# 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

(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?

• What are your definitions of $\sup S$ and $\inf S$? Apr 20, 2022 at 0:32
• @RobertShore its the same in the "here" link, I can add it in to the question Apr 20, 2022 at 0:34

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 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

• Thank you so much, its a massive relief that I got a few thing right. Thanks again. Apr 20, 2022 at 1:20
• This is a useful result, (ii) especially. Anytime you have $\sup S \in \mathbb R$, it assures you elements $s$ near to the supremum. Consider the case when $\sup S \not\in S.$ Now you can use (ii) to create a sequence of distinct points of $S$ that converges to $\sup S.$ Apr 20, 2022 at 1:57