TITLE: Proof of $S \subset \mathbb{R}$, $ \inf(S)\leq \sup(S)$
QUESTION [5 upvotes]: Prove that for any nonempty set $S \subset \mathbb{R}$, $ \inf(S)\leq  \sup(S)$ and give necessary and sufficient conditions for equality.
This is what I have so far but I think I am on the wrong track:
Since set S is contained in R, we have four options:
$$S=(a,b) ; S=[a,b) ; S=(a,b] ; S=[a,b]$$
 for some $ a \text{ and } b \in \mathbb{R}$
via the ordering of interval notation
$\inf(S)=a$ and $\sup(S)=b$ and $a\leq b$ by definition of interval notation. Hence $\inf(S) \leq \sup(S)$.

REPLY [6 votes]: I'm assuming you want to show that $\inf S\leq \sup S$, whenever $S\subseteq \mathbb{R}$ is a non-empty subset, and not the reverse inequality. 
A part of the definition of the supremum is that it is an upper bound for $S$, i.e. $$s\leq \sup S\quad \text{for all }\;s\in S.$$ 
The same goes for the infimum, it is a lower bound for $S$: $$\inf S\leq s\quad\text{for all }\; s\in S.$$
Now use the fact that $S$ is non-empty to deduce the inequality.