The Supremum and Infimum of a Bounded Set

# The Supremum and Infimum of a Bounded Set

Recall that a nonempty subset $S$ of $\mathbb{R}$ is bounded above if there exists an $M$ such that for all $x \in \mathbb{S}$, $x \leq M$. And similarly, we say that $S$ is bounded below if there exists an $m$ such that for all $x \in \mathbb{S}$, $m \leq x$. We should note that these bounds are not unique. Recall the set of natural numbers $\mathbb{N} = \{ 1, 2, 3, ... \}$. We could choose $m = 1$ to be our lower bound, or $m = 0$, etc… So if a lower or upper bound exists for a set, then we can say that there is an infinite number of lower or upper bounds for that set. What we are sometimes more interested in is the least upper bound or the greatest lower bound. These bounds have a very special name which we will define as follows.

 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 $t \leq w$.

Looking at the example of $\mathbb{N}$, we note that since this subset of $\mathbb{R}$ is bounded below, then $\inf \mathbb{N}$ exists. Namely, $\inf \mathbb{N} = 1$. We can show this (rather tediously) with the definition of the infimum of a set. We will first show that $1$ is a lower bound for $\mathbb{N}$, that is for all $x \in \mathbb{N}$, $1 \leq x$. We note that $\mathbb{N} = \{ 1, 2, 3, ... \}$, so clearly $1 \leq x$ for all $x \in \mathbb{N}$. So we have shown that $1$ is a lower bound. Now let's show that $1$ is the greatest lower bound.

Suppose there exists a greatest lower bound $t$. That is $1 \leq t$. But then the element $1 \in \mathbb{N}$ is less or equal to $t$ so then $t = 1$. Therefore $\inf \mathbb{N} = 1$.

Now let's look at some other examples of sets. For example, consider the set $S := \{ x \in \mathbb{R} : -2 \leq x \leq 2 \}$. This set is clearly bounded above and bounded below. We note that $\inf S = -2$ and $\sup S = 2$. In this case the infimum and supremum of this set are in the set $S$ itself. This is not always the case though. For example consider the set $T = \{ x \in \mathbb{R} : -2 < x < 2 \}$. In this case $\inf T = -2$ and $\sup T = 2$, but note that $-2, 2 \not \in T$ due to the strict inequalities defining $T$.

Yet another example is the set $\mathbb{Z}$. We noted that this set is not bounded below or bounded above, so in fact this subset does not contain an infimum or supremum.

 Lemma 1: Let $S$ be a nonempty subset of $\mathbb{R}$. Then either $S$ has both a supremum and infimum, $S$ has one or the other, or $S$ does not have either.

We have already looked at an example that contains both a supremum and infimum, an example that contains only an infimum, and an example that contains neither. We just need to provide an example of a subset of $\mathbb{R}$ that contains only a supremum. Consider the set of negative integers $\mathbb{Z}^{-} = \{ ... , -3, -2, -1 \}$. This set is bounded above (but not bounded below), and in fact $\sup \mathbb{Z}^{-} = -1$.

We will now look at another way to describe the supremum of a set that is bounded above, and the infimum of a set that is bounded below.

 Lemma 2: Let $S \subset \mathbb{R}$ where $S \neq \emptyset$. Then $\sup S = u$ if and only if the number $u$ satisfies the properties that for all $x \in S$, $x \leq u$ and if $v < u$ then there exists an $x' \in S$ such that $v < x'$
• Proof: $\Rightarrow$ Suppose that $\sup S = u$. Then for all $x \in S$, $x \leq u$ by the definition of a supremum. Let $v < u$. If there does not exist an $x' \in S$ such that $v < x'$ then $v$ is an upper bound of $S$ and since $v < u$ this contradicts the fact that $u = \sup S$, so such an $x' \in S$ exists.
• $\Leftarrow$ Suppose that for all $x \in S$, $x \leq u$ and if $v < u$ then there exists an $x' \in S$ such that $v < x'$. Therefore the first condition for $u$ to be the supremum of $S$ is satisfied. Since $v < u$, there exists an $x' \in S$ such that $v < x' ≤ u$. But this is true for all $v < u$, so then $u = \sup S$. $\blacksquare$

Lemma 3 is analogous to lemma 2 but pertains to the infimum of a set that is bounded below.

 Lemma 3: Let $S \subset \mathbb{R}$ where $S \neq \emptyset$. Then $\inf S = w$ if and only if the number $w$ satisfies the properties that for all $x \in S$, $w \leq x$ and if $w \leq t$ then there exists an $x' \in S$ such that $x' \leq t$.

Now consider $S \subset \mathbb{R}$ where $\sup S = u$. If for all $x \in S$, $x \leq u$ then $u$ is an upper bound for $S$. Now the second property says that if $v \leq u$ then there exists an element $x' \in S$ such that $v \leq x'$. If such an element $x'$ did not exist, then $\sup S = v$.