The Supremum and Infimum of the Sum of Nonempty Subsets of Real Numbers

# The Supremum and Infimum of the Sum of Nonempty Subsets of Real Numbers

Definition: Let $A, B \subseteq \mathbb{R}$ be nonempty subsets of real numbers. The Sum $A + B$ is defined as the set $A + B = \{ a + b : a \in A, b \in B \}$. |

For example, consider the sets $A = \{ 1, 2, 3 \}$ and $B = \{ 4, 5 \}$. Then:

(1)\begin{align} \quad A + B = \{ 1 + 4, 1 + 5, 2 + 4, 2 + 5, 3 + 4, 3 + 5 \} = \{ 5, 6, 7, 8 \} \end{align}

We now state some important theorems regarding the supremum and infimum of the sum of two nonempty subsets of real numbers.

Theorem 1: Let $A, B \subseteq \mathbb{R}$ be nonempty subsets of real numbers. Then:a) $\sup (A + B) = \sup A + \sup B$.b) $\inf (A + B) = \inf A + \inf B$. |

**Proof of a):**Let $\sup A = M_A$ and $\sup B = M_B$. Let $M = M_A + M_B$. Then $a \leq M_A$ for all $a \in A$ and $b \leq M_B$ for all $b \in B$. Therefore, for all $a \in A$ and $b \in B$ we have that:

\begin{align} \quad a + b \leq M_A + M_B = M \end{align}

- So $M$ is an upper bound for $A + B$. We now establish $M$ to be the least upper bound of $A + B$.

- Let $\epsilon > 0$ be given. Since $\sup A = M_A$ there exists an element $x_A \in A$ such that $\displaystyle{M_A - \frac{\epsilon}{2} < x_A}$, and since $\sup B = M_B$ there exists an element $x_B \in B$ such that $\displaystyle{M_B - \frac{\epsilon}{2} < x_B}$. So:

\begin{align} \quad \left ( M_A - \frac{\epsilon}{2} \right ) + \left ( M_B - \frac{\epsilon}{2} \right ) < x_A + x_B \quad \Leftrightarrow \quad M - \epsilon < x_A + x_B \end{align}

- Since $x_A \in A$ and $x_B \in B$ we see that $x_A + x_B \in A + B$. So for all $\epsilon > 0$ there exists an element $(x_A + x_B) \in (A + B)$ such that $M - \epsilon < x_A + x_B$. Thus by the theorem on the Properties of the Supremum and Infimum of a Nonempty Subset of Real Numbers page we must have that:

\begin{align} \quad \sup (A + B) = M = M_A + M_B = \sup A + \sup B \quad \blacksquare \end{align}

**Proof of b):**Let $\inf A = m_A$ and $\inf B = m_B$. Let $m = m_A + m_B$. Then $m_A \leq a$ for all $a \in A$ and $m_B \leq b$ for all $b \in B$. Therefore, for all $a \in A$ and $b \in B$ we have that:

\begin{align} \quad m = m_A + m_B \leq a + b \end{align}

- So $m$ is an lower bound for $A + B$. We now establish $m$ to be the greatest lower bound of $A + B$.

- Let $\epsilon > 0$ be given. Since $\inf A = m_A$ there exists an element $x_A \in A$ such that $\displaystyle{x < m_a + \frac{\epsilon}{2}}$, and since $\sup B = m_B$ there exists an element $x_B \in B$ such that $\displaystyle{x_B < m_B + \frac{\epsilon}{2}}$. So:

\begin{align} \quad x_A + x_B < \left ( m_A + \frac{\epsilon}{2} \right ) + \left ( m_B + \frac{\epsilon}{2} \right ) \quad \Leftrightarrow \quad x_A + x_B < m + \epsilon \end{align}

- Since $x_A \in A$ and $x_B \in B$ we see that $x_A + x_B \in A + B$. So for all $\epsilon > 0$ there exists an element $(x_A + x_B) \in (A + B)$ such that $x_A + x_B < m + \epsilon$. Thus by the theorem on the Properties of the Supremum and Infimum of a Nonempty Subset of Real Numbers page we must have that:

\begin{align} \quad \inf (A + B) = m = m_A + m_B = \inf A + \inf B \quad \blacksquare \end{align}