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;
 * $$-\sup_{x \in S} x = \inf_{x \in S} \left({-x}\right)$$.

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 $$\forall x \in T: -x \ge -B$$.

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.