About the $\sup$ of a subset of $\mathbb R$

Question

Suppose that $S$ is a nonempty bounded subset of $\mathbb R$ and let $T = \{\vert x\vert : x\in S\}$. Prove that $\sup T = \max\{\sup S,-\inf S\}$.

Answer 1

Suppose first that $\sup S > -\inf S$ and assume, by contradiction, that there is a $y \in S$ such that $y \in T$ and $y > a$.

By definition of $T$ there is an $x \in S$ such that $|x| = y > \sup S$, but either

$1)$, $|x| = x$, which giver $x \in S$ greater than $\sup S$, a contradiction, or

$2)$, $|x| = -x$, which gives $-\inf S \geq -x > \sup S$, another contradiction.

So either way, we have that there can`t be any such $y$, and therefore, $\sup T \leq \sup S$.

The argument for the case $-\inf S > \sup S$ is entirely analogous.