1. ## supremum

Let $\displaystyle A$ be a nonempty set of real numbers which is bounded below. Let $\displaystyle -A = \{-x: x \in A \}$. Prove that $\displaystyle \inf A = -\sup(-A)$.

So I want to show that (i) $\displaystyle \inf A \leq -\sup(-A)$ and (ii) $\displaystyle \inf A \geq -\sup(-A)$. Now $\displaystyle \inf A$ exists because $\displaystyle A \neq \emptyset$ and $\displaystyle A$ is bounded below. Also, $\displaystyle \sup(-A)$ exists since $\displaystyle -A$ is non-empty and bounded above.

Now $\displaystyle -x \leq \sup(-A) \leq \inf A$ and $\displaystyle \sup(-A) \leq \inf A \leq x$. I took $\displaystyle A$ to have a lower bound greater than $\displaystyle 0$. You can apply the same argument if you set $\displaystyle A$ to have a lower bound less than $\displaystyle 0$ (e.g. break it into cases)? From these inequalities, we can conclude that $\displaystyle \inf A = -\sup(-A)$? Adding them we get $\displaystyle \sup(-A)-x \leq \sup(-A)+ \inf A \leq \inf A + x$.

$\displaystyle x \in A\, \Rightarrow \,\inf (A) \leqslant x\, \Rightarrow \, - x \leqslant - \inf (A)\, \Rightarrow \,\sup ( - A) \leqslant - \inf (A)$.

Now suppose that $\displaystyle sup ( - A) < - \inf (A)$, then $\displaystyle \inf (A) < - \sup ( - A)$.

By definition of inf, $\displaystyle \left( {\exists a \in A} \right)\left[ {\inf (A) \leqslant a < - \sup ( - A)} \right]$.

But that means that $\displaystyle \sup ( - A) < - a$, a contradiction.

3. Originally Posted by Plato
$\displaystyle x \in A\, \Rightarrow \,\inf (A) \leqslant x\, \Rightarrow \, - x \leqslant - \inf (A)\, \Rightarrow \,\sup ( - A) \leqslant - \inf (A)$.
Now suppose that $\displaystyle sup ( - A) < - \inf (A)$, then $\displaystyle \inf (A) < - \sup ( - A)$.
By definition of inf, $\displaystyle \left( {\exists a \in A} \right)\left[ {\inf (A) \leqslant a < - \sup ( - A)} \right]$.
But that means that $\displaystyle \sup ( - A) < - a$, a contradiction.
Or $\displaystyle x \in A \Rightarrow x \geq \inf A$. Then $\displaystyle -x \leq -\inf A$. Thus $\displaystyle \sup(-A)$ exists. Now suppose there is some $\displaystyle p$ such that $\displaystyle -x \leq p < -\inf A$. Then $\displaystyle \inf A \leq -p < x$. This contradicts the definition of infimum. Hence $\displaystyle \sup(-A) = - \inf A$. Or $\displaystyle \inf A = -\sup(-A)$.