Prove that $Sup(A + B) = Sup(A) + Sup(B)$

Earlier on in the book it showed that to prove $a = b$ it is often best to show that $a \leq b$ and that $b \leq a$. This is the way I want to go about the proof. I am sure there is an easier way but I am new to this and I am trying to get a basic grasp on everything before I start class in the fall. Forgive my terrible MathJax.

Question If $A + B := \{a + b | a \in A \land b \in B\}$ Where $A \space and \space B$ are non-empty, bounded subsets of $\mathbb{R}$ then $\sup(A+B) = \sup(A)+\sup(B)$

Since both A and B are non-empty bounded subsets of $\mathbb{R}$ if follows that both A and B have a finite Supremum. Thus $\sup(A)$ and $\sup(B)$ exist. Obviously $\sup(A) \geq a \space \space \forall a \in A$ and $\sup(B) \geq b \space \space \forall \space b \in B$. So $\sup(A) + \sup(B) \geq A + B$ Which means $\sup(A) + \sup(B)$ is an upper bound of $A + B$ and therefore it follows that $\sup(A) + \sup(B) \geq \sup(A + B)$

Likewise since A and B are non-empty bounded subsets of $\mathbb{R}$ It follows that the set $A + B := \{a + b| a \in A \land b \in B\}$ is also a finite bounded subset of $\mathbb{R}$ Therefore the set $A + B$ has a finite \supremum so $\sup(A + B)$ exists. Since $\sup(A) + \sup(B) \in A + B$. It then directly follows that $\sup(A+B)$ is an upperbound of $\sup(A) + \sup(B)$ and then it directly follows that $\sup(A+B) \geq \sup(A) + \sup(B)$

Which completes the proof that $\sup(A) + \sup(B) = \sup(A + B)$.

Solution 1:

You have a couple of things wrong: first you write $\sup(A) + \sup(B) \ge A+B$, and second claim that $\sup(A) + \sup(B) \in A+B$.

Your approach is on the right track, though. To get the first inequality you select an arbitrary element $c \in A+B$. Then $c$ may be expressed as $c = a+b$, where $a \in A$ and $b \in B$. Thus $$c = a + b \le \sup(A) + \sup(B).$$ This means that $\sup(A) + \sup(B)$ is an upper bound of $A+B$, hence $$\sup(A+B) \le \sup(A) + \sup(B).$$

On the other hand, for any $\epsilon > 0$ you may select numbers $a \in A$ and $b \in B$ with $\sup(A) < a + \epsilon$ and $\sup(B) < b + \epsilon$. This means that $$\sup(A) + \sup(B) < a + b + 2\epsilon \le \sup(A+B) + 2\epsilon.$$ The fact that this is valid for arbitrary $\epsilon > 0$ gives you that $$\sup(A) + \sup(B) \le \sup(A+B).$$