Help with a simple theorem proof involving supremum(part 2)

Question

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?

Answer 1

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