Let $A$ and $B$ be sets of real numbers. Define a set $A+B$ by $A+B =\{a+b|a \in A, b \in B\}$.
Show that if $A$ and $B$ are bounded sets, then $g.l.b.(A+B) = (g.l.b. A)+(g.l.b.B)$.
(The g.l.b. being the greatest lower bound.)
So I'm just really confused at how to prove this.
So far I have,
$Proof. $ Let $A,B$ be sets of real numbers and bounded sets. Let $\alpha = glbA$ and $\beta = glb B$.
$\forall x \in A,$ there exists an $x \geq \alpha$ and $\forall y \in B,$ there exists $y \geq \beta$.
Not sure where I want to go from here. Please help!
By contradiction:
Assume $$glb(A+B) \neq glb(A) + glb(B)$$ Then either $glb(A+B) > glb(A) + glb(B)$ or $<$.
Assume the former. Since $A$ and $B$ are bounded, $A+B$ must be bounded (can you see this?). Then, it exist some $c \in A+B$ such that:
On the other hand, since $A$ and $B$ are bounded, there are some $\alpha$ and $\beta$ such that $\alpha = glb(A)$ and $\beta = glb(B)$
What is the relation between $a'$ and $\alpha$. and $b'$ and $\beta$? That answered, what is the relation between $a' + b'$ and $\alpha + \beta$?
Do you find anything weird, or something impossible (given the assumptions we agreed on) between the former relations and the definition of our $c$?
This is halfway done, you also need to see what happens when you assume $glb(A+B) < glb(A) + glb(B)$