Help with a simple theorem proof involving supremum

Question

Here is the theorem(just how they gave it, no details left out):

(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 $\beta \le w<\beta$

My first question is that for part (a) of the theorem, are they saying:

$\alpha = sup(S) \iff ( \alpha \ge x$ $\forall x \in S$ and $\forall a < \alpha , \exists s \in S$ such that $a<s\le\alpha$)

and not

1. $\alpha = sup(S) \iff \alpha \ge x$ $\forall x \in S$

2. $\alpha = sup(S) \iff \forall a < \alpha , \exists s \in S$ such that $a<s\le\alpha$

and similar for part (b)?

My second question is that they defined supremum/infimum as follows:

This came just before the proof, but can't supremum/infimum $\alpha$ and $\beta$ be $\pm\infty$? or is that a different case?

My Third question is that they did not state what set of numbers $\alpha$ and $\beta$ belong to and did not define what S was in the theorem so are we to assume that its the same as stated in the definitions of supremum/infimum?

I then tried to prove this but didn't get very far and would really appreciate it if you could offer some advice.

Answer 1

(1) For your first question, your first interpretation is correct. With the definition you provided, you can try to see why the second interpretation would not be correct.

(2) With regards to your second question, if $S$ is nonempty and bounded above, then $\alpha = \sup(S)$ cannot be $\pm \infty$. To see why, note that since $\alpha$ is nonempty, we know that there must be some real number $x_0 \in S$. By definition of supremum, $x_0 \leq \alpha$.

Furthermore, since $S$ is bounded from above, there must be some real number $M$ such that for all $s \in S$, $s \leq M$. Once again, by definition of supremum, $\alpha \leq M$.

Putting both inequalities, we obtain $x_0 \leq \alpha \leq M$ and this means that $\alpha$ can never be $\pm \infty$ in this case.

The same explanation goes for the case where $\beta = \inf(S)$ where $S$ is nonempty and bounded from below.

(3) For the third question, $\alpha$ and $\beta$ are just real numbers, that is, they belong to the set $\mathbb{R}$. In particular, $\alpha$ and $\beta$ may not be elements of the set itself. As an example, consider the set $S = (1,2)$. You can verify that $\sup(S) = 2$ and $\inf(S) = 1$ but neither of them are in $S$.

Answer 2

1. The former interpretation, i.e. $$(\alpha = \sup S) \iff \big( (\alpha \ge x\;\forall x \in S )\text{ and } (\forall a < \alpha ,\,\exists s \in S \text{ such that } a<s\le\alpha)\big).$$

2. Generally speaking, the supremum of $S$ can be $\pm \infty$. But the definition 02.4 rules out the case $S=\emptyset$, which is the only way to have $\sup S = -\infty.$ Easy examples of the situation $\sup S=+\infty$ are seen by letting $S$ be the integers, or simply $S=\mathbb{R}$. But again your author rules out these types of examples by requiring that $S$ be bounded (from) above. Anyway, you are correct that the idea of supremum is easily extended to apply to the set of all subsets of the real line.

3. I think the author did say so, but the notification occurs in your definitions 02.4 and 02.5. Note that the words "a real number" appear in each definition. The author is implicitly referring you to the definition, merely by using the symbol $\sup$ in your theorem. Similarly for $S$, I would presume $S$ is a real, non-empty set.

As for your proof, I have offered some feedback for you: https://math.stackexchange.com/a/4431587