$\overline S = \{ 1-s: s \in S \}$. 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're not using at all the fact that it is bounded, so I don't really see where you are trying to go.
Simply use the definition of the supremum :
$\forall s \in S, s \le M \Leftrightarrow 1-s\ge1-M$
$\forall s' \in \overline S, \exists s \in S \ s.t.\ s'=1-s$, and $s'=1-s \ge 1-M$
So $\overline S$ is bounded below.
