Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A = -\sup(-A)$.
We know that $-A$ is bounded above. Hence $\sup(-A)$ exists. So to proceed, one would show that $\inf A \leq -\sup(-A)$ and $\inf A \geq -\sup(-A)$? Or $-\inf A > \sup(-A)$ and $-\inf A < \sup(-A)$.
I get the sense that one can use the following fact as well: (i) Let $\alpha = \sup S$. If $\beta < \alpha$ then $\beta$ is not an upper bound for $S$.
Best Answer
This is just a matter of using the definitions of supremum and infimum.
Just let $x = \inf{A}$ and $y = \sup{(-A)}$. So since $x = \inf{A}$ then for every $a \in A$ you know that $x \leq a$. But then this implies that $-x \geq -a$ for all $a \in A$, so $-x = - \inf{A}$ is an upper bound for the set $-A$. But since $y = \sup{(-A)}$ is the least upper bound for $-A$, then this means that $y \leq -x$ or that $\sup{(-A)} \leq - \inf{A}$.
You can prove the other inequality.