The Real Numbers ($\mathbb{R}$)

Introduction

The real numbers are the order completion of the rationals: the unique complete ordered field, obtained by filling the Dedekind cuts of $\mathbb{Q}$ and, equivalently, as the metric completion of $\mathbb{Q}$ for the usual absolute value. The two constructions agree, and the result is unique: any two complete ordered fields are isomorphic by a unique order-preserving isomorphism, so $\mathbb{R}$ is determined by its order and its field structure together, not by either alone. The least upper bound property, the existence of the supremum of every nonempty set that is bounded above, is the single axiom that separates $\mathbb{R}$ from $\mathbb{Q}$, and completeness forces the field to be real closed.

This article carries out the completion of $\mathbb{Q}$, records the complete ordered field and its suprema, proves the uniqueness of the complete ordered field, and describes its real closedness and its subfields. The automorphisms and derivations of $\mathbb{R}$ belong to Real Automorphisms and Derivations, its topology to Real Topology, and the analytic theory of functions on it to Real Analysis and Real Integration. The construction carried out here is algebraic and order-theoretic: the order completion and the least upper bound property need no distance. The metric reading of the same construction — the Cauchy completion, metric completeness and the topology it induces — is the topological development of Real Topology, and is named here only as the equivalent construction that completes the ordered field. Real closedness and the uniqueness of the complete ordered field are developed from completeness in Real-Closed and Complete Ordered Fields and are used here.

The Completion of $\mathbb{Q}$

Dedekind Cuts

Definition. A Dedekind cut of $\mathbb{Q}$ is a subset $A \subseteq \mathbb{Q}$ such that

(C1) $A \neq \emptyset$ and $A \neq \mathbb{Q}$;

(C2) if $a \in A$ and $q \in \mathbb{Q}$ with $q < a$, then $q \in A$;

(C3) $A$ has no greatest element.

The set of all cuts is denoted $\mathcal{R}$.

Theorem. With the order $A \leq B \iff A \subseteq B$ and the operations

$$ A + B = \{a + b : a \in A,\ b \in B\}, \qquad A \cdot B = \{ab : a \in A,\ b \in B,\ a, b \geq 0\} \cup \{q \in \mathbb{Q} : q < 0\} \ (\text{for } A, B \geq 0), $$

and with negative elements handled by sign, $\mathcal{R}$ is an order-complete ordered field. The map $q \mapsto \{a \in \mathbb{Q} : a < q\}$ embeds $\mathbb{Q}$ as an ordered subfield, and $\mathcal{R}$ is the order completion of $\mathbb{Q}$.

Proof sketch. $\mathcal{R}$ is totally ordered by inclusion, since cuts are downward closed subsets of $\mathbb{Q}$. A nonempty family of cuts that is bounded above has a supremum: every upper bound is a cut and hence is not all of $\mathbb{Q}$, the family is contained in such a bound, so the union of the family is a downward closed subset of $\mathbb{Q}$ that is neither empty nor all of $\mathbb{Q}$, and it has no greatest element because no member of the family has one; hence the union is a cut, and it is the least upper bound. A nonempty family that is bounded below has an infimum, namely the intersection of the family with the greatest element removed if the intersection has one: the intersection is downward closed and nonempty because it contains a lower bound, and a cut contained in every member of the family has no greatest element, so it lies in the intersection with the greatest element removed, which is therefore the largest such cut. Addition of cuts is well defined and makes $\mathcal{R}$ an ordered abelian group whose zero is the cut of negative rationals, and multiplication is defined on nonnegative cuts by the display and extended by the sign rule; the field axioms are verified case by case, distributivity being the only substantive one. The embedding $q \mapsto \{a < q\}$ preserves the order and the operations, and the least upper bound property is the sup-of-unions statement.

Remark. The cut construction is the one explained in Real-Closed and Complete Ordered Fields as the order completion $\widehat{F}$ of an Archimedean ordered field $F$, applied to $F = \mathbb{Q}$.

Cauchy Sequences

Definition. A sequence $(a_n)$ in $\mathbb{Q}$ is Cauchy if for every rational $\epsilon > 0$ there is $n_0$ with $\lvert a_n - a_m \rvert < \epsilon$ for all $n, m \geq n_0$. Two Cauchy sequences are equivalent if $\lvert a_n - b_n \rvert \to 0$, and the completion is

$$ \mathcal{C} = \{\text{Cauchy sequences in } \mathbb{Q}\}/\sim . $$

Theorem. With termwise addition and multiplication, $\mathcal{C}$ is an order-complete ordered field, and the map $q \mapsto (q, q, q, \dots)$ embeds $\mathbb{Q}$ as an ordered subfield; $\mathcal{C}$ is the metric completion of $\mathbb{Q}$ for the usual absolute value, and it is complete as a metric space.

Proof. Termwise sums and products of Cauchy sequences are Cauchy, and the operations respect the equivalence relation, so $\mathcal{C}$ is a commutative ring with $1$; it is a field because a Cauchy sequence that does not converge to $0$ is eventually bounded away from $0$ in absolute value, so a sufficiently far tail can be inverted termwise, producing an inverse class. Completeness of $\mathcal{C}$ is proved by the diagonal argument on Cauchy sequences of Cauchy sequences, and order-completeness by identifying each class with the cut of rationals below it.

Agreement

Theorem. The two constructions produce isomorphic ordered fields, canonically: the map

$$ \Phi : \mathcal{C} \to \mathcal{R}, \qquad \Phi[(a_n)] = \left\{q \in \mathbb{Q} : q + \epsilon < a_n \text{ for all sufficiently large } n \text{ and some rational } \epsilon > 0\right\}, $$

is an order-preserving field isomorphism.

Proof sketch. The set displayed is a cut: it is nonempty and not all of $\mathbb{Q}$ because $(a_n)$ is Cauchy and hence bounded; it is downward closed, and it has no greatest element, because $q$ lies in it with gap $\epsilon$ precisely when $q + \epsilon/2$ lies in it with gap $\epsilon/2$; and it depends only on the class of $(a_n)$, since two equivalent Cauchy sequences are eventually within any prescribed rational distance of one another. The element $q$ lies in $\Phi[(a_n)]$ exactly when $q < a$, where $a$ is the limit of $(a_n)$ in the complete field $\mathcal{C}$: if $q < a$ then eventually $a_n \geq q + \epsilon$ for a rational $\epsilon$ with $q + \epsilon < a$, and conversely $q + \epsilon < a_n$ eventually gives $q < a$. Conversely every cut arises from a Cauchy sequence, for instance from an increasing sequence of rationals converging to the cut, which exists by density. The map is additive and multiplicative by the arithmetic of cuts, and it preserves the order by construction.

Corollary. Any two order-complete ordered fields are isomorphic by a unique order-preserving isomorphism, by Real-Closed and Complete Ordered Fields; hence $\mathcal{R} \cong \mathcal{C} \cong \mathbb{R}$ and $\mathbb{R}$ is unique up to a unique ordered-field isomorphism.

Theorem. $\mathbb{R}$ is the completion of $\mathbb{Q}$ at the Archimedean place: it is the metric completion of $\mathbb{Q}$ for $\lvert \cdot \rvert_\infty$ and the order completion of $\mathbb{Q}$ for its unique order. The other completions of $\mathbb{Q}$ are the fields $\mathbb{Q}_p$, and no $\mathbb{Q}_p$ is isomorphic to $\mathbb{R}$ as a field.

Proof. The two completions of $\mathbb{Q}$ that give $\mathbb{R}$ are the theorems of the first section and of Real-Closed and Complete Ordered Fields; the classification of completions is Ostrowski's theorem from Absolute Values, Valuations and Completions. Finally $\mathbb{R}$ is real closed and hence formally real, while $\mathbb{Q}_p$ is not formally real, so no field isomorphism can exist.

Remark (a warning about "the" completion). A field can have several inequivalent completions, and a field isomorphism between two completions need not respect the topology or the valuation. As abstract fields, $\mathbb{C}$ and the completed algebraic closure $\mathbb{C}_p$ of $\mathbb{Q}_p$ are isomorphic, because both are algebraically closed fields of characteristic $0$ with transcendence degree $2^{\aleph_0}$ and hence fall in the same Steinitz class of Algebraically Closed Fields; they are not isomorphic as topological or as valued fields. The real numbers, by contrast, are rigid in both senses: they have no nontrivial field automorphism, so no such phenomenon occurs for $\mathbb{R}$.

The Complete Ordered Field

Definition. A field $F$ carrying a total order compatible with its operations is an ordered field; it is complete, or order-complete, if every nonempty subset of $F$ that is bounded above has a least upper bound in $F$.

Theorem. $\mathbb{R}$ is a complete ordered field, and it is the unique one up to a unique order-preserving isomorphism.

The least upper bound property — the existence of the supremum of every nonempty set that is bounded above — is the single axiom that distinguishes $\mathbb{R}$ from $\mathbb{Q}$, and it is the source of every further property of the line. It is developed for its own sake, with the Archimedean property and the density of the rationals, in Suprema and Order-Completeness below; the uniqueness of the complete ordered field is the subject of the next section, and real closedness, which follows from completeness, is treated after that.

Theorem. $\mathbb{R}$ is uncountable, of cardinality $$ \lvert \mathbb{R} \rvert = 2^{\aleph_0}, $$

and the set of irrationals is uncountable, of the same cardinality; in particular $\mathbb{R}$ is neither countable nor discrete.

Proof. $\mathbb{R}$ has a Hamel basis of cardinality continuum, or directly: $\mathbb{R}$ is uncountable by Cantor's diagonal argument applied to the interval $(0,1)$, it has cardinality at most $2^{\aleph_0}$ as the set of Dedekind cuts of $\mathbb{Q}$, which is a subset of the power set of $\mathbb{Q}$, and at least $2^{\aleph_0}$ because the subsets of $\mathbb{N}$ inject into $\mathbb{R}$ by mapping a subset to the real whose ternary expansion has digit $2$ exactly at the positions in that subset. The irrationals are $\mathbb{R} \setminus \mathbb{Q}$, the complement of a countable set in an uncountable one.

Uniqueness of the Complete Ordered Field

The Order Type of $\mathbb{R}$

Definition. A linear order is dense if between any two distinct points there is a third, without endpoints if it has neither least nor greatest element, and complete if every nonempty subset that is bounded above has a least upper bound.

Theorem (Cantor). Any two countable dense linear orders without endpoints are order-isomorphic. In particular every such order is isomorphic to $\mathbb{Q}$.

Proof. The back-and-forth argument enumerates both orders as $(a_n)$ and $(b_n)$ and builds an increasing bijection by alternating the extension of the partial isomorphism on the side with the smaller index, using density and the absence of endpoints to place each new point.

Theorem. Any two complete dense linear orders without endpoints that contain a countable dense subset are order-isomorphic; in particular any such order is isomorphic to $\mathbb{R}$.

Proof. The same back-and-forth construction on the countable dense subsets, with completeness and density extending the partial isomorphism to a map of the whole order: given $a$ in the first order, send it to the supremum in the second order of the images of the dense elements below $a$, which exists by completeness and is strictly increasing by density.

Corollary. The order type of $\mathbb{R}$ is the unique complete dense order type without endpoints that has a countable dense subset; $\mathbb{Q}$ is the unique countable dense order type without endpoints, and it is the dense subset of $\mathbb{R}$ in the above.

The Archimedean Criterion for Embedding into $\mathbb{R}$

Theorem. An ordered field $F$ admits an order-preserving field embedding into $\mathbb{R}$ if and only if $F$ is Archimedean, and the embedding is then unique.

Proof sketch. If $F$ embeds in $\mathbb{R}$ it is Archimedean, because $\mathbb{R}$ is. Conversely, if $F$ is Archimedean then the map

$$ \varphi : F \to \mathbb{R}, \qquad \varphi(a) = \sup\{q \in \mathbb{Q} : q < a\}, $$

is well defined because $\{q \in \mathbb{Q} : q < a\}$ is nonempty and bounded above in $F$, hence bounded by a real after the identification of the rationals; it is order-preserving by construction, additive because the rationals below $a + b$ are exactly the sums of rationals below $a$ and below $b$ up to arbitrarily small error, and multiplicative by the analogous squeezing argument using the Archimedean property; it fixes $\mathbb{Q}$ and is therefore the unique such embedding.

Corollary. $\mathbb{R}$ contains an isomorphic copy of every Archimedean ordered field, and it is the order-complete field generated by $\mathbb{Q}$; a non-Archimedean ordered field, such as $\mathbb{Q}(t)$ with $t$ infinite, admits no order-preserving embedding into $\mathbb{R}$, though its underlying field may still embed by a non-order-preserving map.

Real Closedness

Real Closedness and the Algebraic Closure

Definition. A field $F$ is real closed if every positive element of $F$ is a square in $F$ and every polynomial of odd degree over $F$ has a root in $F$.

Theorem. $\mathbb{R}$ is real closed, and it is not algebraically closed. Its algebraic closure is $\mathbb{C} = \mathbb{R}(i)$ of degree $2$ over $\mathbb{R}$, and every polynomial over $\mathbb{R}$ factors into linear factors together with irreducible quadratic factors.

Proof. Every non-negative real $a$ has a square root, by completeness, so every positive element of $\mathbb{R}$ is a square. A polynomial of odd degree over $\mathbb{R}$ has a real root, because its values at $+\infty$ and $-\infty$ have opposite signs and it is continuous; this is the intermediate value theorem of Real Analysis, and its algebraic content is the real closedness of $\mathbb{R}$. A real-closed field $F$ that is not algebraically closed becomes algebraically closed on adjoining a square root of $-1$: the field $F(i)$ with $i^2 = -1$ is algebraically closed, so the algebraic closure of $\mathbb{R}$ is $\mathbb{R}(i) = \mathbb{C}$, of degree $2$.

The Real Algebraic Numbers and Transcendence

Definition. The real algebraic numbers are the elements of $\mathbb{R}$ that are algebraic over $\mathbb{Q}$, that is, the set $\overline{\mathbb{Q}} \cap \mathbb{R}$.

Theorem. The set $\overline{\mathbb{Q}} \cap \mathbb{R}$ is a countable subfield of $\mathbb{R}$, it is real closed, and it is not order-complete. The field $\mathbb{R}$ has characteristic $0$ and prime field $\mathbb{Q}$, and its transcendence degree over $\mathbb{Q}$ is

$$ \operatorname{tr.deg}_{\mathbb{Q}} \mathbb{R} = 2^{\aleph_0}. $$

Proof. The elements algebraic over $\mathbb{Q}$ form a subfield, closed under sums, products and inverses. It is countable, being the union over $n$ of the finite sets of roots of the degree-$n$ polynomials with rational coefficients. It is real closed: a positive algebraic $a$ has a square root in the real-closed field $\mathbb{R}$, and that root satisfies the quadratic $X^2 - a$ over $\overline{\mathbb{Q}} \cap \mathbb{R}$, hence is algebraic over $\mathbb{Q}$; and a polynomial of odd degree over $\overline{\mathbb{Q}} \cap \mathbb{R}$ has a root in $\mathbb{R}$ which is algebraic over $\overline{\mathbb{Q}} \cap \mathbb{R}$, hence over $\mathbb{Q}$. It is not order-complete, since the set of its elements with square less than $2$ is bounded above in it but has no least upper bound in it, its supremum in $\mathbb{R}$ being the irrational $\sqrt2$. The field $\mathbb{R}$ has characteristic $0$, its prime field being the image of $\mathbb{Q}$; and $\operatorname{tr.deg}_{\mathbb{Q}} \mathbb{R} = 2^{\aleph_0}$, because $\mathbb{R}$ has cardinality $2^{\aleph_0}$ while every extension of $\mathbb{Q}$ generated by fewer than $2^{\aleph_0}$ elements is of smaller cardinality.

Remark. Characteristic $0$ makes $\mathbb{R}$ a perfect field, so every algebraic extension of $\mathbb{R}$ is separable. Combined with real closedness, it also makes the order the only order compatible with the field structure, since the positive elements are exactly the nonzero squares. This is why $\mathbb{R}$ is not isomorphic as an ordered field to any non-Archimedean completion $\mathbb{Q}_p$ of $\mathbb{Q}$.

Subfields and the Real Closure

Theorem. Every subfield $F \subseteq \mathbb{R}$ is ordered by the restriction of the order of $\mathbb{R}$, and this order is Archimedean. Let

$$ M = \{a \in \mathbb{R} : a \text{ is algebraic over } F\}. $$

Then $M$ is a subfield of $\mathbb{R}$, it is real closed, and it is the smallest real-closed subfield of $\mathbb{R}$ containing $F$; it is the real closure of $F$ inside $\mathbb{R}$.

Proof. The restriction of the order satisfies the ordered-field axioms on any subfield and is Archimedean because $\mathbb{R}$ is. The set of elements algebraic over $F$ is a subfield, being closed under the field operations by the algebraic dependence of sums, products and inverses. It is real closed: a positive $a \in M$ has a square root in the real-closed field $\mathbb{R}$, and $\sqrt a$ satisfies the quadratic $X^2 - a$ over $M$, hence is algebraic over $F$ and lies in $M$; and a polynomial of odd degree over $M$ has a root in the real-closed field $\mathbb{R}$, and that root is algebraic over $M$ and hence over $F$, so it lies in $M$. Minimality is because any real-closed subfield of $\mathbb{R}$ containing $F$ must contain every real root of a polynomial over $F$, hence every element algebraic over $F$.

Corollary. The real closure of an ordered field is unique up to an order-preserving isomorphism, by Real-Closed and Complete Ordered Fields, and inside $\mathbb{R}$ it is the literal set $M$ above; in particular $\overline{\mathbb{Q}} \cap \mathbb{R}$ is the real closure of $\mathbb{Q}$.

Examples. The subfields $\mathbb{Q} \subset \mathbb{Q}(\sqrt2) \subset \overline{\mathbb{Q}} \cap \mathbb{R} \subset \mathbb{R}$ show the tower from the prime field to the real algebraic numbers, which are real closed and countable and not order-complete; adjoining a transcendental $t$ gives $\mathbb{Q}(t) \subset \mathbb{R}$ with the induced Archimedean order, in which $t$ is a finite real number rather than an infinite element; and the real closure of $\mathbb{Q}(t)$ inside $\mathbb{R}$ consists of the real numbers algebraic over $\mathbb{Q}(t)$.

Suprema and Order-Completeness

The supremum of a set $S \subseteq \mathbb{R}$ is its least upper bound and the infimum its greatest lower bound; both exist whenever $S$ is nonempty and bounded, above and below respectively. The following properties of $\mathbb{R}$ are the consequences of the completeness axiom.

Theorem. $\mathbb{R}$ has the following properties.

(a) Every nonempty subset of $\mathbb{R}$ that is bounded above has a least upper bound, and every nonempty subset bounded below has a greatest lower bound.

(b) $\mathbb{R}$ is Archimedean, and $\mathbb{Q}$ is dense in $\mathbb{R}$; every real is the supremum of the rationals below it.

(c) $\mathbb{R}$ is order-complete, and therefore no proper subfield of $\mathbb{R}$ is order-complete.

Proof. (a) is the least upper bound property, proved in the cut model. (b) is the theorem on order-complete fields from Real-Closed and Complete Ordered Fields. (c) Let $F \subsetneq \mathbb{R}$ be a subfield and let $a \in \mathbb{R} \setminus F$. The set $S = \{q \in \mathbb{Q} : q < a\}$ is a nonempty subset of $F$, since $\mathbb{Q} \subseteq F$, and it is bounded above in $F$ because $\mathbb{R}$ is Archimedean; its least upper bound in $\mathbb{R}$ is $a$. If $b \in F$ is an upper bound of $S$ then $b \geq a$, and if $b > a$ then, $\mathbb{Q}$ being dense in $\mathbb{R}$, there is a rational $q$ with $a < q < b$, and $q \in F$ is an upper bound of $S$ smaller than $b$. Hence a least upper bound of $S$ in $F$, if it existed, would have to equal $a$ and lie in $F$, a contradiction. Hence no proper subfield is order-complete.

Remark. Property (b) is what makes the order of $\mathbb{R}$ usable: every real is determined by the rationals below it, which is the content of the cut construction, and by Cantor's theorem this determination is what makes $\mathbb{R}$ unique.

Summary

$\mathbb{R}$ is the order completion of $\mathbb{Q}$ and the metric completion of $\mathbb{Q}$ for the usual absolute value; the two constructions, by Dedekind cuts and by Cauchy sequences, give canonically isomorphic ordered fields, and any two order-complete ordered fields are isomorphic by a unique order-preserving isomorphism, so $\mathbb{R}$ is the unique complete ordered field. It is Archimedean, the rationals are dense in it, every real is the supremum of the rationals below it, and no proper subfield of $\mathbb{R}$ is order-complete. It has cardinality $2^{\aleph_0}$, and the set of irrationals is uncountable.

The field $\mathbb{R}$ has characteristic $0$ and prime field $\mathbb{Q}$; it is real closed, so every positive real is a square and every odd-degree real polynomial has a real root, and it is not algebraically closed, its algebraic closure being $\mathbb{C} = \mathbb{R}(i)$ of degree $2$ over it. The real algebraic numbers $\overline{\mathbb{Q}} \cap \mathbb{R}$ form the real closure of $\mathbb{Q}$ and are countable and not complete, while $\operatorname{tr.deg}_{\mathbb{Q}} \mathbb{R} = 2^{\aleph_0}$. And $\mathbb{R}$ is not isomorphic to any $\mathbb{Q}_p$, since it is formally real and they are not; the triviality of $\operatorname{Aut}(\mathbb{R})$ is treated in Real Automorphisms and Derivations.

In the order-theoretic terms of Cantor's theorems, $\mathbb{Q}$ is the unique countable dense linear order without endpoints and $\mathbb{R}$ is the unique complete dense linear order without endpoints with a countable dense subset. An ordered field embeds order-preservingly into $\mathbb{R}$ exactly when it is Archimedean, the embedding being $a \mapsto \sup\{q \in \mathbb{Q} : q < a\}$, and it is then unique; so $\mathbb{R}$ is the order-complete field universal for Archimedean ordered fields. Every subfield of $\mathbb{R}$ inherits an Archimedean order and has a real closure inside $\mathbb{R}$, namely its relative algebraic closure.

Property	Value for $\mathbb{R}$
Order	complete, Archimedean, $\mathbb{Q}$ dense
Uniqueness	unique order-complete ordered field
Cardinality	$2^{\aleph_0}$
Characteristic	$0$, prime field $\mathbb{Q}$
Real closed	yes
Algebraically closed	no; closure $\mathbb{C} = \mathbb{R}(i)$
$\operatorname{tr.deg}_{\mathbb{Q}} \mathbb{R}$	$2^{\aleph_0}$
Real algebraic subfield	$\overline{\mathbb{Q}} \cap \mathbb{R}$, countable, real closed, not complete
Completions of $\mathbb{Q}$ giving $\mathbb{R}$	$\lvert \cdot \rvert_\infty$, order

Summary of Notation

symbol	meaning
$\mathbb{R}$	The real numbers, the complete ordered field
$\mathbb{Q}$	Rational numbers, prime field
$\mathcal{R}$, $\mathcal{C}$	Cut model and Cauchy-sequence model of $\mathbb{R}$
$\lvert \cdot \rvert_\infty$	Usual absolute value
$\widehat{F}$	Order or metric completion of $F$
$A \leq B \iff A \subseteq B$	Order on cuts
$i$	Square root of $-1$
$\mathbb{C} = \mathbb{R}(i)$	Algebraic closure of $\mathbb{R}$
$\overline{\mathbb{Q}} \cap \mathbb{R}$	Real algebraic numbers, real closure of $\mathbb{Q}$
$\operatorname{tr.deg}_{\mathbb{Q}} \mathbb{R}$	Transcendence degree, $2^{\aleph_0}$
$\mathbb{Q}_p$, $\mathbb{C}_p$	Non-Archimedean completions of $\mathbb{Q}$
$\varphi : F \to \mathbb{R}$	Archimedean embedding $a \mapsto \sup\{q \in \mathbb{Q} : q < a\}$
$M$	Relative algebraic closure of a subfield $F \subseteq \mathbb{R}$