Set of Real Numbers is Bounded Below iff Set of Negatives is Bounded Above/Corollary
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Corollary to Set of Real Numbers is Bounded Below iff Set of Negatives is Bounded Above
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
Let $V$ be the set defined as:
- $V = \set {x \in \R: -x \in T}$
From Set of Real Numbers is Bounded Below iff Set of Negatives is Bounded Above:
- $T$ is bounded below
- $V$ is bounded above.
Then we have:
\(\ds V\) | \(=\) | \(\ds \set {x \in \R: -x \in T}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {x \in \R: -\paren {-x} \in S}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {x \in \R: x \in S}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds S\) |
The result follows.
$\blacksquare$