Prove that if \(a\) is an upper bound for \(A\), and if \(a\) is also an element of \(A\), then it must be that \(a=\sup A\)
Proof
Let \(a\) be an upper bound for \(A\) where \(a \in A\). Let \(b = \sup A\). Then by definition, \(b\) is the least upper bound. Since \(a\) is an upper bound of \(A\), then
$$
\begin{align*}
b \leq a
\end{align*}
$$
Now, since \(a \in A\), then by definition of an upper bound, we must have \(a \leq b\).
$$
\begin{align*}
a \leq b
\end{align*}
$$
But this implies that \(a = b\) and so \(a = \sup A\). \(\ \blacksquare\)