For $A \subseteq \mathbb{R}$, we denote $-A$ to be the set obtained by taking the opposite of everything in $A$. That is, $- A : = \{-x: x \in A\}.$ Suppose that $A \neq \emptyset $ and that $A$ is bounded below. Prove that (1) $-A \neq \emptyset$ , (2) $-A$ is bounded above, and (3) $\sup(- A) = - \inf(A)$.
Here are my attempts: (1): Let $x\in A$ Then by the definition of -A we have $-x \in -A$. Therefore $-A \neq \emptyset$
(2): There exists an $a \in \mathbb{R}$ such that $a \leq x $ $\forall x \in A$. Then by the definition of -A we have $-x \leq -a$. Thus -A is bounded below by -x.
(3): Let $y=\sup(-A)$. Then Then $b \leq y $ $\forall b \in -A$. Then by the definition of -A it follows that $-y \leq a$ $\forall a \in A$, So $-y$ is a lower bound of A. Suppose there exists some $x \in A$ such that $x <-y+\varepsilon$ for all $\varepsilon>0$. Choose $\varepsilon=y+x$. Then we have the contradiction $x<x$. Finally $-y=\inf(A)$, so $\sup(-A)=-\inf(A)$.
I'm really not sure if that proof is correct. Can you spot a mistake?