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?
Best Answer
Here is a direct proof:
let $\alpha =\sup A.$ Since $A$ is bounded above, $\alpha\in \mathbb R.$ It is enough to show that $\sup(c+A)=c+\alpha.$
If $a\in A$ then since $a\le \alpha,$ we have $c+a\le c+\alpha$ so
$\tag 1c+\alpha\ \text{is a upper bound for}\ c+A.$
Suppose $\beta $ is another upper bound for $c+A.$ Then, $\beta-c$ is an upper bound for $A$ which means that $\beta-c\ge \alpha$ (since $\alpha = \sup A$). It follows that
$\tag2 \beta\ge c+\alpha.$
To finish, combine $(1)$ and $(2).$