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

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:

$U$ is an upper bound of $S$

if and only if:

$-U$ is a lower bound of $T$.

Proof

Let $V$ be the set defined as:

$V = \set {x \in \R: -x \in T}$

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

$B$ is a lower bound of $T$

if and only if:

$-B$ is an upper bound of $V$

Then we have:

\(\ds V\)

\(=\)

\(\ds \set {x \in \R: -x \in T}\)

\(\ds \)

\(=\)

\(\ds \set {x \in \R: -\paren {-x} \in S}\)

\(\ds \)

\(=\)

\(\ds \set {x \in \R: x \in S}\)

\(\ds \)

\(=\)

\(\ds S\)

That is:

$B$ is a lower bound of $T$

if and only if:

$-B$ is an upper bound of $S$

The result follows by setting $U = -B$.

$\blacksquare$

Also see

• Negative of Lower Bound of Set of Real Numbers is Upper Bound of Negatives