Supremum and Infimum
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 $S$.
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.
You can do it by inequalities, or you can use the definitions of supremum and infimum.
Remember that a real number $S$ is the supremum of $A$ if and only if
- $a\leq S$ for all $a\in A$ ($S$ is an upper bound for $A$); and
- For all $\epsilon\gt 0$ there exists $a_{\epsilon}\in A$ such that $S-\epsilon \lt a_{\epsilon}$ (no number strictly smaller than $S$ is an upper bound for $A$).
(2 above can be written slightly differently so that it applies to arbitrary partially ordered sets: you require that for all $s$, if $s\lt S$, then there exists $a\in A$ such that $s\lt a$).
Dually, a real number $T$ is the infimum of $A$ if and only if
- $a\geq T$ for all $a\in A$ ($T$ is a lower bound for $A$); and
- For all $\epsilon \gt 0$ there exists $a_{\epsilon}\in A$ such that $T+\epsilon \gt a_{\epsilon}$ (no number strictly larger than $T$ is a lower bound for $A$).
So, you can use the fact that $T=\inf(A)$ is the infimum to show that $-T$ is the supremum of $-A$ by verifying the properties of the supremum. It should be straightforward.