1.2 Least upper bounds

Going back to $A:=\{a\in\mathbb{R}:0\leq a\leq 1\}$ from Example 1.2 with upper bound $1$ , notice that there are many other upper bounds for this set. For instance, $2$ is also an upper bound for $A$ since it is also true that $a\leq 2$ for all $a\in A$ . Indeed, any number $x\geq 1$ is an upper bound for $A$ .

Exercise 1.8.

Let $A\subseteq\mathbb{R}$ . Show that upper bounds are never unique: in particular, if $x\in\mathbb{R}$ is an upper bound for $A$ , then there exists some $y\in\mathbb{R}$ with $y\neq x$ such that $y$ is also an upper bound for $A$ .

Continuing with the set $A:=\{a\in\mathbb{R}:0\leq a\leq 1\}$ from Example 1.2, note that any number $x\in\mathbb{R}$ satisfying $x<1$ is not an upper bound for $A$ , since $1\in A$ . From these observations, we see that the upper bounds for $A$ in $\mathbb{R}$ are precisely those numbers $x\geq 1$ .

Exercise 1.9.

Find all the upper bounds for the following sets:

(i)

$\{a\in\mathbb{R}:3\leq a\leq 4\}$ ;
(ii)

$\{a\in\mathbb{R}:0\leq a\leq 1\}\cup\{a\in\mathbb{R}:3\leq a\leq 4\}$ .

In Example 1.2 the upper bound $1$ is special: it is the least possible (one might say most efficient) upper bound for $A$ . This leads to the following important definition.

Definition 1.10.

Let $A\subseteq\mathbb{R}$ . We say $s\in\mathbb{R}$ is a least upper bound (or supremum) for $A$ if both

(i)

$s$ is an upper bound for $A$ , and
(ii)

if $x\in\mathbb{R}$ is an upper bound for $A$ , then $x\geq s$ .

In this case, we write $s=\sup A$ .

Exercise 1.11.

Let $A\subseteq\mathbb{R}$ . Show that if a least upper bound for $A$ exists, then it must be unique. In other words, if $s_{1}$ , $s_{2}\in\mathbb{R}$ are least upper bounds for $A$ , then $s_{1}=s_{2}$ .

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

Example 1.12.

For the set $A:=\{a\in\mathbb{R}:0\leq a\leq 1\}$ , we know from Example 1.2 that

1.

The number $1$ is an upper bound for $A$ ;
2.

If $x\in\mathbb{R}$ is an upper bound for $A$ , then $x\geq 1$ .

Hence the least upper bound for $A$ is $1$ : in other words, $\sup A=1$ .

Exercise 1.13.

Find the least upper bound of each of the sets from Exercise 1.9.

So far our examples have been fairly straightforward and do not provide a particularly new insight into numbers. However, the following example is going to change all that!

Example 1.14 (Important example!).

Suppose we temporarily restrict ourselves to the world of rational numbers. Consider the set of rational numbers $S:=\{a\in\mathbb{Q}:a^{2}\leq 2\}$ .

If $a\geq 2$ , then $a^{2}\geq 4$ and so $a\notin S$ . It therefore follows that $2$ is an upper bound for $S$ (why?), so $S$ is bounded above.

However, there are smaller upper bounds. For instance, $1.5$ is an upper bound. So too is $1.42$ and $1.415$ and $1.4143$ (can you see where these numbers are coming from?). We can in fact form a whole sequence of upper bounds which get successively smaller.

What we would like to say is that $\sqrt{2}$ is the least upper bound for $S$ . However, we know that $\sqrt{2}$ does not exist in $\mathbb{Q}$ . Using these observations, one can show that, although the set is bounded above, $S$ has no least upper bound in $\mathbb{Q}$ . You are asked to make this line of reasoning precise later in Exercise 1.48. Thus, although $S\subseteq\mathbb{Q}$ , to find the supremum of $S$ we need to leave the world of rationals, and enter the reals.

Example 1.14 is very important, since it helps motivate the completeness axiom.