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

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

Let $T = \set {x \in \R: -x \in S}$ be the set of negatives of the elements of $S$.

Then:
 * $B$ is a lower bound of $S$


 * $-B$ is an upper bound of $T$.
 * $-B$ is an upper bound of $T$.

Proof
Let $B$ be a lower bound of $S$.

That is:
 * $\forall x \in S: x \ge B$

Let $x \in T$ be arbitrary.

As $x$ is arbitrary it follows that:
 * $\forall x \in T: x \le -B$

That is, $-B$ is an upper bound for $T$.

Necessary Condition
Let $U$ be an upper bound for $T$.


 * $\forall x \in T: x \le U$

Let $x \in S$ be arbitrary.

As $x$ is arbitrary it follows that:
 * $\forall x \in S: x \ge -U$

That is, $-U$ is a lower bound for $S$.

Also see

 * Negative of Upper Bound of Set of Real Numbers is Lower Bound of Negatives