Prove that if s is the supremum of $X$, then $z + s$ is the supremum of $X + z$ I am having a bit of trouble trying to solve this problem.
Question: For any $X \subseteq \mathbb{R}$ and $z\in\mathbb{R}$ define $X + z$ to be the set of all $x + z$ where $x\in X$. Prove that if $s$ is the supremum of $X$, then $z + s$ is the supremum of $X + z$
My attempt is that you're given that the supremum, $s$, is the supremum of $X$. Then that should mean that $s \geq x, \forall x \in X$, then that should mean that $z + s \geq x + z, \forall x \in X$.
Any help or tips will be useful!

Let be $\varepsilon>0$. Since $s$ is the supremum of $X$, then there exists $x_n\in X$ such that :
$$s-\varepsilon < x_n \le s$$
Now
$$z+s-\varepsilon < z+x_n \le z+s$$
let $z_n = z+x_n$ and $w=z+s$. We have showed that for all $\varepsilon>0$, there exists a $z_n = z+x_n \in z+X$ such that:
$$w-\varepsilon < z_n \le w$$
therefore $w = z+s$ is the supremum of $z+X$
