Negative of Supremum is Infimum of Negatives

Theorem
Let $S$ be a non-empty 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
As $S$ is non-empty and bounded above, it follows by the Continuum Property that $S$ admits a supremum.

Let $B = \sup S$.

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

By definition, $B$ is an upper bound for $S$.

From Negative of Upper Bound of Set of Real Numbers is Lower Bound of Negatives:


 * $-B$ is a lower bound for $T$.

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

Let $C$ be the infimum of $T$.

Then by definition of infimum:
 * $C \ge -B$

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.

Also see

 * Negative of Infimum is Supremum of Negatives