The Supremum and Infimum of The Bounded Set (aS)

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

Theorem 1: Let $S$ be a nonempty subset of $\mathbb{R}$ and let $a \in \mathbb{R}$ such that $a > 0$. Then $\sup (aS) = a\sup S$.

• Proof: Let $S \subset \mathbb{R}$ that is nonempty and bounded and let $u = \sup S$. Then $\forall x \in S$ $x ≤ u$. Now since $a > 0$, we can multiply both sides of this inequality to get that $ax ≤ au$. Therefore $au$ is an upper bound to the set $(aS)$ and so $\sup (aS) ≤ au$.

• Now suppose that $v$ is any upper bound to the set $(aS)$. Then $\forall x \in S$ it follows that $ax ≤ v$ and since $a > 0$ there exists a multiplicative inverse $a^{-1} > 0$ such that $a^{-1}ax ≤ a^{-1}v$ which when simplified gives us that $x ≤ a^{-1}v$. So $a^{-1}v$ is an upper bound to the set $S$ and so $\sup S = u ≤ a^{-1}v$ which implies that $au ≤ v$. Since $v$ is any upper bound to the set $(aS)$ we thus have that $au ≤ \sup (aS)$.

• Since $\sup (aS) ≤ au$ and $au ≤ \sup (aS)$ then $\sup (aS) = au = a \sup S$. $\blacksquare$

Theorem 2: Let $S$ be a nonempty subset of $\mathbb{R}$ and let $a \in \mathbb{R}$ such that $a > 0$. Then $\inf (aS) = a \inf S$.

• Proof: Let $S \subset \mathbb{R}$ that is nonempty and bounded and let $w = \inf S$. Then $\forall x \in S$ $w ≤ x$. Now since $a > 0$, we can multiply both sides of this inequality to get that $aw ≤ ax$. Therefore $aw$ is a lower bound to the set $(aS)$ and so $aw ≤ \inf (aS)$.

• Now suppose that $t$ is any lower bound to the set $(aS)$. Then $\forall x \in S$ it follows that $t ≤ ax$ and since $a > 0$ there exists a multiplicative inverse $a^{-1} > 0$ such that $a^{-1}t ≤ a^{-1}ax$ which when simplified gives us that $a^{-1}t ≤ x$. So $a^{-1}t$ is a lower bound to the set $S$ and so $a^{-1}t ≤ w = \inf S$ which implies that $t ≤ aw$. Since $t$ is any lower bound to the set $(aS)$ we thus have that $\inf (aS) ≤ aw$.

• Since $aw ≤ \inf (aS)$ and $\inf (aS) ≤ aw$ then $\inf (aS) = aw = a \inf S$. $\blacksquare$

Theorem 3: Let $S$ be a nonempty subset of $\mathbb{R}$ and let $a \in \mathbb{R}$ such that $a < 0$. Then $\sup (aS) = a \inf S$.

• Proof: Let $S \subset \mathbb{R}$ that is nonempty and bounded and let $u = \sup S$ and let $w = \inf S$. Then $\forall x \in S$ we have that $w ≤ x$ and since $a ≤ 0$ we have that $aw ≥ ax$. Therefore $aw$ is an upper bound to the set $(aS)$ and so $aw ≥ \sup (aS)$.

• Now suppose that $v$ is any upper bound to the set $(aS)$, and so $\forall x \in S$ it follows that $ax ≤ v$. Since $a < 0$, then we have that $a^{-1} < 0$ and so $a^{-1}ax ≥ a^{-1}v$ which when simplified gives us that $x ≥ a^{-1}v$. So $a^{-1}v$ is a lower bound to the set $S$ and so $\inf S = w ≥ a^{-1}v$ which implies that $aw ≤ v$. Since $v$ is any upper bound of the set $(aS)$ then $aw ≤ \sup (aS)$.

• Since $aw ≥ \sup (aS)$ and $aw ≤ \sup (aS)$ we thus get that $\sup (aS) = aw = a \inf S$. $\blacksquare$

Theorem 4: Let $S$ be a nonempty subset of $\mathbb{R}$ and let $a \in \mathbb{R}$ such that $a < 0$. Then $\inf (aS) = a \sup S$.

• Proof: Let $S \subset \mathbb{R}$ that is nonempty and bounded and let $u = \sup S$ and let $w = \inf S$. Then $\forall x \in S$ we have that $x ≤ u$ and since $a ≤ 0$ we have that $ax ≥ au$. Therefore $au$ is a lower bound to the set $(aS)$ and so $\inf (aS) ≥ au$.

• Now suppose that $t$ is any lower bound to the set $(aS)$, and so $\forall x \in S$ it follows that $t ≤ ax$. Since $a < 0$, then we have that $a^{-1} < 0$ and so $a^{-1}t ≥ a^{-1}ax$ which when simplified gives us that $a^{-1}t ≥ x$. So $a^{-1}t$ is an upper bound to the set $S$ and so $a^{-1}t ≥ u = \sup S$ which implies that $t ≤ au$. Since $t$ is any lower bound of the set $(aS)$ then $\inf (aS) ≤ au$.

• Since $\inf (aS) ≥ au$ and $\inf (aS) ≤ au$ we thus get that $\inf (aS) = au = a \sup S$. $\blacksquare$