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 $S$

if and only if:

$-B$ 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 $S$.

$\blacksquare$


Also see