Showing a sequence is Cauchy with metric $d(x,y)=|\arctan(x)-\arctan(y)|$ I am trying to prove that the sequence $(x_n)=n$ in $\mathbb{R}$ is Cauchy with respect to the metric $d(x,y)=|\arctan(x)-\arctan(y)|$. 
Is it enough to say for some every $N<n<m$ we have $d(x_n,x_m)=|\arctan(n)-\arctan(m)|<|\arctan(n)-\arctan(N)|$ which goes to zero as $n$ goes to infinity? If not, then where does this fall apart, and how would I go about fixing it? 
 A: By definition: a sequence $(x_n)$ is Cauchy if for every $\varepsilon>0$ there is some $N\in\mathbb{N}$ such that 
$$d(x_m,x_n)<\varepsilon$$
whenever $m,n\geqslant N$. In your case we have 
$$d(x_m,x_n):=|\arctan x_m-\arctan x_n|<\varepsilon$$
whenever $m,n\geqslant N$. In the special case $x_n:=n$ notice that $\arctan(m)\leqslant\arctan (n)$ since $\arctan(\cdot)$ is an increasing function on $\mathbb{R}$. Moreover $|\arctan x|<\pi/2$ for all $x\in\mathbb{R}$. This implies that your sequence $\arctan(x_n)=\arctan(n)$ is a bounded monotone sequence hence it has a limit point by Bolzano-Weierstrass (the monotone convergence theorem for real sequences). In other words there is an $L>0$ such that $\lim_n\arctan(n)=L$ in fact $L=\pi/2$. This in particular implies that $(\arctan(n))$ is a Cauchy sequence i.e. for every $\varepsilon>0$ there is some $N\in\mathbb{N}$ such that 
$$|\arctan(m)-\arctan(n)|<\varepsilon$$
whenever $m,n\geqslant N$. This is equivalent to saying whenever $m,n\geqslant N$ 
$$d(x_m,x_n)<\varepsilon$$
by the definition of your metric. Hence $(x_n)$ is a Cauchy sequence. 
