$\overline S = \{ 1-s: s \in S \}$. Prove that $\overline S$ is bounded below. Let $S$ be a nonempty subset of $\mathbb R$ that is bounded above. Set
$\overline S = \{ 1-s: s \in S \}$. Prove that $\overline S$ is bounded below.
Prove that $\overline S$ is bounded below.
I'm just not entirely sure how I should go about doing this proof.
My attempt:
Since $S$ is bounded above, $S$ has upper bounds and a least upper bound (i.e. the supremum).
Let $M = \text{Sup}S$
Where $s < 0 => \overline S = 1-s > 0$
Where $s=0 => \overline S = 1-s = 1-0 = 1$
Where $s > 0 => \overline S = 1-s \leq 1$
Where $s >0$, $\overline S$ decreases as $s$ increases.
Now $s$ is bounded above but it's supremum can be $<,=$ or $> 0$
Do you think I'm on the right track or is there a better way to set this out?

You got off to a great start! Indeed, $S$ had a least upper bound, say $M.$ In particular, then, $$s\le M\tag{$\heartsuit$}$$ for all $s\in S.$
Now, to show that $\overline S$ is bounded below, we need to show that there is some $m$ such that $m\le t$ for all $t\in \overline S,$ or equivalently that $$m\le 1-s\tag{$\star$}$$ for all $s\in S.$ As a hint for how to find an $m$ that satisfies $(\star)$ for all $s\in S,$ note that you haven't yet used the fact that $(\heartsuit)$ holds for all $s\in S.$
