Proving sup(A + B) = sup A + sup B There are two sets $A$ and $B$ which are bound and are not empty. Now we'll define another set as:
$A + B = \{a + b | a \in A, b \in B\}.$
I need to prove that $\sup(A + B) = \sup A + \sup B.$ 
I don't know if the way in which I proved it is enough or is missing something. This is my proof:
I know that for every $a \in A, b \in B$: $\sup A \geq a$ and $\sup B \geq b.$  Therefore $\sup A + sup B \geq a + b.$
I'll take a $c \in A+B$ and by $A+B$ definition: $c = a + b$.
Then, it is safe to say that $c = a + b \leq \sup A + \sup B$ which gives:
$c \leq \sup A + \sup B.$
I know that $c \leq \sup A + \sup B$ is true for every general $c \in A+B$, so the supremum of $A + B$ is: $\sup(A + B) = \sup A + \sup B.$
Is this a way to prove it or is it not enough? Thanks.
 A: I think you actually mean set instead of group. And no, that is not a correct proof. E.g., $x \le 1 \forall x \in \{0\}$ but $\sup \{0\} = 0$.
Hint:
As @Gibbs pointed out in the comments, you showed $\sup (A + B) \le \sup A + \sup B$. Now for the equality you need to show $\sup (A + B) \ge \sup A + \sup B$ also holds. Take $\epsilon > 0$, 
$$ a \in A : |a - \sup A| < \frac \epsilon 2 $$
$$ b \in B : |b - \sup B| < \frac \epsilon 2 $$
which exist by characterization of supremum. Can you take it from here?
A: As Gibbs pointed out, till now we have:
$sup(A+B)\leq supA + supB$
We'll proceed by Contradiction. 
Suppose, $sup(A+B)=M, M<supA+supB$
Using Denseness of $\mathbb{Q}$,
$\exists r\in \mathbb{Q}: M<r<supA+supB$
$ supA+supB \geq sup(A+B)\geq a+b$ $\forall a\in A, \forall b\in B$ 
$supA+supB\geq a+b>r\implies r\in A+B$
But, $r>M \implies M \neq sup(A+B) \implies sup(A+B)\geq supA+supB$
But, $supA+supB\geq sup(A+B) \implies sup(A+B)=supA+supB$
A: That's not really enough as you didn't really relate $c$ to $\sup (A+ B)$.
First the boring part: showing $A+B$ is non-empty and bounded above. That's boring but needs to be done but is trivially done by noting if $a\in A, b\in B$, $A$ bounded by $c$ and $B$ bounded by $d$ then $a+b \in A+B$ and $A+B$ bounded by $c + d$.
Actually that last clause needs a little exposition.  If $f \in A+B$ there are $a \in A; b\in B$ so that $f = a + b$.  As $a\le \sup A$ and $b\le \sup B$ we know $f = a+b \le \sup A + \sup B$ so $A+B$ is bounded above by $\sup A + \sup B$ so $\sup (A+B) \le \sup A + \sup B$.
We need to show that if $g < \sup A + \sup B$ then $g$ can not be an upper bound of $A + B$.  Or in other words there exists an $f = a+b; a\in A; b\in B$ so that $g < f=a+b \le \sup A + \sup B$.  Intuitively we want $g= a' + b'$ where $a'<\sup A$ and $b' < \sup B$ so that, there will be $a,b,$ so that $a' < a\le \sup A$ and $b' < b \le \sup B$.  But can we split $g$ like that?
Let $\epsilon = (\sup A + \sup B)$ then $g = [\sup A - \frac {\epsilon}2] + [\sup B -\frac {\epsilon}2]$.  By def of $\sup$ we know there is an $a\in A; [\sup A - \frac {\epsilon}2] < a \le \sup A$ and a $b\in B;[\sup B - \frac {\epsilon}2] < b \le \sup B$.  So $g < a + b \le \sup A + \sup B$ and, of course, $a+b \in A+B$.
So $g$ is not an upper bound of $\sup A + \sup B$.  So $\sup A + \sup B = \sup (A+B)$.
