Set of Real Numbers is Bounded Below iff Set of Negatives is Bounded Above
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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.
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
- $T$ 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 $S$
- $-B$ 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 $T$
- $-U$ 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