Method 2 — Alternative Characterisation of Suprema

Solution.

First, note for \(n\in\mathbb{N}\): \[\frac{2n-1}{n+1} = \frac{2n+2-3}{n+1} = 2 - \frac{3}{n+1} < 2.\] Hence \(B\) is bounded above by \(2\). Therefore as \(B \neq \emptyset,\) the completeness axiom says that \(\sup(B)\) exists.

We claim that \(\sup(B) = 2.\) Fix \(\epsilon > 0.\) Then, for \(n \in \mathbb{N}:\) \[\begin{align*} 2 - \frac{3}{n+1} &> 2-\epsilon,\\ \Leftrightarrow \epsilon &> \frac{3}{n+1},\\ \Leftrightarrow n\epsilon &> 3 - \epsilon,\\ \Leftrightarrow n &> \frac{3-\epsilon}{\epsilon}. \end{align*}\] Now, by Archimedes Postulate, \(\exists N \in \mathbb{N}\) such that \(N > \frac{3-\epsilon}{\epsilon}\), from which \[2 - \frac{3}{N+1} > 2- \epsilon.\] At this stage, take \(b = 2 - \frac{3}{N+1} \in B\). Since \(\epsilon > 0\) was arbitrary, we have that \(\forall \epsilon > 0, \exists b \in B\) such that \(b > 2-\epsilon.\) So, by the alternative characterisation of suprema (Theorem 3.2), \(\sup(B) = 2.\)