The Supremum and Infimum of The Bounded Set (S+T)

# The Supremum and Infimum of The Bounded Set (S+T)

Recall from The Supremum and Infimum of a Bounded Set page the following definitions:

Definition: Let $S$ be a set that is bounded above. We say that the supremum of $S$ denoted $\sup S = u$ is a number $u$ that satisfies the conditions that $u$ is an upper bound of $S$ and $u$ is the least upper bound of $S$, that is for any $v$ that is also an upper bound of $S$ then $u \leq v$. |

Definition: Let $S$ be a set that is bounded below. We say that the infimum of $S$ denoted $\inf S = w$ is a number $w$ that satisfies the conditions that $w$ is a lower bound of $S$ and $w$ is the greatest lower bound of $S$, that is for any $t$ that is also a lower bound of $S$ then $t \leq w$. |

Now let $S$ and $T$ be bounded sets and define the set $(S + T) := \{ x + y : x \in \mathbb{S}, y \in T \}$. We will now look at some important supremum and infimum theorems regarding this set.

Theorem 1: Let $S$ and $T$ be nonempty subsets of $\mathbb{R}$. Then $\sup (S + T) = \sup S + \sup T$. |

**Proof:**Let both $S$ and $T$ be nonempty subsets of $\mathbb{R}$, and let $u_1 = \sup S$ and $u_2 = \sup T$. Since $u_1 = \sup S$ then $\forall x \in S$ we have that $x ≤ u_1$ and similarly, since $u_2 = \sup T$ then $\forall y \in T$ we have that $y ≤ u_2$. Adding these inequalities together we get that $\forall x \in S$ and $\forall y \in T$ then $x + y ≤ u_1 + u_2$. So $u_1 + u_2$ is an upper bound to the set $(S + T)$ and so $\sup (S + T) ≤ u_1 + u_2$, that is $\sup (S + T) ≤ \sup S + \sup T$.

- Let $\epsilon > 0$. Now since $u_1 = \sup S$ then $u_1 - \frac{\epsilon}{2}$ is not an upper bound to the set $S$ and so $\exists x' \in S$ such that $x' > u_1 - \frac{\epsilon}{2}$. Similarly, since $u_2 = \sup T$ then $u_2 - \frac{\epsilon}{2}$ is not an upper bound to the set $T$ and so $\exists y' \in T$ such that $y' > u_2 - \frac{\epsilon}{2}$. Adding these inequalities together we get for all $\epsilon > 0$ that $x' + y' > u_1 + u_2 - \epsilon$, and since $\epsilon > 0$ is arbitrary we have that $x' + y' ≥ u_1 + u_2$. But $\forall x \in S$ and $\forall y \in T$ we have that $\sup (S + T) ≥ x + y$, so $\sup (S + T) ≥ u_1 + u_2$ or rather $\sup (S + T) ≥ \sup S + \sup T$.

- Whence we have that since $\sup S + \sup T ≤ \sup (S + T)$ and $\sup (S + T) ≤ \sup S + \sup T$ then $\sup (S + T) = \sup S + \sup T$. $\blacksquare$

Theorem 2: Let $S$ and $T$ be nonempty subsets of $\mathbb{R}$. Then $\inf (S + T) = \inf S + \inf T$. |

**Proof:**Let both $S$ and $T$ be nonempty subsets of $\mathbb{R}$, and let $w_1 = \inf S$ and $w_2 = \inf T$. Since $w_1 = \inf S$ then $\forall x \in S$ we have that $w_1 ≤ x$ and similarly, since $w_2 = \inf T$ then $\forall y \in T$ we have that $w_2 ≤ y$. Adding these inequalities together we get that $\forall x \in S$ and $\forall y \in T$ then $w_1 + w_2 ≤ x + y$. So $w_1 + w_2$ is a lower bound to the set $(S + T)$ and so $w_1 + w_2 ≤ \inf (S + T)$, that is $\inf S + \inf T ≤ \inf (S + T)$.

- Let $\epsilon > 0$. Now since $w_1 = \inf S$ then $w_1 + \frac{\epsilon}{2}$ is not a lower bound to the set $S$ and so $\exists x' \in S$ such that $x' < w_1 + \frac{\epsilon}{2}$. Similarly, since $w_2 = \inf T$ then $w_2 + \frac{\epsilon}{2}$ is not a lower bound to the set $T$ and so $\exists y' \in T$ such that $y' < w_2 + \frac{\epsilon}{2}$. Adding these inequalities together we get for all $\epsilon > 0$ that $x' + y' < w_1 + w_2 + \epsilon$, and since $\epsilon > 0$ is arbitrary we have that $x' + y' ≤ w_1 + w_2$. But $\forall x \in S$ and $\forall y \in T$ we have that $\inf (S + T) ≤ x + y$ and so $\inf (S + T) ≤ w_1 + w_2$ or rather $\inf (S + T) ≤ \inf S + \inf T$.

- Whence we have that since $\inf S + \inf T ≤ \inf (S + T)$ and $\inf (S + T) ≤ \inf S + \inf T$ then $\inf (S + T) = \inf S + \inf T$. $\blacksquare$