Assume that \(A\) and \(B\) are nonempty, bounded above, and satisfy \(B \subseteq A\). Show \(\sup B \leq \sup A\).
Proof
Since \(A\) and \(B\) are bounded above, then there must exists least upper bounds for \(A\) and \(B\). We claim that \(\sup B \leq \sup A\). To show this, let \(b\) be an element in \(B\). Since \(B \subseteq A\), then \(b \in A\). Therefore,
$$
\begin{align*}
b \leq \sup A.
\end{align*}
$$
Since this holds for all \(b \in B\), then this means that \(\sup A\) is an upper bound for \(B\). But \(\sup B\) is the least upper bound of \(B\). By the definition of the least upper bound, therefore
$$
\begin{align*}
\sup B \leq \sup A.
\end{align*}
$$
as desired. \(\ \blacksquare\)