Prove that $\sup(A+B) = \sup(A) + \sup(B)$ and why does $\sup(A+B)$ exist?

Question

We want to show that $\sup(A)+\sup(B)$ is the least upper bound of the set $A + B$. First, we need to show that $\sup(A) + \sup(B)$ is an upper bound for the set $A + B$. Indeed, if $z\in A + B$, then there exists $a\in A$ and $b\in B$ such that $z = a + b$. But by definition of $\sup(A)$ and $\sup(B)$, $a \leq \sup(A)$ and $b \leq \sup(B)$, so $z = a + b \leq \sup(A) + \sup(B)$. So, $\sup(A) + \sup(B)$ is an upper bound for $A + B$.

We now wish to show that $\sup(A) + \sup(B)$ is the least upper bound for the set $A + B$. So, if $u$ is an upper bound for $A + B$, we need to show that $\sup(A)+\sup(B) \leq u$. We will use part (i): that is, we need to show that there exists $\varepsilon > 0$, $\sup(A) + \sup(B) < u + \varepsilon$.

To do this, note that since $\sup(A)$ is the least upper bound for $A$, $\sup(A) - \varepsilon/2$ is not an upper bound for A, so there exists an $a\in A$ so that $\sup(A) - \varepsilon/2 < a$. Similarly, there is a $b\in B$ so that $\sup(B) - \varepsilon/2 < b$.

Adding these two inequalities gives $$ \sup(A) + \sup(B) - \varepsilon < a + b; $$ in other words $$ \sup(A) + \sup(B) < a + b + \varepsilon. $$ But $u$ is an upper bound for the set $A + B$, so $a + b \leq u$, and hence we have $$ \sup(A) + \sup(B) < u + \varepsilon. $$ Thus, by part (i), $\sup(A) + \sup(B) \leq u$, so $\sup(A) + \sup(B)$ is the least upper bound for $A + B$, as required.

Now how do I show that $\sup(A+B)$ exists?

Answer 1

Show $A+B$ is nonempty and bounded above.

Answer 2

Assume $A,B$ are both nonempty. Otherwise you can run into some $\infty-\infty$ strangeness.

You have $a \le \sup A$ for all $a \in A$, and similarly $b \le \sup B$ for all $b \in B$. Hence we have $a+b \le \sup A + \sup B$ for all $a \in A, b \in B$. Hence we have $\sup (A+B) = \sup_{a \in A, b \in B} a+b \le \sup A + \sup B$.

Now let $\epsilon >0$, and choose $a \in A, b \in B$ such that $a > \sup A -\frac{\epsilon}{2}$ and $b > \sup B -\frac{\epsilon}{2}$. Then $\sup(A+B) \ge a+b > \sup A + \sup B - \epsilon$. Since $\epsilon>0$ was arbitrary, we are have the desired result.