# $\overline S = \{ 1-s: s \in S \}$. 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?

• $$\forall s \in S, s \leq M \Longrightarrow \forall s \in S, 1 - s \geq 1 - M \Longrightarrow \forall (1 - s) \in \bar{S}, 1 - s \geq 1 - M$$ – Jose Arnaldo Bebita-Dris Apr 16 '15 at 18:31
• @ArnieDris that is really confusing to read. OP: why don't you try seeing what happens when $S$ is an interval, say $[a,b]$? What would $\bar{S}$ be? How does an upper bound for $S$ relate to a lower bound for $\bar{S}$ in this case? – Cameron Williams Apr 16 '15 at 18:32
• Actually I like @Arnie Dris's way tbh – StephanCasey Apr 16 '15 at 18:35

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.
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 $s\in S.$