I need to prove that $\sup \{-x \mid x \in A\} = -\inf\{x \mid x \in A\}$ and am having trouble moving the $-x$ out of the $\sup$ to $\inf$. Another thing is that I don't quite know how to prove $b = \inf$ (see below). Any help would be appreciated! Here my attempt:
Let $a = \sup\{-x \mid x \in A\}$
Then by definition $a \geq -x$ for $x \in A \implies -a\leq x$ for $x \in A$
Thus $-a$ is a lower bound of $A$. Assume $b$ is a lower bound of $A$
Then $b \leq -a \implies a \geq -b$ and $a = \sup \{−x \mid x\in A\} = -\inf(A)$
After you say that $-a\leq x$ for all $x\in A$, you can then conclude that $$ -a \leq b:= \inf A $$ and hence $$ a \geq -b $$ Similarly, $b\leq x$ for all $x\in A$, and hence $-b \geq -x$ for all $x\in A$, and so $$ -b \geq a $$