Negative of Supremum is Infimum of Negatives

Theorem
Let $S$ be a subset of the real numbers $\R$.

Let $S$ be bounded above.

Then:
 * $\left\{{x \in \R: -x \in S}\right\}$ is bounded below;
 * $\displaystyle -\sup_{x \in S} x = \inf_{x \in S} \left({-x}\right)$.

Proof
Let $B = \sup S$.

Let $T = \left\{{x \in \R: -x \in S}\right\}$.

Since $\forall x \in S: x \le B$ it follows that $\forall x \in S: -x \ge -B$.

Hence $-B$ is a lower bound for $T$, and so $\left\{{x \in \R: -x \in S}\right\}$ is bounded below.

If $C$ is the largest lower bound for $T$, it follows that $C \ge -B$.

On the other hand, $\forall y \in T: y \ge C$.

Therefore $\forall y \in T: -y \le -C$.

Since $S = \left\{{x \in \R: -x \in T}\right\}$ it follows that $-C$ is an upper bound for $S$.

Therefore $-C \ge B$ and so $C \le -B$.

So $C \le -B$ and $C \ge -B$ and the result follows.