Inf and Supremum of $\{\arctan(x) \; : \; x \in \mathbb R\}$

Question

I had trouble with this, but I think it's because there is no minimum and no maximum, correct?

For the $\inf$ I got $-\frac{\pi}{2}$ and for the $\sup$ I got $\frac{\pi}{2}$

But on second thought I feel like there is no $\inf$, $\min$, $\sup$, or $\max$ because $\arctan(x)$ is a subset of $\mathbb R$, however, there is a limit. So I'm confused.

Hints?

Answer 1

You have $\arctan x \le {\pi \over 2} $ for all $x$, hence ${\pi \over 2}$ is an upper bound. Furthermore, $\arctan ( \tan ({\pi \over 2} - {1 \over n})) = {\pi \over 2} - {1 \over n}$, hence ${\pi \over 2} = \sup_x \arctan x$.