Negative of Infimum is Supremum of Negatives

Theorem

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

Then:

$(1): \quad \set {x \in \R: -x \in T}$ is bounded above

$(2): \quad \ds -\inf_{x \mathop \in T} x = \map {\sup_{x \mathop \in T} } {-x}$

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

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

Let $B = \inf T$.

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

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 $\set {x \in \R: -x \in T}$ is bounded above.

Let $C$ be the supremum of $S$.

Then by definition of supremum:

$C \le -B$

On the other hand:

$\forall y \in S: y \le C$

Therefore:

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

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

$\blacksquare$