1.1 Upper bounds for sets

To introduce the completeness axiom, we need some preliminary definitions concerning the order structure $\leq$ on the real line.

Definition 1.1.

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

1

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

We say $A$ is bounded above if there exists an upper bound $x\in\mathbb{R}$ for $A$ .

To understand this definition, we consider some examples.

Example 1.2.

Consider the set $A:=\{a\in\mathbb{R}:0\leq a\leq 1\}$ of all numbers lying between $0$ and $1$ . The number $1$ is an upper bound for $A$ since $a\leq 1$ for all $a\in A$ . Thus, since there exists an upper bound, the set $A$ is bounded above.

Example 1.3.

The set $\displaystyle A:=\bigg{\{}\frac{6n^{2}+n+4}{n^{2}+2n+1}:n\in\mathbb{N}\bigg{\}}$ is bounded above.

Proof.

Let $n\in\mathbb{N}$ . Since $n\geq 1$ , we have $n^{2}\geq n$ . Thus,

0\leq 6n^{2}+n+4\leq 6n^{2}+n^{2}+4n^{2}=11n^{2}\quad\text{and}\quad n^{2}+2n+% 1\geq n^{2}>0.

Combining these observations,

0\leq\frac{6n^{2}+n+4}{n^{2}+2n+1}\leq\frac{11n^{2}}{n^{2}}=11

and so $11$ is an upper bound for $A$ . Since an upper bound exists, $A$ is bounded above. ∎

Exercise 1.4.

Show that $\displaystyle A:=\bigg{\{}\frac{n+5}{8n^{2}-7}:n\in\mathbb{N}\bigg{\}}$ is bounded above.

Example 1.5.

The set of natural numbers $\mathbb{N}$ is not bounded above.

Proof.

If $x\in\mathbb{R}$ with $x\leq 0$ , then $1\in\mathbb{N}$ and $1>x$ , so $x$ is not an upper bound for $\mathbb{N}$ .

Given $x\in\mathbb{R}$ with $x>0$ , we may consider the number $a:=\lceil x\rceil+1\in\mathbb{N}$ , where $\lceil x\rceil$ is the least integer $m$ satisfying $m\geq x$ . Then $a$ satisfies $a>x$ . Thus, $x$ is not an upper bound for $\mathbb{N}$ .

Since we have shown every $x\in\mathbb{R}$ is not an upper bound for $\mathbb{N}$ , we conclude that $\mathbb{N}$ is not bounded above. ∎

Remark 1.6 (Quantifiers).

Definition 1.1 involves the expressions for all and there exists, which we have underlined for emphasis. These expressions have a very precise meaning in mathematics, and are called quantifiers. Sometimes quantifiers are written in shorthand: the symbol $\forall$ is used to denote for all and the symbol $\exists$ is used to denote there exists.¹¹ 1 For instance, we could rewrite the first part of Definition 1.1 as: $x\in\mathbb{R}$ is an upper bound for $A$ if $a\leq x$ $\forall$ $a\in A$ . However, in these notes we shall tend to be explicit and avoid the use of such symbols.

When we negate a for all statement, it becomes a there exists statement. For instance, to show $1/2$ is not an upper bound for $A$ in Example 1.2, we just need to show there exists some $a\in A$ with $a\not\leq 1/2$ (in particular, $a=1$ works).

When we negate a there exists statement, it becomes a for all statement. For instance, to show $\mathbb{N}$ is not bounded above in Example 1.5 we need to show that for all $n\in\mathbb{N}$ there exists some $a\in\mathbb{N}$ such that $n<a$ .

We shall work with quantifiers a lot throughout this course, and one of the main difficulties students often have is to understand how strings of quantifiers work together. For this reason, we shall often highlight the role of for all and there exists in our definitions and theorems and provide exercises to help build quantifier fluency.

Exercise 1.7.

Considering the following two statements:

(I)

For all students $S$ , the height of student $S$ is at least $1.7$ meters;
(II)

There exists a student $S$ such that the height of student $S$ is at least $1.7$ meters.

Which of (I) and (II) is true for students in this class? What are the negations of these statements? Do the negations hold for the class?