Absolute Value of a Bounded Set Let $S \subseteq \mathbb{R}$ be non-empty. Suppose that $S$ is bounded. Let $|S| = \{|s| \, | \, s \in S\}$. Let $c = \max\{|\inf(S)|, \, |\sup(S)|\}$. Prove that $|S|$ is bounded above and $\sup(|S|) = c$.
How can I interpret the absolute value of a set? Since we know that the set S is complete by the completeness axiom, can we also conclude that |S| is also complete? 
And can someone give me some insight on how to start this proof?
Thanks for your help!
 A: Generally we say that $S$ is bounded if there is some $B$ such that $|s| \le B$ for all $s \in S$. Hence bounded implies that $|S|$ is bounded above (by the same constant).
(Note that in general, $|x| \le L$ iff $-L \le x \le L$.)
Note that $-|S| \le s \le |S|$ for all $s \in S$, hence 
$-|S| \le \inf S \le |S|$ and $-|S| \le \sup S \le |S|$ and so
$|\inf S| \le |S|$ and $|\sup S| \le |S|$ from which we get $c \le |S|$.
Note that $-c \le \inf S \le s \le \sup |S| \le c$ for all $s$ and so $|s| \le c$ and
so $|S|=\sup |s| \le c$.
A: Let do this:
$$s \in S \Longrightarrow \inf(S) \leqslant s \leqslant \sup(S) \Longrightarrow -|\inf(S)| \leqslant \inf(S) \leqslant s \leqslant \sup(S) \leqslant \sup(S)| \Longrightarrow$$
$$-|\inf(S)| \leqslant s \leqslant |\sup(S)| \Longrightarrow |s| \leqslant \max(|\inf(S)|,|\sup(S)|) \Longrightarrow |s| \leqslant c \Longrightarrow \sup(|S|) \leqslant c$$
For equality first notice that:
$$\left\{\begin{array}{c} c & = & |\inf(S)| & \Longrightarrow & \inf(S) & \leqslant & 0 \\ c & = & |\sup(S)| & \Longrightarrow & \sup(S) & \geqslant & 0 \\ \end{array}\right\}$$
Since $|S|$ has no negative elements, we have:
$$0\leqslant d<c \Longrightarrow \left\{\begin{array}{c} \inf(S) & < & -d & \leqslant & 0 \\ 0 & \leqslant & +d & < & \sup(S) \\ \end{array}\right\} \Longrightarrow$$
$$\Longrightarrow \left\{\begin{array}{c} \exists i \in S: & \inf(S) & \leqslant & i & <& -d & \leqslant & 0 \\ \exists s \in S:&  0 & \leqslant & +d & < & s & \leqslant & \sup(S) \\ \end{array}\right\}\Longrightarrow \left\{\begin{array}{c} |-d| < |i| \in |S| \\ |+d| < |s| \in |S| \\ \end{array}\right\}\Longrightarrow$$
$$\Longrightarrow d<\sup(|S|)$$
And this by first result show:
$$c=\sup(|S|)$$
