# Proof of $S \subset \mathbb{R}$, $\inf(S)\leq \sup(S)$

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 $\inf(S) \leq \sup(S)$.

• Question about sets? Definitely [set-theory]... – Asaf Karagila Jan 21 '13 at 0:23
• $S$ need not be an interval. – Shaun Ault Jan 21 '13 at 0:24
• Surely you mean that $\inf S\le\sup S$. – Brian M. Scott Jan 21 '13 at 0:29
• @Asaf: Nah, it must be [tennis]. – Brian M. Scott Jan 21 '13 at 0:31
• The vote to close this as not a real question is incomprehensible. Not only is the question perfectly clear (apart from the unfortunate typo), but the OP has provided an attempt at a solution. – Brian M. Scott Jan 21 '13 at 0:36

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.
Hint: Suppose that $S$ has more than one point. If $a,b \in S$ and $a<b$ what can you deduce about the $\inf$ and $\sup$ of $S$? What happens if $S$ has one point?