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

Question

This is another probably heavily searched for problem from Rudin (Proof that $\inf A = -\sup(-A)$) But, my proof differs and I'd just like some verification and critique of proof writing and style.

Let $A$ be a nonempty subset of $\mathbb{R}$. Prove that $\inf A = - \sup(-A)$

Let $\alpha = \inf A$ and $\beta = - \sup B$ where $B = \{-x | x\in A\}$

WLOG assume $\alpha \neq \beta$ and $\alpha > \beta$.

$\exists q \in\mathbb{Q}$ such that $\alpha > q > \beta$.

So $-q < -\beta = \sup B$ Therefore $ \exists b \in B$ where $b > -q$ by definition of $\sup$.

Since $\exists a \in A$ where $-a = b$ implies $a < q$. But that leads to $a < q < \alpha$.

This is a contradiction since $\alpha$ is the $\inf$.

Thus $\inf A = - \sup(-A)$

Answer 1

Globally, your proof is correct. I have only two minor remarks:

1. Why do you introduce $\mathbb Q$? It is not wrong, but it sounds a bit arbitrary.
2. Writing $\sup-A$ looks strange. It would be beter to write $\sup(-A)$.