I'm having trouble understanding part of this proof:
Take a sequence $x_n=n$. Since $arctan(n)\to \pi/2$, for give $\epsilon>0$, there is N so that $|arctan(n)-\pi/2|<\epsilon/2$ for $n\geq N.$ Hence
$d(x_n,x_m)=|arctan(n) - arctan(m)|\leq|arctan(n) - \pi/2|+|arctan(m) - \pi/2|<\epsilon$.
for $n,m\geq N$. Assume $d(x,y)\to 0$ for some $x \in R$. Then $pi/2=arctan x$ which is impossible.
What I don't understand is the first part where we say that since $arctan(n) \to \pi/2$ we can have:
$|arctan(n)-\pi/2|<\epsilon/2$
After that I see we use the triangle inequality using this same expression. Why do we substitute the second arctan with $pi/2$ since arctan will never become $\pi/2$ and what is the sole purpose of having this expression $<$ than $\epsilon$/2 and not just $\epsilon$.
Any hint is appreciated, regards.
This is just the definition of convergence: to say that $\arctan(n)$ converges to $\pi/2$ means that for any $\delta>0$, there exists $N$ such that $|\arctan(n)-\pi/2|<\delta$ for all $n\geq N$. Now given $\epsilon>0$, define $\delta=\epsilon/2$ and use the definition to obtain $N$ such that $|\arctan(n)-\pi/2|<\epsilon/2$ for all $n\geq N$.
The reason to choose $\epsilon/2$ instead of $\epsilon$ here is that you eventually want to show that $d(x_n,x_m)<\epsilon$ whenever $n,m\geq N$ (to satisfy the definition of a Cauchy sequence) and in the course of proving that our $\delta$ will end up getting doubled.