nhodges
Let $S$ be nonempty and bounded above and $S+c$ = {${x=s+c:s \in S ,c \in R}$}. Since $s\in S$ and $c\in R$ $s+c\ne \emptyset$. Let $T$ be the supremum of $S$. Then $s\le T$ and $s+c \le T+c$, thus $S+c$ is bounded above. Let $y$ be any upperbound for $S+c$. Then $$x \le y$$ $$s+c \le y$$ $$s \le y-c$$ Thus $y-c$ is an upperbound for $S$ and by definition $$T \le y-c$$ and $$T+c \le y$$ Since $y$ is any upperbound for $S+c$, $T+c$ is the supremum of $S+c$. and $$Sup(S+c)= T+c= Sup(s) +c$$
