Let \(A\) be a nonempty subset of \(\mathbb{R}\) that is bounded above. Define \(S = \{ x \) is an upper bound for \(A\}\). Prove that \(\inf S = \sup A\).
Proof
Let \(A\) be a nonempty subset of \(\mathbb{R}\) that is bounded above. Let \(S\) be defined as above. We want to show that \(\inf S = \sup A\). To see this, observe that since \(A\) is bounded above, then \(\sup A\) exists by the axiom of completeness. Furthermore, every element \(x \in S\) is an upper bound of \(A\) and since \(\sup A\) is the least upper bound, then
$$
\begin{align*}
\sup A \leq x
\end{align*}
$$
That is \(\sup A\) is a lower bound for \(S\). But \(\sup A\) is also an element of \(S\) itself. Therefore, \(\sup A\) is a lower bound of \(S\) and belongs to \(S\), therefore, it must be the greatest lower bound of \(S\). Thus
$$
\begin{align*}
\inf S = \sup A
\end{align*}
$$
as desired. \(\ \blacksquare\)
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