Negative of Infimum is Supremum of Negatives

Theorem
Let $T$ be a non-empty subset of the real numbers $\R$.

Let $T$ be bounded below.

Then:
 * $(1): \quad \left\{{x \in \R: -x \in T}\right\}$ is bounded above
 * $(2): \quad \displaystyle -\inf_{x \mathop \in T} x = \sup_{x \mathop \in T} \left({-x}\right)$

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

Proof
As $T$ is non-empty and bounded below, it follows by the Continuum Property that $T$ admits an infimum.

Let $B = \inf T$.

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

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

Hence $-B$ is an upper bound for $S$, and so $\left\{{x \in \R: -x \in T}\right\}$ is bounded above.

If $C$ is the supremum of $S$, it follows that $C \le -B$.

On the other hand, $\forall y \in S: y \le C$.

Therefore $\forall y \in S: -y \ge -C$.

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

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

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