1.3 The completeness axiom

In the system of rational numbers, the least upper bound for $S:=\{a\in\mathbb{Q}:a^{2}\leq 2\}$ is missing. In order to find it, we need to pass to a larger system of numbers: this is where the completeness axiom and the real numbers $\mathbb{R}$ come in.

Axiom 1.15 (Completeness).

Every nonempty subset $A\subseteq\mathbb{R}$ which is bounded above has a least upper bound in $\mathbb{R}$ .

Since this is an axiom, it does not need to be proved. We will simply define the real numbers $\mathbb{R}$ to be a system of numbers which satisfy this property. In Section 1.6, we will derive consequences from this axiom to establish facts about the real numbers.

Note that we cannot replace $\mathbb{R}$ with $\mathbb{Q}$ in 1.15; that is, the number system $\mathbb{Q}$ does not satisfy the completeness axiom. We can see this from Example 1.14, which shows that the set $S:=\{a\in\mathbb{Q}:a^{2}\leq 2\}\subset\mathbb{Q}$ is bounded above and nonempty but does not have a least upper bound in the rational number system $\mathbb{Q}$ .

Remark 1.16.

Note that the completeness axiom applies only to $A$ which are nonempty. This is because every number $x\in\mathbb{R}$ is an upper bound for $\emptyset$ in $\mathbb{R}$ , and so $\emptyset$ cannot possibly have a least upper bound. Indeed, given any $x\in\mathbb{R}$ , it vacuously holds that $a\leq x$ for all $a\in\emptyset$ , since there are no elements in the empty set to check the condition on!

We shall invest some time becoming acquainted with the completeness axiom, studying sets where the supremum (i.e., least upper bound) can be explicitly computed.

Intervals and their suprema

In Example 1.2 and Example 1.12, we considered the set $A:=\{a\in\mathbb{R}:0\leq a\leq 1\}$ and showed $\sup A=1$ . This set is an example of an interval, with the endpoints 0 and 1, and we could represent it as $[0,1]$ , using the interval notation introduced in IMU. In this section, we will look at other examples of intervals, and consider their suprema.

Given $a,b\in\mathbb{R}$ with $a\leq b$ , we define the sets

(1.1)

\displaystyle\begin{split}[a,b]&:=\{t\in\mathbb{R}:a\leq t\leq b\},\\ (-\infty,b]&:=\{t\in\mathbb{R}:t\leq b\},\\ [a,\infty)&:=\{t\in\mathbb{R}:a\leq t\}.\end{split}

We say $I\subseteq\mathbb{R}$ is a closed interval if it is of the form of one of the three sets in (1.1), $I=\mathbb{R}$ or $I=\emptyset$ .

Warning 1.17.

Here we just think of ‘ $\pm\infty$ ’ as symbols used in the notation rather than any actual object. In particular, $\infty$ and $-\infty$ do not represent numbers in $\mathbb{R}$ .

The argument used in Example 1.2 can be applied to solve the following exercise.

Exercise 1.18.

For $a,b\in\mathbb{R}$ with $a\leq b$ , show that

\sup\,[a,b]=b\quad\text{and}\quad\sup\,(-\infty,b]=b.

What about the remaining closed intervals $[a,\infty)$ , $\mathbb{R}$ and $\emptyset$ : do they have suprema?

It is useful to visualise closed intervals as line segments on the real line: see Figure 1.1. It is customary to draw vertical straight lines at the endpoints to signify that the endpoints are included in the interval.

Figure 1.1: A closed interval.

Given $a,b\in\mathbb{R}$ with $a<b$ , we define the sets

(1.2)

\displaystyle\begin{split}(a,b)&:=\{t\in\mathbb{R}:a<t<b\},\\ (-\infty,b)&:=\{t\in\mathbb{R}:t<b\},\\ (a,\infty)&:=\{t\in\mathbb{R}:a<t\}.\end{split}

We say $I\subseteq\mathbb{R}$ is an open interval if it is of the form of one of the sets in (1.2), $I=\mathbb{R}$ or $I=\emptyset$ .²² 2 Note that, by definition, $\mathbb{R}$ and $\emptyset$ are both open intervals and closed intervals. It is again useful to visualise open intervals as line segments on the real line: see Figure 1.2. Here it is customary to draw round brackets at the endpoints to signify that the endpoints are not included in the interval.

Figure 1.2: An open interval.

In addition, we define the half-open intervals,

(1.3)

\displaystyle\begin{split}[a,b)&:=\{t\in\mathbb{R}:a\leq t<b\},\\ (a,b]&:=\{t\in\mathbb{R}:a<t\leq b\}.\end{split}

We say $I\subseteq\mathbb{R}$ is an interval if it is either an open interval, a closed interval, or of the form of one of the sets in (1.3).

Example 1.19.

Consider the open interval $(0,1):=\{a\in\mathbb{R}:0<a<1\}$ , consisting of all real numbers lying strictly between $0$ and $1$ . We claim that $\sup\,(0,1)=1$ . This looks similar to the set $[0,1]$ from Example 1.12, but is in fact harder since now $1\notin(0,1)$ . We need to check both parts of Definition 1.10:

1.

Since $a<1$ for all $a\in(0,1)$ , we see that $1$ is an upper bound for $(0,1)$ .
2.
Suppose $x\in\mathbb{R}$ and $x<1$ .
- •
  
  If $x\leq 0$ , then $a=1/2\in(0,1)$ and $a>x$ , so that $x$ is not an upper bound for $(0,1)$ .
- •
  
  If $0<x<1$ , then the midpoint $t:=\frac{x+1}{2}$ between $x$ and $1$ satisfies $x<t<1$ . Moreover, $t\in(0,1)$ and $t>x$ , so that $x$ is not an upper bound for $(0,1)$ .
This shows that if $x\in\mathbb{R}$ is an upper bound for $(0,1)$ , then $x\geq 1$ .

Thus, by definition, $\sup\,(0,1)=1$ .

Exercise 1.20.

Draw a sketch to illustrate the proof used in Example 1.19, including the interval $(0,1)$ , the point $x$ and the midpoint $t:=\frac{1+x}{2}$ .

Exercise 1.21.

For $a,b\in\mathbb{R}$ with $a<b$ , show that

\sup\,(a,b)=\sup\,(a,b]=\sup\,[a,b)=b\quad\text{and}\quad\sup\,(-\infty,b)=b.

What about the remaining open intervals $(a,\infty)$ , $\mathbb{R}$ and $\emptyset$ : do they have suprema?

Maxima

The concept of a maximum of a set is closely related to the supremum, although – as we will see – the two are not the same!

Definition 1.22.

Let $A\subseteq\mathbb{R}$ . We say $M\in\mathbb{R}$ is a maximum of $A$ if $M\in A$ and $x\leq M$ for all $x\in A$ . In this case we write $M=\max A$ .

In other words, $M$ is maximum of $A$ if $M\in A$ and $M$ is an upper bound of $A$ .

Exercise 1.23.

Let $A\subseteq\mathbb{R}$ . Show that if a maximum for $A$ exists, then it must be unique. In other words, if $M_{1}$ , $M_{2}\in A$ are maxima for $A$ , then $M_{1}=M_{2}$ .

In light of the previous exercise, if $A\subseteq\mathbb{R}$ has a maximum, then we can refer to it as the (definite article) maximum of $A$ .

Example 1.24.

We claim that $[0,1]$ has a maximum and, moreover, $\max\,[0,1]=1$ . Indeed, this follows directly from the definition, since $1\in[0,1]$ and $a\leq 1$ for all $a\in[0,1]$ . Recall from Example 1.12 that $\sup\,[0,1]=1$ , so in this case we have $\sup\,[0,1]=\max\,[0,1]=1$ .

Exercise 1.25.

For each of the following subsets of $\mathbb{R}$ , show that both the maximum and the supremum exists and compute their values.

(i)

$\{1,2,3\}$ ;
(ii)

$\{2/n:n\in\mathbb{N}\}$ .

The above observations suggest that suprema and maxima are closely related. However, the following important example shows that the two concepts are different.

Example 1.26.

We claim that the set $(0,1)$ does not have a maximum. This follows by the same argument we used to compute the supremum in Example 1.19. Indeed, for any $M\in(0,1)$ , the midpoint $t:=\frac{M+1}{2}$ satisfies $M<t<1$ . Moreover, $t\in(0,1)$ and $M\not\geq t$ and so $M$ cannot be a maximum for $(0,1)$ .

On the other hand, recall from Example 1.19 that $\sup\,(0,1)=1$ . Thus, here the supremum exists but no maximum exists for the set.

Exercise 1.27.

Using the fact that $\sup\,(0,1)=1$ , give another proof that $(0,1)$ does not have a maximum as follows.

(i)

Suppose $(0,1)$ does have a maximum $M$ . Using $\sup\,(0,1)=1$ , show that $M\geq 1$ .
(ii)

Use part (i) to obtain a contradiction.

Suprema: further examples

We round off this section with some more involved examples of least upper bounds / suprema, to allow us to further master the concept.

Example 1.28.

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

1.

Since $2-1/n\leq 2$ for all $n\in\mathbb{N}$ , it follows that $2$ is an upper bound for $A$ ;
2.

Suppose $x\in\mathbb{R}$ with $x<2$ . Our aim is to show that $x$ cannot be an upper bound for $A$ , so we need to find an element of $A$ that is larger than $x$ .

Let $r:=2-x$ . Since $\mathbb{N}$ is not bounded above (see Example 1.5), there exists some $N\in\mathbb{N}$ such that $N>1/r$ and therefore $1/N<r$ . We have $2-1/N\in A$ and $2-1/N>2-r=x$ , so that $x$ is not an upper bound for $A$ .

This shows that if $x\in\mathbb{R}$ is an upper bound for $A$ , then $x\geq 2$ .

Thus, by definition, $\sup A=2$ .

Exercise 1.29.

Does the set $A:=\{2-1/n:n\in\mathbb{N}\}$ from Example 1.28 have a maximum? Either prove a maximum exists and compute its value, or prove that no maximum exists.

Exercise 1.30.

For each of the following subsets of $\mathbb{R}$ , show that the supremum exists and compute its value. Determine whether a maximum exists.

(i)

$\displaystyle\Big{\{}\frac{3n^{2}-1}{n^{2}}:n\in\mathbb{N}\Big{\}}$ ;
(ii)

$\mathbb{Q}\cap(0,1)$ .