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
(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 $z\in \mathbb R.$