Boundedness in the real numbers Let $r \in R$ and $S \subset R$ that is bounded above. Define $rS = \\{rs : s \in S\\}$. Prove that $rS$ is bounded above and that sup $rS$ =$r$ sup$S$. Okay I think I have this for the case where $r \gt 0$ but I'm stuck otherwise. Let $u \in sup S$ and $a \in rS$. Then there exists an $s \in S$ such that $a=ts$. Since $u$ is an upper boun for $S$, $s \leq u$ and $a=rs \leq ru$ IF r is POSITIVE. Then since a was arbitrary, $ru = r$ sup$S$ is an upper bound for $rS$. I need to prove this for all $r \in R$, how can I get around this constraint?

If $r = 0$ then the result is trivial. If $r < 0$ then it is false, for example for $S = \\{-1,1\\}$ and $r = -1$, $\sup rS = 1$ while $r\sup S = -1$. In fact, for $r < 0$ we have $\sup rS = r \inf S$.