Real-Closed and Complete Ordered Fields

Introduction

Two independent conditions single out the field of real numbers among ordered fields. The first is real closedness, an algebraic condition: the field is ordered, every positive element is a square, and every polynomial of odd degree has a root. The second is order completeness, a condition on the ordering: every nonempty subset that is bounded above has a least upper bound. Real closedness describes the algebraic shape of $\mathbb{R}$ and is inherited by the field of real algebraic numbers, which is countable; order completeness describes its analytic shape and is not inherited by any proper subfield. Each condition characterises $\mathbb{R}$ together with an additional hypothesis: an order-complete ordered field is $\mathbb{R}$ up to isomorphism, and a real-closed field whose ordering is order-complete and whose rationals are order-dense in it is also $\mathbb{R}$.

This article develops both notions, proves that a real-closed field has a unique ordering determined by its field structure, constructs the real closure of an ordered field and proves its uniqueness, and proves the uniqueness of $\mathbb{R}$ as the order-complete ordered field. The relation between real closedness and algebraic closedness is settled by showing that $F(i)$ is algebraically closed whenever $F$ is real closed.

Throughout, $F$ is an ordered field with positive cone $P$ as in Ordered Fields, and $F^{rc}$ denotes a real closure. The reader should not confuse the real closure $F^{rc}$, which is an algebraic extension, with the topological completion of $F$, which is not; the two constructions are compared in the final section.

---

Real-Closed Fields

Definition

Definition. An ordered field $F$ is real closed if

(RC1) every positive element of $F$ is a square in $F$, and

(RC2) every polynomial in $F[x]$ of odd degree has a root in $F$.

Theorem. A field $F$ is real closed if and only if it is formally real and admits no proper algebraic extension that is formally real. Equivalently, $F$ is real closed if and only if $F$ has an ordering and no proper algebraic extension of $F$ can be ordered.

Proof sketch. If $F$ is real closed and $K/F$ is a proper algebraic extension, then $K$ contains an element $\alpha$ of degree $> 1$ over $F$; its minimal polynomial has even degree (otherwise (RC2) gives a root in $F$, contradicting the choice of $\alpha$), and an argument with (RC1) shows that $-1$ is a sum of squares in $K$, so $K$ is not formally real. Conversely, if $F$ is formally real and has no proper formally real algebraic extension, then every positive element of $F$ has a square root (else adjoining a square root is a proper formally real extension by the standard analysis of $x^2 - a$ with $a > 0$), and every odd-degree polynomial has a root (else an odd-degree extension of $F$ can be ordered, by the standard ordering-extension lemma for an irreducible polynomial of odd degree).

Equivalent Conditions

Theorem. For an ordered field $F$ the following are equivalent.

(a) $F$ is real closed.

(b) $F(i)$ is algebraically closed, where $i^2 = -1$.

(c) $F$ has no proper algebraic extension that is formally real, and $F$ is formally real.

(d) $F$ is elementarily equivalent to $\mathbb{R}$: every first-order statement in the language of ordered fields is true in $F$ if and only if it is true in $\mathbb{R}$.

Proof sketch. (a) $\Leftrightarrow$ (c) is the theorem above. (a) $\Rightarrow$ (b) is the algebraic-closedness theorem proved below. (b) $\Rightarrow$ (a): if $F(i)$ is algebraically closed, then $F$ is not algebraically closed itself (as $x^2+1$ has no root in $F$) and $F$ is real closed by the standard result that a field whose extension by a square root of $-1$ is algebraically closed is real closed. (d) is Tarski's theorem on the completeness of the theory of real-closed fields: the axioms of an ordered field together with (RC1) and (RC2) are first-order and true in every real-closed field, and any two real-closed fields are elementarily equivalent, so a first-order sentence holds in one if and only if it holds in all.

Proposition. If $F$ is real closed, then its positive cone is the set of nonzero squares, $P = \{a^2 : a \in F^\times\}$; consequently a real-closed field carries a unique ordering, and that ordering is determined by the field structure. Every field isomorphism between real-closed fields is order-preserving.

Proof. (RC1) says that every positive element is a square, so $P \subseteq \{a^2 : a \in F^\times\}$, and every nonzero square is positive by the ordered-field axioms, so equality holds. The cone is therefore intrinsic to the field, and there is only one ordering of $F$; an isomorphism of fields carries squares to squares, hence carries the cone of one field onto the cone of the other, and so preserves the order.

Remark (the square condition is not equivalent to real closedness). An ordered field in which every positive element is a square is called Euclidean, and every real-closed field is Euclidean by the proposition above, but the converse fails. The field of real numbers constructible from $\mathbb{Q}$ by straightedge and compass is Euclidean, since a square root of a positive constructible number is constructible; it is not real closed, because every constructible number has degree a power of $2$ over $\mathbb{Q}$, while $\sqrt[3]{2}$ has degree $3$, so $\sqrt[3]{2}$ is not constructible and $x^3 - 2$ has no root in the field. What the equivalence (a) $\Leftrightarrow$ (b) above adds to the square condition is exactly the algebraic closedness of $F(i)$, which fails here.

Examples

The table separates the two phenomena: $\mathbb{R}$ and the real algebraic numbers are both real closed, one uncountable and one countable, so real closedness does not determine the cardinality.

---

The Real Closure

Definition

Definition. Let $F$ be an ordered field. A real closure of $F$ is an ordered field $F^{rc}$ together with an order-preserving embedding $F \hookrightarrow F^{rc}$ such that $F^{rc}$ is algebraic over $F$ and real closed.

Theorem (Artin–Schreier). Every ordered field has a real closure, and any two real closures of $F$ are isomorphic by a unique isomorphism fixing $F$ pointwise and preserving the order.

Proof sketch. By Zorn's lemma take an algebraic extension of $F$ that is maximal with respect to being orderable, which exists because the union of a chain of orderable algebraic extensions is orderable; a maximal such extension has no proper orderable algebraic extension, hence is real closed by the characterization above. Uniqueness: given two real closures, the isomorphism between the algebraic extensions is constructed by the usual step-by-step extension of the identity on $F$ to the roots of irreducible polynomials, and it is order-preserving because the order in a real-closed field is determined by the field structure, by the corollary above.

Example. The real closure of $\mathbb{Q}$ is the field of real algebraic numbers $\overline{\mathbb{Q}} \cap \mathbb{R}$, the field of real roots of polynomials with rational coefficients. It is countable, Archimedean, and real closed, but it is not order-complete: the set

$$ A = \{x \in \overline{\mathbb{Q}} \cap \mathbb{R} : x < \pi\} $$

is nonempty and bounded above by $4$, and its supremum in $\mathbb{R}$ is $\pi$, since $\mathbb{Q} \subseteq A$ is order-dense in $\mathbb{R}$; if $A$ had a supremum $s$ in $\overline{\mathbb{Q}} \cap \mathbb{R}$, then $s$ would equal the real supremum $\pi$, contradicting the transcendence of $\pi$. So $A$ has no least upper bound in the real algebraic numbers.

Example. The real closure of $\mathbb{R}(t)$ with $t$ infinite is a real-closed field of generalized Puiseux series; it is not a subfield of $\mathbb{R}$, in contrast to the real closure of any ordered subfield of $\mathbb{R}$.

Real Closure versus Algebraic Closure

Theorem. Let $F$ be an ordered field with real closure $F^{rc}$ and algebraic closure $\overline{F}$. Then

$$ \overline{F} = F^{rc}(i), \qquad i^2 = -1, \qquad [\overline{F} : F^{rc}] = 2 . $$

Thus the algebraic closure of a real-closed field is obtained by adjoining a single square root of $-1$. In particular, for $F = \mathbb{R}$ one has $\overline{F} = \mathbb{C} = \mathbb{R}(i)$.

Proof. The field $F^{rc}$ is real closed, so $F^{rc}(i)$ is algebraically closed by the theorem on algebraic closedness of $F(i)$ below, and it is algebraic over $F$ because $F^{rc}$ is. Since it contains $F$, the embedding $F^{rc}(i) \to \overline{F}$ fixing $F$ is an isomorphism, and $i \notin F^{rc}$ because a formally real field contains no square root of $-1$. Hence the degree is $2$.

---

Order Completeness

Dedekind Completeness

Definition. An ordered field $F$ is order-complete (or Dedekind-complete) if every nonempty subset $A \subseteq F$ that is bounded above has a least upper bound $\sup A$ in $F$. Equivalently (applying the condition to $-A$), every nonempty subset bounded below has a greatest lower bound.

Proposition. Every order-complete ordered field is Archimedean.

Proof. Suppose $F$ is not Archimedean. Then the set $A = \{n \cdot 1 : n \geq 1\}$ is nonempty and bounded above, so it has a supremum $s = \sup A$. Since $n \cdot 1 \leq s$ for all $n$, also $(n+1) \cdot 1 \leq s$ for all $n$, hence $n \cdot 1 \leq s - 1$ for all $n$, so $s - 1$ is an upper bound of $A$ strictly smaller than $s$, contradicting the definition of $s$. Hence $F$ is Archimedean.

Theorem (equivalent formulations). For an ordered field $F$ the following are equivalent.

(a) $F$ is order-complete: every nonempty subset bounded above has a supremum.

(b) Every nonempty subset bounded below has an infimum.

(c) Every Dedekind cut of $F$ is realized in $F$: for every partition $F = A \cup B$ with both parts nonempty and $a < b$ for all $a \in A$, $b \in B$, either $A$ has a greatest element or $B$ has a least element.

Proof. (a) $\Leftrightarrow$ (b): negating all elements turns upper bounds into lower bounds and suprema into infima. (a) $\Rightarrow$ (c): given a cut $(A,B)$, the set $A$ is nonempty and bounded above by any element of $B$, so $s = \sup A$ exists; then either $s \in A$ and $s$ is the greatest element of $A$, or $s \in B$, and $s$ is a lower bound of $B$ while no element of $B$ is smaller, so $s$ is the least element of $B$. (c) $\Rightarrow$ (a): given a nonempty $A$ bounded above, if $A$ has a greatest element then that element is $\sup A$. Otherwise let $B$ be the set of upper bounds of $A$; no element of $A$ is an upper bound, so $A \cap B = \emptyset$, both parts of the partition $(F \setminus B, B)$ are nonempty, and every element of $F \setminus B$ lies below every element of $B$. So (c) applies. No element of $F \setminus B$ is greatest: if $x$ is not an upper bound of $A$ then some $a \in A$ satisfies $x < a$, and that $a$ is again not an upper bound. Hence $B$ has a least element, which is the smallest upper bound of $A$, that is, $\sup A$.

Density and the Least Upper Bound Property

Theorem. Let $F$ be an order-complete ordered field. Then $\mathbb{Q}$ is order-dense in $F$, and every element of $F$ is both the supremum of the rationals below it and the infimum of the rationals above it.

Proof. $F$ is Archimedean by the proposition, so $\mathbb{Q}$ is order-dense by the characterization of Archimedean fields in Ordered Fields. For $x \in F$, the set $A_x = \{q \in \mathbb{Q} : q < x\}$ is nonempty (as $F$ is Archimedean, $-n < x$ for $n$ large) and bounded above by $x$, so it has a supremum $s \leq x$. If $s < x$, order-density gives a rational $q$ with $s < q < x$, hence $q \in A_x$ and $q > s$, contradicting $s = \sup A_x$; so $s = x$. The statement for the infimum follows by applying the result to $-x$.

Corollary. In an order-complete ordered field, the rationals separate points and the order is the order generated by the embedding of $\mathbb{Q}$; consequently, if $F$ and $F'$ are order-complete ordered fields, any order-preserving isomorphism between their prime fields extends to at most one isomorphism $F \to F'$.

Proof. An order-preserving field isomorphism is determined by its values on $\mathbb{Q}$ together with the description of each element as a supremum of rationals, which is preserved.

---

The Uniqueness of $\mathbb{R}$

The Isomorphism Theorem

Theorem (Cantor). Any two order-complete ordered fields are isomorphic by a unique order-preserving field isomorphism.

Proof sketch. Let $F$ and $F'$ be order-complete ordered fields. Their prime fields are the images of $\mathbb{Q}$, and the unique ordering of $\mathbb{Q}$ is preserved, giving an isomorphism $\varphi_0$ of the prime subfields that is order-preserving. Extend $\varphi_0$ to a map $\varphi : F \to F'$ by

$$ \varphi(x) = \sup\{ \varphi_0(q) : q \in \mathbb{Q},\ q < x \}, $$

which is defined because $\varphi_0(A_x)$ is nonempty and bounded above in $F'$. The map is order-preserving and additive and multiplicative by the arithmetic of suprema, and its kernel is trivial; it is surjective because, given $y \in F'$, the element $x = \sup\{q \in \mathbb{Q} : \varphi_0(q) < y\}$ satisfies $\varphi(x) = y$ by the order-density of $\mathbb{Q}$ in both fields. Uniqueness holds because an order-preserving field map is determined by its restriction to $\mathbb{Q}$, hence to the order-dense subfield $\mathbb{Q}$, hence to $F$.

Corollary (the real numbers are unique). Up to a unique order-preserving field isomorphism there is exactly one order-complete ordered field, namely $\mathbb{R}$. In particular every complete ordered field is $\mathbb{R}$, and every order-complete ordered field contains $\mathbb{Q}$ as an order-dense subfield and is isomorphic to $\mathbb{R}$.

Corollary (characterisation by order). $\mathbb{R}$ is Archimedean and order-complete; conversely an Archimedean order-complete ordered field is $\mathbb{R}$. Hence among ordered fields the field of real numbers is characterised by the least upper bound property.

Remark (completeness does not follow from Archimedean). The field $\mathbb{Q}$ is Archimedean but not order-complete: the set $\{q \in \mathbb{Q} : q > 0,\ q^2 < 2\}$ is nonempty and bounded above and has no supremum in $\mathbb{Q}$. The real closure $\overline{\mathbb{Q}} \cap \mathbb{R}$ is Archimedean and real closed but still not order-complete, for the same reason. Order completeness is a strictly stronger condition than Archimedean real closedness, and it is the condition that makes $\mathbb{R}$ unique.

Real Closedness of $\mathbb{R}$

Theorem. $\mathbb{R}$ is real closed; equivalently, every positive real number is a square and every real polynomial of odd degree has a real root.

Pro. That every positive real has a square root is the completeness of $\mathbb{R}$ applied to the set $\{y \geq 0: y^2 \leq x\}$, which is nonempty and bounded above and whose supremum $\sqrt x$ satisfies $(\sqrt x)^2 = x$; the verification that the supremum has this property uses the order and the Archimedean property. For odd degree, a real polynomial $p$ of odd degree takes both signs, say $p(a) < 0 < p(b)$ with $a < b$ chosen far enough out, and the set $S = \{x \in (a,b): p < 0 \text{ on } (a,x]\}$ is nonempty and bounded above; its supremum $s \in (a,b]$ satisfies $p(s) = 0$, because a value $p(s) \neq 0$ persists on both sides of $s$: factoring $p(x) - p(s) = (x-s)q(x)$ with $q$ a polynomial, the finitely many coefficients of $q$ are bounded, so for $|x - s|$ small enough $|(x-s)q(x)| < |p(s)|$ and $p(x)$ has the sign of $p(s)$; a value $p(s) < 0$ would put $s+\delta$ in $S$ for small $\delta > 0$, so $s$ would not bound $S$, and $p(s) > 0$ would put $s-\delta$ above every element of $S$ for small $\delta > 0$, so $s$ would not be the least upper bound. The case of negative leading coefficient is analogous.

Theorem (fundamental theorem of algebra, real-closed form). Let $F$ be a real-closed field and let $i^2 = -1$ in an algebraic closure. Then $F(i)$ is algebraically closed.

Proof sketch. One shows that every nonconstant polynomial over $F(i)$ has a root, reducing by conjugation to a real polynomial $p \in F[x]$ and then to $p\bar p$. By the odd-degree condition (RC2) applied to an auxiliary polynomial, one exhibits a root. The argument is the classical proof of the fundamental theorem of algebra in the real-closed setting and is given in the references.

Corollary. $\mathbb{C} = \mathbb{R}(i)$ is algebraically closed, and it is an algebraic closure of $\mathbb{R}$; the algebraic closure of the real algebraic numbers $\overline{\mathbb{Q}} \cap \mathbb{R}$ is the field $\overline{\mathbb{Q}}$ of algebraic numbers, which is a proper subfield of $\mathbb{C}$ because $\mathbb{C}$ is transcendental over $\mathbb{Q}$.

---

Real Closure, Completion and $\mathbb{R}$

The two constructions of this article differ in a way worth recording.

Proposition. For an Archimedean ordered field $F$ the order completion $\widehat{F}$ is an order-complete ordered field containing $F$ as an order-dense subfield; it is unique up to a unique order-preserving $F$-isomorphism, and $\widehat{\mathbb{Q}} = \mathbb{R}$.

Proof sketch. The completion is constructed from the Dedekind cuts of $F$, ordered by inclusion, with the field operations defined by the arithmetic of cuts; the same construction applied to $\mathbb{Q}$ produces $\mathbb{R}$. Uniqueness is Cantor's theorem.

Remark. The Archimedean hypothesis is not removable. A non-Archimedean ordered field has a positive infinitesimal $\epsilon$, so $\{n\epsilon : n \geq 1\}$ is bounded above by $1$ and has a supremum $u$ in the Dedekind completion $\widehat{F}$. If $\widehat{F}$ were an ordered field, multiplication by $2$ would preserve suprema, and since $\{2n\epsilon : n \geq 1\}$ and $\{n\epsilon : n \geq 1\}$ are cofinal in one another,

$$ 2u = \sup_{n \geq 1} 2n\epsilon = \sup_{n \geq 1} n\epsilon = u, $$

forcing $u = 0$, although $u \geq \epsilon > 0$. Hence $\widehat{F}$ is a complete linear order containing $F$ as an order-dense subset but carries no field structure making it an ordered field extension of $F$: the completion row of the table above is an ordered field only in the Archimedean case.

Corollary. $\mathbb{R}$ is simultaneously the order completion of $\mathbb{Q}$ and the order completion of the real algebraic numbers. The real closure of $\mathbb{Q}$ is the smaller field $\overline{\mathbb{Q}} \cap \mathbb{R}$, and $\mathbb{R}$ is the order completion of that real closure.

Remark. The distinction is exactly the distinction between the algebraic and the order-theoretic content of the real numbers. The real closure of $\mathbb{Q}$ contains every real root of every rational polynomial and is countable; the completion adds the transcendental limits, and its construction requires the supremum principle. The companion articles The Rational Numbers and The Real Numbers carry out both constructions in full.

---

The Cardinality of $\mathbb{R}$

Theorem. $|\mathbb{R}| = |\mathcal{P}(\mathbb{N})| = 2^{\aleph_0}$; in particular $\mathbb{R}$ is uncountable, and the cardinal $\mathrm{C}$ of Cardinality and the Axiom of Choice is its cardinality.

Proof sketch. The identification $|\mathbb{R}| = |\mathcal{P}(\mathbb{N})|$ is the binary expansion: a real number has a binary expansion, which is a function $\mathbb{N} \to \{0,1\}$ modulo the ambiguity of the expansions ending in $1$s, and the standard reduction to the non-ambiguous expansions gives a bijection between $\mathbb{R}$ and a subset of $\mathcal{P}(\mathbb{N})$; order-completeness supplies the expansion, since each digit is decided by a bounded-above set, and Schröder–Bernstein, applied to the inclusion and to a suitable injection the other way, gives the equality. Uncountability is then immediate from Cantor's theorem.

Remark. The powers $\mathbb{R}^n$ are equipotent to $\mathbb{R}$, by interleaving binary expansions of the coordinates; the interval $(0,1)$ is equipotent to $\mathbb{R}$ by the rational function $x \mapsto (2x-1)/(x(1-x))$, which is strictly increasing on $(0,1)$ and carries it onto $\mathbb{R}$. The limiting language that the description of that map invites is only a way of saying that the values become arbitrarily large in absolute value near the endpoints; the rigorous statement of that fact belongs to Part II, where the order is enriched by a distance, and the cardinality result itself uses only the explicit formula. The arithmetic of the cardinal $\mathrm{C} = 2^{\aleph_0}$ is treated in Cardinality and the Axiom of Choice.

Summary

An ordered field is real closed when every positive element is a square and every polynomial of odd degree has a root. This is equivalent to $F$ being formally real with no proper formally real algebraic extension, to $F(i)$ being algebraically closed, and, by Tarski's theorem, to $F$ being elementarily equivalent to $\mathbb{R}$; a real-closed field carries a unique ordering, determined by the field structure through the identity $P = \{a^2 : a \neq 0\}$, so all isomorphisms of real-closed fields preserve the order. Every ordered field has a real closure, unique up to order-preserving isomorphism fixing the base field: the real closure of $\mathbb{Q}$ is the countable field of real algebraic numbers, while the real closure of $\mathbb{R}(t)$ with $t$ infinite is a field of generalized Puiseux series that does not sit inside $\mathbb{R}$.

An ordered field is order-complete when every nonempty bounded-above subset has a supremum; order-complete fields are Archimedean, they contain $\mathbb{Q}$ as an order-dense subfield, and they are all isomorphic to $\mathbb{R}$ by Cantor's uniqueness theorem. Among Archimedean ordered fields, real closedness is strictly weaker than order completeness: the real algebraic numbers are real closed but not order-complete, and no proper subfield of $\mathbb{R}$ is order-complete, while completeness does imply real closedness, because every order-complete ordered field is isomorphic to $\mathbb{R}$. The field $\mathbb{R}$ is real closed, and for a real-closed field $F$ the extension $F(i)$ is algebraically closed, so $\mathbb{C} = \mathbb{R}(i)$ is algebraically closed and is the algebraic closure of $\mathbb{R}$.

Summary of Notation

Further Reading

• Emil Artin and Otto Schreier, Algebraische Konstruktion reeller Körper (Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg 5, 1927), for the existence and uniqueness of the real closure.
• Serge Lang, Algebra (Springer, 3rd ed. 2002), for real-closed fields, real closures and the Artin–Schreier theory.
• Alfred Tarski, A Decision Method for Elementary Algebra and Geometry (University of California Press, 1951), for the completeness and decidability of the theory of real-closed fields.
• Nathan Jacobson, Basic Algebra I (Dover, 2nd ed. 2009), for Dedekind completeness and Cantor's uniqueness theorem for ordered fields.
• Walter Rudin, Principles of Mathematical Analysis (McGraw-Hill, 3rd ed. 1976), for the construction of $\mathbb{R}$ by cuts, the least upper bound property and the intermediate value theorem.
• Alexander Prestel and Charles N. Delzell, Positive Polynomials (Springer, 2001), for real-closed fields and their orderings in the general theory.