TITLE: Prove that $\sup(A+B) = \sup(A) + \sup(B)$ and why does $\sup(A+B)$ exist?
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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.