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.

Let $s=\sup S$ and $t= \sup T$
$$\forall x \in S \Rightarrow x \leq s \Rightarrow cx \leq cs \Rightarrow t \leq cs$$
Now $$\forall x \in S \rightarrow x=\frac{cx}{c}\leq \frac{t}{c} \Rightarrow s \leq \frac{t}{c} \Rightarrow cs \leq t $$
Thus  $t=s$
