The question is:
"Let $A ⊆ R$ be nonempty and bounded above, and let $c ∈ R$. This time
define the set $cA = \left\{ca : a ∈ A \right\}$. Show that If $c \geq 0$, $\sup(cA)=c\sup A$."
The way I tried to prove this is like the following.
$Proof.$ We know that $\sup A \geq a$ $, \forall a \in A .$ Then for any $c \geq 0$ , $c\sup A \geq ca$. So $c\sup A$ is an upper bound for $cA$. Now since $\sup A \in A$, $c\sup A \in cA$ and $c\sup A \geq ca$, it's clear that $c\sup A$ is the least upper bound for $cA$. Hence $\sup(cA)=c\sup A$.
Is that the right way to prove it?
Best Answer
Assuming $c>0$. The inequality $\sup (cA) \leqslant c\sup A$ is clear. Let $t$ be such that $ca\leqslant t$ for all $a\in A$. Then $a\leqslant c^{-1}t$ for all $a\in A$ implies $\sup A \leqslant c^{-1}t$. Conclude that $c\sup A= \sup (cA)$.