Supremum of $B$, where $\sup B = \lambda + \sup A $

Question

Question Setup:

Let $A$ be a set that is bounded above, where $\sup A = \lambda $ $A$ is a subset of $\mathbb{R}$ and $\lambda \in \mathbb{R} \\ $. A set $B$ is defined as $B = \{\, x + k \ | \ x \in A, k \in \mathbb{R}\,\}$

Question: Prove that $\sup B = \lambda + k $

My attempt: Let $\mu \in \mathbb{R}$ $$\forall x + k \in B, \mu \geq x + k $$ i.e. $\mu$ is an upper bound of $B$ Select $\mu$ such that , $$\forall m \in B, m \leq \mu \implies \exists n \in B, n \geq m $$ i.e $\mu = \sup B $

We know, (everything from here on is shaky) $$\forall x + k \in B, \mu - k \geq x $$ $$\forall x \in A, \mu - k \geq x $$ i.e $$\mu - k \geq \sup A $$ As $ \mu = \sup B$ $$\mu - k = \sup A $$ $$\sup B = \lambda + k$$

Answer 1

If $b\in B$, then $b=a+k$, for some $a\in A$. So, since $a\leqslant\sup A$, $b\leqslant k+\sup A$. Therefore, $k+\sup A$ is an upper bound of $B$.

And if $\mu<k+\sup A$, then $\mu-k<\sup A$. So, there is some $a\in A$ such that $a>\mu-k$, and therefore $a+k>\mu$. But $a+k\in B$. So, $\mu$ is not an upper bound of $B$. It follows that $k+\sup A$ is the least upper bound of $B$. In other words, $k+\sup B=\sup A$.

Answer 2

Note that $\sup B$ does not exist because $B = \mathbb{R}$, i. e., $B$ is unbunded. I think he lacks the hypothesis that B is bounded or something similar.