Proof that $\inf(A)=-\sup(-A)$

Question

Prepping for a master's program in pure mathematics. I'm working on my problem-solving skills and was hoping someone would kindly verify my proof.

Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A = \{-x| x \in A\}$. Prove that $$ \inf A = -\sup (-A) $$

My attempt: Lets begin with a lemma.

Lemma: $x$ is a lower bound of $A$ if and only if $-x$ is an upper bound of $-A$.

proof of lemma: For each $a \in A$, \begin{align*} x \text{ is a lower bound of } A &\iff x \leq a \\ &\iff -x \geq -a \\ &\iff -x \text{ is an upper bound of } -A. \end{align*}

Now, let $\inf A = \alpha$. Then $- \alpha$ is an upper bound of $-A$. Let $y \in \mathbb{R}$ and suppose $y < -\alpha$. So, $-y > \alpha$ which implies $-y$ not a lower bound $A$. Hence, $y$ is an upper bound of $A$. Therefore, $-\alpha = \sup (-A)$ and $\inf (A) = - \sup (-A)$.

Answer 1

You wrote: For each $a \in A$: $$ x \text{ is a lower bound of } A \iff x \leq a$$

This is wrong because the quantifier "For each $a \in A$" is in the wrong place: it should not apply to the whole statement, because there may be lots of $a \in A$ where $x \leq a$ without $x$ being a lower bound of $A$.
It should be instead

$$ x \text{ is a lower bound of } A \iff \text{ for each }a \in A,\ x \leq a$$