Prove that inf(A+B) = infA + infB

A+B={a+b} I proved that the set A+B is bounded below. Now I'm stuck on how to prove that inf(A+B) = infA + infB

Solution 1:

For all $a\in A$ and $b\in B$ we have $$a+b\ge \inf A+\inf B$$ so we deduce that $$\inf(A+B)\ge \inf A+\inf B$$ for the other inequality we have $$\inf(A+B)\leq a+b\quad \forall a\in A, b\in B$$ so $$\inf(A+B)-a\leq b\quad \forall a\in A, b\in B$$ so $$\inf(A+B)-a\leq\inf B \quad \forall a\in A\iff \inf(A+B)-\inf B\leq a \quad \forall a\in A$$ hence $$\inf(A+B)-\inf B \leq \inf A\iff \inf(A+B) \leq \inf A+\inf B$$

Solution 2:

To show that $\inf (A+B) \leq \inf A+\inf B.$

For any $\epsilon>0$ there exist $a \in A$ such that $a<\inf A+\frac{\epsilon}{2}$

and also there exist $b \in B$ such that $b<\inf B+\frac{\epsilon}{2}.$

Hence $a+b<\inf A+\inf B+\epsilon.$ This is true for any $\epsilon>0.$

Therefore, $\inf(A+B) \leq \inf A+\inf B. ------(1)$

Now, $\inf A \leq a ~~\forall a \in A$ also $\inf B \leq b ~~\forall b \in B. $

So, $\inf A+\inf B \leq a+b ~~\forall a \in A ~~and~~ b\in B.$

Therefore, $\inf A+\inf B \le \inf(A+B).-------(2)$

From (1) and (2) result follows.

Solution 3:

Hint. It is easy to show that $\inf (A+B)\ge \inf A +\inf B$ : because $a+b\ge \inf A+\inf B$ for all $a\in A$, $b\in B$. To prove the reverse, use the following fact : for all $\varepsilon>0$ there is $a\in A$ s.t. $a-\varepsilon\le\inf A$.