# Dedekind Cuts

As we have seen, the set \(\mathbb{Q}\) of rational numbers contains gaps at numbers such as \(\sqrt{2}\) and \(\pi.\) We have previously gained some preliminary insight into real numbers and how these numbers allowed us to fill in the gaps. In this section, we give a rigorous mathematical construction of the real numbers.

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## Real Numbers as Dedekind Cuts

A

Dedekind cut\(x = (A, B)\) in \(\mathbb{Q}\) is a pair of subsets \(A,B\) of \(\mathbb{Q}\) satisfying

- \(A \cup B = \mathbb{Q}, A \cap B = \emptyset, A \not= \emptyset, B \not= \emptyset\)
- If \(a \in A\) and \(b \in B\), then \(a < b\)
- \(A\) contains no largest element.
A

real numberis a Dedekind cut in \(\mathbb{Q}\) and the set of real numbers is denoted \(\mathbb{R}\).

Note that the cut is *ordered* and the elements of \(A\) are all smaller than the elements of \(B\). In the above definition, for a cut \( x = (A,B),\) we have \(B = \mathbb{Q} \backslash A\). Given any rational number \(r,\) an example of a cut is one of the form

\[ [r] = ( \{ q \in \mathbb{Q} : q < r \} , \{ q \in \mathbb{Q} : q \geq r \} ), \]

which is called a **rational cut**. This gives an interpretation of rational numbers as cuts and therefore, every rational number is also a real number. The real number zero is defined as the rational cut

\[ [0] = ( \{ q \in \mathbb{Q} : q < 0 \}, \{ q \in \mathbb{Q} : q \geq 0 \} ). \]

Note that in a rational cut \([r]\), the set \(B\) contains a smallest element, namely \(r\). A real number is **irrational** if the set \(B\) contains no smallest element. An example of a cut that defines an irrational number is

\[ ( \{ q \in \mathbb{Q} : q \leq 0 \text{ or } q^2 < 2 \}, \{ q \in \mathbb{Q} : q > 0 \text{ and } q^2 > 2 \} ) \]

We now use cuts to define arithmetic operations and order relations.

## Arithmetic Operations

We first use cuts to define the familiar arithmetic operations for real numbers, including addition, subtraction, multiplication, and division. For cuts \(x=(A,B)\) and \(y= (C,D)\), define addition as follows:

\[x + y = (E, \mathbb{Q} \backslash E), \text{ where } E = \{ q \in \mathbb{Q} : q = a+c \text{ for some } a \in A, c \in C \} \]

Observe that cut addition is well defined and satisfies \(x + [0] = x, x+y=y+x \) (commutativity), and \((x+y)+z = x + (y+z)\) (associativity). The additive inverse of \(x=(A , B)\) is \(-x = (C, \mathbb{Q} \backslash C )\) with

\[ C = \{ q \in \mathbb{Q} : q =-b \text{ for } b \in B \text{ not the smallest element} \}.\]

Then subtraction is defined by \(x -y = x + (-y),\) and for any cut \(x\), we have \( x + (-x) = (-x) + x = [0].\) The absolute value \( \vert x \vert \) is defined by

\[ \lvert x \rvert = \begin{cases} x, & \mbox{ if } x \geq 0, \\ -x & \mbox{ otherwise} \end{cases}\]

and multiplication is defined by

\[ x \times y = \begin{cases} 0 & \mbox{ if } x = 0 \mbox{ or } y = 0\\ \lvert x \rvert \times \lvert y \rvert &\mbox{ if } x >0, y> 0 \mbox{ or } x < 0, y < 0\\ - \left( \lvert x \rvert \times \lvert y \rvert \right) &\mbox{ if } x >0, y< 0 \mbox{ or } x < 0, y > 0. \end{cases}\]

Then addition and multiplication satisfy the properties \(x \times [1] = x, x \times y=y \times x \) (commutativity), \((x \times y) \times z = x \times (y \times z)\) (associativity), \( (x + y) \times z = (x \times z) + (y \times z)\) (distributivity).

For a real number \(x = (A, \mathbb{Q} \backslash A) \) with \(x > 0\), the multiplicative inverse is defined by \(x^{-1} = (A' , \mathbb{Q} \backslash A' ) \), where

\[ A' = \left\{ q \in \mathbb{Q} : q \leq 0 \mbox{ or } q > 0 \mbox{ and } \frac{1}{q} \in \mathbb{Q} \backslash A \text{ not the smallest element} \right\} . \]

For \(x< 0\), \(x^{-1} = - \left( \lvert x \rvert^{-1} \right) \).

For rational cuts, all of the above arithmetic operations are consistent with arithmetic operations over the rationals.

## Order relations

Given real numbers \(x = (A, B)\) and \(y = (C, D),\) \(x\) is less than or equal to \(y,\) denoted \(x \leq y,\) if \(A \subseteq C.\) The inequality is strict if \(A \subset C.\)

This ordering on the real numbers satisfies the following properties:

- \(x < y\) and \(y< z\) implies \(x < z\)
- Exactly one of \(x < y, y < x,\) or \(x=y\) holds
- \(x < y\) implies \(x + z < y + z\).

All of the above show that \(\mathbb{R}\) is an **ordered field**. Note that according to our definitions, \(\infty\) is not a real number because \( ( \mathbb{Q} , \emptyset ) \) is not a cut. Furthermore, there are infinitely many elements in the set \(\mathbb{R} \backslash \mathbb{Q}\), which is the set of irrational numbers.

## Upper Bound Properties

\(B \in \mathbb{R}\) is an **upper bound** for a set \(S \subset \mathbb{R}\) if each \(s \in S\) satisfies \(s \leq B\). If \(B\) is an upper bound for \(S\) and every upper bound \(B'\) satisfies \(B' \geq B\), then \(B\) is the **least upper bound** for \(S\). Note that if a least upper bound exists, then it is unique.

## Worked Examples

## Let \(x = (A,B)\) be a Dedekind cut. Show that given any rational number \(\epsilon>0\), there exist rational numbers \(a \in A\) and \(b \in B\) such that \(b - a < \epsilon\).

Solution:By definition, \(A\) and \(B\) are nonempty, so there exist rational numbers \(a_0 \in A\) and \(b_0 \in B\). Consider the sequence of rational numbers \( (a_i, b_i) , i = 1, 2, 3 \ldots\) defined by\[ a_i = \begin{cases} \frac{ a_{i-1} + b_{i-1}}{2} & \text{ if } \frac{ a_{i-1} + b_{i-1}}{2} \in A\\ a_{i-1} & \text{ otherwise} \end{cases}\\ b_i = \begin{cases} \frac{ a_{i-1} + b_{i-1}}{2} & \text{ if } \frac{ a_{i-1} + b_{i-1}}{2} \in B\\ b_{i-1} & \text{ otherwise.} \end{cases} \]

Note that \(a_i \in A, b_i \in B\) for \(i = 0, 1, 2, \ldots\) and \(b_n - a_n \leq \frac{1}{2^{n}} (b_0-a_0) \). Then by choosing \(n\) such that \(2^{n} > \frac{b_0 - a_0}{\epsilon}, \) the rational numbers \(a_n, b_n\) satisfy the conditions \(a_n \in A, b_n \in B\) and \(b_n - a_n < \epsilon\).

## Show that for any nonempty subset \(S\) of \(\mathbb{R}\) with an upper bound, there exists a least upper bound for \(S\).

Solution:A nonempty subset \(S\) of \(\mathbb{R}\) is a set of cuts \( \{ (C_1,D_1), ( C_2, D_2), (C_3, D_3), \ldots \} \). Let\[ C= \cup_i C_i , D = \mathbb{Q} \backslash C.\]

Then \( (C, D) \) is a cut and is an upper bound since \(C_i \subseteq C\) implies \( (C_i, D_i ) \leq (C, D)\) for all \(i\) . Since an upper bound for \(S\) exists, let \( (C',D')\) be any upper bound. Then by definition of an upper bound, \(C'\) contains \( \cup_i C_i = C ,\) implying \( (C,D) \leq (C', D')\). Therefore, \( (C, D) \) is the least upper bound for \(S\).

## Suppose \(x\) and \(y\) are real numbers. Prove the following statements

- If \( x \leq y + \epsilon\) for every \(\epsilon > 0,\) then \(x \leq y.\)
- If \( \lvert x - y \rvert < \epsilon \) for every \(\epsilon>0,\) then \(x = y.\)

Solution:

By the ordering principle, we have either \(x \leq y\) or \(x > y\). If \(x > y\), then let \(\epsilon\) be any number in the interval \(0 < \epsilon < x-y\). This gives \[ \epsilon < x- y \leq \epsilon,\] a contradiction. Therefore, \(x \leq y\).

If \(x \ne y\), then let \(\epsilon\) be any number in the interval \( 0 < \epsilon < \lvert x - y \rvert\). This gives \[ \epsilon < \lvert x - y \rvert \leq \epsilon,\] a contradiction. Therefore, \(x = y.\)