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

if and only if:

$T$ is bounded above.

Corollary

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 above

if and only if:

$T$ is bounded below.

Proof

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 $S$

if and only if

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

It follows that $T$ is bounded above.

$\Box$

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 $T$

if and only if:

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

It follows that $S$ is bounded below.

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 5$: Limits