3  Real Numbers

In this chapter we introduce the system of Real Numbers \(\mathbb{R}\) and study some of its properties. We will follow quite an abstract approach, which requires the definition of ordered field. The fact that \(\mathbb{R}\) is a continuum with no gaps (unlike \(\mathbb{Q}\)) will be consequence of the Axiom of Completeness.

Therefore \(\mathbb{R}\) will be defined as an ordered field in which the Axiom of Completeness holds.

3.1 Fields

In order to introduce \(\mathbb{R}\), we need the concepts of binary operation and field. We proceed in a general setting, starting from a set \(K\). We will assume that there is an equivalence relation on \(K\) denoted by \(=\).

Definition 1: Binary operation

A binary operation on a set \(K\) is a function \[ \circ \ \colon K \times K \to K \] which maps the ordered pair \((x,y)\) into \(x \circ y\).

Notation 2

We use the special symbols of \(+\) and \(\cdot\) to refer to addition and multiplication.

• Addition: The addition, or sum of \(x,y \in K\) is denoted by \[ x + y \]
• Multiplication: The multiplication, or product of \(x,y \in K\) is denoted by \[ x \cdot y \,\, \mbox{ or } \,\, xy \]

Warning

When we talk about \(+\) and \(\cdot\) in a set \(K\) we are not necessarily thinking about the usual sum and multiplication in \(\mathbb{Q}\). This is the case in the next Example, where we define \[ 1 + 1 = 0 \]

Example 3: of binary operation

Let \(K=\{0,1\}\). We can for example define operations of sum and product on \(K\) according to the tables \[ \begin{array}{c|cc} + & 0 & 1 \\ \hline 0 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} \qquad \qquad \begin{array}{c|cc} \cdot & 0 & 1 \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{array} \] The above mean that \[ \begin{gathered} 0+0 = 1+1 = 0 \,, \quad 0 + 1 = 1 + 0 = 1\,, \\ 0 \cdot 0 = 0 \cdot 1 = 1 \cdot 0 = 0 \,, \quad 1 \cdot 1 = 1 \,. \end{gathered} \tag{3.1}\]

Note that we have defined (rather controversially!) \[ 1+1 = 0 \] The other option would have been to define \[ 1 + 1 = 1 \] This would have been absolutely fine. However we could not have defined \[ 1 + 1 = 2 \] because \(2 \notin K\).

Note: The operations in (3.1) look more natural if you identify \(0\) with the even numbers and \(1\) with the odd numbers. In that case \(1+1 = 0\) reads \[ \mathrm{Odd} + \mathrm{Odd} = \mathrm{Even} \] Indeed all the definitions given in (3.1) agree with such identification, see tables below: \[ \begin{array}{c|cc} + & \mathrm{Even} & \mathrm{Odd} \\ \hline \mathrm{Even} & \mathrm{Even} & \mathrm{Odd} \\ \mathrm{Odd} & \mathrm{Odd} & \mathrm{Even} \\ \end{array} \qquad \qquad \begin{array}{c|cc} \cdot & \mathrm{Even} & \mathrm{Odd} \\ \hline \mathrm{Even} & \mathrm{Even} & \mathrm{Even} \\ \mathrm{Odd} & \mathrm{Even} & \mathrm{Odd} \\ \end{array} \]

Binary operations take ordered pairs of elements of \(K\) as input. Therefore the operation \[ x \circ y \circ z \] does not make sense, since we do not know which one between \[ x \circ y \quad \mbox{ or } \quad y \circ z \] has to be performed first. Moreover the outcome of an operation depends on order: \[ x \circ y \neq y \circ x \,. \] This motivates the following definition.

Definition 4

Let \(K\) be a set and \(\circ \ \colon K \times K \to K\) be a binary operation on \(K\). We say that:

1. \(\circ\) is commutative if \[ x \circ y = y \circ x \,, \quad \, \forall \, x,y \in K \]
2. \(\circ\) is associative if \[ (x \circ y) \circ z = x \circ (y \circ z) \,, \quad \, \forall \, x,y,z \in K \]
3. An element \(e \in K\) is called neutral element of \(\circ\) if \[ x \circ e = e \circ x = x \,, \quad \, \forall \, x \in K \]
4. Let \(e\) be a neutral element of \(\circ\) and let \(x \in K\). An element \(y \in K\) is called an inverse of \(x\) with respect to \(\circ\) if \[ x \circ y = y \circ x = e \,. \]

Example 5

Consider \(\mathbb{Q}\) with the usual operations of sum and multiplication

• We already know that \(+\) and \(\cdot\) are commutative and associative

• The neutral element of the sum is \(e = 0\)

• The neutral element of the product is \(e = 1\)

• The additive inverse of \(x\) is \(-x\)

• The multiplicative inverse of \(x \neq 0\) is \(1/x\).

Example 6

Let \(K\) with \(+\) and \(\cdot\) be as in Example 3. The sum satisfies:

• \(+\) is commutative, since \[ 0 + 1 = 1 + 0 = 0 \,. \]
• \(+\) is associative, since for example \[ (0 + 1) + 1 = 1 + 1 = 0 \,, \quad 0 + (1 + 1) = 0 + 0 = 0 \,, \] and therefore \[ (0 + 1) + 1 = 0 + (1 + 1) \,. \] In general one can show that \(+\) is associative by checking all the other permutations.
• The neutral element of \(+\) is \(0\), since \[ 0 + 0 = 0 \,, \quad 1 + 0 = 0 + 1 = 1 \,. \]
• Every element has an inverse. Indeed, the inverse of \(0\) is \(0\), since \[ 0 + 0 = 0\,, \] while the inverse of \(1\) is \(1\), since \[ 1 + 1 = 1 + 1 = 0 \,. \]

The multiplication satisfies:

• \(\cdot\) is commutative, since \[ 1 \cdot 0 = 0 \cdot 1 = 0 \,. \]
• \(\cdot\) is associative, since for example \[ (0 \cdot 1) \cdot 1 = 0 \cdot 1 = 0 \,, \quad 0 \cdot (1 \cdot 1) = 0 \cdot 1 = 0 \,, \] and therefore \[ (0 \cdot 1) \cdot 1 = 0 \cdot (1 \cdot 1) \,. \] By checking all the other permutations one can show that \(\cdot\) is associative.
• The neutral element of \(\cdot\) is \(1\), since \[ 0 \cdot 1 = 1 \cdot 0 = 0 \,, \quad 1 \cdot 1 = 1 \,. \]
• The element \(0\) has no inverse, since
  \[ 0 \cdot 0 = 0 \cdot 1 = 1 \cdot 0 = 0\,, \] and thus we never obtain the neutral element \(1\). The inverse of \(1\) is given by \(1\), since \[ 1 \cdot 1 = 1 \,. \]

Example 7

Let \(K=\{0,1\}\) be a set with binary relation \(\circ\) defined by the table \[ \begin{array}{c|cc} \circ & 0 & 1 \\ \hline 0 & 1 & 1 \\ 1 & 0 & 0 \\ \end{array} \] In this case \(\circ\) is not commutative since \[ 0 \circ 1 = 1 \, , \quad 1 \circ 0 = 0 \] and therefore \[ 0 \circ 1 \neq 1 \circ 0 \,. \] Moreover \(\circ\) is not associative, since \[ (0 \circ 1) \circ 1 = 1 \circ 1 = 0 \,, \] while \[ 0 \circ (1 \circ 1) = 0 \circ 0 = 1 \,, \] so that \[ (0 \circ 1) \circ 1 \neq 0 \circ (1 \circ 1)\,. \]

We are ready to define fields.

Definition 8: Field

Let \(K\) be a set with binary operations of addition \[ +\ \colon K \times K \to K \,, \quad (x,y) \mapsto x + y \] and multiplication \[ \cdot\ \colon K \times K \to K \,, \quad (x,y) \mapsto x \cdot y = xy \,. \] We call the triple \((K, + , \cdot)\) a field if:

1. The addition \(+\) satisfies: \(\,\forall \, x,y,z \in K\)
   • (A1) Commutativity and Associativity: \[ x+y = y+x \] \[ (x+y)+z = x+(y+z) \]
   • (A2) Additive Identity: There exists a neutral element in \(K\) for \(+\), which we call \(0\). It holds: \[ x + 0 = 0 + x = x \]
   • (A3) Additive Inverse: There exists an inverse of \(x\) with respect to \(+\). We call this element the additive inverse of \(x\) and denote it by \(-x\). It holds \[ x + (-x) = (-x) + x = 0 \]
2. The multiplication \(\cdot\) satisifes: \(\,\forall \, x,y,z \in K\)
   • (M1) Commutativity and Associativity: \[ x \cdot y = y \cdot x \] \[ (x \cdot y) \cdot z = x \cdot (y \cdot z) \]
   • (M2) Multiplicative Identity: There exists a neutral element in \(K\) for \(\cdot\), which we call \(1\). It holds: \[ x \cdot 1 = 1 \cdot x = x \]
   • (M3) Multiplicative Inverse: If \(x \neq 0\) there exists an inverse of \(x\) with respect to \(\cdot\). We call this element the multiplicative inverse of \(x\) and denote it by \(x^{-1}\). It holds \[ x \cdot x^{-1} = x^{-1} \cdot x = 1 \]
3. The operations \(+\) and \(\cdot\) are related by
   • (AM) Distributive Property: \(\,\forall \, x,y,z \in K\) \[ x \cdot (y + z) = (x \cdot y) + (y \cdot z) \,. \]

Example 9

\((\mathbb{Q},+,\cdot)\) is a field, where \(+\) and \(\cdot\) are the usual operations of addition and multiplication. This is trivially true, as the field axioms clearly encode properties that \(\mathbb{Q}\) already has.

Remark 10

Since fields are an abstract model for \(\mathbb{Q}\), we have chosen to denote the two field operations with \(+\) and \(\cdot\). We could have denoted them with other symbols, e.g. \(\clubsuit\) and \(\heartsuit\). In this case (AM) would read \[ x \, \clubsuit \, (y \, \heartsuit \, z) = (x \, \clubsuit \, y ) \,\heartsuit \, (y \, \clubsuit \, z ) \] This would of course still make sense. It would just be a lot harder to write calculations and to remember which operation is which!

Warning

It might be tempting to dismiss the definition of field, and to think that all fields look like \(\mathbb{Q}\). This is not the case, as seen in the following Example.

Example 11

Let \(K\) with \(+\) and \(\cdot\) be as in Example 3. We can show that \((K,+,\cdot)\) is a field. Indeed we have already shown in Example 6 that:

• (A1) and (M1) hold,
• (A2) holds with neutral element \(0\),
• (M2) holds with neutral element \(1\),
• (A3) every element has an additive inverse, with \[ -0 = 0 \,, \quad - 1 = 1 \,, \]
• (M3) every element which is not \(0\) a multiplicative inverse, with \[ 1^{-1} = 1\,. \]

We are left to show the Distributive Property (AM). Indeed:

• (AM) For all \(y,z \in K\) we have \[ 0 \cdot (y + z) = 0 \,, \quad (0 \cdot y) + (0 \cdot z) = 0 + 0 = 0\,, \] and also \[ 1 \cdot (y + z) = y + z \,, \quad (1 \cdot y) + (1 \cdot z) = y + z \,. \] Thus (AM) holds.

Definition 12: Subtraction and division

Let \((K,+,\cdot)\) be a field. We define:

• Subtraction as the operation \(-\) defined by \[ x - y := x + (-y) \,, \quad \forall \, x , y \in K \,, \] where \(-y\) is the additive inverse of \(y\).
• Division as the operation \(/\) defined by \[ x/y := x \cdot y^{-1}\,, \quad \forall \, x , y \in K \,, \,\, y \neq 0 \,, \] where \(y^{-1}\) is the multiplicative inverse of \(y\).

Warning

One might wonder why we are defining subtraction and division. Are they not already defined in \(\mathbb{Q}\)? Yes they are. But they make no sense in a general field \((K,+,\cdot)\), unless we define them explicitly.

Using the field axioms we can prove the following properties.

Proposition 13: Uniqueness of neutral elements and inverses

Let \((K,+,\cdot)\) be a field. Then

1. There is a unique element in \(K\) with the property of \(0\),
2. There is a unique element in \(K\) with the property of \(1\),
3. For all \(x \in K\) there is a unique additive inverse \(-x\),
4. For all \(x \in K\), \(x \neq 0\), there is a unique multiplicative inverse \(x^{-1}\).

Proof

1. Suppose that \(0 \in K\) and \(\widetilde{0} \in K\) are both neutral element of \(+\), that is, they both satisfy (A2). Then \[ 0 + \widetilde{0} = 0 \] since \(\widetilde{0}\) is a neutral element for \(+\). Moreover \[ \widetilde{0} + 0 = \widetilde{0} \] since \(0\) is a neutral element for \(+\). By commutativity of \(+\), see property (A1), we have \[ 0 = 0 + \widetilde{0} = \widetilde{0} + 0 = \widetilde{0} \,, \] showing that \(0 = \widetilde{0}\). Hence the neutral element for \(+\) is unique.
2. Exercise.
3. Let \(x \in K\) and suppose that \(y, \widetilde{y} \in K\) are both additive inverses of \(x\), that is, they both satisfy (A3). Therefore \[ x + y = 0 \] since \(y\) is an additive inverse of \(x\) and \[ x + \widetilde{y} = 0 \] since \(\widetilde{y}\) is an additive inverse of \(x\). Therefore we can use commutativity and associativity and of \(+\), see property (A1), and the fact that \(0\) is the neutral element of \(+\), to infer \[\begin{align*} y & = y + 0 = y + (x + \widetilde{y}) \\ & = (y + x) + \widetilde{y} = (x + y) + \widetilde{y} \\ & = 0 + \widetilde{y} = \widetilde{y} \,, \end{align*}\] concluding that \(y = \widetilde{y}\). Thus there is a unique additive inverse of \(x\), and \[ y = \widetilde{y} = -x \,, \] with \(-x\) the element from property (A3).
4. Exercise.

Using the properties of field we can also show that the usual properties of sum, subtraction, multiplication and division still hold in any field. We list such properties in the following proposition.

Proposition 14: Properties of field operations

Let \((K,+,\cdot)\) be a field. Then for all \(x,y,z \in K\),

• \(x + y = x + z \,\, \implies \,\, y = z\)
• \(x \cdot y = x \cdot z \,\) and \(\,x \neq 0 \,\, \implies \,\, y = z\)
• \(- 0 = 0\)
• \(1^{-1} = 1\)
• \(x \cdot 0 = 0\)
• \(-1 \cdot x = -x\)
• \(-(-x) = x\)
• \((x^{-1})^{-1} = x \,\) if \(\, x \neq 0\)
• \((x \cdot y)^{-1} = x^{-1} \cdot y^{-1}\)

The above properties can be all proven with elementary use of the field properties (A1)-(A3), (M1)-(M3) and (AM). This is an exercise in patience, and is left to the reader.

Let us conclude with examining the sets of numbers introduced in Chapter 1.

Theorem 15

Consider the sets \(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\) with the usual operations \(+\) and \(\cdot\). We have:

• \((\mathbb{N}, + , \cdot)\) is not a field:
  It satisfies properties (A1), (A2), (M1), (M2), (AM) of fields. It is missing properties (A3) and (M3), the additive and multiplicative inverse properties, respectively.
• \((\mathbb{Z}, + , \cdot)\) is not a field:
  It satisfies properties (A1), (A2), (A3), (M1), (M2), (AM) of fields. Thus it is only missing (M3), the multiplicative inverse property.
• \((\mathbb{Q}, + , \cdot)\) is a field.

The proof is omitted.

3.2 Ordered fields

Definition 16

Let \(K\) be a set with binary operations \(+\) and \(\cdot\), and with an order relation \(\leq\). We call \((K,+,\cdot,\leq)\) an ordered field if:

1. \((K,+,\cdot)\) is a field

2. There \(\leq\) is of total order on \(K\): \(\, \forall \, x, y, z \in K\)

   • (O1) Reflexivity: \[ x \leq x \]
   • (O2) Antisymmetry: \[ x \leq y \, \mbox{ and } \, y \leq x \,\, \implies \,\, x = y \]
   • (O3) Transitivity: \[ x \leq y \,\, \mbox{ and } \,\, y \leq z \,\, \implies \,\, x = z \]
   • (O4) Total order:
     \[ x \leq y \,\, \mbox{ or } \,\, y \leq x \]

3. The operations \(+\) and \(\cdot\), and the total order \(\leq\), are related by the following properties: \(\, \forall x, y, z \in K\)

   • (AM) Distributive: Relates addition and multiplication via \[ x \cdot (y + z) = x \cdot y + x \cdot z \]
   • (AO) Relates addition and order with the requirement: \[ x \leq y \,\, \implies \,\, x + z \leq y + z \]
   • (MO) Relates multiplication and order with the requirement: \[ x \geq 0, \, y \geq 0 \,\, \implies \,\, x \cdot y \geq 0 \]

Example 17

\((\mathbb{Q},+,\cdot,\leq)\) is an ordered field.

3.3 Cut Property

We have just introduced the notion of fields and ordered fields. We noted that the set of rational numbers with the usual operations of sum and multiplication \[ (\mathbb{Q}, + , \cdot) \] is a field. Moreover it is an ordered field with the usual order \(\leq\).

We now need to address the key issue we proved in Chapter 1, namely, the fact that \[ \sqrt{2} \notin \mathbb{Q}\,. \] Intuitively, this means that \(\mathbb{Q}\) has gaps, and cannot be represented as a continuous line. The rigorous definition of lack of gaps needs the concept of cut of a set. This, in turn, needs the concept of partition.

Definition 18: Partition of a set

Let \(S\) be a non-empty set. The pair \((A,B)\) is a partition of \(S\) if \[ A , B \subseteq S\,, \quad A \neq \emptyset\,, \quad B \neq \emptyset \,, \] and \[ S = A \cup B \,, \quad A \cap B = \emptyset \, . \]

Schematic picture of a partition \((A,B)\) of the set \(K\).

Definition 19: Cut of a set

Let \(S\) be a non-empty set with a total order relation \(\leq\). The pair \((A,B)\) is a cut of \(S\) if

1. \((A,B)\) is a partition of \(S\),
2. We have \[ a \leq b \,, \quad \forall \, a \in A \,, \,\, \forall \, b \in B \,. \]

The cut of a set is often called Dedekind cut, named after Richard Dedekind, who used cuts to give an explicit construction of the real numbers \(\mathbb{R}\), see Wikipedia page.

Definition 20: Cut property

Let \(S\) be a non-empty set with a total order relation \(\leq\). We say that \(S\) has the cut property if for every cut \((A,B)\) of \(S\) there exists some \(s \in S\) such that \[ a \leq s \leq b \,, \quad \forall \, a \in A\,, \,\, \forall \, b \in B \,. \] We call \(s\) the separator of the cut \((A,B)\).

Example 21

Let \(S = \mathbb{Q}\) and consider the sets \[ A = (-\infty,s] \cap \mathbb{Q}\,, \quad B = (s,\infty) \cap \mathbb{Q}\,. \] for some \(s \in \mathbb{Q}\). Then the pair \((A,B)\) is a cut of \(\mathbb{Q}\), and \(s\) is the separator.

\((A,B)\) is a cut of \(\mathbb{Q}\) with separator \(s\).

Question 22

Do all ordered fields have the Cut Property? Does \(\mathbb{Q}\) have the Cut Property?

The answer to the above question is NO. For example the pair \[ A = (-\infty,\sqrt{2}) \cap \mathbb{Q}\,, \quad B = (\sqrt{2},\infty) \cap \mathbb{Q}\,. \tag{3.2}\] is a cut of \(\mathbb{Q}\), since \(\sqrt{2} \notin \mathbb{Q}\). However what is the separator? It should be \(s = \sqrt{2}\), given that clearly \[ a \leq \sqrt{2} \leq b \,, \quad \forall \, a \in A\,, \,\, \forall \, b \in B \,. \] However \(\sqrt{2} \notin \mathbb{Q}\), so we are NOT ALLOWED to take it as separator. Indeed, we can show that \((A,B)\) defined as in (3.2) has no separator.

\((A,B)\) is a cut of \(\mathbb{Q}\) which has no separator

Theorem 23: \(\mathbb{Q}\) does not have the cut property.

\(\mathbb{Q}\) does not have the cut property. More explicitly, there exist a cut \((A,B)\) of \(\mathbb{Q}\) which has no separator.

Remark 24: Ideas for the proof of Theorem 23

Before proceeding with the proof, let us summarize the ideas behind it:
We will consider the cut \((A,B)\) in (3.2). We then assume by contradiction that \((A,B)\) admits a separator \(L \in \mathbb{Q}\), so that \[ a \leq L \leq b \,, \quad \forall \, a \in A\,, \,\, \forall \, b \in B \,. \tag{3.3}\] Since \((A,B)\) is a partition of \(\mathbb{Q}\), then either \(L \in A\) or \(L \in B\). These will both lead to a contradiction:

• If \(L \in A\), by definition of \(A\) we have \[ L < \sqrt{2} \,. \] We want to contradict the fact that \(L\) is a separator for the cut \((A,B)\). The idea is that, since \(\sqrt{2} \notin \mathbb{Q}\), it is possible to squeeze a rational number \(\widetilde{L} \in \mathbb{Q}\) in between \(L\) and \(\sqrt{2}\), i.e. \[ L < \widetilde{L} < \sqrt{2} \,. \] How do we find such \(\widetilde{L}\) in practice? We look for a number \(\widetilde{L}_n\) of the form \[ \widetilde{L}_n = L + \frac{1}{n} \] for some \(n \in \mathbb{N}\) to be suitably chosen later. Clearly \(\widetilde{L}_n \in \mathbb{Q}\) and \[ L < \widetilde{L}_n \] for all \(n \in \mathbb{N}\). We will then be able to find \(n_0 \in \mathbb{N}\) such that \[ L < \widetilde{L}_{n_0} < \sqrt{2} \,. \tag{3.4}\] Now comes the contradiction: From (3.4) we see that \(\widetilde{L}_{n_0} \in A\). However \(L\) is a separator, and so from (3.3) we have \[ \widetilde{L}_{n_0} \leq L \,, \] which contradicts (3.4).

• If \(L \in B\), by definition of \(B\) we have \[ \sqrt{2} < L \,. \] The idea is the same as above: Since \(\sqrt{2} \notin \mathbb{Q}\), we can squeeze a rational number \(\widetilde{L} \in \mathbb{Q}\) between \(\sqrt{2}\) and \(L\), i.e., \[ \sqrt{2} < \widetilde{L} < L \,. \] Since we want \(\widetilde{L}\) to be a rational number smaller than \(L\), we look for \(\widetilde{L}\) of the form \[ \widetilde{L}_n := L - \frac{1}{n} \,, \] for a suitable \(n \in \mathbb{N}\). Clearly \(\widetilde{L}_n \in \mathbb{Q}\) and \[ \widetilde{L}_n < L \,, \] for all \(n \in \mathbb{N}\). We will be able to find \(n_0 \in \mathbb{N}\) such that \[ \sqrt{2} < \widetilde{L}_{n_0} < L \,. \tag{3.5}\] Therefore \(\widetilde{L}_{n_0} \in B\). Again, contradiction: \(L\) is a separator and so \[ L \leq \widetilde{L}_{n_0} \,, \] which contradicts (3.5).

Both cases \(L \in A\) or \(L \in B\) lead to a contradiction. Since these are all the possibilities, we conclude that the cut \((A,B)\) has no separator in \(\mathbb{Q}\).

Time to make the ideas in the above remark rigorous. Two main issues need fixing:

1. \(\sqrt{2}\) is just a symbol for a number \(x\) such that \(x^2 = 2\). As \(\sqrt{2} \notin \mathbb{Q}\), what is the meaning of the expression \[ a < \sqrt{2} < b \] when \(a,b \in \mathbb{Q}\)? We can only compare rational numbers with rational numbers, so the above inequalities are meaningless. We need a more clever way to write down the sets \(A\) and \(B\) so that they make sense as objects in \(\mathbb{Q}\).

2. We said it is possible to find \(n_0 \in \mathbb{N}\) such that \[ L < \widetilde{L}_{n_0} < \sqrt{2} \quad \text{ or } \quad \sqrt{2} < \widetilde{L}_{n_0} < L \] We need to prove it!

Proof: Proof of Theorem 23

Let \(A\) and \(B\) be the sets defined in (3.2). It is useful to rewrite \(A\) and \(B\) in the form \[ A = A_1 \cup A_2 \,, \] where \[ A_1 = \{ q \in \mathbb{Q}\, \colon \,q< 0 \}\,, \] \[ A_2 =\{ q \in \mathbb{Q}\, \colon \,q \geq 0\,, \,\, q^2 < 2 \} \,, \] and \[ B = \{ q \in \mathbb{Q}\, \colon \,q > 0, \,\, q^2 > 2 \} \,. \]

Step 1. \((A,B)\) is a cut of \(\mathbb{Q}\):
We need to prove the following:

1. \((A,B)\) is a partition of \(\mathbb{Q}\). This is because \(A , B \subseteq \mathbb{Q}\) with \(A \neq \emptyset\) and \(B \neq \emptyset\). Moreover \(A \cap B = \emptyset\) and \[ A \cup B = \mathbb{Q}\,, \] given that \(\sqrt{2} \notin \mathbb{Q}\), and so there is no element \(q \in \mathbb{Q}\) such that \(q^2 = 2\).
2. It holds \[ a \leq b \,, \quad \forall a \in A \,, \,\, \forall \, b \in B \,. \] Indeed, suppose that \(a \in A\) and \(b \in B\). We have two cases:
   • \(a \in A_1\): Therefore \(a<0\). In particular \[ a < 0 < b\,, \] given that \(b > 0\) for all \(b \in B\). Thus \(a<b\).
   • \(a \in A_2\): Therefore \(a \geq 0\) and \(a^2 < 2\). In particular \[ a^2 < 2 < b^2 \,, \] since \(b^2>2\) for all \(b \in B\). In particular \[ a^2 < b^2 \,. \] Since \(b>0\) for all \(b \in B\), from the above inequality we infer \(a<b\), concluding.

Step 2. \((A,B)\) has no separator:
Suppose by contradiction that \((A,B)\) admits a separator \[ L \in \mathbb{Q}\,. \] By definition this means \[ a \leq L \leq b \,, \quad \forall a \in A \,, \,\, \forall \, b \in B \,. \tag{3.6}\] Since \[ L \in \mathbb{Q}\,, \quad \mathbb{Q}= A \cup B \,, \quad A \cap B = \emptyset \,, \] then either \(L \in A\) or \(L \in B\). We will see that both these possibilities lead to a contradiction:

Case 1: \(L \in A\).
By (3.6) we know that \[ a \leq L \,, \quad \forall \, a \in A \,. \tag{3.7}\] In particular the above implies \[ L \geq 0 \tag{3.8}\] since \(0 \in A\). Therefore we must have \(L \in A_2\), that is, \[ L \geq 0 \,\, \mbox{ and } \,\, L^2 < 2 \,. \tag{3.9}\] Set \[ \widetilde{L} := L + \frac1n \] for \(n \in \mathbb{N}\), \(n \neq 0\) to be chosen later. Clearly we have \[ \widetilde{L} \in \mathbb{Q}\,\,\, \mbox{ and } \,\,\, L < \widetilde{L} \,. \tag{3.10}\] From (3.8) and (3.10) we have also \[ \widetilde{L}>0 \,. \tag{3.11}\] We now want to show that there is a choice of \(n\) such that \(\widetilde{L}^2 < 2\), which will lead to a contradiction. Indeed, we can estimate \[\begin{align*} \widetilde{L}^2 & = \left( L + \frac1n \right)^2 \\ & = L^2 + \frac{1}{n^2} + 2 \frac{L}{n} \\ & < L^2 + \frac{1}{n} + 2 \frac{L}{n} \qquad \left(\mbox{using } \, \frac{1}{n}<\frac{1}{n^2} \right) \\ & = L^2 + \frac{2L + 1}{n} \,. \end{align*}\] If we now impose that \[ L^2 + \frac{2L + 1}{n} < 2 \,, \] we can rearrange the above and obtain \[ n(2 - L^2) > 2L + 1 \,. \] Now note that \(L^2 < 2\) by assumption (3.9). Thus we can divived by \((2 - L^2)\) and obtain \[ n >\frac{2L + 1}{2 - L^2} \,. \] Therefore we have just shown that \[ n >\frac{2L + 1}{2 - L^2} \,\, \implies \,\, \widetilde{L}^2 < 2 \,. \] Together with (3.11) this implies \(\widetilde{L} \in A\). Therefore we have \[ \widetilde{L} \leq L \] by (3.7). On the other hand it also holds \[ \widetilde{L} > L \] by (3.10), and therefore we have a contradiction. Thus \(L \notin A\).

Case 2: \(L \in B\).
As \(L \in B\), we have by definition \[ L > 0 \,, \quad L^2 > 2 \,. \tag{3.12}\] Moreover since \(L\) is a separator, see (3.6), in particular \[ L \leq b \,, \,\, \forall \, b \in B \,. \tag{3.13}\] Define now \[ \widetilde{L} := L - \frac1n \] with \(n \in \mathbb{N}\), \(n \neq 0\) to be chosen later. Clearly we have \[ \widetilde{L} \in \mathbb{Q}\,, \quad \widetilde{L} < L \,. \tag{3.14}\] We now show that \(n\) can be chosen so that \(\widetilde{L} \in B\). Indeed \[\begin{align*} \widetilde{L}^2 & = \left( L - \frac1n \right)^2 \\ & = L^2 + \frac{1}{n^2} - 2 \frac{L}{n} \\ & > L^2 - \frac{1}{n^2} - 2 \frac{L}{n} \qquad \left(\mbox{using } \, \frac{1}{n^2} > - \frac{1}{n^2} \right) \\ & > L^2 - \frac{1}{n} - 2 \frac{L}{n} \qquad \left(\mbox{using } \, -\frac{1}{n^2} > - \frac{1}{n} \right) \\ & = L^2 - \frac{1 + 2L}{n} \,. \end{align*}\] Now we impose \[ L^2 - \frac{1 + 2L}{n} > 2 \] which is equivalent to \[ n(L^2 - 2) > 1 + 2L \,. \] Since we are assuming \(L \in B\), then \(L^2 > 2\), see (3.12). Therefore we can divide by \((L^2 -2)\) and get \[ n > \frac{1+2L}{L^2 - 2} \,. \] In total, we have just shown that \[ n > \frac{1+2L}{L^2 - 2} \quad \implies \quad \widetilde{L}^2 > 2\,, \] proving that \(\widetilde{L} \in B\). Therefore by (3.13) we get \[ L \leq \widetilde{L} \,. \] This contradicts (3.14).

Conclusion:
We have seen that assuming that \((A,B)\) has a separator \(L \in \mathbb{Q}\) leads to a contradiction. Thus the cut \((A,B)\) has no separator.

Remark 25

The above proof is also suggesting the following observations:

• The set \[ A = (-\infty, \sqrt{2}) \cap \mathbb{Q} \] does not admit a largest element in \(\mathbb{Q}\)

• The set \[ B = (\sqrt{2}, \infty) \cap \mathbb{Q} \] does not admit a lowest element in \(\mathbb{Q}\).

It turns out that the largest and lowest element play a crucial role in analysis. We will give precise definitions in the next section.

3.4 Supremum and infimum

A crucial definition in Analysis is the one of supremum or infimum of a set. This is also another way of studying the gaps of \(\mathbb{Q}\).

Example 26: Intuition about supremum and infimum

Consider the set \[ A = [0,1) \cap \mathbb{Q}\,. \] Intuitively, we understand that \(A\) is bounded, i.e. not infinite. We also see that

• \(0\) is the lowest element of \(A\)
• \(1\) is the highest element of \(A\)

However we see that \(0 \in A\) while \(1 \notin A\). We will see that

• \(0\) can be defined as the infimum and minimum of \(A\).
• \(1\) can be defined as the supremum, but not maximum, of \(A\).

3.4.1 Upper bound, supremum, maximum

We start by defining the supremum. First we need the notion of upper bound of a set.

Definition 27: Upper bound and bounded above

Let \((K,+,\cdot,\leq)\) be an ordered field and \(A \subseteq K\):

1. We say that \(b \in K\) is an upper bound for \(A\) if \[ a \leq b \,, \quad \forall \, a \in A \,. \]
2. We say that \(A\) is bounded above if there exists and upper bound \(b \in K\) for \(A\).

Definition 28: Supremum

Let \((K,+,\cdot,\leq)\) be an ordered field and \(A \subseteq K\). A number \(s \in K\) is called least upper bound or supremum of \(A\) if:

1. \(s\) is an upper bound for \(A\),
2. \(s\) is the smallest upper bound of \(A\), that is, \[ \mbox{If } \, b \in K \, \mbox{ is upper bound for } \, A \, \mbox{ then } \, s \leq b \,. \]

Notation 29

We will almost always prefer the name supremum to least upper bound. For \(A \subseteq K\) the supremum is denoted by \[ s := \sup \ A \,. \]

Remark 30

Note that if a set \(A \subseteq K\) in NOT bounded above, then the supremum does not exist, as there are no upper bounds of \(A\).

Proposition 31

Let \((K,+,\cdot,\leq)\) be an ordered field and \(A \subseteq K\). If \[ \sup A \] exists, then it is unique.

Proof

Suppose there exist \(s_1, s_2 \in K\) such that \[ s_1 = \sup A, \,\, s_2 = \sup A \,. \] Then:

• Since \(s_2 = \sup A\), in particular \(s_2\) is an upper bound for \(A\). Since \(s_1 = \sup A\) then \(s_1\) is the lowest upper bound. Thus we get \[ s_1 \leq s_2 \,. \]
• Exchanging the roles \(s_1\) and \(s_2\) in the above reasoning we also get \[ s_2 \leq s_1 \,. \]

This shows \(s_1 = s_2\).

Warning

In general:

• A set can have infinite upper bounds,
• The supremum does not belong to the set.

For example \[ A = [0,1) \cap \mathbb{Q} \] has for upper bounds all the numbers \(b \in \mathbb{Q}\) with \(b>1\). Moreover one can show that \[ \sup A = 1\,, \] and so \[ \sup A \notin A \,. \]

Warning

The supremum does not exist in general. For example let \[ A = [0,\sqrt{2}) \cap \mathbb{Q}\,. \] We will show that \(\sup A\) does not exist in \(\mathbb{Q}\). Indeed we will have that \[ \sup A = \sqrt{2} \in \mathbb{R}\,. \]

Supremum and upper bounds of a set \(A\) in the field \(K\)

Definition 32: Maximum

Let \((K,+,\cdot,\leq)\) be an ordered field and \(A \subseteq K\). A number \(M \in K\) is called the maximum of \(A\) if: \[ M \in A \,\, \mbox{ and } \,\, a \leq M \,, \, \forall a \in A \,. \] We denote the maximum by \[ M = \max A \,. \]

Proposition 33

Let \((K,+,\cdot,\leq)\) be an ordered field and \(A \subseteq K\). If the maximum of \(A\) exists, then also the supremum exists, and \[ \sup A = \max A \,. \]

Proof

Let \[ M = \max A \,. \] Then:

• By definition we have \(M \in A\) and \[ a \leq M \,, \quad \forall \, a \in A \,. \] In particular the above tells us that \(M\) is an upper bound of \(A\).
• We claim that \(M\) is the least upper bound. Indeed, suppose \(b\) is an upper bound of \(A\), that is,
  \[ a \leq b \,, \quad \forall \, a \in A \,. \] In particular, since \(M \in A\), by the above condition we have \[ M \leq b \,. \]

Therefore \(M\) is the least upper bound of \(A\), meaning that \(M = \sup A\).

Warning

The converse of the above statement is not true: In general the \(\sup\) might exist while the \(\max\) does not. For example \[ A = [0,1) \cap \mathbb{Q} \] is such that \[ \sup A = 1 \] but \(\max A\) does not exist. Instead for the set \[ B = [0,1] \cap \mathbb{Q} \] we have that \[ \max A = \sup A = 1 \,. \]

3.4.2 Lower bound, infimum, minimum

We now introduce the definitions of lower bound, infimum, minimum. These are the counterpart of upper bound, supremum and maximum, respectively.

Definition 34: Upper bound, bounded below, infimum, minimum

Let \((K,+,\cdot,\leq)\) be an ordered field and \(A \subseteq K\):

1. We say that \(l \in K\) is a lower bound for \(A\) if \[ l \leq a \,, \quad \forall \, a \in A \,. \]

2. We say that \(A\) is bounded below if there exists a lower bound \(l \in K\) for \(A\).

3. We say that \(i \in K\) is the greatest lower bound or infimum of \(A\) if:

   • \(i\) is a lower bound for \(A\),
   • \(i\) is the largest lower bound of \(A\), that is, \[ \mbox{If } \, l \in K \, \mbox{ is a lower bound for } \, A \, \mbox{ then } \, l \leq i \,. \] If it exists, the infimum is denoted by \[ i = \inf A \,. \]

4. We say that \(m \in K\) is the minimum of \(A\) if: \[ m \in A \,\, \mbox{ and } \,\, m \leq a \,, \, \forall a \in A \,. \] If it exists, we denote the minimum by \[ m = \min A \,. \]

Infimum and lower bounds of a set \(A\) in the field \(K\)

Proposition 35

Let \((K,+,\cdot,\leq)\) be an ordered field and \(A \subseteq K\):

1. If \(\inf A\) exists, then it is unique.
2. If the minimum of \(A\) exists, then also the infimum exists, and \[ \inf A = \min A \,. \]

The proof uses similar arguments to the one employed in the previous section, and is left to the reader as an exercise.

Warning

We have

• A set can have infinite lower bounds,
• The infimum does not belong to the set.

For example \[ A = (0,1) \cap \mathbb{Q} \] has for lower bounds all the numbers \(b \in \mathbb{Q}\) with \(b<1\). Moreover we will show that \[ \inf A = 0\,, \] and so \[ \inf A \notin A \,. \]

Warning

The infimum does not exist in general. For example let \[ A = (\sqrt{2},5] \cap \mathbb{Q}\,. \] We will show that \(\inf A\) does not exist in \(\mathbb{Q}\). Indeed we will have that \[ \inf A = \sqrt{2} \in \mathbb{R}\,. \]

Warning

In general the \(\inf\) might exist while the \(\min\) does not. For example \[ A = (0,1) \cap \mathbb{Q} \] is such that \[ \inf A = 0 \] but \(\min A\) does not exist. Instead for the set \[ B = [0,1] \cap \mathbb{Q} \] we have that \[ \inf A = \min A = 0 \,. \]

Proposition 36

Let \((K,+,\cdot,\leq)\) be an ordered field and \(A \subseteq K\). If \(\inf A\) and \(\sup A\) exist, then \[ \inf A \leq a \leq \sup A \,, \quad \forall a \in A \,. \]

The proof is simple, and is left as an exercise. We now have a complete picture about supremum and infimum, see figure below.

Supremum, upper bounds, infimum and lower bounds of a set \(A\) in \(K\)

We conclude with another simple proposition. The proof is again left to the reader.

Proposition 37: Relationship between sup and inf

Let \((K,+,\cdot,\leq)\) be an ordered field and \(A \subseteq K\). Define \[ - A := \{ - a \, \colon \,a \in A \} \,. \] It holds:

• If \(\sup A\) exists, then \(\inf A\) exists and \[ \inf(-A) = - \sup A \,. \]
• If \(\inf A\) exists, then \(\sup A\) exists and \[ \sup(-A) = - \inf A \,. \]

3.5 Completeness

We have introduced the concepts of supremum and infimum on an ordered field \(K\).

Question 38

Suppose \((K,+,\cdot, \leq)\) is an ordered field, and that \(A \subseteq K\) is non-empty and bounded above. Does \[ \sup A \] always exist?

The answer to the above question is NO. Like we did with the Cut Property, the counterexample can be found in the set of rational numbers \(\mathbb{Q}\). A set bounded above for which the supremum does nor exist is, for example, \[ A = [0, \sqrt{2}) \cap \mathbb{Q}\,. \tag{3.15}\]

Theorem 39

There exists a set \(A \subseteq \mathbb{Q}\) such that

• \(A\) is non-empty,
• \(A\) is bounded above,
• \(\sup A\) does not exist in \(\mathbb{Q}\).

The proof uses the same ideas we used for showing that \(\mathbb{Q}\) does not have the Cut Property.

Proof

Define the set \(A\) as in (3.15). Equivalently, this can be written as \[ A = \{ q \in \mathbb{Q}\, \colon \,q \geq 0 \,, \,\, q^2 < 2 \} \,. \]

Step 1. \(A\) is bounded above.
Take \(b:=9\). Then \(b\) is an upper bound for \(A\). Indeed by definition \[ q^2 < 2 \,, \,\, q \geq 0 \,, \,\,\, \forall q \in A \,. \] Therefore \[ q^2 < 2 < 9 \implies q^2 < 9 \implies q < 3 = b \, . \]

Step 2. \(\sup A\) does not exist.
Assume by contradiction that \[ s = \sup A \in \mathbb{Q} \] exists. By definition it holds \[ s \geq q \,, \quad \forall \, q \in A \tag{3.16}\] \[ b \geq q \,, \, \forall \, q \in A \,\, \implies \,\, s \leq b \tag{3.17}\] There are two possibilities: \(s \in A\) or \(s \notin A\):

• Case 1. \(s \in A\).
  If \(s \in A\) by definition \[ s \geq 0\,, \quad s^2 < 2 \,. \tag{3.18}\] Define \[ \widetilde{s} := s + \frac{1}{n} \] with \(n \in \mathbb{N}\), \(n \neq 0\) to be chosen later. Then \[\begin{align*} \widetilde{s}^2 & = \left( s + \frac1n \right)^2 \\ & = s^2 + \frac{1}{n^2} + 2 \frac{s}{n} \\ & < s^2 + \frac{1}{n} + 2 \frac{s}{n} \qquad \left(\mbox{using } \, \frac{1}{n}<\frac{1}{n^2} \right) \\ & = s^2 + \frac{2s + 1}{n} \,. \end{align*}\] If we now impose that \[ s^2 + \frac{2s + 1}{n} < 2 \,, \] we can rearrange the above and obtain \[ n(2 - s^2) > 2s + 1 \,. \] Now note that \(s^2 < 2\) by assumption (3.18). Thus we can divide by \((2 - s^2)\) and obtain \[ n >\frac{2s + 1}{2 - s^2} \,. \] To summarize, we have just shown that \[ n >\frac{2s + 1}{2 - s^2} \,\, \implies \,\, \widetilde{s}^2 < 2 \,. \] Moreover \(\widetilde{s} := (s + 1/n) \in \mathbb{Q}\). Therefore \[ \widetilde{s} \in A \,. \] Since \(s = \sup A\), we then have \[ \widetilde{s} \leq s \,. \] However \[ \widetilde{s} := s + \frac{1}{n} > s \,, \] yielding a contradiction. Thus \(s \in A\) is not possible.

• Case 2. \(s \notin A\).
  If \(s \notin A\), by the fact that \(s = \sup A\) and by definition of \(A\) we get \[ s > 0 \,, \quad s^2 > 2 \,. \tag{3.19}\]
  Define \[ \widetilde{s} := s - \frac{1}{n} \,. \] We have \[\begin{align*} \widetilde{s}^2 & = \left( s - \frac1n \right)^2 \\ & = s^2 + \frac{1}{n^2} - 2 \frac{s}{n} \\ & > s^2 - \frac{1}{n^2} - 2 \frac{s}{n} \qquad \left(\mbox{using } \, \frac{1}{n^2} > - \frac{1}{n^2} \right) \\ & > s^2 - \frac{1}{n} - 2 \frac{s}{n} \qquad \left(\mbox{using } \, -\frac{1}{n^2} > - \frac{1}{n} \right) \\ & = s^2 - \frac{1 + 2s}{n} \,. \end{align*}\] Now we impose \[ s^2 - \frac{1 + 2s}{n} > 2 \] which is equivalent to \[ n(s^2 - 2) > 1 + 2s \,. \] By (3.19) we have \(s^2 > 2\). Therefore we can divide by \((s^2 -2)\) and get \[ n > \frac{1+2s}{s^2 - 2} \,. \] In total, we have just shown that \[ n > \frac{1+2s}{s^2 - 2} \quad \implies \quad \widetilde{s}^2 > 2\,. \] Therefore \(\widetilde{s} \notin A\), and by definition of \(A\) we have \[ \widetilde{s} \geq q \,, \quad \forall q \in A\,. \] Moreover \(\widetilde{s} := (s - 1/n) \in \mathbb{Q}\). Therefore \(\widetilde{s}\) is an upper bound of \(A\) in \(\mathbb{Q}\). Since \(s=\sup A\) is the smallest upper bound, see (3.17), it follows \[ s \leq \widetilde{s} \,. \] However \[ \widetilde{s} := s - \frac{1}{n} < s \,, \] obtaining a contradiction. Then \(s \notin A\).

Conclusion.
We have assumed by contradiction that \(s = \sup A\) exists in \(\mathbb{Q}\). In this case either \(s \in A\) or \(s \notin A\). In both cases we found a contradiction. Therefore \(\sup A\) does not exist.

The above theorem shows that the supremum does not necessarily exist. What about the infimum?

Question 40

Suppose \((K,+,\cdot, \leq)\) is an ordered field, and that \(A \subseteq K\) is non-empty and bounded below. Does \[ \inf A \] always exist?

The answer to the above question is again NO. A set bounded below for which the infimum does nor exist is, for example, \[ A = (\sqrt{2}, 10] \cap \mathbb{Q}\,. \] The proof of this fact is, of course, very similar to the one of Theorem 39, and is therefore omitted.

Thus infimum and supremum do not exist in general. The fields for which all the bounded sets admit supremum or infimum are called complete.

Definition 41: Completeness

Let \((K,+,\cdot,\leq)\) be an ordered field. We say that \(K\) is complete if it holds the property:

• (AC) For every \(A \subseteq K\) non-empty and bounded above \[ \sup A \in K \,. \]

Notation 42

We have that:

• Property (AC) is called Axiom of Completeness
• If \(K\) is an ordered field in which (AC) holds, then \(K\) is called a complete ordered field

Notice that if the Axiom of Completeness holds, then also the infimum exists. This is shown in the following proposition.

Proposition 43

Let \((K,+,\cdot,\leq)\) be a complete ordered field. Suppose that \(A \subseteq K\) is non-empty and bounded below. Then \[ \inf A \in K\,. \]

Proof

Suppose that \(A \subseteq K\) is non-empty and bounded below. Then \[ -A:= \{ -a \, \colon \,a \in A \} \] is non-empty and bounded above. By completeness we have that \(\sup(-A)\) exists in \(K\). But then Proposition 37 implies that \(\inf A\) exists in \(K\), with \[ \inf A = - \sup (-A)\,. \]

3.6 Equivalence of Completeness and Cut Property

We can show that Completeness is equivalent to the Cut Property. Such result is not essential, but its proof is very instructive.

Theorem 44: Equivalence of Cut Property and Completeness

Let \((K,+,\cdot, \leq)\) be an ordered field. Then they are equivalent:

1. \(K\) has the Cut Property
2. \(K\) is Complete

Remark 45: Ideas for proving Theorem 44

The proof of Theorem 44 is rather long, but the ideas are simple:

Step 1. Cut Property \(\implies\) Completeness. Suppose \(K\) has the Cut Property. To prove that \(K\) is Complete, we need to:

• Consider an arbitrary set \(A \subseteq K\) such that \(A \neq \emptyset\) and \(A\) is bounded above.

• Show that \(A\) has a supremum.

To achieve this, consider the set \[ B := \{ b \in K \, \colon \,b \geq a \,, \,\, \forall a \in A \} \,, \] which is the set of Upper Bounds of \(A\). We can show that the pair \[ (B^c,B) \] is a Cut of \(K\). As \(K\) has the Cut Property, then there exists \(s \in K\) separator of \((B^c,B)\). We will show that the separator \(s\) is the supremum of \(A\) \[ s = \sup A \,. \] Thus \(K\) is complete. See Figure 3.1 for a schematic picture of the above construction.

Step 2. Completeness \(\implies\) Cut Property. Conversely, suppose that \(K\) is Complete. To prove that \(K\) has the Cut Property, we need to:

• Consider a cut \((A,B)\) of \(K\).
• Show that \((A,B)\) has a separator \(s \in K\).

This implication is easier. Indeed, since \(A\) is non-empty and bounded above, by Completeness there exists \[ \sup A \in K \,. \] We will show that \[ s := \sup A \] is a separator for the cut \((A,B)\). See Figure 3.2 for a schematic picture of the above construction.

Figure 3.1: Let \(s\) be the separator of the cut \((B^c,B)\), with \(B\) the set of upper bounds of \(A\). Then \(s = \sup A\).

Figure 3.2: Let \((A,B)\) be a cut of \(K\) and let \(s = \sup A\). Then \(s\) is the separator of the cut \((A,B)\).

Keeping the above ideas in mind, let us proceed with the proof.

Proof: Proof of Theorem 44

Step 1. Cut Property \(\implies\) Completeness.
We need to prove that \(K\) is complete. To this end, consider \(A \subseteq K\) non-empty and bounded above. Define the set of upper bounds of \(A\): \[ B := \{ b \in K \, \colon \,b \geq a \,, \,\, \forall a \in A \} \,. \]

Claim. The pair \((B^c,B)\) is a cut of \(K\).

Proof of Claim. We have to prove two points:

• \((B^c,B)\) forms a partition of \(K\).

Indeed, we have \(B \neq \emptyset\), since \(A\) is bounded above. Further, we have \(B^c \neq \emptyset\), since \(A\) is non-empty. Thus \[ K = B^c \cup B \,, \quad B^c \cap B = \emptyset \,. \] Then \((B^c,B)\) is a partition of \(K\).

• We have \[ x \leq y \,, \quad \forall \, x \in B^c\,, \,\forall \, y \in B \,. \tag{3.20}\]

To show the above, let \(x \in B^c\) and \(y \in B\). By definition of \(B\) we have that elements of \(B^c\) are not upper bounds of \(A\). Therefore \(x\) is not an upper bound. This means there exists \(\widetilde{a} \in A\) which is larger than \(x\), that is, \[ x \leq \widetilde{a} \,. \] Since \(y \in B\), then \(y\) is an upper bound for \(A\), so that \[ a \leq y \,, \,\, \forall a \in A\,. \] Therefore \[ x \leq \widetilde{a} \leq y \,, \] concluding (3.20).

Thus \((B^c,B)\) is a cut of \(K\) and the claim is proven.

Since \((B^c,B)\) is a cut of \(K\), by the Cut Property there exists a separator \(s \in K\) such that \[ x \leq s \leq y \,, \quad \forall \, x \in B^c\,, \,\forall \, y \in B \,. \tag{3.21}\]

Claim. \(s\) is an upper bound for \(A\).

Proof of Claim.
Suppose by contradiction that \(s\) is not an upper bound for \(A\). Therefore by definition of upper bound, there exists \(\widetilde{a} \in A\) such that \[ s < \widetilde{a} \,. \] Consider the mid-point between \(s\) and \(\widetilde{a}\), that is, \[ m:=\frac{s + \widetilde{a}}{2} \in K \,. \] Since \(m\) is the mid-point between \(s\) and \(\widetilde{a}\), and \(s < \widetilde{a}\), it holds \[ s < m < \widetilde{a} \,. \]

Indeed, since \(s < \widetilde{a}\) then \[ s = \frac{2s}{2} < \frac{s + \widetilde{a}}{2} < \frac{2 \widetilde{a}}{2} = \widetilde{a} \,. \]

In particular the above tells us that \(m\) is not an upper bound for \(A\), given that \(\widetilde{a} \in A\) and \(m < \widetilde{a}\). Therefore \(m \in B^c\), by definition of \(B^c\). Therefore(3.21) implies \[ m \leq s \,, \] which contradicts \(s < m\). Hence \(s\) is an upper bound of \(A\), concluding the proof of Claim.

Conclusion. We have shown that \(s\) is an upper bound of \(A\). Condition
(3.21) tells us that \[ s \leq y \,, \,\, \forall y \in B \,. \] Recalling that \(B\) is the set of upper bounds of \(A\), this means that \(s\) is the smallest upper bound of \(A\), that is, \[ s = \sup A \in K \,. \]

Step 2. Completeness \(\implies\) Cut Property.
Suppose \(K\) is complete. We need to show that \(K\) has the Cut Property. Therefore assume \((A,B)\) is a cut of \(K\), that is, \[ A \neq \emptyset\,, \quad B \neq \emptyset \,, \] \[ K = A \cup B\,, \quad A \cap B = \emptyset \,, \] \[ a \leq b \,, \quad \forall a \in A \,, \,\, \forall \, b \in B \,. \tag{3.22}\] Since \(B \neq \emptyset\), from (3.22) it follows that \(A\) is bounded above: indeed, every element of \(B\) is an upper bound for \(A\), thanks to (3.22). Since \(A \neq \emptyset\), by the Axiom of Completeness we have \[ s = \sup A \in K \,. \] In particular, by definition of supremum, we have \[ a \leq s \,, \,\, \forall \, a \in A \,. \] Let now \(b \in B\) be arbitrary. From (3.22) we have that \[ a \leq b \,, \,\, \forall \, a \in A \,. \tag{3.23}\] Therefore \(b\) is an upper bound of \(A\). Since \(s = \sup A\), we have that \(s\) is the smallest upper bound, and so \[ s \leq b \,. \] Given that \(b \in B\) was arbitrary, it actually holds \[ s \leq b \,, \,\, \forall \, b \in B \,. \tag{3.24}\] From (3.23) and (3.24) we therefore have \[ a \leq s \leq b \,, \,\, \forall a \in A \,, \, \forall \, b \in B \,, \] showing that \(s\) is a separator of \((A,B)\). Thus \(K\) has the Cut Property.

3.7 Axioms of Real Numbers

We now have all the key elements to introduce the Real Numbers \(\mathbb{R}\). These ingredients are:

• Definition of ordered field,
• The Cut Property or Axiom of Completeness.

The definition of \(\mathbb{R}\) is given in an axiomatic way.

Definition 46: System of Real Numbers \(\mathbb{R}\)

A system of Real Numbers is a set \(\mathbb{R}\) with two operations \(+\) and \(\cdot\), and an order relation \(\leq\), such that

• \((\mathbb{R},+,\cdot, \leq)\) is an ordered field

• \(\mathbb{R}\) sastisfies the Axiom of Completeness

For reader’s convenience we explicitly state the above mentioned properties.

1. There is an operation \(+\) of addition on \(\mathbb{R}\) \[ +\ \colon \mathbb{R}\times \mathbb{R}\to \mathbb{R}\,, \quad (x,y) \mapsto x + y \] The addition satisifes: \(\,\forall \, x,y,z \in \mathbb{R}\)
   • (A1) Commutativity and Associativity: \[ x+y = y+x \] \[ (x+y)+z = x+(y+z) \]
   • (A2) Additive Identity: \(\,\exists \, 0 \in \mathbb{R}\, \text{ s.t. } \, \) \[ x + 0 = 0 + x = x \]
   • (A3) Additive Inverse: \(\, \exists \, (-x) \in \mathbb{R}\, \text{ s.t. } \, \) \[ x + (-x) = (-x) + x = 0 \]
2. There is an operation \(\cdot\) of multiplication on \(\mathbb{R}\) \[ \cdot\ \colon \mathbb{R}\times \mathbb{R}\to \mathbb{R}\,, \quad (x,y) \mapsto x \cdot y = xy \] The multiplication satisifes: \(\, \forall \, x,y,z \in \mathbb{R}\)
   • (M1) Commutativity and Associativity: \[ x \cdot y = y \cdot x \] \[ (x \cdot y) \cdot z = x \cdot (y \cdot z) \]
   • (M2) Multiplicative Identity: \(\, \exists \, 1 \in \mathbb{R}\, \text{ s.t. } \, \) \[ x \cdot 1 = 1 \cdot x = x \]
   • (M3) Multiplicative Inverse: If \(x \neq 0 \,, \,\, \exists \, x^{-1} \in \mathbb{R}\, \text{ s.t. } \, \) \[ x \cdot x^{-1} = x^{-1} \cdot x = 1 \]
3. There is a relation \(\leq\) of total order on \(\mathbb{R}\). The order satisfies: \(\, \forall \, x, y, z \in \mathbb{R}\)
   • (O1) Reflexivity: \[ x \leq x \]
   • (O2) Antisymmetry: \[ x \leq y \, \mbox{ and } \, y \leq x \,\, \implies \,\, x = y \]
   • (O3) Transitivity: \[ x \leq y \,\, \mbox{ and } \,\, y \leq z \,\, \implies \,\, x = z \]
   • (O4) Total order:
     \[ x \leq y \,\, \mbox{ or } \,\, y \leq x \]
4. The operations \(+\) and \(\cdot\), and the total order \(\leq\), are related by the following properties: \(\, \forall x, y, z \in \mathbb{R}\)
   • (AM) Distributive: Relates addition and multiplication via \[ x \cdot (y + z) = x \cdot y + x \cdot z \]
   • (AO) Relates addition and order with the requirement: \[ x \leq y \,\, \implies \,\, x + z \leq y + z \]
   • (MO) Relates multiplication and order with the requirement: \[ x \geq 0, \, y \geq 0 \,\, \implies \,\, x \cdot y \geq 0 \]
5. Axiom of Completeness holds:
   • (AC) For every \(A \subseteq \mathbb{R}\) non-empty and bounded above \[ \sup A \in \mathbb{R} \]

Remark 47

Since Axiom of Completeness and Cut Property are equivalent by Theorem 44, one can replace the Axiom of Completeness in Definition 46 Point 5 with:

5. Cut Property holds:
   • (CP) Every cut \((A,B)\) of \(\mathbb{R}\) admits a separator \(s \in \mathbb{R}\, \, \text{ s.t. } \, \) \[ a \leq s \leq b \,, \quad \forall \, a \in A \,, \,\forall \, b \in B \]

Notation 48

For \(x \in \mathbb{R}\), \(x \neq 0\), the multiplicative inverse is also denoted by \[ x^{-1} = \frac{1}{x} \,. \]

Remark 49

Recall that

• \((K,+,\cdot)\) satisfying \[ \mbox{(A1)-(A3), (M1)-(M3), (AM)} \] is a field
• \((K,+,\cdot, \geq)\) satisfying \[ \mbox{(A1)-(A3), (M1)-(M3), (O1)-(O4) ,(AM), (AO), (MO)} \] is an ordered field

In particular we have that \[ (\mathbb{R},+, \cdot,\leq) \] is a complete ordered field: that is, an ordered field in which the Cut Property (CP) or Axiom of Completeness (AC) hold

Important

It can be shown that \((\mathbb{R},+, \cdot,\leq)\) is the only complete ordered field.

The above has to be intended in the following sense: if \((K,+, \cdot,\geq)\) is another complete ordered field, then \(K\) looks like \(\mathbb{R}\). Mathematically this means that there exists an invertible map \(\Psi \ \colon \mathbb{R}\to K\), called isomorphism of fields, which preserves the operations \(+\), \(\cdot\) and the order \(\leq\).

Question 50

We have only postulated the existence of \(\mathbb{R}\). Does such complete ordered field actually exist?

The answer is YES. There are several equivalent models for the system \(\mathbb{R}\). If time allows, we will look into one of these models at the end of the module.

3.8 Special subsets of \(\mathbb{R}\)

In Definition 46 we have introduced \(\mathbb{R}\) as a complete ordered field. This was done axiomatically and in a non constructive way. What happens now to the sets \(\mathbb{N}, \mathbb{Z}, \mathbb{Q}\)? Are they well defined? Does it still hold that \[ \mathbb{N}\,, \mathbb{Z}\,, \mathbb{Q}\, \subseteq \mathbb{R}\,? \] The definitions that we gave in Chapter 1 for \(\mathbb{N}, \mathbb{Z}\) and \(\mathbb{Q}\) are not related to the system of real numbers \(\mathbb{R}\) we just introduced. To overcome this problem, we will have to define new sets \[ {\mathbb{N}}_{\mathbb{R}}, \,\, {\mathbb{Z}}_{\mathbb{R}}\,, \,\, {\mathbb{Q}}_{\mathbb{R}} \] from scracth, starting from the axioms of \(\mathbb{R}\). Note that we are using the subscript \(\mathbb{R}\) to distinguish these new sets from the old ones.

3.8.1 Natural numbers

Let us start with the definition of \({\mathbb{N}}_{\mathbb{R}}\). We would like \({\mathbb{N}}_{\mathbb{R}}\) to be \[ {\mathbb{N}}_{\mathbb{R}} = \{ \mathbf{1}, \mathbf{2} , \mathbf{3} , \ldots \} \,. \] We are denoting the above numbers with bold symbols in order to distinguish them from the elements of \(\mathbb{R}\). The key property that we would like \({\mathbb{N}}_{\mathbb{R}}\) to have is the following: \[ \mbox{Every } \mathbf{n} \in {\mathbb{N}}_{\mathbb{R}} \, \mbox{ has a successor } \, (\mathbf{n+1}) \in {\mathbb{N}}_{\mathbb{R}} \,. \] How do we ensure this property? We could start by defining \[ \mathbf{1}:=1\,, \] with \(1\) the neutral element of the multiplication in \(\mathbb{R}\), which exists by the field axiom (M2) in Defintion 46. We could then define \(\mathbf{2}\) by setting \[ \mathbf{2} := 1 + 1 \,. \] We need a formal definition to capture this idea. This is the concept of inductive set.

Definition 51: Inductive set

Let \(S \subseteq \mathbb{R}\). We say that \(S\) is an inductive set if they are satisfied:

• \(1 \in S\),
• If \(x \in S\), then \((x + 1) \in S\).

Note that in the above definition we just used:

• The existence of the neutral element \(1\), given by axiom (M2).
• The operation of sum in \(\mathbb{R}\), which is again given as an axiom.

Example 52

We have that

• \(\mathbb{R}\) is an inductive set.

Indeed we have \(1 \in \mathbb{R}\) by axiom (M2). Moreover \((x + 1) \in \mathbb{R}\) for every \(x \in \mathbb{R}\), by definition of sum \(+\).

• The set \(A=\{0,1\}\) is not an inductive set.

This is because \(1 \in A\), but \((1 + 1) \notin A\), since \(1 + 1 \neq 0\).

Therefore \(\mathbb{R}\) is an inductive set, showing that the definition of inductive set is not sufficient to fully describe our intuitive idea of \({\mathbb{N}}_{\mathbb{R}}\). The right way to define \({\mathbb{N}}_{\mathbb{R}}\) is as follows: \[ {\mathbb{N}}_{\mathbb{R}} \mbox{ is the smallest inductive subset of } \mathbb{R}\,. \] To make the above definition formal we need a few observations.

Proposition 53

Let \(\mathcal{M}\) be a collection of inductive subsets of \(\mathbb{R}\). Then \[ S := \bigcap_{M \in \mathcal{M}} \, M \] is an inductive subset of \(\mathbb{R}\).

Proof

We have to show that the two properties of inductive sets hold for \(S\):

• We have \(1 \in M\) for every \(M \in \mathcal{M}\), since these are inductive sets. Thus \[ 1 \in \bigcap_{M \in \mathcal{M}} \, M = S \,. \]

• Suppose that \(x \in S\). By definition of \(S\) this implies that \(x \in M\) for all \(M \in \mathcal{M}\). Since \(M\) is an inductive set, then \((x + 1) \in M\). Therefore \((x+1) \in M\) for all \(M \in \mathcal{M}\), showing that \((x+1) \in S\).

Therefore \(S\) is an inductive set.

We are now ready to define the natural numbers \({\mathbb{N}}_{\mathbb{R}}\).

Definition 54: Set of Natural Numbers

Let \(\mathcal{M}\) be the collection of all inductive subsets of \(\mathbb{R}\). We define the set of natural numbers in \(\mathbb{R}\) as \[ {\mathbb{N}}_{\mathbb{R}} := \bigcap_{M \in \mathcal{M}} \, M \,. \]

Therefore \({\mathbb{N}}_{\mathbb{R}}\) is the intersection of all the inductive subsets of \(\mathbb{R}\). From this definition it follows that \({\mathbb{N}}_{\mathbb{R}}\) is the smallest inductive subset of \(\mathbb{R}\), as shown in the following proposition.

Proposition 55: \({\mathbb{N}}_{\mathbb{R}}\) is the smallest inductive subset of \(\mathbb{R}\)

Let \(C \subseteq \mathbb{R}\) be an inductive subset. Then \[ {\mathbb{N}}_{\mathbb{R}} \subseteq C \,. \] In other words, \({\mathbb{N}}_{\mathbb{R}}\) is the smallest inductive set in \(\mathbb{R}\).

Proof

Let \(\mathcal{M}\) be the collection of all inductive subsets of \(\mathbb{R}\). By definition \[ {\mathbb{N}}_{\mathbb{R}} = \bigcap_{M \in \mathcal{M}} \, M \,. \] Let \(x \in {\mathbb{N}}_{\mathbb{R}}\), then \(x \in M\) for all \(M \in \mathcal{M}\). Since \(C \in \mathcal{M}\) then \(x \in C\). This shows \({\mathbb{N}}_{\mathbb{R}} \subseteq C\).

The definition of \({\mathbb{N}}_{\mathbb{R}}\) guarantees that all numbers in \({\mathbb{N}}_{\mathbb{R}}\) are larger than \(1\).

Theorem 56

Let \(x \in {\mathbb{N}}_{\mathbb{R}}\). Then \[ x \geq 1 \,. \]

Proof

Define the set \[ C := \{ x \in \mathbb{R}\, \colon \,x \geq 1 \} \,. \] We have that \(C\) is an inductive subset of \(\mathbb{R}\).

By definition \(1 \in C\). Suppose now that \(x \in C\), so that \(x \geq 1\). Since \(1 \geq 0\) as a consequence of the field axioms, we deduce that \[ x + 1 \geq x + 0 = x \geq 1 \,, \] showing that \(x + 1 \geq 1\). Thus \((x + 1) \in C\).

By Proposition 55 we conclude that \[ {\mathbb{N}}_{\mathbb{R}} \subseteq C \,, \] showing that \(x \geq 1\) for all \(x \in {\mathbb{N}}_{\mathbb{R}}\).

Notation 57

We have just shown that all the numebers \(x \in {\mathbb{N}}_{\mathbb{R}}\) satisfy \[ x \geq 1 \,. \] Moreover by the fact that \({\mathbb{N}}_{\mathbb{R}}\) is an inductive set, we know that \[ 1 + 1 \in {\mathbb{N}}_{\mathbb{R}} \,, \] since \(1 \in {\mathbb{N}}_{\mathbb{R}}\). We denote \[ 2 := 1 + 1\,. \] Similarly, we will have that \[ 2 + 1 \in {\mathbb{N}}_{\mathbb{R}}\,, \] since \(2 \in {\mathbb{N}}_{\mathbb{R}}\). We denote \[ 3 := 2 + 1\,. \] In this way we give a name to all the numbers in \({\mathbb{N}}_{\mathbb{R}}\).

3.8.2 Principle of induction

The Principle of Induction is a consequence of the definition of \({\mathbb{N}}_{\mathbb{R}}\), see Definition 54, and of the field axioms of \(\mathbb{R}\) in Definition 46.

Theorem 58: Principle of Induction

Let \(\alpha(n)\) be a statement depending on \(n \in {\mathbb{N}}_{\mathbb{R}}\). Assume that

1. \(\alpha(1)\) is true.
2. If \(\alpha(n)\) is true then also \(\alpha(n+1)\) is true.

Then \(\alpha(n)\) is true for all \(n \in {\mathbb{N}}_{\mathbb{R}}\).

Proof

Define the set \[ C:=\{ x \in {\mathbb{N}}_{\mathbb{R}} \, \colon \,\alpha(n) \mbox{ is true} \} \,. \] We have that \(C\) is an inductive subset of \(\mathbb{R}\).

Indeed:

• \(1 \in C\) since \(\alpha(1)\) is true by assumption.
• If \(n \in C\) then \(\alpha(n)\) is true. By assumption \(\alpha(n+1)\) is true. Therefore \((n+1) \in C\).

By Proposition 55 we conclude that \[ {\mathbb{N}}_{\mathbb{R}} \subseteq C\,. \] As by definition \(C \subseteq {\mathbb{N}}_{\mathbb{R}}\), we have proven that \[ {\mathbb{N}}_{\mathbb{R}} = C \,, \] showing that \(\alpha(n)\) is true for all \(n \in {\mathbb{N}}_{\mathbb{R}}\).

As a consequence of the principle of induction, we can prove that \({\mathbb{N}}_{\mathbb{R}}\) is closed under the field operations of sum and multiplication.

Theorem 59

For all \(n,m \in {\mathbb{N}}_{\mathbb{R}}\) we have:

1. \({\mathbb{N}}_{\mathbb{R}}\) is closed under addition, that is, \[ m + n \in {\mathbb{N}}_{\mathbb{R}} \,. \]

2. \({\mathbb{N}}_{\mathbb{R}}\) is closed under multiplication, that is, \[ m \cdot n \in {\mathbb{N}}_{\mathbb{R}} \,, \]

3. If \(m > n\) there exists \(k \in {\mathbb{N}}_{\mathbb{R}}\) such that \[ m = n + k \,. \]

Proof

We only prove the first point, the other statements are left as an exercise. Fix \(m \in {\mathbb{N}}_{\mathbb{R}}\). We prove that \[ m + n \in {\mathbb{N}}_{\mathbb{R}} \,, \quad \forall \, n \in {\mathbb{N}}_{\mathbb{R}} \,, \tag{3.25}\] by using induction.

• Induction base: We have \(m+1 \in {\mathbb{N}}_{\mathbb{R}}\), since \(m \in {\mathbb{N}}_{\mathbb{R}}\) and \({\mathbb{N}}_{\mathbb{R}}\) is an inductive set.
• Inductive hypothesis: Suppose \(m+n \in {\mathbb{N}}_{\mathbb{R}}\). Since \({\mathbb{N}}_{\mathbb{R}}\) is an inductive set, we have \((m+n)+ 1 \in {\mathbb{N}}_{\mathbb{R}}\). By associativity of the sum, see axiom (A1), we get \[ m + (n + 1) = ( m + n ) + 1 \in {\mathbb{N}}_{\mathbb{R}} \,, \] which is the desired theis.

By the Induction Principle of Theorem 58 we conclude (3.25).

As a consequence of the above theorem, we see that the restriction of the operations of sum and multiplication to \({\mathbb{N}}_{\mathbb{R}}\) are still binary operations: \[ + \, \colon {\mathbb{N}}_{\mathbb{R}} \times {\mathbb{N}}_{\mathbb{R}} \to {\mathbb{N}}_{\mathbb{R}} \,, \quad \cdot \, \colon {\mathbb{N}}_{\mathbb{R}} \times {\mathbb{N}}_{\mathbb{R}} \to {\mathbb{N}}_{\mathbb{R}} \,. \]

Equipped with the above operations, \({\mathbb{N}}_{\mathbb{R}}\) satisfies the following properties.

Theorem 60

\(({\mathbb{N}}_{\mathbb{R}}, + , \cdot, \leq )\) satisfies the following axioms from Definition 46:

• (A1).
• (M1), (M2).
• (O1)-(O4).
• (AM), (AO), (MO).

The proof is trivial, as it follows immediately from the inclusion of \({\mathbb{N}}_{\mathbb{R}}\) in \(\mathbb{R}\).

3.8.3 Integers

We have seen in Theorem 59 that \({\mathbb{N}}_{\mathbb{R}}\) is closed under addition. However \({\mathbb{N}}_{\mathbb{R}}\) is not closed under subtraction. We therefore define the set of integers \({\mathbb{Z}}_{\mathbb{R}}\) in a way that we can perform subtraction of any two natural numbers.

Definition 61: Set of Integers

The set of integers in \(\mathbb{R}\) is defined by \[ {\mathbb{Z}}_{\mathbb{R}} := \{ m - n \, \colon \,n,m \in {\mathbb{N}}_{\mathbb{R}} \} \,. \]

In the definition of \({\mathbb{Z}}_{\mathbb{R}}\) we denote by \(-n\) the inverse of \(n\) in \(\mathbb{R}\), which exists by the field axiom (A3) in Definition 46. The following characterization explains the relationship between \({\mathbb{Z}}_{\mathbb{R}}\) and \({\mathbb{N}}_{\mathbb{R}}\).

Theorem 62

It holds \[ {\mathbb{Z}}_{\mathbb{R}} = \{ -n \, \colon \,n \in {\mathbb{N}}_{\mathbb{R}} \} \cup \{0\} \cup {\mathbb{N}}_{\mathbb{R}} \,. \]

Proof

Define the set \[ M:= \{ -n \, \colon \,n \in {\mathbb{N}}_{\mathbb{R}} \} \cup \{0\} \cup {\mathbb{N}}_{\mathbb{R}} \,. \]

• \(M \subseteq {\mathbb{Z}}_{\mathbb{R}}\): Suppose \(m \in M\). We have 3 cases:

  • If \(m \in \{ -n \, \colon \,n \in {\mathbb{N}}_{\mathbb{R}}\}\) then there exists \(n \in {\mathbb{N}}_{\mathbb{R}}\) such that \(m=-n\). Thus \[ m = -n = 1 - (n+1) \in {\mathbb{Z}}_{\mathbb{R}} \,, \] since \(1 \in {\mathbb{N}}_{\mathbb{R}}\) and \(n+1 \in {\mathbb{N}}_{\mathbb{R}}\) because \(n \in {\mathbb{N}}_{\mathbb{R}}\).
  • If \(m=0\) then \[ m = 0 = 1 - 1 \in {\mathbb{Z}}_{\mathbb{R}}\,, \] as \(1 \in {\mathbb{N}}_{\mathbb{R}}\).
  • If \(m \in {\mathbb{N}}_{\mathbb{R}}\) then \[ m = (m + 1) - 1 \in {\mathbb{Z}}_{\mathbb{R}} \,, \] since \(1 \in {\mathbb{N}}_{\mathbb{R}}\) and \(m+1 \in {\mathbb{N}}_{\mathbb{R}}\), given that \(m \in {\mathbb{N}}_{\mathbb{R}}\).

  In all 3 cases we have shown that \(m \in {\mathbb{Z}}_{\mathbb{R}}\), proving that \(M \subseteq {\mathbb{Z}}_{\mathbb{R}}\).

• \({\mathbb{Z}}_{\mathbb{R}} \subseteq M\): Let \(z \in {\mathbb{Z}}_{\mathbb{R}}\). Then \(z = m - n\) for some \(n,m \in {\mathbb{N}}_{\mathbb{R}}\). We have 3 cases:

  • If \(m=n\) then \[ z = m - n = m - m \stackrel{(A3)}{=} 0 \in M \,. \]
  • If \(m>n\), by Theorem 59 there exists \(k \in {\mathbb{N}}_{\mathbb{R}}\) such that \(m = k + n\). Therefore \[\begin{align*} z & = m - n = (k + n) - n \\ & \stackrel{(A1)}{=} k + (n-n) \stackrel{(A3)}{=} k + 0 \\ & \stackrel{(A2)}{=} k \in M \,, \end{align*}\] since \(k \in {\mathbb{N}}_{\mathbb{R}}\).
  • If \(m<n\), by Theorem 59 there exists \(k \in {\mathbb{N}}_{\mathbb{R}}\) such that \(n = k + m\). Therefore \[ z = m - n = - k \in M \,, \] since \(k \in {\mathbb{N}}_{\mathbb{R}}\), where again we have used (implicitly) the field axioms (A1), (A2) and (A3).

Therefore \({\mathbb{Z}}_{\mathbb{R}} = M\).

Like we did with \({\mathbb{N}}_{\mathbb{R}}\), we can also show that \({\mathbb{Z}}_{\mathbb{R}}\) is closed under the operations of sum and multiplication.

Theorem 63

For all \(n,m \in {\mathbb{Z}}_{\mathbb{R}}\) we have:

1. \({\mathbb{Z}}_{\mathbb{R}}\) is closed under addition, that is, \[ m + n \in {\mathbb{Z}}_{\mathbb{R}} \,. \]

2. \({\mathbb{Z}}_{\mathbb{R}}\) is closed under multiplication, that is, \[ m \cdot n \in {\mathbb{Z}}_{\mathbb{R}} \,, \]

The proof is left as an exercise. As a consequence of Theorem 63 we have that the restriction of the operations of sum and multiplication to \({\mathbb{Z}}_{\mathbb{R}}\) are still binary operations: \[ + \, \colon {\mathbb{Z}}_{\mathbb{R}} \times {\mathbb{Z}}_{\mathbb{R}} \to {\mathbb{Z}}_{\mathbb{R}} \,, \quad \cdot \, \colon {\mathbb{Z}}_{\mathbb{R}} \times {\mathbb{Z}}_{\mathbb{R}} \to {\mathbb{Z}}_{\mathbb{R}} \,. \]

Equipped with the above operations, \({\mathbb{Z}}_{\mathbb{R}}\) satisfies the following properties.

Theorem 64

\(({\mathbb{Z}}_{\mathbb{R}}, + , \cdot, \leq )\) satisfies the following axioms from Definition 46:

• (A1), (A2), (A3).
• (M1), (M2).
• (O1)-(O4).
• (AM), (AO), (MO).

Proof

The fact that \[ \mbox{(A1), (A2), (M1), (M2), (O1)-(O4), (AM), (AO), (MO)} \] are satisfied descends immediately from the inclusion \[ {\mathbb{Z}}_{\mathbb{R}} \subseteq \mathbb{R}\,. \] We are left to prove (A3). This is non-trivial because a priori the additive inverse \(-z\) of some \(z \in {\mathbb{Z}}_{\mathbb{R}}\) belongs to \(\mathbb{R}\). We need to check that \(- z \in {\mathbb{Z}}_{\mathbb{R}}\). Indeed, since \(z \in {\mathbb{Z}}_{\mathbb{R}}\), there exist \(n,m \in {\mathbb{N}}_{\mathbb{R}}\) such that \(z = m - n\). Define \(y:= n - m\). We have that \(y \in {\mathbb{Z}}_{\mathbb{R}}\) and \[ z + y = (m - n) + (n - m) = (m - m) + (n - n) = 0 \,. \] Therefore \(y\) is the inverse of \(z\) and \(y \in {\mathbb{Z}}_{\mathbb{R}}\), proving that the sum in \({\mathbb{Z}}_{\mathbb{R}}\) satisfies (A3).

Remark 65

\({\mathbb{Z}}_{\mathbb{R}}\) does not satisfy (M3).

For example, let us show that \(2 \in {\mathbb{Z}}_{\mathbb{R}}\) has no inverse in \({\mathbb{Z}}_{\mathbb{R}}\). Indeed, let \(m \in {\mathbb{Z}}_{\mathbb{R}}\). By Theorem 62 we have 3 cases:

• \(m \in {\mathbb{N}}_{\mathbb{R}}\): Since \(2 > 1\) we have \[ 2 \cdot m > 1 \cdot m \geq 1 \] where in the last inequality we used that \(m \geq 1\) for all \(m \in {\mathbb{N}}_{\mathbb{R}}\), as shown in Theorem 56. The above shows that \[ 2 \cdot m > 1\,, \] and therefore \(m\) cannot be the inverse of \(2\).

• \(m=0\): Then \(2 \cdot m = 0\), so that \(m\) cannot be the inverse of \(2\).

• \(m = -n\) with \(n \in {\mathbb{N}}_{\mathbb{R}}\). Then \[ 2 \cdot m = 2 \cdot (-n) < 0 \,, \] so that \(m\) cannot be the inverse of \(2\).

As we have exhausted all the possibilities, we conclude that \(2\) does not have a multiplicative inverse in \({\mathbb{N}}_{\mathbb{R}}\).

3.8.4 Rational numbers

In Theorem 64 and 65 we have seen that \({\mathbb{Z}}_{\mathbb{R}}\) satisfy all the field axiom, except for (M3). We therefore extend \({\mathbb{Z}}_{\mathbb{R}}\) in a way that the extension contains multiplicative inverses. The extension is the set of rational numbers \({\mathbb{Q}}_{\mathbb{R}}\).

Definition 66: Set of Rational Numbers

The set of rational numbers in \(\mathbb{R}\) is \[ {\mathbb{Q}}_{\mathbb{R}} := \left\{ \frac{m}{n} \, \colon \,m \in {\mathbb{Z}}_{\mathbb{R}}\,, \,\, n \in {\mathbb{N}}_{\mathbb{R}} \right\} \,. \]

Notice that in the above definition we are just using the field axiom (M3), with \[ \frac{m}{n} := m \cdot n^{-1} \,. \] The inverse of \(n\) exists because we are assuming \(n \in {\mathbb{N}}_{\mathbb{R}}\), and therefore \(n\) cannot be \(0\), as a consequence of Theorem 56.

The set \({\mathbb{Q}}_{\mathbb{R}}\) is closed under addition and multiplication (exercise). Therefore they are well defined the operations: \[ + \, \colon {\mathbb{Q}}_{\mathbb{R}} \times {\mathbb{Q}}_{\mathbb{R}} \to {\mathbb{Q}}_{\mathbb{R}} \,, \quad \cdot \, \colon {\mathbb{Q}}_{\mathbb{R}} \times {\mathbb{Q}}_{\mathbb{R}} \to {\mathbb{Q}}_{\mathbb{R}} \,. \]

Theorem 67

\(({\mathbb{Q}}_{\mathbb{R}}, + , \cdot, \leq)\) is an ordered field.

Proof

All the field properties, except for (M3), follow from the inclusion \[ {\mathbb{Q}}_{\mathbb{R}} \subseteq \mathbb{R} \] and from the field properties of \(\mathbb{R}\). To check (M3), let \(q \in {\mathbb{Q}}_{\mathbb{R}}\) with \(q \neq 0\). Therefore \(q = m/n\) for \(m \in {\mathbb{Z}}_{\mathbb{R}}\), \(n \in {\mathbb{N}}_{\mathbb{R}}\). As \(q \neq 0\) and \(n \neq 0\), see Theorem 56, we deduce that \(m \neq 0\). We have two cases:

• \(m>0\): In this case \(m \in {\mathbb{N}}_{\mathbb{R}}\) by Theorem 62. Therefore \[ p = \frac{n}{m} \in {\mathbb{Q}}_{\mathbb{R}} \] by definition, since \(n,m \in {\mathbb{N}}_{\mathbb{R}}\). By commutativity we have \[ q \cdot p = \frac{m}{n} \cdot \frac{n}{m} = 1 \,. \]
• \(m<0\): Then \(m=-x\) with \(x \in {\mathbb{N}}_{\mathbb{R}}\) by Theorem 62. Therefore \[ p = \frac{-n}{x} \in {\mathbb{Q}}_{\mathbb{R}} \] by definition, since \(-n \in {\mathbb{Z}}_{\mathbb{R}}\) and \(x \in {\mathbb{N}}_{\mathbb{R}}\). By commutativity we have \[ q \cdot p = \frac{m}{n} \cdot \frac{-n}{x} = \frac{m}{n} \cdot \frac{-n}{-m} = 1 \,. \]

Therefore \(q\) always admits a multiplicative inverse \(q^{-1}\) belonging to \({\mathbb{Q}}_{\mathbb{R}}\), proving (M3).

The set \({\mathbb{Q}}_{\mathbb{R}}\) does not have the Cut Property or the Axiom of Completeness.

Theorem 68

\({\mathbb{Q}}_{\mathbb{R}}\) is not complete.

The proof of the above Theorem replicates exactly the proof of Theorem 39: This is because the proof of Theorem 39 only makes use of field axioms, and thus it applies to \({\mathbb{Q}}_{\mathbb{R}}\).

Notation 69

From now on we denote \[ \mathbb{N}:= {\mathbb{N}}_{\mathbb{R}} \,, \quad \mathbb{Z}:= {\mathbb{Z}}_{\mathbb{R}} \,, \quad \mathbb{Q}:= {\mathbb{Q}}_{\mathbb{R}} \,, \quad \] dropping the subscript \(\mathbb{R}\).