Elementary Classical Analysis, Marsden, 2nd ed, Chapter 1, Problem 7

Problem

For sets $A,B \subset \mathbb{R}$, let $A+B=\{x+y | \text{$x \in A$ and  $y \in B$}\}$. Show that $\sup(A+B) = \sup(A) + \sup(B)$. Make a similar statement for inf's.

Answer

Let $a = \sup(A), b= \sup(B)$. Then for $z \in A+B$, $z = x+y$ for some $x \in A, y \in B$ and $z = x+y \leq a+b$, which means $a+b$ is an upper bound of $A+B$.

For $\epsilon > 0$, there exists $x_\epsilon \in A, y_\epsilon \in B$ such that $x_\epsilon > a - \epsilon/2, y_\epsilon > b - \epsilon/2$  $\Rightarrow$  $x_\epsilon + y_\epsilon > a+b - \epsilon$. This implies $a+b = \sup(A+B)$.

For inf's we can also prove the following in similar way.

$$ \inf(A+B) = \inf(A) + \inf(B)$$

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