Let \(A \subseteq \mathbf{R}\) be nonempty and bounded above, and let \(c \in \mathbf{R}\). Define the set \(cA = \{ca : a \in A\}\). Then if \(c \geq 0\), show that \(\sup cA = c\sup A\).


For the definitions of an upper bound and the least upper bound of a set. See This.

Proof:

Let \(A \subseteq \mathbf{R}\) be nonempty and bounded above, let \(c \in \mathbf{R}\) and let \(cA = \{ca : a \in A\}\). We’ll verify both conditions of the least upper bound. Let \(a \in A\). We know that \(a \leq \sup A\). Multiply both sides by \(c\) to get

$$ \begin{align*} ca \leq c\sup A. \end{align*} $$

This means that \(c\sup A\) is an upper bound for \(cA\) and this verifies condition one. To verify condition two, let \(b\) be an arbitrary upper bound for \(cA\). This means that

$$ \begin{align*} ca &\leq b \\ a &\leq b/c. \end{align*} $$

This means that \(b/c\) is an upper bound for \(A\). But \(A\) has a least upper bound \(\sup A\) and we know by condition two that any upper bound is greater than or equal to \(\sup A\) so

$$ \begin{align*} b/c &\leq \sup A \\ b &\leq c \sup A. \end{align*} $$

Recall that \(b\) was an arbitrary upper bound of \(cA\). Therefore, we can conclude from this that \(c \sup A\) is a least upper bound for \(cA\) as required. \(\blacksquare\)

References: