1. ## supremum

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

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

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

2. Here is an outline that may help you get some ideas on your proof.

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

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

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

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

3. Originally Posted by Plato
Here is an outline that may help you get some ideas on your proof.

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

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

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

But that means that $\sup ( - A) < - a$, a contradiction.
Or $x \in A \Rightarrow x \geq \inf A$. Then $-x \leq -\inf A$. Thus $\sup(-A)$ exists. Now suppose there is some $p$ such that $-x \leq p < -\inf A$. Then $\inf A \leq -p < x$. This contradicts the definition of infimum. Hence $\sup(-A) = - \inf A$. Or $\inf A = -\sup(-A)$.