Dedekind Cuts
Abstract
In the first chapter of Walter Rudin’s Principles of Mathematical Analysis [Rud76], he omits a complete proof of the Dedekind construction of the real field \(\R\). This article borrows previous definitions and lemmas from the first chapter and will detail a rigorous construction and elaboration of the real field \(\R\) and go into further detail regarding the Archimedean property of ordered fields with the leastupperbound property and the topological denseness of \(\Q\) on \(\R\). The consequences of the completeness of \(\R\), due to its leastupperbound property, will be explored through discovering that reconstructing \(\R\) again by Dedekind completion yields an ordered field isomorphic to \(\R\). Illustrations are provided for the motivation of Rudin’s “secret formula” and the definition/order of Dedekind cuts.^{1} The contents assumes familiarity with the rational numbers \(\Q\) and settheoretic principles.
Table of Contents:
 A Discussion of the Limitations of ℚ.
 Order, Fields, and Ordered Fields.
 Outline of the Construction.
 ℚ as a Subfield of ℝ
 Properties of ℝ.
List of Figures:
 A first approximation of √2 + 1.
 Final approximation of √2.
 Illustration of a Dedekind Cut on a Number Line
 Illustration of the Order of Dedekind Cuts on a Number Line
Tip: Mobile users, this document should be viewed in landscape orientation for a better reading experience because some of the equations are too long to fully render in portrait orientation.
1. A Discussion of the Limitations of ℚ.
In other words, why do we need real numbers? The short answer is that there are “gaps” in the set of rational numbers \(\mathbb{Q},\) even though between every two rationals is another rational. We will inquire about such “gaps” in \(\mathbb{Q}\) from a settheoretic perspective and construct a new number system which fills in these “gaps.”
To illustrate this gap, we will consider the number whose square is \(2.\) This number is denoted as the square root of \(2,\) or \(\sqrt{2}.\) A standard proof relying on the ability to represent a rational number as the ratio of a pair of coprime integers shows that \(\sqrt{2} \notin \mathbb{Q}\) and could be found here [Alf96]. It turns out that \(\sqrt{2}\) is a real number (as we will see later in our construction), and since it is not a rational number, we call it irrational. We will dive into this “gap” from a the perspective of sets by defining: \[A = \{x \in \mathbb{Q} : x > 0 \wedge x^2 < 2\}, \\ B = \{x \in \mathbb{Q} : x > 0 \wedge x^2 > 2\}.\]
The following technique inspired by Professor George Bergman’s exercise packet [Ber06, p.3], we will assume that \(A\) has a largest member or \(B\) has a smallest member and reach a contradiction. First, we observe that \(s \in A\) if and only if \(2/s \in B.\) If \(s \in A,\) then \(s > 0,\) and thus \(2/s > 0.\) Also, since \(s^2 < 2,\) \[\left( \frac{2}{s} \right)^2 = \frac{4}{s^2} > \frac{4}{2} = 2,\] and hence \(2/s \in B.\)
Conversely, if \(2/s \in B,\) put \(t = 2/s\) and hence \(s = 2/t.\) Since \(t > 0,\) \(s = 2/t > 0.\) And since \(t^2 > 2,\) \[s^2 = \left( \frac{2}{t} \right)^2 = \frac{4}{t^2} < \frac{4}{2} = 2,\] and hence \(s \in A.\)
Then, if we assume that the set \(A\) has a largest number \(p,\) then the number \(q = 2/p\) is a member of \(B.\) If there is a rational number \(r \in B\) such that \(r < q,\) then \(2/r \in A\) and \(2/r > 2/q = p,\) contradicting the assumption that \(p\) is the largest member. Therefore, \(q\) must also be the smallest member of \(B.\)
Conversely, if we assume that the set \(B\) has a smallest number \(q,\) then the number \(p = 2/q\) must be a member of \(A.\) If there is a rational number \(r \in A\) such that \(r > p,\) then \(2/r \in A\) and \(2/r < 2/p = q,\) contradicting the assumption that \(q\) is the smallest member. Therefore, \(p\) must also be the largest member of \(B.\)
Now, we have shown that \(p\) is the largest member of \(A\) if and only if \(q\) is the smallest member of \(B.\) Here is where the magic happens with our assumption of \(p\) and \(q.\) Let \[\eta = \frac{p + q}{2}.\] We see that \(p < \eta < q.\) If \(\eta^2 < 2,\) then \(\eta \in A,\) contradicting the choice of \(p\); if \(\eta^2 > 2,\) then \(\eta \in B,\) contradicting the choice of \(q.\) Finally, if \(\eta^2 = 2,\) then
\[\eta^2 = \frac{(p + \frac{2}{p})^2}{4} = \frac{(p^2 + 2)^2}{4p^2} = 2, \] and thus \[(p^2 + 2)^2 = 8p^2.\]
Since \(p > 0,\) we can ignore the negative solution when taking the square root of the right hand side.
\[p^2 + 2 = 2\sqrt{2}p,\] and thus \[(p  \sqrt{2})^2 = 0,\]
hence we see that \(p = \sqrt{2},\) which is not a member of \(\mathbb{Q}\) and thus not a member of \(A,\) contradicting our choice of \(p\) once again.
It seems that no matter what the number \(\eta\) is, we reach some kind of contradiction. Clearly, the set \(A\) cannot have a largest member, and thus the set \(B\) cannot have a smallest member. Either side of the rational number line “gap” at \(\sqrt{2}\) seems to infinitely extend towards the value \(\sqrt{2},\) yet the number \(\sqrt{2}\) itself is missing on the rational number line.
Meanwhile, Rudin suggests a more direct approach, where for any member \(p \in A,\) a number \(q\) could be constructed explicitly such that \(q > p\) and \(q \in A.\) Similarly, for any member \(p \in B,\) a number \(q\) could be found such that \(q < p\) and \(q \in B.\) At the moment, we are just making observations and trying to find that generating formula. Once we are able to obtain such a formula, we will give a rigorous proof. To find such a formula, we first observe that \[\frac{1}{\sqrt{2} + 1} = \frac{\sqrt{2}  1}{2  1} = \sqrt{2}  1. \] Setting \(u = \sqrt{2} + 1,\) we see that \[u = \frac{1}{u} + 2.\] If we define \[g(x) = \frac{1}{x} + 2,\] we see that for any \(x > 1,\) \(g(x)\) seems to be a better approximation (that is, closer) to \(\sqrt{2}\) than \(x.\) Notably, \(g(u) = u.\) We will graph these functions below. Observing from Figure 1, we see that the slope of \(g(x)\) for \(x > 1\) seems to have an absolute value less than \(1\) (the function is somewhat “flat”). Hence, as \(x\) moves away from \(u,\) \(g(x)\) moves away from \(u\) at a slower pace. Hence for any \(x > 1,\) \(g(x)\) yields a desirable value that is closer to \(u.\)
We then see that we could salvage this graph for our use since \(\sqrt{2} = u  1.\) Essentially, we want to move the point \( (u, u) \) on \(g(x)\) to \( (\sqrt{2}, \sqrt{2}) \) on a new function \(g^*(x).\) We can accomplish this by transposing \(g(x)\) one unit to the left and one unit down. \[g^*(x) = \frac{1}{x + 1} + 1 = \frac{x + 2}{x + 1}.\]
For any \(x > 0,\) it seems that \(g^*(x)  \sqrt{2} < x  \sqrt{2}.\) However, for \(p < \sqrt{2},\) \(g^*(p) > \sqrt{2},\) and for \(p > \sqrt{2},\) \(g^*(p) < \sqrt{2}.\) Recall that for rational numbers \(a\) and \(b,\) if \(0 < a < b,\) then \(0 < 1/b < 1/a.\) Then notice that \(\sqrt{2} = g^*(\sqrt{2}) = 2/g^*(\sqrt{2}).\) We will take the reciprocal of \(g^*(x)\) while preserving the property \(g^*(\sqrt{2}) = \sqrt{2}.\) For \(x > 0,\) let \[f(x) = \frac{2}{g^*(x)} = \frac{2x + 2}{x + 2}.\] The new function \(f(x)\) (see Figure 2 apparently satisfies all the properties we desire. We have motivated the discovery of Rudin’s “secret formula” [Rud76, p.2] which he uses to demonstrate that \(A\) has no largest member. We will now construct the proof using the formula we attained with \(f.\)
Proof. We first observe that we can factor a term of \(x\) out of the formula. For the convenience of notation, denote \(f(x)\) as \(y.\) \[y = \frac{2x + 2}{x + 2} = \frac{x(x + 2)  (x^2  2)}{x + 2} = x  \frac{x^2  2}{x + 2}.\] This makes comparing the size of \(x\) and \(y\) more convenient. Then,
\[\begin{align} y^2 &= \bigg( \frac{2x + 2}{x + 2} \bigg)^2 = \frac{4x^2 + 8x + 4}{(x + 2)^2}\\\\ &= \frac{2(x^2  2) + 2(x + 2)^2}{(x + 2)^2} = \frac{2(p^2  2)}{(x + 2)^2} + 2, \end{align}\]and hence \[y^2  2 = \frac{2(x^2  2)}{(x + 2)^2}.\] This makes comparing the size of \(y^2\) and \(2\) more convenient. If \(x \in A,\) then \(x^2  2 < 0,\) and we have \[y = x  \frac{x^2  2}{x + 2} > x,\] and \[y^2  2 = \frac{2(p^2  2)}{(p + 2)^2} < 0,\] hence \(y^2 < 2\) and \(y \in A.\) Next, if \(x \in B,\) then \(x^2  2 > 0,\) and we have \[y = x  \frac{x^2  2}{x + 2} < x,\] and \[y^2  2 = \frac{2(p^2  2)}{(p + 2)^2} > 0,\] hence \(y^2 > 2\) and \(y \in B.\) We have demonstrated that
\[\begin{gather} (\forall x \in A)(\exists y \in A) y > x, \\ (\forall x \in B)(\exists y \in B) y < x. \end{gather}\]Which implies that \(A\) has no largest member and \(B\) has no smallest member.
$\blacksquare$
The set of rational numbers \(\mathbb{Q}\) is clearly missing some numbers, we say that it is not complete. We could, of course, demonstrate many other instances of numbers that exist “around” but not within \(\mathbb{Q}.\) For the sake of brevity, the following example demonstrates that there exist infinitely many irrational numbers based on the irrationality of \(\sqrt{2}\) because the sum and product of a rational number and an irrational number are both irrational.
Example 1. Suppose \(x \notin \mathbb{Q}\) and \(r \in \mathbb{Q}.\) Then, there is an integer \(p\) and a positive integer \(q > 0\) such that \(r = p/q.\)
We want to show that \(r + x \notin \mathbb{Q}.\) Seeking a contradiction, assume \(r + x \in \mathbb{Q}.\) Hence, there is an integer \(m\) and a positive integer \(n > 0\) such that \(r + x = m/n.\) And it follows that
\[x = r + x  r = \frac{m}{n}  \frac{p}{q}  \frac{mq  np}{nq} \in \mathbb{Q},\]
leading to a contradiction.
Similarly, we want to show that given \(r \neq 0,\) \(rx \notin \mathbb{Q}.\) Seeking a contradiction, assume \(rx = \mathbb{Q}.\) Hence, there is an integer $m$ and a positive integer \(n > 0\) such that \(rx = m/n.\) Thus, we have
\[x = \frac{rx}{r} = \frac{mq}{np} \in \mathbb{Q},\]
again, leading to a contradiction.
It follows that we can find an infinite number of such irrational numbers simply by adding and multiplying rational numbers with \(\sqrt{2},\) and the set of rational numbers \(\mathbb{Q}\) has infinitely many such gaps. We wish to construct a real number system which includes all these gaps, that is, we wish for it to be complete. But why can’t we just declare that \(\mathbb{R} = \mathbb{Q}\cup \{\sqrt{2}\}\) and keep appending to this set \(\mathbb{R}\) every time we find an irrational number? Because the arithmetic properties of \(\Q\) which we are familiar with follows from the fact that \(\mathbb{Q}\) forms an ordered field under our conventional definitions of addition, multiplication, and order. Under such informal appendages we do not know whether \(\mathbb{R}\) is still an ordered field, and we cannot even embed \(\mathbb{Q}.\) Therefore, we need a formal method to construct a complete real number system that is an ordered field, for which we will need to define what it means to be an ordered field.
2. Order, Fields, and Ordered Fields.
We will reinterpret the “gaps” in \(\Q\) after a brief definition of order and fields. These definitions chiefly come from Rudin’s Principles of Mathematical Analysis [Rud76, pp.3–8].
Definition 1. (Ordered Set) An ordered set $(S, <)$ is a set $S$ that is ordered by a binary relation $<$. The relation $<$ is said to order $S$ if it satisfies the following axioms.
(O1). Trichotomy:
\[\begin{gathered}(\forall x, y \in S)(x < y \vee x = y \vee x > y) \\\\ \wedge \neg ((x < y \wedge x = y) \vee (x < y \wedge x > y) \vee (x = y \wedge x > y)).\end{gathered}\](O2). Transitivity:
\[(\forall x, y, z \in S)((x < y \wedge y < z) \implies x < z).\]
The order axiom of trichotomy states that exactly one (one and only one) of the following is true:
\[x < y, \qquad x = y, \qquad x > y.\]
The expression \(x \leq y\) states that \(x < y \vee x = y,\) it is the negation of \(x > y.\) Similarly, \(x \geq y\) is the negation of \(x < y\) and \(x \neq y\) is the negation of \(x = y.\)
If we make the following definition: \((\forall x, y \in \mathbb{Q})(x < y \iff x  y < 0),\) then \( (\mathbb{Q}, <) \) is an ordered set.
Definition 2. (Bounds) Let \(S\) be an ordered set. Suppose \(E \subset S\) is any subset. \(\beta\) is an upper bound of \(E\) if \[(\forall x \in E)x \leq \beta.\] And \(\alpha\) is a lower bound of \(E\) if \[(\forall x \in E)\alpha \leq x.\] We say \(E\) is bounded above if it has an upper bound and bounded below if it has a lower bound. We say that \(E\) is bounded if it is bounded both above and below.
Suppose there is an \(\alpha \in S\) with the following properties:
(i). \(\alpha\) is an upper bound of \(E.\)
(ii). For each \(\gamma < \alpha,\) \(\gamma\) is not an upper bound of \(E.\)
then \(\alpha\) is called the least upper bound or supremum of \(E.\) From (ii) it is clear that at most one such \(\alpha \in S\) can exist. This is denoted as \[\alpha = \sup{E}.\] The contrapositive of (ii) is sometimes used: if \(\gamma\) is an upper bound of \(E,\) then \(\gamma \geq \alpha.\)
Similarly, if
(i). \(\beta\) is a lower bound of \(E.\)
(ii). For each \(\gamma > \beta,\) \(\gamma\) is not an lower bound of \(E.\)
then \(\alpha\) is the greatest lower bound or infimum of E. This is denoted as \[\alpha = \inf{E}.\] Again, the contrapositive of (ii) is sometimes used: if \(\gamma\) is an upper bound of \(E,\) then \(\gamma \geq \alpha.\)
It is clear from the definition that the least upper bound and the greatest lower bound are both unique.
Definition 3. (Leastupperbound Property) (sometimes denoted LUB Property) An ordered set \(S\) has the leastupperbound property if:
\[\begin{aligned} (\forall E \subseteq S) \Big[ &(E \neq \varnothing \wedge \exists \alpha ((\forall x \in E) x \leq \alpha)) \\\\ &\implies (\exists \beta \in S) \beta = \sup{E} \big]. \end{aligned}\]In other words, if every nonempty subset \(E \subseteq S\) that is bounded above has a least upper bound in \(S.\)
The leastupperbound property on \(\mathbb{R}\) is often referred to as the completeness axiom. The set \(A \in \mathbb{Q}\) presented in the last section is a counterexample to the definition above and shows that \(\mathbb{Q}\) does not have the leastupperbound property: there are sets of rationals that do not have a rational supremum.
In the next theorem, we will explore the close relationship between having the leastupperbound property and the greatestlowerbound property
Theorem 1. An ordered set \(S\) has the leastupperbound property if and only if it has the greatestlowerbound property.
Proof. For each nonempty subset \(B \subseteq S\) that is bounded below, let \(L\) be the set of all lower bounds of \(B.\) \[L = \big\{y \in S : (\forall x \in B)y \leq x\big\}]
\(B\) is bounded below; it has at least one lower bound, hence \(L\) is nonempty. And since every \(x \in B\) is an upper bound of \(L,\) thus \(L\) is bounded above. The set \(S\) has the leastupperbound property and \(L \subseteq S\) is bounded above (See LUB and the LUB Property); therefore, \(L\) has a leastupperbound in \(S.\) Let \[\alpha = \sup{L} \in S\]
For each \(\gamma < \alpha,\) \(\gamma\) is not an upper bound of \(L,\) hence \(\gamma \notin B.\) Since \(\alpha\) is also the least upper bound, it follows that =\(\alpha\) is less than every member of \(B,\) and thus \(\alpha \in L.\)
And for each \(\beta > \alpha,\) \(\beta \notin L\) since \(\alpha\) is an upper bound of \(L.\) Since \(L\) is the set of all lower bounds of \(B,\) we know that \(\alpha\) is a lower bound of \(B,\) and if \(\beta > \alpha,\) then \(\beta\) is not a lower bound of \(B.\) Therefore, we conclude that \[\alpha = \inf{B}.\]
■
Next, we will look at fields, and we will go with Rudin’s definition of a Field in his book Principles of Mathematical Analysis.
Definition 4. (Field) A field \((\mathbb{F}, +, \cdot, e, f)\) is a set \(\mathbb{F}\) with two operations called addition (\(+\)) and multiplication (\(\cdot\)) (multiplication is usually written without the dot, by concatenating the two elements), which satisfy the axioms listed below known as field axioms. The field axioms (A4) and (M4) also imply the existence of the additive identity \(e\) and the multiplicative identity \(f.\) These identities \(0\) and \(1\) respectively in the fields concerned within this article.
The Field Axioms:
(A) Axioms for Addition
(A1) \( (\forall x, y \in \mathbb{F})(x + y \in \mathbb{F}) \) (Closure under Addition)
(A2) \( (\forall x, y \in \mathbb{F})(x + y = y + x) \) (Commutativity of Addition)
(A3) \( (\forall x, y, z \in \mathbb{F})[(x + y) + z = x + (y + z)] \) (Associativity of Addition)
(A4) \( (\exists e \in \mathbb{F}) (\forall x \in \mathbb{F}) e + x = x \) (Existence of Additive Identity)
(A5) \( (\forall x \in \mathbb{F}) (\exists (x) \in \mathbb{F}) x + (x) = e \) (Existence of Additive Inverse)
(M) Axioms for Multiplication
(M1) \( (\forall x, y \in \mathbb{F})(x + y \in \mathbb{F}) \) (Closure under Multiplication)
(M2) \( (\forall x, y \in \mathbb{F})(xy = yx) \) (Commutativity of Multiplication)
(M3) \( (\forall x, y z \in \mathbb{F})[(xy)z = x(yz)] \) (Associativity of Multiplication)
(M4) \( (\exists f \in \mathbb{F}) (f \neq e \wedge (\forall x \in \mathbb{F}) fx = x) \)
(Existence of Nonzero Multiplicative Identity)
(M5) \( (\forall x \in \mathbb{F}) (x \neq e \implies (\exists x^{1} \in \mathbb{F}) x \cdot x^{1} = f) \)
(Existence of Multiplicative Inverse of Nonzero Members)
(D) The Distributive Axiom
\[ (\forall x, y, z \in \mathbb{F}) x(y + z) = xy + xz \]
If addition and multiplication have their customary meaning, then the field axioms hold in \(\mathbb{Q},\) and \(\mathbb{Q}\) is a field.
Note that \(x^{1},\) in (M5), is also denoted as \(1/x.\) And we usually omit the bracket in cases like \( (x + y) + z \) due to the associative property; however, the convention is that expressions associate to the left unless grouped otherwise by parentheses. For example, \(w + x + y + z\) denotes \( ((w + x) + y) + z .\) Although we will try to show the field axioms explicitly, redundant steps involving the commutative property (especially when applying the identity and inverse properties) will be omitted.
Definition 5. (Ordered Field) Ordered fields are more than fields with an order.
Consider the complex numbers \(\mathbb{C}\) as a counterexample: \(\mathbb{C}\) forms a field if we define addition and multiplication customarily, \(\mathbb{C}\) can also be ordered; however, the field it forms cannot be ordered due to the number \(1\) being the square of the imaginary unit (Since \(1 > 0\) is the multiplicative identity, we cannot have \(i^2 = 1\) and thus \(1 > 0\)).
An ordered field is a field \(\mathbb{F}\) which also can be ordered, written \( (\mathbb{F}, +, \cdot, 0, 1, <) ,\) in a way such that it that satisfies (on top of the order axioms and the field axioms) the following two ordered fields axioms.
(OF1) Addition:
\[ (\forall x, y, z \in \mathbb{F}) (y < z \implies x + y < x + z). \]
(OF2) Multiplication:
\[ (\forall x, y \in \mathbb{F}) ((x > 0 \wedge y > 0) \implies xy > 0). \]
If \(x > 0,\) we call \(x\) positive; if \(x < 0,\) we call \(x\) negative. nonnegative and nonpositive are the negations of positive and negative respectively, hence they are both inclusive to \(0.\)
We now denote \(\mathbb{Q}^+\) to be positive rational numbers \( \{p \in \mathbb{Q} : p > 0\} \)
and \(\mathbb{Q}^\) to be the negative rational numbers \( \{p \in \mathbb{Q} : p < 0\} .\)
We also say that an ordered field is complete if it satisfies the completeness axiom, which requires the leastupperbound property.
Definition 6. (Archimedean Property) An ordered field \((S, +, \cdot, e, f, <)\) is Archimedean or has the Archimedean Property if for all \(x, y \in S\) such that \(x > 0,\) there is an integer (n) such that (nx > y).
In Lemma 2 we will prove that \(\mathbb{Q}\) is Archimedean. Furthermore, we will show that \(\mathbb{R}\) is Archimedean in Theorem 4
3. Outline of the Construction.
We will construct the real numbers via Dedekind cuts, which are sets of rational numbers that are open below, like a ray on the number line. Essentially, a real number \(\alpha\) is represented by the cut \( (\infty, \alpha) \cap \mathbb{Q} ,\)^{2} the set of all rational numbers smaller than itself. For example, the set \(A\) from the first section is a Dedekind cut (see Figure 1). Then, we will verify that the set of Dedekind cuts \\(\mathbb{R}\\) is an ordered field that has the leastupperbound property.
It is also possible to construct real numbers through sequences of rational numbers using Cauchy sequences. Rudin began the first chapter of Principles of Mathematical Analysis by asking to what value does the sequence of rational numbers \[1, 1.4, 1.41, 1.414, 1.4142, \ldots\] really tend [Rud76, p.1]. An infinite decimal expansion that looks like \(n_0 . n_1 n_2 n_3\cdots,\) where \(0 \leq n_k < 10\) for all \(k \geq 1,\) could be approximated by of the set of partial decimal expansions where digits after a certain point and defined as its supremum. The number, expressed as an infinite decimal, is equal to \[\sum_{k = 0}^{\infty} \frac{n_k}{10^k}.\] Hence, putting \[E = \left\{\sum_{0 \leq k \leq m} \frac{n_k}{10^k} : m \in \mathbb{N} \right\},\] \(n_0 . n_1 n_2 n_3\cdots = \sup{E}.\) However, a notable problem with Cauchy sequences is that different decimal expansions can approach the same value. Consider the following sequences \[1, 1, 1, 1, 1, \ldots\] and \[0.9, 0.99, 0.999, 0.9999, 0.99999, \ldots.\] These two sequences both approach the value \(1.\) We resolve this issue by creating an equivalence class where two sequences are equal if the sequence constructed by subtracting corresponding terms of the those two sequences approaches \(0\) [Kra14, p.18 definition 3.17]. There is no issue with equivalence for Dedekind cuts as they are sets and their equivalence follows from the Zermelo–Fraenkel Axiom of Extentionality [Lia11, p.2].
Furthermore, the settheoretic construction of the real numbers fits in with the constructions of the number systems on which the real numbers’ construction is based. For instance, the set of natural numbers is the smallest inductive set and also the unique inductive set that is a subset of all other inductive sets. The integers are constructed from the natural numbers as the difference of two natural numbers. Since the differences are not unique to any given pair of numbers, an equivalence class was defined. Then, the rational numbers are constructed in much the same way: as ratios of integers (where, of course, the denominator has to be nonzero), and a similar equivalence class was constructed for distinct ratios that have the same value. The construction of real numbers follow the various other constructions: based on sets of rational numbers [Kra14, pp.3–12].
3.1. Defining a Cut.
We will now define Dedekind cuts rigorously. The members of \(\mathbb{R}\) will be certain subsets of \(\mathbb{Q},\) called cuts. A cut is defined to be any set \(\alpha \subset \mathbb{Q}\) with the following three properties.
Note: we will use the latin alphabet to represent rational numbers and greek letters to represent rational cuts.
(I). Nontrivial: \(\alpha\) is nonempty and \(\alpha \neq \mathbb{Q}.\)
(II). Open Below: \(\forall p \in \alpha, \forall q \in \mathbb{Q}, q < p \implies q \in \alpha.\)
(III). Has No Maximal Element: \(\forall p \in \alpha, \exists r \in \alpha : p < r.\)
Property (III) states that the cut \(\alpha\) has no largest member, not to be confused with a supremum; (II) implies that \(\alpha\) is not bounded below and the following two facts [Rud76, p.18]
 \(p \in \alpha \wedge q \notin \alpha \implies p < q\);
 \(r \notin \alpha \wedge r < s \implies s \notin \alpha.\)
3.2. The Leastupperbound Property.
With the definition of a cut, we can now define order on \(\mathbb{R}\) and show that it has the leastupperbound property.
For two cuts $\alpha, \beta \in \mathbb{R}$ define \[\alpha <_{\mathbb{R}}\beta :\Longleftrightarrow\alpha \subset \beta,\] ($\alpha$ is a proper subset). We now check to see that it follows the order axioms.
Let \(\alpha, \beta \in \mathbb{R}.\) To show that at least one of the three conditions of trichotomy hold, without loss of generality, assume that neither \(\alpha <_{\mathbb{R}} \beta\) nor \(\alpha = \beta,\) i.e. \(\alpha \nsubseteq \beta.\) Thus, there exists a \(q \in \alpha\) such that \(q \notin \beta\) (otherwise \(alpha \subseteq \beta,\) contradicting our assumption). By property (II), every element \(p \in \alpha\) satisfies \(p < q.\) And since \(q \in \beta,\) thus all \(p \in \beta.\) Therefore, \(\alpha \supset \beta\) (\(\alpha >_{\mathbb{R}} \beta\)).
Then, assume neither \(\alpha >_{\mathbb{R}} \beta\) nor \(\alpha <_{\mathbb{R}}\beta.\) If \(\alpha \neq \beta,\) then there must exist some element \(q \in \alpha\) such that \(q \notin \beta\) (or vice versa), which contradicts the assumption by property (II). Therefore, \(\alpha = \beta.\)
Now we want to show that \(\mathbb{R}\) has the leastupperbound property. Let \(A \subset \mathbb{R}\) be a nonempty subset of cuts, and let \(A\) be bounded above with an upper bound \(\beta \in \mathbb{R}.\) We need to show that \(A\) has a supremum \(\gamma\) in \(\mathbb{R}.\) Let \[\gamma = \bigcup_{\alpha \in A} \alpha\]
We will now verify that \(\gamma\) is indeed a bonafide cut. Since \(A\) is nonempty, there is a cut \(\alpha_0 \in A.\) Since \(\alpha_0\) is a cut, \(\alpha_0\) is not empty. Since \(\alpha_0 \subseteq \gamma,\) \(\gamma\) is also nonempty. Next, since \(\beta \geq_{\mathbb{R}} \alpha\) (\(\beta \supseteq \alpha\)) for all \(\alpha \in A,\) it is clear that \(\gamma \subseteq \beta,\) and therefore \(\gamma \neq \mathbb{Q}\) because \(\beta \neq \mathbb{Q}.\) Thus \(\gamma\) satisfies property (I). For any \(p \in \gamma,\) there is an \(\alpha_1 \in A\) such that \(p \in \alpha_1.\) As \(\alpha_1\) is a cut, \(q \in a_1 \subseteq \gamma\) for every \(q < p,\) and (II) follows. Again, since \(\alpha_1\) is a cut, there is an \(r \in \alpha_1\) such that \(r > p,\) and \(r \in \gamma.\) Thus, (III) follows, and we conclude that \(\gamma \in \mathbb{R}.\)
Now we need to verify that \(\gamma\) is the supremum of \(A.\) It is clear that \(\gamma\) is an upper bound since \(\alpha \leq_{\mathbb{R}} \gamma\) for all \(\alpha \in A.\) Then, suppose \(\delta <_{\mathbb{R}} \gamma,\) since \(\delta\) is a cut, there is an \(s \in \gamma\) such that \(s \notin \delta.\) But since \(s \in \gamma,\) \(s \in \alpha’\) for some \(\alpha’ \in A,\) and \(\delta <_{\mathbb{R}} \alpha’.\) Thus, \(\delta\) is not an upper bound of \(A.\) By the definition of the least upper bound, \(\gamma = \sup{A}.\) Therefore, \( (\mathbb{R}, <_{\mathbb{R}}) \) has the leastupperbound property.
3.3. Verifying the Field Axioms (Part 1).
We will now move on to gradually verifying that \(\mathbb{R}\) satisfies the field axioms. We define the addition of two cuts \(\alpha, \beta \in \mathbb{R}\) as \[\alpha +_{\mathbb{R}} \beta := \{r + s : r \in \alpha, s \in \beta\}.\]
Define \(0^* = \mathbb{Q}^,\) it is clear that \(0^*\) is a cut. Let \(0^*\) be the additive identity, and we shall verify that the Axioms for Addition hold in \(\mathbb{R}.\)
We will also find the Archimedean property of the rational numbers \(\mathbb{Q}\) helpful in the verification of (A5) on \(\mathbb{R}.\) We call an ordered field Archimedean if it has the Archimedean property. In Section 5, we demonstrated that the Archimedean Property of \(\mathbb{R}\) follows from its leastupperbound property.
Lemma 2. \(\Q\) is Archimedean.
Proof. Let \(x, y \in \mathbb{Q}\) such that \(x > 0.\) If \(y < 0,\) then the Archimedean property is automatically satisfied by all positive integers. Otherwise, if \(y \geq 0,\) put \(k = y/x,\) we know that \(k\) is a nonnegative rational number, which means it could be expressed as \(p/q\) where \(p\) is a nonnegative integer and \(q\) is a positive integer. Since \(q \geq 1,\) it follows that \[p = qk \geq k.\] Let \(n = p + 1,\) hence \(n > k.\) Since \(x > 0,\) it follows from (OF1) and (M5) of \(\mathbb{Q}\) that \[nx > n \frac{y}{x} \cdot x = y\] This proves the Archimedean property on \(\mathbb{Q}.\)
$\blacksquare$
The following proofs are due to Rudin [Rud76, pp.1819]
(A1) Closure Under Addition.
Let \(\gamma = \alpha +_{\mathbb{R}} \beta,\) and we will verify that \(\gamma\) is cut.
 (I) We shall first establish that \(\gamma\) is a nonempty proper subset of \(\mathbb{Q}.\) Since \(\alpha\) and \(\beta\) are cuts, there exist \(r \in \alpha\) and \(s \in \beta.\) It follows from the definition of addition that \(r + s \in \gamma,\) and hence \(\gamma\) is not empty. Furthermore, there also exist \(r’ \notin \alpha\) and \(s’ \notin \beta.\) Since \(r’ > r\) and \(s’ > s\) for all \(r \in \alpha\) and \(s \in \beta,\) \(r’ + s’ > r + s,\) and thus \(r’ + s’ \notin \gamma\) so \(\gamma \neq \mathbb{Q}.\) Therefore the cut \(\gamma\) is nontrivial.
 (II) Pick \(p \in \gamma.\) Then there must be some \(r \in \alpha\) and \(s \in \beta\) such that \(p = r + s.\) For all \(q < p,\) \(q  s < r,\) and thus \(q  s \in \alpha.\) Hence, \[q = (q  s) + s \in \gamma.\]
 (III) Taking \(p, r, s\) from above, since \(\alpha\) is a cut, there is a \(t \in \alpha\) such that \(t > r.\) Thus, \(p = r + s < t + s\) and \(t + s \in \gamma.\)
Therefore, \(\gamma = \alpha +_{\mathbb{R}} \beta \in \mathbb{R}.\)
(A2) Commutativity of Addition
This follows from the commutative property (A2) of \(\mathbb{Q}.\)
\[\alpha +_{\mathbb{R}} \beta = \{r + s : r \in \alpha, s \in \beta\}, \quad \beta +_{\mathbb{R}} \alpha = \{s + r : r \in \alpha, s \in \beta\},\] \[r, s \in \mathbb{Q}, r + s = s + r \implies \alpha +_{\mathbb{R}} \beta = \beta +_{\mathbb{R}} \alpha.\]
(A3) Associativity of Addition
Similar to the proof for (A2), this follows from the associative property (A3) of \(\mathbb{Q}.\)
(A4) Existence of Additive Identity
If \(r \in \alpha, s \in 0^* ,\) then \(r + s \in \alpha +_{\mathbb{R}} 0^* \) and \[r + s < r + 0 = r.\] Therefore, \(\alpha +_{\mathbb{R}} 0^* \subseteq \alpha.\) Pick \(p \in \alpha,\) then since \(\alpha\) has no largest member, there is an \(r \in \alpha\) such that \(r > p.\) It follows that \(p  r \in 0^* .\) Therefore, \[p = r + (p  r) \in \alpha +_{\mathbb{R}} 0^*,\] and \(\alpha \subseteq \alpha +_{\mathbb{R}} 0^* .\) We conclude that \(\alpha = \alpha +_{\mathbb{R}} 0^* .\)
(A5) Existence of Additive Inverse
Let \(\alpha \in \mathbb{R}\) and let \(\beta\) be the set of all \(p\) for which there exists an \(r > 0\) such that \( p  r \notin a\) (some rational number smaller than \(p\) fails to be in \(\alpha\)). \[\beta := \{p \in \mathbb{Q} : (\exists r \in \mathbb{Q}) p  r \notin \alpha\}.\] We will show that \(\beta\) is a cut.
 (I) Since \(\alpha \in \mathbb{R}\) is a cut, there is an \(s \in \mathbb{Q}\) such that \(s \notin \alpha.\) Put \(p = s  1.\) It follows that \(p  1 = s \notin \alpha,\) and hence \(p \in \beta.\) Thus, \(\beta\) is nonempty. Next, pick \(q \in \alpha.\) It is clear that \(q \notin \beta,\) and thus \(\beta \neq \mathbb{Q}\) and is nontrivial.
 (II) Let \(p \in \beta\) and pick \(r > 0\) such that \(p  r \notin \alpha.\) Then, for each \(q < p,\) \[q  r > p  r,\] thus \(q  r \notin \alpha\) and \(q \in \beta.\)
 (III) Using \(p\) and \(r\) from above, put \(t = p + r/2,\) then \(t > p.\) We have that \[t  \frac{r}{2} = p  r \notin \alpha,\] and thus \(t \in \beta.\)
We conclude that \(\beta \in \mathbb{R}.\)
Next, we wish to show that \(\alpha + \beta = 0^* .\) Notice that for each \(r \in \alpha\) and \(s \in \beta,\) \(s \notin \alpha,\) and thus \(r < s.\) Which means that \(r + s < 0\) and \(r + s \in 0^* .\) Hence, \(\alpha +_{\mathbb{R}} \beta \subseteq 0^* .\)
Then, for each \(v \in 0^* ,\) let \(w = v/2\) and \(w > 0.\) By the Archimedean property of \(\mathbb{Q}\), there exist an integer \(n\) such that \(nw \in \alpha.\) Pick \(n\) such that \((n + 1)w \notin \alpha.\) Let \(p = (n + 2)w,\) then \[p  w = (n + 2)w  w = (n + 1)w \notin \alpha,\] and \[nw + p = nw  (n+2)w = 2w = 2\Big(\frac{v}{2}\Big) = v.\] By the definition of the addition of cuts, \(v = nw + p \in \alpha +_{\mathbb{R}} \beta\) and therefore \(0^* \subseteq \alpha +_{\mathbb{R}} \beta.\)
Therefore, we conclude that \(\alpha +_{\mathbb{R}} \beta = 0^* \) and denote \(\beta\) as \(\alpha.\)
3.4. Verifying the Field Axioms (Part 2).
We will now move on to multiplication. We wish to define multiplication in a similar way to addition; however, since products of negative rationals are positive, we shall restrict ourselves to the set \(\mathbb{R}^+ = \{\alpha \in \mathbb{R} : \alpha >_{\mathbb{R}} 0^* \}\) first.
We will use the following facts throughout the demonstration. If \(\alpha \in \mathbb{R}^+,\) \(p \in \alpha\) for all \(p \leq 0,\) and there must be an \(r \in \alpha\) such that \(r \geq 0.\) And as \(\alpha\) has no largest member, there also exists an \(s \in \alpha\) such that \(s > r \geq 0\) (so \(s > 0\)).
For \(\alpha, \beta \in \mathbb{R}^+\) define \[\alpha \cdot_{\mathbb{R}} \beta = \alpha\beta := \{p : p \leq rs, r \in \alpha, s \in \beta\}.\] Hence, for any \(r \in \alpha, s \in \beta,\) \(rs \in \alpha\beta\) because \(rs \leq rs.\)
Define \(1^* = \{p \in \mathbb{Q} : p < 1\},\) it is clear that \(1^* \) is a cut. Let \(1^* \) be the multiplicative identity, and we shall verify that the Axioms for Multiplication and the distributive law hold in \(\mathbb{R}^+.\)
(M1) Closure Under Multiplication.
We will verify that for any \(\alpha, \beta \in \mathbb{R}^+,\) \(\alpha\beta\) is a cut in \(\mathbb{R}^+.\)
 (I) We shall first establish that \(\alpha\beta\) is a nonempty proper subset of \(\mathbb{Q}.\) Since \(\alpha\) and \(\beta\) are cuts, there exist \(r \in \alpha\) and \(s \in \beta.\) Hence \(rs \in \alpha\beta,\) and \(\alpha\beta\) is not empty. Furthermore, there also exist \(r’ \notin \alpha\) and \(s’ \notin \beta.\) Since \(r’ > r\) and \(s’ > s\) for all \(r \in \alpha\) and \(s \in \beta,\) \(r’s’ > rs,\) and thus \(r’s’ \notin \alpha\beta\) so \(\alpha\beta \neq \mathbb{Q}.\) Therefore the cut \(\alpha\beta\) is nontrivial.
 (II) If \(p \in \alpha\beta,\) there must be some positive rational numbers \(r \in \alpha\) and \(s \in \beta\) such that \(p \leq rs.\) It follows that for each \(q < p,\) \[q < p \leq rs,\] and thus \(q \in \alpha\beta.\)
 (III) Pick \(p \in \alpha\beta.\) If \(p \leq 0,\) pick any \(r \in \alpha\) and \(s \in \beta\) such that \(r > 0\) and \(s > 0,\) then \(rs \in \alpha\beta\) and \(p \leq 0 < rs.\) And if \(p > 0,\) there are positive rational numbers \(r \in \alpha\) and \(s \in \beta\) such that \(p \leq rs.\) and since \(\alpha\) is a cut, there is a \(t \in \alpha,\) \(t > r.\) Hence \(ts \in \alpha\beta,\) \[ts > rs \geq p.\]
Therefore, taking \(rs \in \alpha\beta\) from above, \(rs > 0,\) and so we conclude that \(\alpha\beta \in \mathbb{R}^+.\)
(M2) Commutativity of Multiplication.
This follows from the commutative property of \(\mathbb{Q},\) since \(rs = sr.\)
(M3) Commutativity of Multiplication.
Suppose \(\alpha, \beta, \gamma \in \mathbb{R}^+.\) Following the definition of multiplication, we have \[(\alpha\beta)\gamma = \{ p : p \leq qt \wedge q \in \alpha\beta \wedge t \in \gamma\},\] and \[\alpha(\beta\gamma) = \{ p : p \leq rq \wedge r \in \alpha \wedge q \in \beta\gamma\}.\]
Pick \(p \in (\alpha\beta)\gamma,\) then there exist positive rational numbers \(q \in \alpha\beta\) and \(t \in \gamma\) such that \(p \leq qt.\) Since \(q \in \alpha\beta,\) there are positive rational numbers \(r \in \alpha\) and \(s \in \beta\) such that \(q \leq rs.\) Therefore, we have \[p \leq qt \leq (rs)t = r(st)\] by the multiplicative associativity of \(\mathbb{Q}.\) It is clear that \(st \in \beta\gamma,\) and thus \(r(st) \in \alpha(\beta\gamma).\) And since \(p \leq r(st),\) \(p \in \alpha(\beta\gamma),\) whence \((\alpha\beta)\gamma \subseteq \alpha(\beta\gamma).\) We can prove \(\alpha(\beta\gamma) \subseteq (\alpha\beta)\gamma\) in the exact same manner by setting \(q \in \beta\gamma\) instead. Therefore, we conclude that \((\alpha\beta)\gamma = \alpha(\beta\gamma).\)
(M4) Existence of Nonzero Mutliplicative Identity.
Let \(\alpha \in \mathbb{R}^+,\) then \[1^* \cdot_{\mathbb{R}} \alpha = \big\{p \in \mathbb{Q} : s \in (0, 1) \wedge r \in \alpha \wedge r > 0 \wedge p \leq rs\big\}\].
Since for all positive rational numbers \(r \in \alpha\) and \(s \in 1^* ,\) \(s < 1,\) thus \(rs < r\) and so \(rs \in \alpha.\) It follows that \(1^* \alpha \subseteq \alpha.\)
Pick \(p \in \alpha.\) If \(p \leq 0,\) then it is clear that \(p \in 1^* \alpha\) since \((\cdot_{\mathbb{R}})\) is closed under \(\mathbb{R}^+.\) Otherwise, if \(p > 0,\) then since \(\alpha\) has no largest member, there is an \(r \in \alpha\) such that \(r > p.\) Thus \[0 < \frac{r}{p} < 1\] and \(r/p \in 1^* .\) It follows that \[p = \frac{r}{p} \cdot p \in 1^* \alpha,\] and hence \(\alpha \subseteq 1^* \alpha.\)
Therefore, we conclude that \(\alpha = 1^* \alpha.\)
The Evil Fifth Axiom.
Up until this point, the verifications of the previous nine axioms are striaghtforward, or even trivial. However, the proof that this axiom holds is deceptively difficult. It is hard to prove this directly, so this proof relies on a few contradictions, while tying together the concepts from the second section, to show that only the result can hold.
Fun fact about this article.
(M5) Existence of Nonzero Mutliplicative Inverse.^{3}
Suppose \(a \in \mathbb{R}^+.\) Let \(\beta\) be the set of all \(p\) such that either \(p \leq 0\) or \(p > 0\) and there is some rational number \(r > 0\) such that \(1/p  r \notin \alpha\) (some rational number smaller than \(1/p\) fails to be in \(\alpha\)). \[\beta := 0^* \cup \{0\} \cup \{p \in \mathbb{Q}^+ : (\exists r \in \mathbb{Q}^+)\tfrac{1}{p}  r \notin \alpha\}.\] We will show that \(\beta \in \mathbb{R}^+\) is a cut.
 (I) Since \(\alpha\) is a cut, there is an \(s \in \mathbb{Q}^+\) such that \(s \notin \alpha.\) Let \(p = 1/(2s).\) Thus \(p > 0,\) and \[\frac{1}{p}  s = s \notin \alpha.\] Thus, \(p \in \beta\) is nonempty and contains a positive element. Pick \(q \in \alpha\) such that \(q > 0.\) Clearly, \(1/q \notin \beta\) and hence \(\beta \neq \mathbb{Q}\) is nontrivial
 (II) Pick \(p \in \beta\) such that \(p > 0,\) then there is some rational number \(r > 0\) such that \(1/p  r \notin \alpha.\) For all \(q < p,\) if \(q \leq 0,\) then \(q \in \beta\) by the definition of \(\beta.\) Otherwise, if \(q > 0\) (\(0 < q < p\)), then it follows from \Cref{orderedfieldprop}\bracklet{e} that \[0 < \frac{1}{p} < \frac{1}{q}.\] Therefore, since \[\frac{1}{q}  r > \frac{1}{p}  r,\] it follows that \(1/q  r \notin \alpha\) and hence \(q \in \beta.\)
 (III) Using \(p > 0\) and \(r > 0\) from above, let \(t = p(pr + 1).\) Thus we have \[t = p(pr + 1) = p^2 r + p > p.\] Furthermore,
and since \(pr^2/(pr + 1) > 0,\) we have that \(t > p\) and \(t \in \beta.\)
From the proof of the first property we also found that \(\beta >_{\mathbb{R}} 0^* ,\) so we conclude that \(\beta \in \mathbb{R}^+.\)
Next, we wish to show that \(\alpha \cdot_{\mathbb{R}} \beta = 1^* .\) For each \(p \in \alpha\beta,\) it is trivial that \(p \in 1^* \) if \(p \leq 0.\) Otherwise, if \(p > 0,\) there are positive rational numbers \(r \in \alpha\) and \(s \in \beta\) such that \(p \leq rs.\) It is clear that \(1/s \notin \alpha,\) and thus \(1/s > r.\) Therefore, we have that \[p \leq rs < 1.\] And it follows that \(p \in 1^* ,\) whence \(\alpha\beta \subseteq 1^* .\)
However, showing that the converse, \(1^* \subseteq \alpha\beta,\) holds is of much greater difficulty. In fact, we will have to work through a few assumptions to show that the hypothesis is the only possible solution. First, assume the hypothesis is false, therefore \(1^* >_{\mathbb{R}} \alpha\beta,\) which implies the existence of some \(v \in 1^* \) such that \(v \notin \alpha\beta.\) It is clear that \(v \in (0, 1),\) and thus \(1/v > 1.\) We will define the following two sets \[\gamma := \Big\{\frac{p}{v} : p \in \alpha, p > 0 \Big\}, \quad \alpha’ := \alpha \cup \gamma\] We now verify that \(\alpha’\) is a cut.
 (I)Since \(\alpha\) is a cut that contains a positive element, \(\alpha’\) is nonempty. Pick \(s \notin \alpha,\) it is clear that \(s/v > s\) and thus \(s/v \notin \alpha,\) and that \(s \notin \gamma.\) Therefore, \(s/v \notin \alpha’\) and \(\alpha’ \notin \mathbb{Q}\) is nontrivial.
 (II) Pick \(p \in \alpha’.\) If \(p \in \alpha\) then it inherits property (II) from \(\alpha.\) Otherwise it must be the case that \(p \in \gamma,\) which implies that \(vp \in \alpha.\) For each \(q < p,\) we have that \(vq < vp\) and thus \(vq \in \alpha\) by (II) of \(\alpha\) again. Therefore, \(q \in \gamma,\) and hence \(q \in \alpha’.\)
 (III) Similarly, picking \(p \in \alpha’,\) if \(p \in \alpha,\) then it from (III) of \(\alpha.\) Otherwise \(p \in \gamma\) and \(vp \in \alpha.\) There is an \(r \in \alpha\) such that \(r > vp.\) Then, we have that \(r/v > p,\) \(r/v \in \gamma,\) and thus \(r/v \in \alpha’.\)
Therefore, \(\alpha’ \in \mathbb{R}\) is a cut. It is clear that \(\alpha’ \geq_{\mathbb{R}} \alpha\) since \(\alpha \subseteq \alpha’.\) However, we need that \(\alpha’ >_{\mathbb{R}} \alpha,\) so we assume that \(\alpha’ = \alpha\) to reach a contradiction.
Since the cut \(\alpha\) is bounded above, there exists an \(M \in \mathbb{Q}^+\) such that \(M\) is an upper bound of \(\alpha.\) Next, we see that if \(p \in \alpha\) such that \(p > 0,\) then \(p/v \in \alpha’.\) And since \(\alpha’ = \alpha,\) it also holds that \(p/v \in \alpha.\) Repeating this process, we obtain that \(p/v^n \in \alpha\) for any positive integer \(n.\) Put \[m = \Bigg\lceil \frac{M  p}{p(\tfrac{1}{v}  1)} \Bigg\rceil\] Since \(1/v > 1,\) \(1/v^n > 1\) and we obtain the following useful inequality. \[\frac{1}{v^n}  \frac{1}{v^{n1}} = \frac{1}{v^{n1}} \left( \frac{1}{v}  1 \right) > \frac{1}{v}  1 > 0.\] We can then apply this inequality to the following telescoping sum. \[\sum_{k = 1}^{n} \Big( \frac{1}{v^k}  \frac{1}{v^{k1}} \Big) = \frac{1}{v^n}  1 > n \Big( \frac{1}{v}  1 \Big)\] Applying this sum, we see that \[p \left( \frac{1}{v^m}  1 \right) > pm \left( \frac{1}{v}  1 \right).\] Therefore,
\[\begin{aligned} \frac{p}{v^m}  p & > \Bigg\lceil \frac{M  p}{p(\tfrac{1}{v}  1)} \Bigg\rceil \cdot p \Big( \frac{1}{v}  1 \Big) \\ & \geq \frac{M  p}{p(\tfrac{1}{v}  1)} \cdot p \Big( \frac{1}{v}  1 \Big) \\ & = M  p. \end{aligned}\]And thus, \(p/v^m > M.\) But \(p/v^m \in \alpha,\) so we have shown that \(\alpha\) cannot be bounded above by any \(M > 0.\) This contradicts the definition of \(\alpha\) as a cut. Therefore, \(\alpha’\) cannot equal \(\alpha,\) and it must be true that \(\alpha’ > \alpha.\) Hence, there is a \(p \in \alpha’\) such that \(p \notin \alpha,\) which implies that \(p \in \gamma\) and hence \(vp \in \alpha.\)
Since \(\alpha’\) is a cut, \(p\) cannot be its largest member and there is a \(q \in \alpha’\) such that \(q > p.\) It follows that \(q \notin \alpha,\) thus \(q \in \gamma,\) and hence \(vq \in \alpha.\) We then have that \[\frac{v}{vq} = \frac{1}{q} \notin \beta, \quad \frac{v}{vp} = \frac{1}{p} \notin \beta,\] since \(v \notin \alpha\beta.\)^{4} We see that \(p\) is an upper bound of the cut \(\alpha\) (since \(p \notin \alpha\)), but no rational number smaller than \(p\) is member of \(\alpha.\) Therefore, \(p\) is a supremum of the cut \(\alpha.\) The very same argument could be made to show that \(q\) is a supremum of \(\alpha.\) However, since \(q > p,\) \(q  p\) is a positive rational number, we see that \[p = q  (q  p) \notin \alpha,\] which contradicts the definition of \(p.\) In the demonstration above, \(p\) and \(q\) act like distinct least upper bounds (supremums) of \(\alpha,\) but this is not possible because there can only be one least upper bound. Therefore, our initial assumption must be wrong and it has to be the case that \(1^* \leq_{\mathbb{R}} \alpha\beta\) or, equivalently, \(1^* \subseteq \alpha\beta.\)
Therefore, we conclude that \(\alpha\beta = 1^* \) and denote \(\beta\) as \(\frac{1}{\alpha}.\)
Back to proving axioms. We have one left to prove on \(\mathbb{R}^+\): the distributive axiom.
(D) The Distrbibutive Axiom.
Let \(\alpha, \beta, \gamma \in \mathbb{R}^+\). We wish to show that \(\alpha \cdot_{\mathbb{R}} (\beta +_{\mathbb{R}} \gamma) = \alpha \cdot_{\mathbb{R}} \beta +_{\mathbb{R}} \alpha \cdot_{\mathbb{R}} \gamma\), relying on the fact that the distributive axiom holds for \(\mathbb{Q}\).
Pick \(p \in \alpha \cdot_{\mathbb{R}} (\beta +_{\mathbb{R}} \gamma)\). There exist \(r \in \alpha\) and \(q \in \beta +_{\mathbb{R}} \gamma\) such that \(p \leq rq\). Then, \(q = s + t\) for some \(s \in \beta\) and \(t \in \gamma\). We have that \[p \leq rq = r(s + t) = rs + rt.\] Since \(rs \in \alpha \cdot_{\mathbb{R}} \beta\), thus \(rt \in \alpha \cdot_{\mathbb{R}} \gamma\), \(p\) and \(rs + rt\) are members of \(\alpha \cdot_{\mathbb{R}} \beta +_{\mathbb{R}} \alpha \cdot_{\mathbb{R}} \gamma\), and hence \[\alpha \cdot_{\mathbb{R}} (\beta +_{\mathbb{R}} \gamma) \subseteq \alpha \cdot_{\mathbb{R}} \beta +_{\mathbb{R}} \alpha \cdot_{\mathbb{R}} \gamma\]
Next, pick \(p \in \alpha \cdot_{\mathbb{R}} \beta +_{\mathbb{R}} \alpha \cdot_{\mathbb{R}} \gamma\). There exist \(q_1 \in \alpha\beta\) and \(q_2 \in \alpha\gamma\) such that \(p = q_1 + q_2\). Consequently, there are numbers \(r_1, r_2 \in \alpha, s \in \beta, t \in \gamma\) such that \(q_1 \leq r_1 s\) and \(q_2 \leq r_2 t\). Put \[r = \max\{r_1, r_2\}.\] It is clear that \(r \in \alpha\). We have that \[p = q_1 + q_2 \leq r_1 s + r_2 t \leq rs + rt = r(s + t).\] Since \(s + t \in \beta +_{\mathbb{R}} \gamma\), thus \(p\) and \(r(s + t)\) are members of \(\alpha \cdot_{\mathbb{R}} (\beta +_{\mathbb{R}} \gamma)\), and hence \[\alpha \cdot_{\mathbb{R}} \beta +_{\mathbb{R}} \alpha \cdot_{\mathbb{R}} \gamma \subseteq \alpha \cdot_{\mathbb{R}} (\beta +_{\mathbb{R}} \gamma).\] And we conclude that \(\alpha \cdot_{\mathbb{R}} (\beta +_{\mathbb{R}} \gamma) = \alpha \cdot_{\mathbb{R}} \beta +_{\mathbb{R}} \alpha \cdot_{\mathbb{R}} \gamma\).
Completing Multiplication on ℝ.
Now we wish to extend the Axioms for Multiplication and Distribution, which we showed to hold in \(\mathbb{R}^+\), to the set of all cuts \(\mathbb{R}\).
We complete the definition of multiplication by defining multiplication with the additive identity, \[\alpha 0^* = 0^* \alpha = 0^*,\] and extending the definition of multiplication to negative real cuts.
\[\alpha\beta = \begin{cases} \alpha\beta & \alpha > 0^*, \beta > 0^* \\ (\alpha)(\beta) & \alpha < 0^*, \beta < 0^* \\ \left[ (\alpha)\beta \right] & \alpha < 0^*, \beta > 0^* \\ \left[ \alpha(\beta) \right] & \alpha > 0^*, \beta < 0^* \end{cases}\]Since the products on the right are all contained in (\mathbb{R}^+), the multiplication axioms (M) that we proved holds in (\mathbb{R}^+) can be extended to (\mathbb{R}) using the identity (\gamma = (\gamma)) repeatedly from \Cref{fieldadd}\bracklet{d}, which relies on the proofs for the Axioms for Order and Addition. Since we have already shown that those axioms hold on \(\mathbb{R}\), this identity could be used.
The distributive law, which we already know to hold in (\mathbb{R}^+), could be broken into cases and proven using the extended definition of multiplication on (\mathbb{R}).
In conclusion, We have now completed the proof that \((\mathbb{R}, +_{\mathbb{R}}, \cdot_{\mathbb{R}}, 0^*, 1^*, <_{\mathbb{R}})\) is an ordered field with the leastupperbound property (it is complete).
4. ℚ as a Subfield of ℝ.
We associate every rational number \(r \in \mathbb{Q}\) with the cut \(r^* = \{p \in \mathbb{Q} : p < r\}\) which we refer to as a rational cut. It is clear that \(r^* \in \mathbb{R}\) for all \(r \in \mathbb{Q}\). Let \(\mathbb{Q}^* \) be the set of all rational cuts \(r^* \in \mathbb{R}\). There is a bijective mapping from \(\mathbb{Q}\) to \(\mathbb{Q}^* \), and we see from the theorem below that these rational cuts preserve sums, products, and order. In other words, \(\mathbb{Q}\) is isomorphic to the ordered field \(\mathbb{Q}^* \).
Theorem 3. \(\mathbb{Q}^* \) is isomorphic to \(\mathbb{Q}\). In other words, \(\mathbb{Q}^* \) preserves the arithmetic properties of addition, multiplication, and order in \(\mathbb{Q}\); it satisfies the following three properties.
 \(r^* +_{\mathbb{R}} s^* = (r + s)^* \)
 \(r^* \cdot_{\mathbb{R}} s^* = (rs)^* \)
 \(r^* <_{\mathbb{R}} s^* \iff r < s\)
Proof. Choose \(p \in r^* +_{\mathbb{R}} s^* \), then there exist \(u \in r^* \) and \(v \in s^* \) such that \(p = r + s\). It also follows that \(u < r\) and \(v < s\), hence \(p < r + s\) and \(p \in (r + s)^* \). Thus \(r^* +_{\mathbb{R}} s^* \subseteq (r + s)^* \).
Conversely, suppose \( p \in (r + s)^* \), then put \[2t = r + s  p > 0.\] Let \[r’ = r  t, \quad s’ = s  t,\] Then \(r’ \in r*, s’ \in s*\) and \[r’ + s’ = r + s  2t = p.\]
Therefore, \(p \in r^* +_{\mathbb{R}} s^* \). Thus \((r + s)^* \subseteq r^* +_{\mathbb{R}} s^* \). We conclude that \(r^* +_{\mathbb{R}} s^* = (r + s)^* \). This proves (1).
Similarly, for positive \(r^* \) and \(s^* \) choose \(p \in r^* \cdot_{\mathbb{R}} s^* \), then there exist \(u < r, s < v\) such that \(p \leq uv\), hence \(p \leq uv < rs\) and \(p \in (rs)^* \). Thus \(r^* \cdot_{\mathbb{R}} s^* \subseteq (rs)^* \)
And Conversely, suppose \(p \in (rs)^* \). Hence \(p < rs\), which implies that \(prs < r^2 s^2\), then we divide both sides by \(r^2\) or \(s^2\) and take the square root to obtain the following inequalities. Let \[r’ = \frac{\sqrt{prs}}{s} < r, \quad s’ = \frac{\sqrt{prs}}{2} < s.\] Then we see that \[p = \frac{prs}{rs} = r’s’ < rs,\] and \(p \in r^* \cdot_{\mathbb{R}} s^* \). We use the multiplication completion definitions for nonpositive multiplication of rational cuts. Therefore, \((rs)^* \subseteq r^* \cdot_{\mathbb{R}} s^* \). We conclude that \(r^* \cdot_{\mathbb{R}} s^* = (rs)^* \). This proves (2).
If \(r < s\) then \(r \in s^* \), but \(r \notin r^* \), therefore \(r^* <_{\mathbb{R}} s^* \). Conversely, If \(r^* <_{\mathbb{R}} s^* \), then there exists \(p \in s^* \) such that \(p \notin r^* \), and \(r \leq p < s\). Hence \(r < s\). This proves (3).
$\blacksquare$
This identification of \(\mathbb{Q}\) with \(\mathbb{Q}^* \) allows us to regard \(\mathbb{Q}\) as a subfield of \(\mathbb{R}\). They are isomorphic, allowing us to embed and denote rational numbers within the real number system. Of course, the rational numbers \(r \in \mathbb{Q}\) is by no means same as the cuts \(r^* \in \mathbb{Q}^* \) which are sets containing rational numbers, but the arithmetic and order properties that we are concerned with are the same in the two fields.
Note that this phenomenon also occurs, although much more trivially, when the real numbers \(\mathbb{R}\) are regarded as a subfield of the complex numbers \(\mathbb{C}\) and when the integers are identified with certain subsets of \(\mathbb{Q}\). Also, any two ordered fields with the least upper bound property are isomorphic, but the proof of that fact is outside the scope of this article. From now on, further discussions of real numbers will not be denoted by the subscripts (\(+_{\mathbb{R}}, \cdot_{\mathbb{R}}\), and \(<_{\mathbb{R}}\)) and the asterisk denoting rational cuts such as \(0^*\) and \(1^* \).
5. Properties of ℝ.
Earlier, in Lemma 2, we showed that \(\mathbb{Q}\) has the Archimedean property. We will now show that the completeness of \(\mathbb{R}\) also implies the Archimedean property. Theorems 4 and 5 are inspired and originally given by Rudin [Rud76, p.9 Theorem 1.20].
Theorem 4. An ordered field with the leastupperbound property is Archimedean.
Proof. Let \((\mathbb{F}, +, \cdot, 0, 1, <)\) be an ordered field with the leastupperbound property. We wish to show that \(\mathbb{F}\) is Archimedean. We need to show that for all \(x, y \in \mathbb{F}\) with positive \(x\), there exists a positive integer \(n \in \mathbb{Z}^+\) such that \(nx > y\). Define \[A = \{nx : n \in \mathbb{Z}^+\},\] and assume that \(\mathbb{F}\) does not have the Archimedean property. Then, there would be no \(n \in \mathbb{Z}^+\) such that \(nx > y\), and \(y\) would be an upper bound of the set \(A\). Since \(A\) has an upper bound in \(\mathbb{F}\), the leastupperbound property of \(\mathbb{F}\) implies that \(A\) would also have a leastupperbound in \(\mathbb{F}\). Let \[\alpha = \sup{A}.\] (Note that \(\alpha \in \mathbb{F}\) does not represent a cut.) Since \(x > 0\), we have that \(\alpha  x < \alpha\). However, since \(n + 1 \in \mathbb{Z}^+\) for all \(n \in \mathbb{Z}^+\), we have for all \(n\), \[(n + 1)x < \alpha,\] and hence \[nx < \alpha  x.\] It follows that \(\alpha  x\) is also an upper bound of \(A\), contradicting the definition of \(\alpha\). Therefore, \(\mathbb{F}\) is Archimedean.
$\blacksquare$
Corollary. The real numbers \(\mathbb{R}\) is Archimedean.
Since \(\mathbb{R}\) is an ordered field with the leastupperbound property, we can apply Theorem 4. Next, we will look at an application of the Archimedean property of \(\mathbb{R}\).
Theorem 5. \(\mathbb{Q}\) is dense in \(\mathbb{R}\).
Topologically speaking, with a metric \(d(x, y) = x  y\), \(\mathbb{Q}\) is dense in \(\mathbb{R}\) if every point in \(\mathbb{R}\) is a point in \(\mathbb{Q}\) or a limit point in \(\mathbb{Q}\) (it is in the closure of \(\mathbb{Q}\)). It suffices to show that between every two distinct real numbers \(x, y \in \mathbb{R}\) is a rational number for reasons we will demonstrate after the proof.
Proof. Since \(x \neq y\), without loss of generality, put \(x < y\). Therefore, \(y  x > 0\), and by the Archimedean property of \(\mathbb{R}\), there is a positive integer \(n \in \mathbb{Z}^+\) such that \(n(y  x) > 1\), from which we have \[ny > nx + 1.\] Again, by the Archimedean property, there is a positive integers \(p \in \mathbb{Z}^+\) such that \(p > nx\). Hence, we have \[p < nx < nx + 1\] We want to find an integer \(m\) such that \(nx < m \leq nx + 1\). We know that this number is in the interval \([p, nx + 1] \cap \mathbb{Z}\), and we could find it explicitly by setting \(m\) as the maximum (largest member) of a finite set. \[m = \max\{k \in \mathbb{Z} : p \leq k \leq nx + 1\}.\] Combining the inequalities, we obtain \[nx < m \leq nx + 1 < ny,\] so that \[x < \frac{m}{n} < y.\] It is clear that \(m/n\) (\(n > 0\)), the ratio of two integers, is a rational number and the proof is complete.
$\blacksquare$
To elaborate on the sufficiency of this proof on the denseness of \(\mathbb{Q}\) in \(\mathbb{R}\), we shall show that every real number is either a point in \(\mathbb{Q}\) or a limit point of \(\mathbb{Q}\) using Theorem 5. For any \(x \in \mathbb{R}\), the hypothesis is satisfied automatically if \(x \in \mathbb{Q}\). But either way, we could show that \(x\) is a limit point of \(\mathbb{Q}\) by inductively constructing a sequence of rational terms \((s_n)_{n \in \mathbb{Z}^+}\) such that \(s_n \to x\). We will be constructing a sequence that approaches \(x\) from the left side of the number line. By Theorem 5, there exists an \(s_1 \in \mathbb{Q}\) such that \(s_1 \in (x  1, x)\). Then, if \(s_1, \ldots, s_{n1}\) have been chosen, put \(x  1/n < s_n < x\) again by Theorem 5. We now see that given any \(\epsilon > 0\), let \(N_\epsilon = \lceil 1/\epsilon \rceil\), then for any \(n \geq N\), \[s_n  x \leq \Big x  \frac{1}{n}  x \Big = \frac{1}{n} \leq \frac{1}{N_\epsilon} = \epsilon.\] Therefore, \(s_n \to x\) and \(x\) is a limit point of \(\mathbb{Q}\).^{5}
A Second Round of Dedekind Cuts?
One more problem which one would ponder after constructing the real field \(\mathbb{R}\) using the method of Dedekind cuts, or Dedekind completion, is what happens if we attempt to apply the Dedekind completion but start with a different ordered field \(\mathbb{F}\) instead of \(\mathbb{Q}\)? More specifically, what happens when we try to complete the field \(\mathbb{R}\) that we just constructed again? Through Dedekind completion, we construct a set \(\mathbb{R}^* \) of all cuts (which are now subsets of \(\mathbb{R}\)) that satisfy the properties of Dedekind cuts.
It is clear that each \(r \in \mathbb{R}\) could be mapped to an \(r^* \in \mathbb{R}^* \) where \[r^* = (\infty, r) = \{x \in \mathbb{R} : x < r\}.\] And since \(\mathbb{R}\) satisfies the axiom of completeness, meaning it has the leastupperbound property, we have that for each cut \(\alpha \in \mathbb{R}\), there exists one and only one \(r \in \mathbb{R}\) such that \(r = \sup{\alpha}\). Property (II) of a Dedekind cut implies that \(\alpha = r^* \). \(\alpha\), or \(r^* \), is also unique to \(r\), because any other cut \(\beta\) sharing the same supremum would also be equal to \(r^* \) under the same argument. Hence, each \(r \in \mathbb{R}\) corresponds to an unique \(r^* \in \mathbb{R}^* \). Therefore, the mapping \(\mathbb{R} \to \mathbb{R}^* \) is a bijection.
Next, if we define order and the operations of addition and multiplication in the same manner as we did for the construction of \(\mathbb{R}\), we find that the arguments in Theorem 3 still hold for \(\mathbb{R}\) and \(\mathbb{R}^* \) (since they do not depend on the values being rational).^{6} Therefore, we conclude that a Dedekind completion of \(\mathbb{R}\), which we denoted as \(\mathbb{R}^* \), is isomorphic to \(\mathbb{R}\) itself up to isomorphism.
Since \(\mathbb{R}\) is complete, completing \(\mathbb{R}\) again only yields an ordered field isomorphic to \(\mathbb{R}\) (that is also complete). Hence, \(\mathbb{R}\) has achieved the goal on which we set out: to fill in all the “gaps” in \(\mathbb{Q}\). It turns out that there are no more “gaps” to be filled in \(\mathbb{R}\), which is of critical importance to the foundation of calculus on real numbers.
Thanks for reading! This is still a work in progress. If you spot any errors or have any suggestions, I would really appreciated it if you contacted me
References
Notes
 Axioms are not proven. When we say “prove an axiom for X,” we mean to verify that an axiom holds in some object and/or definition X.

Figures 14 generated using $\LaTeX$
tikz
andpgfplots
. ↩ 
Intervals describe real numbers by default, although we have not constructed them yet. This expression of a cut serves to provide a little bit of intuition from familiar concepts to the reader. ↩

The naming for this axiom (M5) is rather inept. It states that there is a multiplicative inverse for every nonzero term, i.e. every term that is not equal to the additive identity. ↩

If the contrary were true, for example, \(1/p \in \alpha\) would imply that \(1/p \cdot vp = v \in \alpha\beta,\) contradicting the definition of \(v.\) The same could be said for \(1/q.\) ↩

The epsilonic inequalities of the form \(< \varepsilon\) and \(\leq \varepsilon\) are equivalent since \(\varepsilon> 0\) is arbitrary, \(x < \varepsilon\) implies \(x \leq \varepsilon,\) and conversely \(x \leq \varepsilon\) implies \(x < 2\varepsilon.\) ↩

To apply part (2) of the proof of Theorem 3 to \(\mathbb{R},\) we first need to rigorously define the square root of any real number, for which one should refer to [Rud76, pp.1011 Theorem 21]. ↩