I am a beginner at Real Analysis and attempting to prove some of the results in Problems in Mathematical Analysis volume - I by Kaczor and Nowak. I would like someone to verify, if my proof to the below question is rigorous and correct. If it isn't, I would like some hints that would lead me to the correct proof.
1.1.2. Let $A \subset \mathbb{R}$ be a non-empty subset. Define $-A=\{x:-x \in A\}$. Show that \begin{align*} \sup (-A)&=-\inf A \\ \inf (-A)&=-\sup A \end{align*}
Proof.
(1) Let $l \in \mathbb{R}$ be the infimum of the set $A$. From the definition of the greatest lower bound for a set:
(a) $l$ is a lower bound for $A$.
So, $l \le x$ for all $x \in A$. Thus, $-l \ge -x$ for all $x$. So, $-l$ is an upper bound for $-A$.
(b) If $b$ is any lower bound for $A$, $b \le l$. Therefore, $-b \ge -x$ for all elements in $-A$ and $-b \ge -l$.
Hence, we can conclude that $-l$ is the supremum of $-A$.
(2) The proof for the infimum of $-A$ follows similarly.