a problem using upper bounds Let S be a non empty subset of the real numbers which is bounded above. Let c be a real number and define T ={cx|x∈S}. Show that,if c > 0,then T is non empty, bounded above and that sup T = c sup S. Give an example of a set S as above, such that with c = −1, the set T is still bounded above but supT ̸= csupS.
I have made several attempts but my proofs are very inelegant and my friend has guaranteed there is a much shorter proof than mine (I have attached my proof) . If someone could please help me with an elegant proof it would be  very helpful.

Since $S$ is bounded above, $\overline s=\sup S<\infty .$ We show that $\sup T=c\overline s:$ 
$a). \ \overline s$ is an upper bound for $S$ by definition of $\sup.$ Then, by definition of $T,$ and the fact that $c>0,\ c\overline s$ is an upper bound for $T$. 
$b).$ Let $t$ be $\textit {any}$ upper bound for $T.$ Then, by defintion of $T,\  t/c$ is an upper bound for $S,$ and then $\overline s<t/c$ and so $c\overline s<t,$ which implies that  $c\overline s$ is the least upper bound for $T$.
For a counterexample in case $c=-1$ take $S=\left \{ -1,1 \right \}.$
