Let $S\subset \mathbb{R}$ which is bounded below. Let $x\in S$ such that $-x\in -S$. Prove $\inf(S)=-\sup(-S).$
By definition a lower bound on $S\subset \mathbb{R}$ is a number $a\in \mathbb{R}$ such that $s\ge a$ for each $s\in S$. Since $S$ is non-empty and is bounded below then there exists an infimum. How can I continue this proof?
We know $\inf(S) \leq s$ for all s. Therefore, $- \inf (S) \geq -s $. In other words, $- \inf(S) $ is an upper bound for $-S$. Hence, $ \sup(-S) \leq - \inf(S) \implies \inf(S) \leq - \sup (-S)$. This is half of the problem. But, with this in mind, you can figure out the other direction. In other words, you want to show $\inf(S) \geq -\sup(-S)$. Ok, so you can say that since $-S$ is bounded above, then $-s \leq \sup(-S) \implies s \geq - \sup (-S)$. Therefore, $- \sup(-S)$ is a lower bound for $S$ which implies $\inf(S) \geq - \sup(-S) $
Finally $\inf(S) \leq - \sup (-S)$ and $\inf(S) \geq - \sup(-S) $ implies $$\inf(S) = - \sup (-S) $$