Show that: $\inf(A+B) = \inf(A)+ \inf(B)$ 
Let $A,B$ be non-empty, bounded subsets of $R$ and $A+B=\{a+b:a \in A,b \in B\}$. Show that: $$\inf(A+B)=\inf(A)+\inf(B)$$

This is what I have tried:
Suppose that $x \in A+B \Rightarrow x=a+b, \ a \in A, \ b \in B$. We know that $a\geq \inf A \text{ and } b\geq \inf(B)$ . So $x \geq \inf(A)+\inf(B) \Rightarrow\inf(A+B) \geq \inf(A)+\inf(B)$
If $x\in A \Rightarrow x \in A+B, \text{ so we conclude that } \inf(A) \leq  \inf(A+B)$ and if $x \in B \Rightarrow x \in A+B,\text{ so we conclude that } \inf(B) \leq  \inf(A+B)$. Therefore, we have that $\displaystyle \frac{\inf(A)+\inf(B)}{2} \leq \inf(A)+\inf(B) \leq \inf(A+B)$.
Could you tell me if it is right?
 A: Your proof for $\inf (A+B)\ge \inf A+\inf B$ is right.
For the other direction. For any $\delta>0$, there exist $a\in A, b\in B$ s.t. $a<\inf A+\delta$ and $b<\inf B+\delta$, then $a+b<\inf A+\inf B+2\delta$, hence $\inf (A+B)\le \inf A+\inf B+2\delta$. Let $\delta$ go to zero.
A: I shall prove a small lemma.

If $x,y$ and $a$ be arbitrary real numbers such that the inequality $a \leq x \leq a+\frac {y}{n}$ holds for all integers $n \geq 1$, then $x=a$.

This is because if $x>a$,  then the number $\frac {y}{x-a}$ would be an upper bound for the set of positive integers, which is not the case.  
Now, you have proved yourself that $\inf(A+B)\geq \inf(A)+\inf(B)$. 
Also observe that for any positive integer $n$, $\exists a \in A$ such that $a \leq \inf(A)+\frac{1}{n}$ and $\exists b \in B$ such that $b\leq \inf(B)+\frac{1}{n}$.(Why?) 
So, we have $\inf(A+B)\leq a+b \leq \inf(A)+\inf(B)+\frac {2}{n}$ $\forall n \in \mathbb N$ 
Thus we have $\inf(A)+\inf(B) \leq \inf (A+B) \leq \inf(A)+\inf(B) + \frac {2}{n}$ for all positive integers $n\geq1$. Thus from the lemma, we are forced to conclude that $\inf(A+B)=\inf(A)+\inf(B)$.
