Set of Real Numbers is Bounded Below iff Set of Negatives is Bounded Above

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:
 * $S$ is bounded below


 * $T$ is bounded above.
 * $T$ is bounded above.

Sufficient Condition
Let $S$ be bounded below.

Then $S$ has a lower bound.

Let $B$ be a lower bound for $S$.

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


 * $B$ is a lower bound for $S$


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

It follows that $T$ is bounded above.

Necessary Condition
Let $T$ be bounded above.

Then $T$ has an upper bound.

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


 * $U$ is an upper bound for $T$


 * $-U$ is a lower bound for $S$.
 * $-U$ is a lower bound for $S$.

It follows that $S$ is bounded below.