Prove that for any nonempty set \(S\) of real numbers that is bounded below. \(\inf S = -\sup(-S)\) where \(-S = \{-s: s \in S\}\).

Proof

Let \(S\) be nonempty and bounded below. We want to show that \(\inf S = -\sup(-S)\) where \(-S = \{-s: s \in S\}\). To do this, we want to show that \(-S\) is bounded above and then we will show that \(-\sup(-S)\) is a lower bound of \(S\) and finally we will show that it is the greatest lower bound of \(S\).

Since \(S\) is bounded below, then there exists some element \(m \in \mathbb{R}\) such that \(m \leq s\) for all \(s \in S\). Thus, for any \(-s \in -S\), \(-m \geq -s\). This means that \(-S\) is bounded above. Therefore, \(\sup(-S)\) exists.

Next, we want to show that \(-\sup(-S)\) is a lower bound on \(S\). Observe that for any \(-s \in -S\), we have

$$ \begin{align*} -s \leq \sup (-S). \end{align*} $$

Since \(-s \in -S\), then \(s \in S\). Multiply both sides by \(-1\) to see that

$$ \begin{align*} s \geq - \sup (-S). \end{align*} $$

Therefore, \(- \sup (-S)\) is a lower bound of \(S\) since every \(s \in S\) satisfies \(s \geq -\sup (-S)\). Finally, we know that \(-\sup(-S)\) is the greatest lower bound of \(S\). To see this, observe that since \(\sup(-S)\) is least upper bound of \(-S\), then this implies that for any \(\epsilon > 0\), there must exists an element \(-s \in -S\) such that

$$ \begin{align*} -s > \sup (-S) - \epsilon \end{align*} $$

But since \(-s \in -S\), then \(s \in S\). So, multiply both sides by \(-1\) to see that

$$ \begin{align*} s < - \sup (-S) + \epsilon \end{align*} $$

But this says that for every \(\epsilon\), there exists an \(s\) such that \(s < -\sup(-S) + \epsilon\). Then, no number greater that \(-\sup(-S)\) can be a lower bound of \(S\). Therefore, \(- \sup (-S)\) is the greatest lower bound of \(S\) as desired. \(\ \blacksquare\)

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