$S\subset \mathbb{R}$ is non-empty. We define $-S:=\{-s:s\in S\}$.
First, I prove $\inf S \geq -\sup(-S)$.
$-s\leq \sup(-S)$ for every $-s\in -S$ because supremum is an upper bound.
$\implies s \geq -\sup(-S)$ for every $s\in S$. So we have a lower bound for $S$.
$\implies \inf S \geq -\sup(-S)$ because infimum is the greatest among lower bounds.
Now, I prove $\inf S \le -\sup(-S)$ and it suffices to conclude the equality.
$s\ge \inf S$ for every $s\in S$ because infimum is an lower bound.
$\implies -s \le -\inf S$ for every $-s\in -S$. So we have an upper bound for $-S$.
$\implies \sup(-S)\le -\inf S$ because supremum is the smallest among upper bounds.
$\implies \inf S \le -\sup(-S)$.
Is this proof acceptable? If it was a quiz or a test, how much score would I get?
And how do I improve it?
Best Answer
Yes, this proof is acceptable. you will get decent score. Be aware that $\Bbb R$ is complete so for a bounded set supremum and infimum always exist by completeness property. If $S$ is unbounded then you have to consider extended real number system.