Proof that $\sup(c+A) = c + \sup(A)$

Question

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?

Answer 1

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 $(2).$

Answer 2

Let $L=\sup(c+A), S=\sup(A), c+A=\{c+a~|~a\in A\}$

First, show $L\le c+S$

$$\forall x\in c+A, \exists a\in A, s.t. x=c+a, \text{and} ~a\le S\Rightarrow x\le c+S\Rightarrow L\le c+S$$

Next, show $L \ge c+S$

Assume the opposite, $L<c+S$, let $\epsilon=\frac{c+S-L}{2}>0$

$$\exists a\in A, s.t. S-a<\epsilon\Rightarrow c+a-L>\epsilon>0\Rightarrow c+a>L$$

Since $c+a\in c+A$, but we get $c+a>L=\sup(c+A)$, which gives a contradiction. Therefore,

$$L\ge c+S$$

Finally, $L=c+S$

Answer 3

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!