Properties of the Sup. and Inf. of a Nonempty Subset of Real Numbers

# Properties of the Supremum and Infimum of a Nonempty Subset of Real Numbers

Recall from The Supremum and Infimum of a Nonempty Subset of Real Numbers page that if $S \subseteq \mathbb{R}$ is a nonempty subset of the real numbers then if $S$ is bounded above there exists a least upper bound to $S$ called the supremum of $S$ and if $S$ is bounded below there exists a greatest lower bound to $S$ called the infimum of $S$.

We now state some important properties of the supremum and infimum of a nonempty subset of real numbers.

Theorem 1: Let $S \subseteq \mathbb{R}$ be a nonempty subset of real numbers that is bounded above. Then $\sup S = M$ if and only if the following two conditions hold:1) $M$ is an upper bound to $S$.2) For all $\epsilon > 0$ there exists an element $x \in S$ such that $M - \epsilon < x$. |

**Proof:**$\Rightarrow$ Let $\sup S = M$. Then for all $x \in S$ we have that $x \leq M$ and if $M^*$ is any other upper bound to $S$ then $M \leq M^*$.

- Since for all $x \in S$ we have that $x \leq M$ this means that $M$ is an upper bound to $S$ so condition (1) is satisfied.

- Suppose that there exists an $\epsilon_0 > 0$ such that for every element $x \in S$ we have that:

\begin{align} \quad x \leq M - \epsilon_0 \end{align}

- Then $M - \epsilon_0$ is an upper bound to $S$ and $M - \epsilon_0 < M$ which contradicts $\sup S = M$. So for all $\epsilon > 0$ there must exist an element $x \in S$ such that $M - \epsilon < x$, so (2) is satisfied.

- $\Leftarrow$ Suppose that (1) and (2) hold. By (1) we have that $M$ is an upper bound to $S$. We now show that $M$ is a least upper bound. Let $M^*$ be an other upper bound to $S$ and assume that $M^* < M$. Then $M - M^* > 0$. Set $\epsilon = M - M^*$. By (2) there must exist an element $x \in S$ such that:

\begin{align} \quad M - \epsilon < x \quad \Leftrightarrow \quad M - (M - M^*) < x \end{align}

- So $M^* < x$. But this inequality contradicts $M^*$ being an upper bound to $S$. Hence the assumption that $M^*< M$ was false. So for every other upper bound $M^*$ of $S$ we must have that $M \leq M^*$. Thus $\sup S = M$. $\blacksquare$

Theorem 2: Let $S \subseteq \mathbb{R}$ be a nonempty subset of real numbers. Then $\sup S = M$ if and only if the following two conditions hold:1) $M$ is an upper bound to $S$.2) For all $t < M$ there exists an element $x \in S$ such that $t < x$. |

**Proof:**$\Rightarrow$ Let $\sup S = M$. Then for all $x \in S$ we have that $x \leq M$ and if $M^*$ is any other upper bound to $S$ then $M \leq M^*$.

- Since for all $x \in S$ we have that $x \leq M$ this means that $M$ is an upper bound to $S$ so condition (1) is satisfied.

- Suppose that there exists a $t_0 < M$ such that for every $x \in S$ we have that:

\begin{align} \quad x \leq t_0 \end{align}

- Then $t_0$ is an upper bound to $S$ and $t_0 < M$ which contradicts $\sup S = M$. So for all $t < M$ there exists an element $x \in S$ such that $t < x$. So (2) is satisfied.

- $\Leftarrow$ Suppose that (1) and (2) hold. By (1) we have that $M$ is an upper bound to $S$. We now show that $M$ is a least upper bound. Let $M^*$ be an upper bound to $S$ and assume that $M^* < M$. Set $t = M^*$. By (2) we have that there exists an element $x \in S$ such that $t = M^* < x$ so $M^*$ is not an upper bound to $S$ which is a contradiction. Therefore the assumption that $M^*$ is an upper bound to $S$ is false. Thus if $M^*$ is an upper bound to $S$ then $M \leq M^*$, so $\sup S = M$. $\blacksquare$

Theorem 3: Let $S \subseteq \mathbb{R}$ be a nonempty subset of real numbers that is bounded below. Then $\inf S = m$ if and only if the following two conditions hold:1) $m$ is a lower bound to $S$.2) For all $\epsilon > 0$ there exists an element $x \in S$ such that $x < m + \epsilon$. |

**Proof:**$\Rightarrow$ Let $\inf S = m$. Then for all $x \in S$ and for all other lower bounds $m^*$ of $S$ we have that $m^* \leq m \leq x$. So (1) holds.

- Suppose that there exists an $\epsilon_0 > 0$ such that for all $x \in S$ we have that:

\begin{align} \quad m + \epsilon_0 \leq x \end{align}

- Then $m + \epsilon_0$ is an upper bound to $S$. But $m < m + \epsilon_0$ which contradicts $\inf S = m$. Therefore for all $\epsilon > 0$ there must exist an element $x \in S$ such that $x < m + \epsilon$ and (2) is satisfied.

- $\Leftarrow$ Suppose that (1) and (2) hold. From (1) we immediately have that $m$ is a lower bound to $S$. We show that $m$ is the greatest lower bound to $S$. Suppose not, i.e., suppose there exists an upper bound $m^*$ such that $m < m^*$. Then $m^* - m > 0$. Set $\epsilon = m^* - m$. Then from (2) we have that there exists an element $x \in S$ such that:

\begin{align} \quad x < m + \epsilon = m + (m^* - m) = m^* \end{align}

- But this is a contradiction as $m^*$ is a lower bound for $S$. Therefore $m$ is a greatest upper bound for $S$, i.e., $\inf S = m$. $\blacksquare$

Theorem 4: Let $S \subseteq \mathbb{R}$ be a nonempty subset of real numbers that is bounded below. Then $\inf S = m$ if and only if the following two conditions hold:1) $m$ is a lower bound to $S$.For all $t > m$ there exists an element $x \in S$ such that $t > x$. |