How do you prove the infimum and supremum?

Question

Suppose that $S$ is nonempty and bounded above. Show that the set:

$$ -S:= \{-x \mid x \in S\} $$

is bounded below and that $\inf(-S) = -\sup(S)$.

Answer 1

Let $M=\sup(S)$. Then, for all $x\in S$ we have $M\geq x$ and for all $\epsilon> 0$ there is $x\in S$ with $x>M-\epsilon$.

Now, the two conditions above imply that, $-M$ satisfies that for all $-x\in -S$ we have $-M\leq -x$. Also, for all $\epsilon >0$ there is $-x\in -S$ such that $-x<-M+\epsilon$, as this follows from the inequality $x>M-\epsilon$ above, for the same $x\in S$.

This is the definition of infimum. Therefore, $-M=\inf(-S)$.