a problem using upper bounds

Question

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.

Answer 1

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 $S=\left \{ -1,1 \right \}.$

Answer 2

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 $t=s$