1.5 Greatest lower bounds

We conclude our discussion of the completeness axiom by introducing the notation of a greatest lower bound (or infimum), which is a close cousin to the notion of a least upper bound.

Definition 1.39.

Let $A\subseteq\mathbb{R}$ .

1

We say $x\in\mathbb{R}$ is a lower bound if $x\leq a$ for all $a\in A$ .
2

We say $A$ is bounded below if there exists a lower bound $x\in\mathbb{R}$ for $A$ .
3
We say $i\in\mathbb{R}$ is a greatest lower bound (or infimum) for $A$ if both
1. (i)
  
  $i$ is a lower bound for $A$ , and
2. (ii)
  
  if $x\in\mathbb{R}$ is a lower bound for $A$ , then $x\leq i$ .
In this case, we write $i=\inf A$ .
4

We say $A$ is bounded if it is both bounded above and bounded below. Otherwise, $A$ is unbounded.

If an infimum for $A$ exists, then it must be unique (why?). Thus, we refer to it as the infimum of $A$ , which is denoted by $\inf A$ .

Example 1.40.

Consider the set $A:=\big{\{}\frac{n}{2n-1}:n\in\mathbb{N}\big{\}}$ . We claim that $\inf A=1/2$ .

Note that

$\displaystyle\frac{n}{2n-1}$	$\displaystyle=\frac{1}{2}\cdot\frac{(2n-1)+1}{2n-1}$
	$\displaystyle=\frac{1}{2}\Big{(}1+\frac{1}{2n-1}\Big{)}$
(1.4)	(1.4)	$\displaystyle\geq\frac{1}{2}\qquad\text{for all $n\in\mathbb{N}$.}$

Hence $1/2$ is a lower bound for $A$ .

We now show $1/2$ satisfies a version of the approximation property for the infimum. For this, we note that

2(2n-1)\geq 2(2n-n)=2n>n\qquad\text{for all $n\in\mathbb{N}$.}

Let $\varepsilon>0$ be given. Let $N\in\mathbb{N}$ satisfy $N>1/\varepsilon$ so that $0<1/N<\varepsilon$ . If we define $a:=\frac{N}{2N-1}$ , then $a\in A$ and

(1.5) (1.5)

a=\frac{1}{2}+\frac{1}{2(2N-1)}<\frac{1}{2}+\frac{1}{N}<\frac{1}{2}+\varepsilon.

Combining (1.40) and (1.5), we have $1/2\leq a<1/2+\varepsilon$ . This tells us that $1/2+\varepsilon$ cannot be a lower bound for $A$ . As $\varepsilon>0$ was chosen arbitrarily, we conclude that $\inf A=1/2$ .

Many properties of infima can be deduced directly from properties of suprema. In particular, the following lemma establishes a version of the completeness axiom in terms of infima.

Lemma 1.41.

Let $A\subseteq\mathbb{R}$ be nonempty and bounded below. Then $\inf A$ exists and, moreover, $\inf A=-\sup(-A)$ where $-A:=\{-x:x\in A\}$ .

Proof.

Since $A$ is nonempty, there exists some $a\in A$ . Hence, by definition $-a\in-A$ so that $-A$ is nonempty. Since $A$ is bounded below, there exists some $m\in\mathbb{R}$ such that $x\geq m$ for all $x\in A$ . Thus, $-x\leq-m$ for all $x\in A$ so that $y\leq-m$ for all $y\in-A$ . Hence, $-m$ is an upper bound for $-A$ so that $-A$ is bounded above.

Since $-A$ is nonempty and bounded above, $\sup(-A)$ exists. We claim that $\inf A$ exists and $\inf A=-\sup(-A)$ .

Since $\sup(-A)$ is, by definition, an upper bound for $-A$ , we have $y\leq\sup(-A)$ for all $y\in-A$ . Thus, $-x\leq\sup(-A)$ for all $x\in A$ and so $x\geq-\sup(-A)$ for all $x\in A$ . Hence $-\sup(-A)$ is a lower bound for $A$ .

It remains to show $-\sup(-A)$ is the greatest lower bound for $A$ . Indeed, suppose $m\in\mathbb{R}$ is any lower bound for $A$ . Then by our earlier arguments, we know $-m$ is an upper bound for $-A$ . Hence, by the definition of the supremum, $\sup(-A)\leq-m$ . But this implies that $m\leq-\sup(-A)$ . Hence $-\sup(-A)$ is indeed the greatest lower bound for $A$ and thus, by definition, $\inf A=-\sup(-A)$ . ∎

Definition 1.42.

Let $A\subseteq\mathbb{R}$ . We say $m\in\mathbb{R}$ is a minimum of $A$ if $m\in A$ and $x\geq m$ for all $x\in A$ .

If a minimum for $A$ exists, then it must be unique (why?). Thus, we refer to it as the minimum of $A$ , which is denoted by $\min A$ .

Exercise 1.43.

Let $A\subset\mathbb{R}$ .

(i)

Show that if a minimum $\min A$ exists, then $\inf A$ exists and $\inf A=\min A$ .
(ii)

Give an example of a nonempty, bounded set $A\subseteq\mathbb{R}$ for which no minimum exists.