Let \(A = \{x \in \mathbf{R}: x < 0\}\). Then \(\sup(A) = 0\).


For the definitions of an upper bound and the least upper bound of a set, see this.

Proof:

Let \(A = \{x \in \mathbf{R}: x < 0\}\). We will use lemma 1.3.8 to prove that \(\sup(A) = 0\). First, 0 is an upper bound of the set \(A\) since by definition the set \(A\) is the set of real numbers less than 0. Next, we need to prove that there exists an element \(a \in A\) such that for every choice of \(\epsilon > 0\), we have

$$ \begin{align*} a &> \sup(A) - \epsilon \\ a &> 0 - \epsilon \\ a &> -\epsilon. \end{align*} $$

Pick \(a = -\epsilon/2\). First note that since \(\epsilon > 0\), then \(-\epsilon/2 < 0\) and therefore \(-\epsilon/2 \in A\). Now let’s prove that this choice will lead to having \(a > -\epsilon\) as required. Start with

$$ \begin{align*} \epsilon &> 0 \\ \epsilon -2\epsilon &> 0 - 2\epsilon \\ -\epsilon &> -2\epsilon \\ -\epsilon/2 &> -\epsilon \end{align*} $$

Therefore, setting \(a = -\epsilon/2\) has the property that \(a \in A\) and also that \(a > 0 - \epsilon\) for any choice of \(\epsilon > 0\) which proves that \(\sup(A) = 0\) as required by lemma 1.3.8. \(\blacksquare\)

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