The characterization of an interval theorem.

Question

Here's the 1.5 pages where my text defines intervals. I'm having a difficult time understanding the characterization theorem.

If we just consider the first case in the theorem that $S$ is bounded. We are given that $S$ contains at least two points. But what if it contains only two points? Then there's no possible way $S$ could be an interval. But the proof opens up with "Let $a = \inf S$ and let $ b = \sup S$. If $a < z$ then $z$ is not a lower bound of $S$... etc. "

So this characterization theorem is basically saying that, for example, the set $S := \{3,5\}$ is an interval? That seems like a definition of sorts not a theorem.

This is how the proof seems to me:

Consider the set $S := \{3,5\}$. $\inf S = 3$ and $\sup S = 5$. Now $3<4$ so $4$ is not a lower bound of $S$. Therefore there exists an element of $S$ in between $3$ and $4$.

Answer 1

No, the theorem do not say that {3,4} is an interval. S={3,4} does not verify the assumption of the theorem : $x=3 \in S$ and $y=4 \in S$, but [x,y]=[3,4] is not included in S.

Remark: Another way to state the charaterization Theorem is to say that the convex subsets of $\mathbb R$ are its convex sets.

Answer 2

No, $S=\{3,5\}$ is not an interval, indeed it does not satisfy the above property (see Paramanand Singh's comment).

The preliminary observation is: Every interval $S\subseteq \mathbf{R}$ with $|S|\ge 2$ is convex.

The characterization theorem states the converse: Let $S\subseteq \mathbf{R}$ be an arbitrary subset with $|S|\ge 2$ such that $S$ is convex. Then $S$ is an interval.