Proof that $\sup(c+A) = c + \sup(A)$ Understanding Analysis: Abbott exercise $1.3.5$:
Let $A \subseteq \mathbb{R}$ be bounded above and let $c \in \mathbb
{R}$. Define the set $c + A = \{c + a : a \in A\} $.
Proof.
Since $A$ is bounded above, there exists an upper bound $x \in \mathbb{R}$ such that $a \le x$, $\forall a \in A$. It follows that $\sup(A) \le x$. Now, $a \le x \implies c+a \le c+x \implies c+a \le c+\sup(A) $. Therefore, $c+\sup(A)$ is an upper bound of $c+A$.
Let x be any upper bound. Since $a \le x \implies a+c \le x+c$, we can call $x+c$ an upper bound of $c+A$ for all $a \in A$. It follows that $c+ \sup(A) \le x+c$ because $\sup(A)\le x$.
Now, because $c+\sup(A)$ is an upper bound of $c+A$ and $c+\sup(A) \le x+c$ which is another upper bound,
$\sup(c+A) = c+\sup(A)$ by definition.
Is this proof correct?

We have $x\leq\, supA$ for all $x\in A$ and hence $c+x\,\leq\,c+supA$
for all $x\in A$. That implies that $sup_{A}(c+x)\leq\,c+supA$ and by
definition of the set $c+A$ we obtain: $sup(c+A)\leq\,c+supA$.
Now assume that we have strict inequality. i.e. there is a $\delta$
such that $sup(c+A)\,<\delta\,<\,c+supA$.
Take a sequence $x_{n} \in A$ converging to $supA$.
Then clearly for $n$ sufficiently large $c+x_{n}>\,\delta$.
But $c+x_{n}\,\in \,c+A$ and hence we have an element of $c+A$
greater than the supremum which is an obvious contradiction. Therefore
we have equality and we get our result!
