If \(\sup A < \sup B\), then show that there exists an element \(b \in B\) that is an upper bound for \(A\).
Proof
By definition, we know that for any \(a \in A\), we must have \(a < \sup A\). But \(\sup A < \sup B\). Therefore, \(a < \sup B\) and \(\sup B\) is an upper bound for \(A\). Let \(\epsilon = \sup B - \sup A\). Then
$$
\begin{align*}
\sup A + \frac{\epsilon}{2} &= \sup A + \frac{\sup B - \sup A}{2} \\
&= \frac{1}{2}(\sup A + \sup B).
\end{align*}
$$
Clearly
$$
\begin{align*}
\sup A < \frac{1}{2}(\sup A + \sup B) < \sup B.
\end{align*}
$$
Therefore,
$$
\begin{align*}
\sup A < \sup A + \frac{\epsilon}{2} < \sup B.
\end{align*}
$$
So \(\sup A + \frac{\epsilon}{2}\) is an upper bound for \(A\). Meanwhile, \(\sup A + \frac{\epsilon}{2}\) is not an upper bound for \(B\). This means that there exists an element \(b \in B\) such that
$$
\begin{align*}
b > \sup A + \frac{\epsilon}{2}.
\end{align*}
$$
Since \(\sup A + \frac{\epsilon}{2}\) is an upper bound for \(A\) and \(b > \sup A + \frac{\epsilon}{2}\), it follows that \(b\) is an upper bound for \(A\) as we wanted to show. \(\ \blacksquare\)
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