Rudin's POMA Chapter 1 exercise 5 Hi I am writing to check if the proof that I wrote is valid, I feel like it seems kind of right to me but as I am only a beginner in writing proofs I feel like I might've missed something.
Question: Let E be a nonempty set of real numbers which is bounded below. Let -A be the set of all numbers -x, where x is an element of A. Prove that $$ infA=-sup(-A)$$ 
My solution is as follows, 
$\text{Let } \alpha \space\text{be a lower bound of A}$
Then
$\alpha = \inf (A)\space \text{if}\space x\geq\alpha\space\text{for every}\space x\in A\space \text{and there does not exist a }\beta\space \text{where}\space  \beta\gt\alpha\space \text{and} $
$\space x\geq\beta\space \text{for}\space x\in A$
As $x \geq \alpha$ for $x \in A$ 
then, $-x \leq -\alpha $ for all $-x \in -A$ 
Since there is no $\beta \gt \alpha$ where x $\geq \beta $ for $x \in A$ 
Then there is no $-\beta \lt \alpha$ where $-x \leq -\beta$ for $-x \in A$
thus $-\alpha = -\sup (-A)$ and $\inf (A) = -\sup (-A)$
Is it right for me to just assume the existence of a greatest lower bound like that?
Also, what are some good ways to self check my work?

If $A$ is bounded below then there exists $M$ such that $a\in A\implies M\le a$ for all $a\in A$. Therefore, $-a\le -M$ for all $-a\in -A$ and so, by the least upper bound property of the real numbers, there exists $U$ such that $U$ is a least upper bound for $-A$.
Since $-x\le U$ for all $-x\in -A$, $-U\le x$ for all $x\in A$ and therefore is a lower bound for $A$. If $L$ is another lower bound for $A$ then $L\le x\implies -x\le -L$ so $-L$ is also an upper bound for $-A$.
Since $U$ is the least upper bound for $-A$, $U\le -L\implies L\le -U$ so $-U$ is the greatest lower bound.
Now $-U= -\text{sup(-A)}=\text{inf(A)}$ so the theorem is proved.
