$A,B ⊂ R$ are non-empty bounded sets, with $α=\sup(A), B=\sup(B)$. $Z ∈ R$ Prove that $Z + A = \{Z + x : x ∈ A\}$ is bounded and $\sup(Z + A)= Z + α$.
So far in my proof I have: $∀ x ∈ A. x ≤ α, Z + x ≤ Z + α$. Since $α = \sup(A)$ and $x ∈ A$, $x$ is bounded.
I need help finishing and adding more to the proof
It seems right. It has to be clear that $Z + A$ fixes according to your definition $x \in A$. It is clear that $\forall y \in Z, $ $y + x \leq y + \sup(A) = y + \alpha $. Since $y$ is arbitrary in $Z$, this means $Z + A \leq Z + \alpha$. The remaining part would be proving that this is the supreme. If it weren't like that, having fixed a certain $z \in Z$, the smallest upper bound for $z + A$ would not be $z + \alpha$, but for example $z + \beta$. This would mean that $\beta $ would be a smaller upper bound for $A$, which means $\alpha \neq \sup(A)$, which is absurd according to the initial asumption.