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

Question

Let $A \subseteq \mathbb{R}$ be bounded above and let $c \in \mathbb {R}$. Define the set $c + A = \{c + a : a \in A\} $

Now since $a \leq \sup A , \forall a\in A$. Then $a + c \leq \sup A + c $. So $A+c$ has an upper bound. Now the claim is that $\sup(c+A) = c + \sup A.$

So now I have to prove that $\forall \varepsilon> 0, c+a > c + \sup A - \varepsilon.$

How can I proceed? Thanks.

Answer 1

If you wish to use $\varepsilon$s instead, suppose $\varepsilon > 0$. I claim that $c + \sup A - \varepsilon$ is not and upper bound for $c + A$. Since you have proven that $c + \sup A$ is an upper bound for $c + A$, this would mean that no lesser number than $c + \sup A$ can be an upper bound of $c + A$. That is $c + \sup A$ is the supremum of $c + A$.

To prove this claim, it suffices to find some $x \in c + A$ such that $x > c + \sup A - \varepsilon$.

Note that, since $\sup A$ is the least upper bound for $A$, it follows that $\sup A - \varepsilon$ is not an upper bound for $A$. Therefore, there must be some $a \in A$ such that $$a > \sup A - \varepsilon.$$ Adding $c$ to both sides, $$c + a > c + \sup A - \varepsilon.$$ Note that $c + a \in c + A$ by definition of $c + A$. Hence, we can choose $x = c + a$, and our proof of the claim is complete.

Answer 2

You have shown that $c + \sup A$ is an upper bound to $c + A$, and hence $\sup(c + A) \le c + \sup A$. Replacing $A$ with $c + A$ and $c$ with $-c$, we obtain $$\sup(-c + (c + A)) \le -c + \sup(c + A) \le -c + c + \sup A = \sup A.$$ Now, note that $A \subseteq -c + (c + A)$, as if $x \in A$, then $c + x \in c + A$, and so $x = -c + (c + x) \in -c + (c + A)$. Therefore, $$\sup A \le \sup(-c + (c + A)).$$ Putting this together with the above inequalities, we see that all the inequalities must be equality, by the antisymmetry of $\le$. Hence, $$-c + \sup(c + A) = \sup A \implies \sup(c + A) = c + \sup A.$$

Answer 3

So now I have to prove that ∀ε>0,c+a>c+supA−ε.

Well if that is what you have to prove, then you'd have $\forall \epsilon > 0, a > \sup A - \epsilon$ so $c+a > c +\sup A - \epsilon$.

But I think you meant $\forall \epsilon >0 \exists a \in A$ so $a > \sup A - \epsilon$, and so for that $a$, $c+a\in c+A$ and $c+a > c+\sup A - \epsilon$.

Basically everything you can say about the set $A$, its elements and any bounds, can be said about $c + A$ by adding $c$ to all pertainent parts.