1.4 The approximation property

Example 1.12, Example 1.19 and Example 1.28 all use the same strategy to compute the value of a supremum of a given set. In particular, to show some number $s$ is the least upper bound, we always argued by showing any $x<s$ is not an upper bound. This process can be repackaged as the following lemma.

Lemma 1.31 (Approximation property for suprema).

Let $A\subseteq\mathbb{R}$ and suppose $s\in\mathbb{R}$ is an upper bound for $A$ . Then the following are equivalent:

1

The supremum $\sup A$ exists and $s=\sup A$ ;
2

For all³³ 3 Here $\varepsilon$ is epsilon, the fifth letter of the Greek alphabet, which is traditionally used in mathematics to denote a small quantity. $\varepsilon>0$ there exists $a\in A$ such that $s-\varepsilon<a\leq s$ .

We refer to property 2 in Lemma 1.31 as the approximation property for suprema. This is an alternative way to express the fact that the supremum is the least upper bound, since it says that for any proposed smaller value ( $s-\varepsilon$ ) we can always find an element of the set ( $a\in A$ ) which is larger than that value – meaning that no smaller value can be an upper bound.

Before giving the proof of the lemma, we illustrate an example of it in action.

Example 1.32.

Consider the set $A:=\big{\{}\frac{5n}{n+1}:n\in\mathbb{N}\big{\}}$ . We claim that $\sup A=5$ .

We shall prove the claim using the approximation property. To do this, given $\varepsilon>0$ , we need to find some $a=\frac{5n}{n+1}\in A$ satisfying $5-\varepsilon<a\leq 5$ . Before writing out the proof, we carry out some rough work to find such an $a$ .

Rough work. Since $\displaystyle a=\frac{5n}{n+1}=5-\frac{5}{n+1}$ , we need $n\in\mathbb{N}$ such that

$5-\frac{5}{n+1}>5-\varepsilon\quad\iff\quad\frac{5}{n+1}<\varepsilon\quad\iff% \quad n+1>\frac{5}{\varepsilon}.$

So we could take $n>\frac{5}{\varepsilon}$ .

With this in hand, we are now ready to write out the details of the proof.

Proof.

Writing $5n=5(n+1)-5$ , it follows that $\frac{5n}{n+1}=5-\frac{5}{n+1}\leq 5$ for all $n\in\mathbb{N}$ . Thus, $5$ is an upper bound for $A$ .

We now show $5$ satisfies the approximation property from Lemma 1.31. Let $\varepsilon>0$ be given. Let $N\in\mathbb{N}$ satisfy $N>5/\varepsilon$ so that $0<\frac{1}{N}<\frac{\varepsilon}{5}$ . If we define $a:=\frac{5N}{N+1}$ , then $a\in A$ and

a=5-\frac{5}{N+1}>5-\frac{5}{N}>5-\varepsilon.

Thus, $5-\varepsilon<a\leq 5$ , which verifies the approximation property. By Lemma 1.31, we conclude that $\sup A=5$ , as claimed. ∎

Exercise 1.33.

Use the approximation property to give an alternative proof that $A:=\{2-1/n:n\in\mathbb{N}\}$ from Example 1.28 has $\sup A=2$ .

Proof (of Lemma 1.31).

We need to show that both 1 $\Rightarrow$ 2 and 2 $\Rightarrow$ 1.

1 $\Rightarrow$ 2.

We leave this as an exercise: see Exercise 1.34 below.
2 $\Rightarrow$ 1.

Suppose property 2 holds. We know by hypothesis that $s\in\mathbb{R}$ is an upper bound for $A$ and property 2 implies that $A$ is nonempty (why?). Thus, by the completeness axiom, $\sup A$ exists.

It remains to verify that $s=\sup A$ . Since, by hypothesis, $s$ is an upper bound for $A$ , all we have to do is show that $s$ is the least upper bound for $A$ . In other words, given $x\in\mathbb{R}$ with $x<s$ , we need to show that $x$ is not an upper bound for $A$ . If we set $\varepsilon:=s-x>0$ , then by property 2 there exists some $a\in A$ such that $s-\varepsilon<a$ . Since $s-\varepsilon=s-(s-x)=x$ , we have $x<a$ . Thus, $x$ cannot be an upper bound for $A$ .

This concludes the proof. ∎

Exercise 1.34.

Complete the proof of Lemma 1.31 by showing 1 $\Rightarrow$ 2.

Exercise 1.35.

Use the approximation property for suprema to show that

\sup\Big{\{}\frac{12n+5}{3n+2}:n\in\mathbb{N}\Big{\}}

exists and compute its value.

Remark 1.36 (More on quantifiers).

The approximation property in Lemma 1.31 involves multiple quantifiers, which are again underlined for emphasis. Throughout the course (and beyond!), we shall see many different statements involving multiple quantifiers and it is important that you take time to process what exactly each such statement means. In particular, the precise order of the quantifiers is very important. We illustrate this by considering the following two statements:

(I)

For all students $S$ there exists a pencil $P$ such that student $S$ writes with $P$ ;
(II)

There exists a pencil $P$ such that for all students $S$ , student $S$ writes with $P$ .

Statement (I) tells us that each student writes with some pencil. On the other hand, statement (II) describes a rather impractical situation (at least for a class of this size) where all the students write with the same pencil $P$ .

We note that (I) allows both for the situation where each student has their own individual pencil and for the situation described in (II) where all the students share the same pencil. In particular, (II) $\Rightarrow$ (I) but (I) $\not\Rightarrow$ (II).

Exercise 1.37.

In addition to (I) and (II) from Remark 1.36, consider the following statements:

(III)

For all pencils $P$ there exists a student $S$ such that student $S$ writes with $P$ ;
(IV)

There exists a student $S$ such that for all pencils $P$ , student $S$ writes with $P$ .

Think about the precise meanings of (III) and (IV) and how they differ from one another and those of statements in (I) and (II). What are the negations of (I), (II), (III) and (IV)?

Exercise 1.38.

In addition to (I) – (IV) from Exercise 1.37, consider the following statements:

(V)

For all pencils $P$ and for all student $S$ , student $S$ writes with pencil $P$ ;
(VI)

There exists a student $S$ and there exists a pencil $P$ such that student $S$ writes with $P$ .

Think about the precise meanings of (V) and (VI) and how they differ from one another and the earlier statements. What are the negations of (V) and (VI)?