Proof Verification: glb(A) = -lub(-A) Rudin proved this differently than I did, but I want to make sure my proof is valid.
Let $A$ be a non-empty set of real numbers which is bounded below. Let $-A = \{-x \mid x \in A\}$. Prove that $$\inf A = -\sup(-A)$$
Proof: As $A$ is non-empty and bounded below, there exists $\inf A$ by the least upper bound property of $\mathbb{R}$. Let $\alpha = \inf A$.
Then $\alpha \leq x \text{ } \forall x \in A \implies \alpha \leq -x \text{ } \forall x \in -A \text{ by definition of } -A \implies -\alpha \geq x \text{ } \forall x \in -A$
$\therefore -\alpha \text{ is an upper bound of } -A$
Suppose $\beta < -\alpha \implies \alpha < -\beta$
Since $\alpha = \inf A, -\beta$ is not a lower bound of A.
$\implies \exists x \in A$ such that $x < -\beta \implies -x < -\beta \text{ with } x \in -A \implies \beta < x \text{ with } x \in -A$
$\therefore \beta$ is not an upper bound of $-A \implies -\alpha = \sup (-A) \implies \inf A = -\sup(-A) \text{           } \Box$
Edit: Typo
 A: You have a typo after the words "by definition of": you say "for all $x \in -A$" when you mean "for all $a \in -A$", to be consistent with the previous line.
The line "Suppose $\beta < -\alpha \Rightarrow \alpha < -\beta$" is confusing. You mean "Suppose $\beta < -\alpha$. Then $\alpha < -\beta$." Then all subsequent occurrences of the word "with" should ideally be replaced with "there exists… such that", since your statements are all of the form "there exists $x \in -A$ such that $-x < -\beta$", for instance.
Basically, the maths is correct; the writeup is a little sloppy. It's usually better to leave new lines when chaining $\Rightarrow$ signs, and I personally prefer a written-out prose style:

Since $\alpha = \text{inf}(A)$, have $-\beta$ is not a lower bound of $A$. Therefore, there exists $x \in A$ such that $x < -\beta$. That is, there is $x \in -A$ such that $-x < -\beta$; equivalently, there is $x \in -A$ such that $\beta < x$. Therefore, $\beta$ is not an upper bound of $-A$.
Therefore, anything smaller than $-\alpha$ fails to be an upper bound of $-A$, so $-\alpha$ is the least upper bound of $-A$.

