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?