Supremum and Infimum absolute values Let $X \subset \mathbb{R}$. Assume $X$ contains positive and negative numbers and assume $X$ is nonempty and bounded. Show $\sup(|X|) = -\inf(X)$ if $\sup(X) < \sup(|X|)$ where $|X| = \{|x| : x \in X\}$.
Any hints on where to even begin? We know $\inf(X) < 0 $ so -$\inf(X) > 0$.   
 A: After your observation we have that if $sup(X)<sup(|X|)$ than $sup(|X|)$ must be different from $sup(X)$, so it must taken by a negative $x\in X$, in particular the "biggest" in modulus, that is $x=inf(X)$.
A: An alternative statement of this theorem is that $\sup|X|=\max(-\inf X,\sup X)$. Then it must equal one of the two values, so if $\sup|X|\ne\sup X$ then $\sup|X|=-\inf X$.
We can substitute $-\inf X=\sup(-X)$, because $x\mapsto-x$ is an order isomorphism from $(\Bbb R,\le)$ to $(\Bbb R,\ge)$.
To show $\max(\sup(-X),\sup X)\le\sup|X|$, we need $\sup(-X)\le\sup|X|$ and $\sup X\le\sup|X|$, which is true because the sets are "pointwise" less, that is $x\le|x|$ and $-x\le|x|$ for each $x\in X$, which implies that each upper bound of $|X|$ is an upper bound of $X$ and $-X$.
Conversely, to show $\sup|X|\le\max(\sup(-X),\sup X)$ we need $|x|\le\max(\sup(-X),\sup X)$ for all $x\in X$. If $x\ge 0$, then $|x|=x\le\sup X\le\max(\sup(-X),\sup X)$; and if $x\le 0$ then $|x|=-x\le\sup(-X)\le\max(\sup(-X),\sup X)$.
A: Let's try to treat this problem by focusing on the most complex part, expanding the various definitions, and then simplifying, and let's see where that leads us.$%
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%$
Working in $\;\mathbb R\;$, our task is to prove for non-empty bounded $\;X\;$ that
$$
\tag{0} \sup(X) \lt \sup(\abs{X}) \;\then\; \sup(\abs{X}) = -\inf(X)
$$
with the additional assumptions that $\;\sup(X) > 0\;$ and $\;\inf(X) < 0\;$.
Now, the simplest definitions of $\;\sup\;$ and $\;\inf\;$ I know are that, for all $\;z\;$,
\begin{align}
\tag{1} \sup(S) \le z &\;\equiv\; \langle \forall s : s \in S : s \le z \rangle \\
\tag{2} z \le \inf(T) &\;\equiv\; \langle \forall t : t \in T : z \le t \rangle \\
\end{align}
for any non-empty upper-bounded $\;S\;$ and non-empty lower-bounded $\;T\;$.  And to prove equalities through this definitions, properties like the following are useful:
$$
\tag{3} x = y \;\equiv\; \langle \forall z :: x \le z \;\equiv\; y \le z \rangle
$$
which says that two numbers are the same if they have the same upper bounds.

Looking at $\Ref{0}$, we see that the most complex expression there is $\;\sup(\abs{X})\;$.  So let's investigate its upper bounds: for any $\;z\;$, we calculate
$$\calc
    \sup(\abs{X}) \le z
\op\equiv\hint{expand definition $\Ref{1}$ of $\;\sup\;$}
    \langle \forall s : s \in \abs{X} : s \le z \rangle
\op\equiv\hint{expand definition of $\;\abs{X}\;$}
    \langle \forall s : \langle \exists x : x \in X : s = \abs{x} \rangle : s \le z \rangle
\op\equiv\hint{logic: merge quantifiers}
    \langle \forall s,x : x \in X \land s = \abs{x} : s \le z \rangle
\op\equiv\hint{logic: one-point rule}
    \langle \forall x : x \in X : \abs{x} \le z \rangle
\op\equiv\hint{expand definition of $\;\abs{x}\;$}
    \langle \forall x : x \in X : -z \le x \land x \le z \rangle
\op\equiv\hint{logic: distribute $\;\land\;$ over $\;\forall\;$}
    \langle \forall x : x \in X : -z \le x \rangle  \;\land\; \langle \forall x : x \in X : x \le z \rangle \tag{*}
\op\equiv\hint{definition $\Ref{2}$ of $\;\inf\;$; definition $\Ref{1}$ of $\;\sup\;$}
    -z \le \inf(X) \;\land\; \sup(X) \le z
\op\equiv\hint{arithmetic}
    -\inf(X) \le z \;\land\; \sup(X) \le z
\op\equiv\hint{property of $\;\max\;$}
    \max \left\{-\inf(X), \sup(X) \right\} \le z
\endcalc$$
Therefore, by $\Ref{3}$, this longish but straightforward calculation shows that
$$
\tag{5} \sup(\abs{X}) \;=\; \max \left\{-\inf(X), \sup(X) \right\}
$$
And from this, with the definition of $\;\max\;$, $\Ref{0}$ immediately follows.

Note how until $\Ref{*}$, the above calculation is only about expanding definitions and simplifying.  The rest is directly inspired by $\Ref{0}$.
Finally, note that we need did not use the fact that "$\;X\;$ contains positive and negative numbers", so that $\;\sup(X) > 0\;$ and $\;\inf(X) < 0\;$.
