1.5: The Completeness Axiom for the Real Numbers

There are many examples of ordered fields. However, we are interested in the field of real numbers. There is an additional axiom that will distinguished this ordered field from all others. In order to introduce our last axiom for the real numbers, we first need some definitions.

Definition \(\PageIndex{1}\): Upper Bound

Let \(A\) be a subset of \(\mathbb{R}\). A number \(M\) is called an upper bound of \(A\) if

\[x \leq M \text { for all } x \in A.\]

If \(A\) has an upper bound, then \(A\) is said to be bounded above.

Similarly, a number \(L\) is a lower bound of \(A\) if

\[L \leq x \text{ for all } x \in A,\]

and \(A\) is said to be bounded below if it has a lower bound. We also say that \(A\) is bounded if it is both bounded above and bounded below.

It follows that a set \(A\) is bounded if and only if there exist \(M \in \mathbb{R}\) such that \(|x| \leq M \text { for all } x \in A\) (see Exercise 1.5.1)

Definition \(\PageIndex{2}\): Least Upper Bound

Let \(A\) be a nonempty set that is bounded above. We call a number \(\alpha\) a least upper bound or supremum of \(A\), if

1. \(x \leq \alpha \text { for all } x \in A\) (that is, \(\alpha\) is an upper bound of \(A\));
2. If \(M\) is an upper bound of \(A\), then \(\alpha \leq M\) (this means \(\alpha\) is smallest among all upper bounds).

It is easy to see that if \(A\) has a supremum, then it has only one (see Exercise 1.5.2). In this case, we denote such a number by \(\sup A\).

Example \(\PageIndex{1}\)

1. \(\sup [0,3)=\sup [0,3]=3\).
2. \(\sup \{3,5,7,8,10\}=10\).
3. \(\sup \left\{\frac{(-1)^{n}}{n}: n \in \mathbb{N}\right\}=\frac{1}{2}\).
4. \(\sup \left\{x^{2}:-2<x<1, x \in \mathbb{R}\right\}=4\).

Solution

a. First consider the set \([0,3]=\{x \in \mathbb{R}: 0 \leq x \leq 3\}\). By its very definition we see that for all \(x \in[0,3]\), \(x \leq 3\). Thus \(3\) is an upper bound. This verifies condition (1) in the definition of supremum. Next suppose \(M\) is an upper bound of \([0,3]\). Since \(3 \in[0,3]\), we get \(3 \leq M\). This verifies condition (2) in the definition of supremum. It follows that \(3\) is indeed the supremum of \([0,3]\).

Consider next the set \([0,3)=\{x \in \mathbb{R}: 0 \leq x<3\}\). It follows as before that \(3\) is an upper bound of \([0,3)\). Now suppose that \(M\) is an upper bound of \([0,3)\) and assume by way of contradiction that \(3>M\). If \(0>M\), then \(M\) is not an upper bound of \([0,3)\) as \(0\) is an element of \([0,3)\). If \(0 \leq M\), set \(x=\frac{M+3}{2}\). Then \(0 \leq x<3\) and \(x>M\), which shows \(M\) is not an upper bound of \([0,3)\). Since we get a contradiction in both cases, we conclude that \(3 \leq M\) and, hence, \(3\) is the supremum of \([0,3)\).

2. Clearly 10 is an upper bound of the set. Moreover, any upper bound \(M\) must satisfy \(10 \leq M\) as \(10\) is an element of the set. Thus \(10\) is the supremum.
3. Note that if \(n \in \mathbb{N}\) is even, then \(n \geq 2\) and

\(\frac{(-1)^{n}}{n}=\frac{1}{n} \leq \frac{1}{2}\).

If \(n \in \mathbb{N}\) is odd, then

\(\frac{(-1)^{n}}{n}=\frac{-1}{n}<0<\frac{1}{2}\)

This shows that \(\frac{1}{2}\) is an upper bound of the set. Since \(\frac{1}{2}\) is an element of the set, it follows as in the previous example that \(\frac{1}{2}\) is the supremum.

d. Set \(A=\left\{x^{2}:-2<x<1, x \in \mathbb{R}\right\}\). If \(y \in A\), then \(y=x^2\) for some \(x\) satisfying \(-2<x<1\) and, hence, \(|x|<2\). Therefore, \(y=x^{2}=|x|^{2}<4\). Thus, \(4\) is an upper bound of \(A\). Suppose \(M\) is an upper bound of \(A\) but \(M<4\). Choose a number \(y \in \mathbb{R}\) such that \(M<y<4\) and \(0<y\). Set \(x=-\sqrt{y}\). Then \(-2<x<0<1\) and, so, \(y=x^{2} \in A\). However, \(y>M\) which contradicts the fact that \(M\) is an upper bound. Thus \(4 \leq M\). This proves that \(4=\sup A\).

The following proposition is convenient in working with suprema.

Proposition \(\PageIndex{1}\)

Let \(A\) be a nonempty subset of \(\mathbb{R}\) that is bounded above. Then \(\alpha=\sup A\) if an only if

(1') \(x \leq \alpha \text { for all } x \in A\);

(2') For any \(\varepsilon>0\), there exists \(\alpha \in A\) such that \(\alpha-\varepsilon<a\).

Proof

Suppose first that \(\alpha=\sup A\). Then clearly (1') holds (since this is identical to condition (1) in the definition of supremum). Now let \(\varepsilon>0\). Since \(\alpha-\varepsilon<\alpha\), condition (2) in the definition of supremum implies that \(\alpha-\varepsilon\) is not an upper bound of \(A\). Therefore, there must exist an element \(a\) of \(A\) such that \(\alpha-\varepsilon<a\) as desired.

Conversely, suppose conditions (1') and (2') hold. Then all we need to show is that condition (2) in the definition of supremum holds. Let \(M\) be an upper bound of \(A\) and assume, by way of contradiction, that \(M<\alpha\). Set \(\varepsilon=\alpha-M\). By condition (2) in the statement, there is \(a \in A\) such that \(a>\alpha-\varepsilon=M\). This contradicts the fact that \(M\) is an upper bound. The conclusion now follows. \(\square\)

The following is an axiom of the real numbers and is called the completeness axiom.

The Completeness Axiom. Every nonempty subset \(A\) of \(\mathbb{R}\) that is bounded above has a least upper bound. That is, \(\sup A\) exists and is a real number.

This axiom distinguishes the real numbers from all other ordered fields and it is crucial in the proofs of the central theorems of analysis.

There is a corresponding definition for the infimum of a set.

Definition \(\PageIndex{3}\)

Let \(A\) be a nonempty subset of \(\mathbb{R}\) that is bounded below. We call a number \(\beta\) a greatest lower bound or infimum of \(A\), denoted by \(\beta=\inf A\), if

1. \(x \geq \beta \text { for all } x \in A\) (that is, \(\beta\) is a lower bound of \(A\));
2. If \(N\) is a lower bound of \(A\), then \(\beta \geq N\) (this means \(\beta\) is largest among all lower bounds).

Using the completeness axiom, we can prove that if a nonempy set is bounded below, then its infimum exists (see Exercise 1.5.5).

Example \(\PageIndex{2}\)

1. \(\inf (0,3]=\inf [0,3]=0\).
2. \(\inf \{3,5,7,8,10\}=3\).
3. \(\inf \left\{\frac{(-1)^{n}}{n}: n \in \mathbb{N}\right\}=-1\).
4. \(\inf \left\{1+\frac{1}{n}: n \in \mathbb{N}\right\}=1\).
5. \(\inf \left\{x^{2}:-2<x<1, x \in \mathbb{R}\right\}=0\).

The following is a basic property of suprema. Additional ones are described in the exercises.

Theorem \(\PageIndex{3}\)

Let \(A\) and \(B\) be nonempty sets and \(A \subset B\). Suppose \(B\) is bounded above. Then \(\sup A \leq \sup B\).

Proof

Let \(M\) be an upper bound for \(B\), then for \(x \in B\), \(x \leq M\). In particular, it is also true that \(x \leq M\) for \(x \in A\). Thus, \(A\) is also bounded above. Now, since \(\sup B\) is an upper bound for \(B\), it is also an upper bound for \(A\). Then, by the second condition in the definition of supremum, \(\sup A \leq \sup B\) as desired. \(\square\)

It will be convenient for the study of limits of sequences and functions to introduce two additional symbols.

Definition \(\PageIndex{4}\)

The extrended real number system consists of the real field \(\mathbb{R}\) and the two symbols \(\infty\) and \(-\infty\). We preserve the original order in \(\mathbb{R}\) and define

\(-\infty<x<\infty\)

for every \(x \in \mathbb{R}\)

The extended real number system does not form an ordered field but it is customary to make the following conventions:

1. If \(x\) is a real number then

\[x+\infty=\infty, \quad x+(-\infty)=-\infty\]

2. If \(x>0\), then \(x \cdot \infty=\infty, \quad x \cdot(-\infty)=-\infty\).
3. If \(x<0\), then \(x \cdot \infty=-\infty, \quad x \cdot(-\infty)=\infty\).
4. \(\infty+\infty=\infty,-\infty+(-\infty)=-\infty, \infty \cdot \infty=(-\infty) \cdot(-\infty)=\infty, \text { and }(-\infty) \cdot \infty=\infty \cdot(-\infty)=-\infty\).

We denote the extended real number set by \(\overline{\mathbb{R}}\). The expressions \(0 . \infty\), \(\infty+(-\infty)\), and \((-\infty)+\infty\) are left undefined.

The set \(\overline{\mathbb{R}}\) with the above conventions will be conventions will be convenient to describe results about limits in later chapters.

Definition \(\PageIndex{5}\)

If \(A \neq 0\) is not bounded above in \(\mathbb{R}\), we will write \(\sup A=\infty\). If \(A\) is not bounded below in \(\mathbb{R}\), we will wrtie \(\inf A=-\infty\).

With this definition, every nonempty subset of \(\mathbb{R}\) has a supremum and an infimum in \(\overline{\mathbb{R}}\). To complete the picture we adopt the following conventions for the empty set: \(\sup \emptyset=-\infty\) and \(\text { inf } \emptyset=\infty\).

Exercise \(\PageIndex{1}\)

Prove that a subset \(A\) of \(\mathbb{R}\) is bounded if and only if there is \(M \in \mathbb{R}\) such that \(|x| \leq M\) for all \(x \in A\).

Exercise \(\PageIndex{2}\)

Let \(A\) be a nonempty set and suppose \(\alpha\) and \(\beta\) satisfy conditions (1) and (2) in Definition 1.5.2 (that is, both are suprema of \(A\)). Prove that \(\alpha=\beta\)

Exercise \(\PageIndex{3}\)

For each subset of \(\mathbb{R}\) below, determine if it is bounded above, bounded below, or both. If it is bounded above (below) find the supremum (infimum). Justify all your conclusions.

1. \(\{1,5,17\}\)
2. \([0,5)\)
3. \(\left\{1+\frac{(-1)^{n}}{n}: n \in \mathbb{N}\right\}\)
4. \((-3, \infty)\)
5. \(\left\{x \in \mathbb{R}: x^{2}-3 x+2=0\right\}\)
6. \(\left\{x^{2}-3 x+2: x \in \mathbb{R}\right\}\)
7. \(\left\{x \in \mathbb{R}: x^{3}-4 x<0\right\}\)
8. \(\{x \in \mathbb{R}: 1 \leq|x|<3\}\)