=|f(x)-f(y)|\ne 0$ because $f$ is $1$-to-$1$.

(ii). $d(x,y)=d(y,x).$ Obvious.

(iii). $d(x,z)=|f(x)-f(z)|=|(f(x)-f(y))+(f(y)-f(z)|\leq$ $\leq |f(x)-f(y)|+|f(y)-f(z)|=d(x,y)+d(y,z).$

The derivative of $\arctan x$ is $1/(1+x^2) ,$ which is always positive. So $\arctan x$ is strictly increasing, and therefore is $1$-to-$1.$

(II). We have $\lim_{n\to \infty}\arctan n=\pi /2,$ so $(\arctan n)_n$ is a Cauchy sequence, that is, $$\lim_{n\to \infty}\sup_{a>b>n}|\arctan a-\arctan b|=0.$$ So with respect to the metric $d(x,y)=|\arctan x-\arctan y|,$ the sequence $(n)_n$ is a $d$-Cauchy sequence.

But there is no $z\in \mathbb R$ such that $\lim_{n\to \infty}d(n,z)=0$, as this would require $\lim_{n\to \infty}|\arctan (n)-\arctan (z)|=0,$ which implies $\arctan z=\lim_{n\to \infty}\arctan n=\pi /2,$ and there is no such $z\in \mathbb R.$


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