Negative of Upper Bound of Set of Real Numbers is Lower 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:
 * $U$ is an upper bound of $S$


 * $-U$ is a lower bound of $T$.
 * $-U$ is a lower bound of $T$.

Proof
Let $V$ be the set defined as:


 * $V = \set {x \in \R: -x \in T}$

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


 * $B$ is a lower bound of $T$


 * $-B$ is an upper bound of $V$
 * $-B$ is an upper bound of $V$

Then we have:

That is:


 * $B$ is a lower bound of $T$


 * $-B$ is an upper bound of $S$
 * $-B$ is an upper bound of $S$

The result follows by setting $U = -B$.

Also see

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