The Supremum and Infimum of a Nonempty Subset of Real Numbers

# The Supremum and Infimum of a Nonempty Subset of Real Numbers

Recall from the Upper and Lower Bounds of Subsets of Real Numbers page that if $S \subseteq \mathbb{R}$ is a subset of real numbers then a real number $M \in \mathbb{R}$ is an upper bound of $S$ if for all $x \in S$ we have that $x \leq M$. Furthermore, $m \in \mathbb{R}$ is a lower bound of $S$ if for all $x \in S$ we have that $m \leq x$.

We will now discuss a very important property of the real numbers known as the axiom of completeness. It will be very important later on.

 Axiom (Completeness of the Real Numbers): If $S \subseteq \mathbb{R}$ is a nonempty subset of real numbers that is bounded above then $S$ has a least upper bound $M \in \mathbb{R}$. That is, if $M^* \in \mathbb{R}$ is also an upper bound of $S$ then $M \leq M^*$.

An analogous result for nonempty subsets of real numbers that are bounded below can be derived from the axiom of completeness.

 Theorem 1: If $S \subseteq \mathbb{R}$ is a nonempty subset of real numbers that is bounded below then $S$ has a greatest lower bound $m \in \mathbb{R}$. That is, if $m^* \in \mathbb{R}$ is also a lower bound of $S$ then $m^* \leq m$.
• Proof: Let $S \subseteq \mathbb{R}$ be such that $S \neq \emptyset$ and assume that $S$ is bounded below. Then there exists an $m' \in \mathbb{R}$ such that for all $x \in S$ we have that $m' \leq x$. Consider the set:
(1)
\begin{align} \quad -S = \{ -x : x \in S \} \end{align}
• Clearly $-S \neq \emptyset$. Furthermore, since $m' \leq x$ for all $x \in S$ we see that $-x \leq -m'$ for all $-x \in -S$. So $-S$ is a nonempty subset of the real numbers that is bounded from above. By the completeness axiom, $-S$ has a least upper bound $M \in \mathbb{R}$ with the property that $-x \leq M \leq -m'$ for all $-x \in -S$ and for all upper bounds $-m'$ of $-S$. But then $m' \leq -M \leq x$ for all $x \in S$ and for all lower bounds $m'$ of $S$. Set $m = -M$. Then $m' \leq m \leq x$ for all $x \in S$ and for all lower bounds $m'$ of $S$ which shows that $m$ is a greatest lower bound for $S$. $\blacksquare$
 Definition: Let $S \subseteq \mathbb{R}$ be nonempty. 1) If $S$ is bounded above then the least upper bound of $S$ is called the Supremum of $S$ and is denoted $\sup S$. 2) If $S$ is bounded below then the greatest lower bound of $S$ is called the Infimum of $S$ and is denoted $\inf S$.

By convention, if $S$ is unbounded from above we say that $\sup S$ exists and that $\sup S = \infty$. If $S$ is unbounded from below we say that $\inf S$ exists and that $\inf S = -\infty$. Therefore the supremum and infimum of a nonempty subset of real numbers always exists in the set of extended real numbers.

Note that if $S$ is a nonempty subset of real numbers that is bounded above then $\sup S$ need not be contained in $S$. Similarly, if $S$ is a nonempty subset of real numbers that is bounded below then $\inf S$ need not be contained in $S$.

For example, consider the set $\mathbb{N} = \{ 1, 2, 3, ... \}$ of natural numbers. Then:

(2)
\begin{align} \quad \sup \mathbb{N} &= \infty \\ \quad \inf \mathbb{N} &= 0 \end{align}

For another example, consider the open interval $(0, 1] = \{ x \in \mathbb{R} : 0 < x \leq 1 \}$. Then:

(3)
\begin{align} \quad \sup (0, 1] &= 1 \\ \quad \inf (0, 1] &= 0 \end{align}

In this case $\sup (0, 1] \in (0, 1]$ but $\inf (0, 1] \not \in (0, 1]$.

We now state a simple proposition regarding the supremum and infimum of a nonempty finite subset of real numbers. We omit the proof but it is rather simple to prove.

 Proposition 1: If $S \subseteq \mathbb{R}$ is a nonempty finite subset of real numbers then: a) $\sup S = \max \{ x : x \in S \}$. b) $\inf S = \min \{ x : x \in S \}$.