Prove that $\sup(a+S)=a+\sup(S)$

Question

We know $a+\sup(S)$ is an upper bound for the set $a+S$ because we have $a+s\leq a+\sup(S)$ for all $s\in S$. Therefore, $\sup(a+S)\leq a+\sup(S)$. But I cannot continue from here, what should I do to show $\sup(a+S)=a+\sup(S)$?

Answer 1

If $x$ is an upper bound for the set $a+S$, you can show that $x-a$ is an upper bound for the set $S$. Then $a+(x-a)$ is bigger than $a+sup(S)$ and so is $x$.