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

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    $\begingroup$ Question about sets? Definitely [set-theory]... $\endgroup$ – Asaf Karagila Jan 21 '13 at 0:23
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    $\begingroup$ $S$ need not be an interval. $\endgroup$ – Shaun Ault Jan 21 '13 at 0:24
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    $\begingroup$ Surely you mean that $\inf S\le\sup S$. $\endgroup$ – Brian M. Scott Jan 21 '13 at 0:29
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    $\begingroup$ @Asaf: Nah, it must be [tennis]. $\endgroup$ – Brian M. Scott Jan 21 '13 at 0:31
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    $\begingroup$ 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. $\endgroup$ – 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.

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    $\begingroup$ I love when math is written this obviously. Nice job, Stefan. $\endgroup$ – 000 Jan 21 '13 at 0:51
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    $\begingroup$ Thanks for your response :) $\endgroup$ – math101 Jan 21 '13 at 0:51

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?


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