Let $S\subseteq R$ be nonempty and bounded above. Let $s$=sup $S$ and define
$$T=(3x\in R : x \in S)$$
Prove that sup $T$ exists and sup $T$=$3s$.
What i tried.
Clearly $T$ is non empty as it contains the element $3x$. $T$ is bounded above by $3S$ (explanation needed). And thus by the axiom of completeness sup $T$ exists.
Next to prove that sup $T$ =$3s$, we have that since the set $S$ is 3 times that of the set $T$ and that since $s$ lies in $S$ and is the lowest upper bound of the set $S$, then $3s$ also lies in $T$ and thus sup $T$=$3s$.
While i know intuitively how to solve the problem and at least have a rough idea of how to solve the problem, my explanation isn't clear enough. Could someone explain how do i solve the problem that is written more clearly. Thanks