Question:
Let $A$ be a set of real numbers that is bounded above and set $\alpha = supA$. Let $B=\{-a:a\in A\}$. Prove that $infB=-\alpha$.
My approach: (edited with the help of a comment by Dave)
For $-\alpha$ to be equal to $infB$, the following two statements should satisfy.
$-\alpha\leq b$ for all $b\in B$.
If there exist $\mu$ which is a lower bound for $B$, then $-\alpha\geq\mu$
We first note that $\alpha$ being $supA$ implies that $a\leq \alpha, \forall a \in A$. This can also be stated as $-a \geq -\alpha$.
Now we prove statement 1 by a contradiction. Assume $\exists b \in B: b< -\alpha$. Now set $b=-a$. This results in $-a<-\alpha$. Which is a contradiction (see the previous paragraph). Hence, statement 1 is true.
Now we prove statement 2. Let $\mu$ be a lower bound on $B$. This implies that, $\mu \leq -a, \forall -a$. By rearranging we get $-\mu \geq a, \forall a$. Therefore, $-\mu$ is an upper bound on $A$. But, $\alpha \leq -\mu$ since $\alpha = supA$. We can rearrange again to get $-\alpha \geq \mu$ as required.
We now conclude our proof is complete.
I need to check whether this proof is right. Especially, the proof of the second statement.