Help with a simple theorem proof involving supremum(part 2) This is similar to the question I asked here, but this time I want to focus on the first part of the theorem(assuming second part is similarly done)
I have been struggling with trying to prove stuff and any advice\hints will help me to get a better understanding.
Here is the  theorem(again):
(a) $\alpha = \sup S \iff (i)\ \alpha \ge x$ $\forall x \in S$; and
$\qquad\qquad\qquad\qquad\ \ \ \ \ (ii)\ \forall a < \alpha , \exists s \in S$ such that $a<s\le\alpha$
(b) $\beta = \inf S \iff (i)\ \beta \le x$ $\forall x \in S$; and
$\qquad\qquad\qquad\qquad\ \ \ \ (ii)\ \forall b > \beta , \exists w \in S$ such that $\beta \le w \lt b$
Here are the definitions


Here is my proof attempt for proving part (a)
Let $\alpha \in \mathbb{R}$ and $S$ be a non empty set in $\mathbb{R}$ (I think that this is correct)
forward
Let $\alpha = sup(S)$
This implies $\alpha$ is an upper bound thus $\alpha \ge x$ $\forall x \in S$
Now  assume by contradiction that$ \ \ \forall a < \alpha$ and  $\forall s \in S$ that $a \ge s$
This implies that $a$ is an upper bound of $S$.
But as $a< \alpha$, this contradicts that $\alpha$ is a supremum.
backward
Assume that (i) and (ii) are true.
Suppose $\exists \beta \ge x$ s.t $\beta < \alpha$
Thus $\beta$ is an upper bound of $S$
But then $\exists s \in S$ s.t $\beta < s \le \alpha$ which is a contradiction to the existence of $\beta$.
thus $sup(S) = \alpha$.
This concludes my proof attempt
Does my proof for part(a) look right?
 A: Your attempt is mostly okay. We can improve it as follows. (My additions in bold.)
(a) $\alpha = \sup S \iff (i)\ \alpha \ge x$ $\forall x \in S$; and
$\qquad\qquad\qquad\qquad\ \ \ \ \ (ii)\ \forall a < \alpha , \exists s \in S$ such that $a<s\le\alpha$

forward
Let $\alpha = \sup(S)$.
This implies $\alpha$ is an upper bound thus $\alpha \ge x$ $\forall x \in S.$
Now  assume by contradiction that $\textbf{there exists } a < \alpha \textbf{ such that } \forall s \in S, \textbf{ we have that } a \ge s.$
This implies that $a$ is an upper bound of $S$.
But as $a< \alpha$, this contradicts that $\alpha$ is a supremum, by the second bullet of the definition.
$\blacktriangledown$
backward
Assume that (i) and (ii) are true.
Suppose $\exists \beta, \textbf{ an upper bound of } S, \text{s.t. } \beta < \alpha.$
But then (by (ii)) $\exists s \in S$ s.t $\beta < s \le \alpha$ which is a contradiction to the existence of $\beta$.
thus $\sup(S) = \alpha$.
$\blacksquare$
You might need some practice in negating logical statements. For example: Elementary logic. Negation
