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

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

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 $a \in \mathbb{R}$ and define the set $(a + S) := \{ a + x : x \in \mathbb{S} \}$. We will now look at some important supremum and infimum theorems regarding this set. Be sure to check out the The Supremum and Infimum of The Bounded Set (aS) as well for similar proofs.

Theorem 1: Let $S$ be a nonempty bounded subset of $\mathbb{R}$. Then $\sup (a + S) = a + \sup S$. |

**Proof:**Let $S \subset \mathbb{R}$ that is nonempty and bounded and let $u = \sup S$. Then we know that $\forall x \in S$, $x ≤ u = \sup S$. By adding $a$ to both sides of this inequality we get that $a + x ≤ a + u$ and so $a + u$ is an upper bound for the set $(a + S)$ and so $\sup (a + S) ≤ a + u$.

- Now suppose that $v$ is any upper bound to the set $(a + S)$ and then so $\forall x \in S$, $a + x ≤ v$ which implies that $x ≤ v - a$. Therefore $v - a$ is an upper bound to the set $S$ and so $\sup S = u ≤ v - a$ or rather just $u ≤ v - a$ which implies that $a + u ≤ v$. Since $v$ is any upper bound of the set $(a + S)$ we can replace it in this inequality to get that $a + u ≤ \sup (a + S)$.

- Since $\sup (a + S) ≤ a + u$ and $a + u ≤ \sup (a + S)$ then it follows that $\sup (a + S) = a + u = a + \sup S$. $\blacksquare$.

Theorem 2: Let $S$ be a nonempty bounded subset of $\mathbb{R}$. Then $\inf (a + S) = a + \inf S$. |

**Proof:**Let $S \subset \mathbb{R}$ that is nonempty and bounded and let $w = \inf S$. Then we know that $\forall x \in S$, $\inf S = w ≤ x$. By adding $a$ to both sides of this inequality we get that $a + w ≤ a + x$ and so $a + w$ is a lower bound for the set $(a + S)$ and so $a + w ≤ \inf (a + S)$.

- Now suppose that $t$ is any lower bound to the set $(a + S)$ and then so $\forall x \in S$, $t ≤ a + x$ which implies that $t - a ≤ x$. Therefore $t - a$ is a lower bound to the set $S$ and so $t - a ≤ w = \inf S$ or rather just $t - a ≤ w$ which implies that $t ≤ a + w$. Since $t$ is any lower bound to the set $(a + S)$ we can replace it in this inequality to get that $\inf (a + S) ≤ a + w$.

- Since $a + w ≤ \inf (a + S)$ and $\inf (a + S) ≤ a + w$ then it follows that $\inf (a + S) = a + w = a + \inf S$. $\blacksquare$