Properties of The Supremum and Infimum Of A Set

# Properties of The Supremum and Infimum Of A Set

We will now look at some important properties given that a set $S$ has a supremum or infimum. Before that though, we will first touch upon a notational matter. Let $S$ be a nonempty set and let $a \in \mathbb{R}$. We will denote the set $(a + S)$ to be defined as $(a + S) := \{ a + x : x \in S \}$. Similarly we will denote the set $(aS)$ to be defined as $(aS) := \{ ax : x \in S \}$. We will now look at some of these important supremum and infimum properties.

Property 1: Let $S$ be a nonempty bounded set and $a \in \mathbb{R}$. Then $\sup (a + A) = a + \sup A$. |

**Proof of Property 1:**First let $u = \sup S$. By the definition of a supremum of a set we know that $\forall x \in S$, $x \leq u$ which implies that $\forall x \in S$, $a + x \leq a + u$. Therefore we note that $a + u$ is an upper bound for the set $(a + S)$. Therefore $\sup (a + S) \leq a + u$.

- Now suppose that $v$ is any upper bound to the set $(a + S)$. Then it follows that $\forall x \in S$, $a + x \leq v$ which implies that $\forall x \in S$, $x \leq v - a$. Therefore we note that $v - a$ must be an upper bound to the set $S$. Therefore $u = \sup S \leq v - a$ and so $u \leq v - a$. So then $a + u \leq v$, and since $v$ is any upper bound of the set $(a + S)$ we get that $a + u \leq \sup (a + S)$.

- Since $\sup (a + S) \leq a + u$ and $a + u \leq \sup (a + S)$ we conclude that $\sup (a + s) = a + u = a + \sup S$. $\blacksquare$

Property 2: Let $S$ be a nonempty bounded set and $a \in \mathbb{R}$. Then $\inf (a + S) = a + \inf S$. |

**Proof of Property 2:**Let $w = \inf S$. By the definition of an infimum of a set we know that $\forall x \in S$, $w \leq x$ which implies that $\forall x \in S$ $a + w \leq a + x$. Therefore we note that $a + w$ is a lower bound for the set $(a + S)$. Therefore $a + w \leq \inf (a + S)$.

- Now suppose that $t$ is any lower bound to the set $(a + S)$. Then it follows that $\forall x \in S$, $t \leq a + x$ which implies that $\forall x \in S$, $t - a \leq x$. Therefore we note that $t - a$ must be a lower bound to the set $S$. Therefore $t - a \leq \inf S = w$ and so $t - a \leq w$. So then $t \leq a + w$, and since $t$ is any lower bound of the set $(a + S)$ we get that $\inf (a + S) \leq a + w$.

- Since $a + w \leq \inf (a + S)$ and $\inf ( a + S) \leq a + w$ we conclude that $\inf (a + S) = a + w = a + \inf S$. $\blacksquare$

Property 3: Let $S$ be a nonempty bounded set and let $a \in \mathbb{R}$ such that $a > 0$. Then $\sup (aS) = a\sup S$. |

**Proof of Property 3:**Let $a > 0$ and denote $u = \sup S$. By the definition of a supremum we note that $\forall x \in S$, $x \leq u$. Therefore since $a > 0$ we get that $ax \leq au$. So $au$ is an upper bound for the set $(aA)$. That is $\sup(aA) \leq au$.

- Now suppose that $v$ is any upper bound to the set $(aS)$. Then it follows that $\forall x \in S$, $ax \leq v$ which implies that $\forall x \in S$, $x \leq a^{-1}v$. Therefore $a^{-1}v$ must be an upper bound to the set $S$. So $u = \sup S \leq a^{-1} v$ or rather $u \leq a^{-1} v$ and so $au \leq v$. And since $v$ is any upper bound to the set $aS$ we have that $au \leq \sup (aS)$.

- Now since $\sup (aA) \leq au$ and $au \leq \sup (aS)$ we conclude that $\sup(aS) = au = a\sup S$. $\blacksquare$

Property 4: Let $S$ be a nonempty bounded set and let $a \in \mathbb{R}$ such that $a > 0$. Then $\inf (aS) = a \inf S$. |

We will not prove property 4 since it is analogous to property 3.

Property 5: Let $S$ be a nonempty bounded set and let $a \in \mathbb{R}$ such that $a < 0$. Then $\sup (aS) = a \inf S$. |

**Proof of Property 5:**Let $a < 0$ and denote $u = \sup S$. By the definition of a supremum we note that $\forall x \in S$, $x \leq u$. Therefore since $a < 0$ we get that $ax \geq au$. So $au$ is a lower bound for the set $(aA)$. that is $\inf (aS) \geq au$.

**Proof of Property 5:**Let $a < 0$ and let $\inf S = i$ so that $i ≤ x$ for all $x \in S$

**Proof of Property 5:**Let $a < 0$ and denote $u \ sup S$ and $w = \inf S$. By the definition of a supremum we note that $\forall x \in S$, $x \ leq u$. Therefore since $a < 0$ we get that $ax \geq au$. So $au$ is a lower bound to the set $(aS)$ and so $ax \geq w$

- Now suppose that $v$ is any upper bound to the set $(aS)$. The it follows that $\forall x \in S$, $ax \leq v$ which implies (since $a < 0$) that $\forall x \in S$ $x \geq a^{-1} v$. Therefore $a^{-1}v$ must a lower bound to the set $S$. Denote $\inf S = w$. Therefore $\inf S = w \geq a^{-1} v$ and so $aw \leq v$. Since $v$ is any upper bound of $(aS)$ we get that $aw \leq \sup S$.