Negative of Lower Bound of Set of Real Numbers is Upper Bound of Negatives
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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:
- $B$ is a lower bound of $S$
- $-B$ 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.
\(\ds x\) | \(\in\) | \(\ds T\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds -x\) | \(\in\) | \(\ds S\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -x\) | \(\ge\) | \(\ds B\) | as $B$ is a lower bound for $S$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\le\) | \(\ds -B\) | Ordering of Real Numbers is Reversed by Negation |
As $x$ is arbitrary it follows that:
- $\forall x \in T: x \le -B$
That is, $-B$ is an upper bound for $T$.
$\Box$
Necessary Condition
Let $U$ be an upper bound for $T$.
- $\forall x \in T: x \le U$
Let $x \in S$ be arbitrary.
\(\ds x\) | \(\in\) | \(\ds S\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds -x\) | \(\in\) | \(\ds T\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -x\) | \(\le\) | \(\ds U\) | as $U$ is an upper bound for $S$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\ge\) | \(\ds -U\) | Ordering of Real Numbers is Reversed by Negation |
As $x$ is arbitrary it follows that:
- $\forall x \in S: x \ge -U$
That is, $-U$ is a lower bound for $S$.
$\blacksquare$