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

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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 ${\pi \over 2} = \sup_x \arctan x$.

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### hax0r_n_code

Updated on August 01, 2022

• hax0r_n_code over 1 year

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?

• JMoravitz over 7 years
The supremum of a set need not necessarily be an element of the set itself. $[0,1)=\{x~:~0\leq x<1\}$ has $1$ for a supremum despite $1\not\in[0,1)$.
• MathematicsStudent1122 over 7 years
The image of $f:\mathbb{R} \to \mathbb{R}$ given by $f(x) = \arctan x$ is the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. The $\sup$ and $\inf$ of an open interval $(a,b)$ is, respectively, $b$ and $a$.
• hax0r_n_code over 7 years
@MathematicsStudent1122 ok, so I got that part right, was I correct in assuming no max and min since it is of an open interval?
• MathematicsStudent1122 over 7 years
@free_mind Yes.
• IAmNoOne over 7 years
Recall the english meaning of sup and inf (least upperbound, greatest lowerbound).