Negative of Supremum is Infimum of Negatives

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

Let $S$ be bounded above.

Then:
 * $(1): \quad \set {x \in \R: -x \in S}$ is bounded below
 * $(2): \quad \ds -\sup_{x \mathop \in S} x = \map {\inf_{x \mathop \in S} } {-x}$

where $\sup$ and $\inf$ denote the supremum and infimum respectively.

Proof
Let $B = \sup S$.

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

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 $T$.

Thus $\set {x \in \R: -x \in S}$ is bounded below.

If $C$ is the infimum of $T$, 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 $S = \set {x \in \R: -x \in T}$ 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.