---
title: Ordered Fields and the Completeness Axiom
module: Foundations and the Real Number System
moduleNumber: 1
lessonNumber: 2
order: 102
summary: >
  The real numbers are the unique ordered field with the least-upper-bound property.
  The field and order axioms, the exact failure of the rationals (no supremum for the
  set of rationals below √2), and completeness as the defining axiom of ℝ lead to the
  first consequences: the existence of √2, the Archimedean property, and the density
  of ℚ in ℝ.
topics: [Foundations and the Real Number System]
sources:
  - book: Lebl
    ref: "Ch. 1 — Real Numbers; §1.1 Basic properties; §1.2 The set of real numbers"
  - book: Rosenlicht
    ref: "Ch. 2 — The real number system"
draft: false
---

Calculus assumes the real numbers behave a certain way: that a bounded increasing sequence
converges, that a continuous function crossing from negative to positive has a root,
that $\sqrt2$ exists. Each of these is false for the rational numbers. What distinguishes
$\mathbb{R}$ from $\mathbb{Q}$ is a single order property, **completeness**, and everything
calculus took for granted is a theorem about it. Rather than construct $\mathbb{R}$ from
$\mathbb{Q}$, we take its existence as given and pin it down by its axioms: an ordered field
with the least-upper-bound property.[^lebl-11]

## Ordered sets, bounds, and suprema

> **Definition (Ordered set).** An **ordered set** is a set $S$ with a relation $<$ satisfying
> - **trichotomy**: for all $x, y \in S$ exactly one of $x < y$, $x = y$, $y < x$ holds;
> - **transitivity**: $x < y$ and $y < z$ imply $x < z$.
>
> We write $x \le y$ for "$x < y$ or $x = y$".

The rationals are ordered by declaring $x < y$ when $y - x$ is a positive rational; $\mathbb{N}$
and $\mathbb{Z}$ inherit the same order. Bounds are defined from the order alone.

> **Definition (Bounds, supremum, infimum).** Let $E \subset S$ be a subset of an ordered set.
> - $b \in S$ is an **upper bound** of $E$ if $x \le b$ for all $x \in E$; then $E$ is
>   **bounded above**. Lower bounds and **bounded below** are defined with $\ge$.
> - An upper bound $b_0$ that satisfies $b_0 \le b$ for every upper bound $b$ is the **least
>   upper bound** or **supremum**, written $\sup E$.
> - A lower bound $b_0$ with $b_0 \ge b$ for every lower bound $b$ is the **greatest lower
>   bound** or **infimum**, $\inf E$.
>
> $E$ is **bounded** if it is bounded above and below.

The supremum, when it exists, is unique: two least upper bounds $b, b'$ each satisfy
$b \le b'$ and $b' \le b$, so $b = b'$. A supremum need not belong to the set and need not
exist at all. The set
$\{x \in \mathbb{Q} : x < 1\}$ has supremum $1$, which is outside it. The set
$\{x \in \mathbb{Q} : x \ge 0\}$ has no upper bound in $\mathbb{Q}$, so no supremum.

$$
% caption: A set bounded above sits below all of its upper bounds; the supremum is
% the leftmost upper bound, the exact boundary between the two regions.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% the number line
\draw[->, black] (-0.4,0) -- (10.2,0);
% the set E as a shaded band on the line
\draw[acc, line width=2.4pt] (0.6,0) -- (4.2,0);
\node[text=acc, anchor=south] at (2.4,0.12) {the set $E$};
% upper bounds region
\draw[black, line width=2.4pt] (4.2,-0.32) -- (9.6,-0.32);
\node[black, anchor=north] at (6.9,-0.42) {upper bounds of $E$};
% the supremum
\fill[acc] (4.2,0) circle (2.2pt);
\draw[acc, thick] (4.2,0.55) -- (4.2,-0.55);
\node[text=acc, anchor=south] at (4.2,0.6) {$\sup E$};
\node[anchor=north, black] at (1.5,-0.05) {smaller};
\node[anchor=north, black] at (8.6,-0.05) {bigger};
\end{tikzpicture}
$$

The property that makes suprema always available is the axiom that defines $\mathbb{R}$.

> **Definition (Least-upper-bound property).** An ordered set $S$ has the **least-upper-bound
> property** if every nonempty subset of $S$ that is bounded above has a supremum in $S$.

This is also called **Dedekind completeness**. It is a property of the _order_ alone, but it is
exactly what the rationals lack.

## The gap in the rationals

> **Worked example.** The set $E = \{x \in \mathbb{Q} : x^2 < 2\}$ is nonempty ($1 \in E$) and
> bounded above (if $x \ge 2$ then $x^2 \ge 4$, so every element is below $2$), yet $E$ has no
> supremum in $\mathbb{Q}$. The only candidate for the boundary is $\sqrt2$, which is not
> rational: suppose $x = m/n$ in lowest terms with $x^2 = 2$; then $m^2 = 2n^2$, so $m^2$ is
> even, so $m$ is even; writing $m = 2k$ gives $2k^2 = n^2$, so $n$ is even too, contradicting
> lowest terms. Around this missing number the rationals close in from both sides: those with
> $x^2 < 2$ can be pushed up, those with $x^2 > 2$ can be pushed down, and no rational sits at
> the boundary to serve as supremum.

$$
% caption: Rationals below and above the value with square two close in from both
% sides by nested intervals, but the boundary point itself is absent from the line
% of rationals.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\definecolor{red}{HTML}{C0392B}
\draw[->, black] (-0.3,0) -- (10.3,0);
\node[anchor=north] at (0.4,-0.08) {1.0};
\node[anchor=north] at (9.6,-0.08) {2.0};
% the gap point
\draw[red, thick] (5.5,0.4) -- (5.5,-0.4);
\node[text=red, anchor=south] at (5.5,0.45) {the gap};
\filldraw[fill=white, draw=red, thick] (5.5,0) circle (2.4pt);
% nested left endpoints (squares below 2) pushing right
\foreach \x/\lab in {2.0/, 3.6/, 4.7/, 5.15/} {
  \fill[acc] (\x,0) circle (1.7pt);
}
\draw[acc, ->] (2.0,0.28) -- (5.05,0.28);
\node[text=acc, anchor=south] at (3.4,0.3) {squares below two};
% nested right endpoints (squares above 2) pushing left
\foreach \x/\lab in {9.0/, 7.3/, 6.3/, 5.85/} {
  \fill[black] (\x,0) circle (1.7pt);
}
\draw[black, ->] (9.0,-0.28) -- (5.95,-0.28);
\node[black, anchor=north] at (7.6,-0.3) {squares above two};
\end{tikzpicture}
$$

So $\mathbb{Q}$ serves algebra well but has holes, and analysis needs a number system with
none — one that keeps the algebraic structure of $\mathbb{Q}$. That structure is a **field**.

## Field and order axioms

> **Definition (Field).** A set $F$ with two operations $x + y$ and $xy$ is a **field** if
> addition and multiplication are each commutative and associative, each has an identity
> ($0$ for $+$, $1 \ne 0$ for $\times$) and inverses ($-x$ for every $x$; $1/x$ for every
> $x \ne 0$), and multiplication distributes over addition: $x(y+z) = xy + xz$.

The rationals $\mathbb{Q}$ form a field; the integers $\mathbb{Z}$ do not, because $2$ has no
multiplicative inverse in $\mathbb{Z}$. The usual arithmetic identities follow from the axioms —
for instance $0x = 0$, since $0x = (0+0)x = 0x + 0x$ forces $0x = 0$ after adding $-(0x)$.

Order and arithmetic must be compatible.

> **Definition (Ordered field).** A field $F$ that is also an ordered set is an **ordered
> field** if
> - $x < y$ implies $x + z < y + z$ for all $z$;
> - $x > 0$ and $y > 0$ imply $xy > 0$.
>
> An $x > 0$ is **positive** and an $x < 0$ is **negative**.

From these two compatibility rules the familiar sign laws follow, the ones used in nearly
every later inequality:[^lebl-11]

> **Proposition (Sign rules).** In an ordered field, for all $x, y$:
> - $x > 0 \iff -x < 0$;
> - $x \ne 0 \implies x^2 > 0$; in particular $1 > 0$;
> - $0 < x < y \implies 0 < 1/y < 1/x$;
> - $xy > 0 \iff x, y$ are both positive or both negative.

Since $x^2 > 0$ for every $x \ne 0$, no ordered field can contain a square root of $-1$. The
complex numbers $\mathbb{C}$ form a field but cannot be ordered as a field, since
$i^2 = -1 < 0$ would violate this rule.

## The definition of ℝ

Every property demanded so far — ordered field, completeness, containing the rationals — is
satisfied by exactly one system, and that system is the definition of the reals.

> **Theorem (Existence and uniqueness of ℝ).** There exists a unique ordered field
> $\mathbb{R}$ with the least-upper-bound property such that $\mathbb{Q} \subset \mathbb{R}$.

Uniqueness is up to relabeling (isomorphism); existence can be established by constructing
$\mathbb{R}$ from $\mathbb{Q}$ via Dedekind cuts or Cauchy sequences, which we take on faith
so as to get to analysis. From now on $\mathbb{R}$ means this system, and $x \ge 0$ or
$\epsilon > 0$ silently means $x, \epsilon \in \mathbb{R}$.

Because $\mathbb{R}$ has the least-upper-bound property, it automatically has the greatest-
lower-bound property as well: if $A$ is nonempty and bounded below, then
$B = \{-x : x \in A\}$ is bounded above, and $\inf A = -\sup B$.

> **Proposition (The $\epsilon$ trick).** If $x \in \mathbb{R}$ satisfies $x \le \epsilon$ for
> all $\epsilon > 0$, then $x \le 0$.

> **Proof.** If instead $x > 0$, then $0 < x/2 < x$, and taking $\epsilon = x/2$ contradicts
> $x \le \epsilon$. $\blacksquare$

Read together with its mirror image, this says: to prove $x = 0$, show $|x| \le \epsilon$ for
every $\epsilon > 0$; to prove $x \le y$, show $x \le y + \epsilon$ for every $\epsilon > 0$.
Nearly every inequality in the coming lessons is proved this way, by driving an error term
below an arbitrary positive bound. It also encodes the fact that between any two reals
$a < b$ lies another, for instance the midpoint $(a+b)/2$; $\mathbb{R}$ has no smallest positive
element.

## The existence of √2

Completeness immediately supplies the number the rationals were missing.

> **Theorem (Existence of √2).** There is a unique positive $r \in \mathbb{R}$ with $r^2 = 2$.

> **Proof sketch.** Let $A = \{x \in \mathbb{R} : x^2 < 2\}$. It is nonempty and bounded above,
> so $r = \sup A$ exists by completeness. The claim is $r^2 = 2$, proved by ruling out
> $r^2 < 2$ and $r^2 > 2$ separately.[^lebl-12]
>
> - If $r^2 < 2$, choose a small $h \in (0,1)$ with $h < \frac{2 - r^2}{2r + 1}$. Then
>   $(r+h)^2 - r^2 = h(2r + h) < h(2r+1) < 2 - r^2$, so $(r+h)^2 < 2$ and $r + h \in A$ with
>   $r + h > r$, contradicting $r = \sup A$.
> - If $r^2 > 2$, choose $h = \frac{r^2 - 2}{2r} > 0$. Then $r^2 - (r-h)^2 < 2rh = r^2 - 2$, so
>   $(r-h)^2 > 2$; hence $r - h$ is an upper bound for $A$ smaller than $r$, again a
>   contradiction.
>
> Both cases fail, so $r^2 = 2$. Uniqueness follows from $0 < s < r \implies s^2 < r^2$.
> $\blacksquare$

The same argument gives a unique positive $n$-th root $x^{1/n}$ for every $x > 0$ and every
$n \in \mathbb{N}$. The set $\mathbb{R} \setminus \mathbb{Q}$ of **irrational** numbers is
therefore nonempty; the [uncountability of $\mathbb{R}$](/real-analysis/foundations/intervals-uncountability)
shows it is in fact far larger than $\mathbb{Q}$.

## The Archimedean property and density of ℚ

Completeness has a second consequence: $\mathbb{R}$ contains no infinitely large or
infinitely small elements.

> **Theorem (Archimedean property; density of ℚ).**
> 1. If $x, y \in \mathbb{R}$ with $x > 0$, there is an $n \in \mathbb{N}$ with $nx > y$.
> 2. If $x < y$ in $\mathbb{R}$, there is an $r \in \mathbb{Q}$ with $x < r < y$.

> **Proof of (1).** Dividing by $x$, the claim is that $\mathbb{N}$ is not bounded above in
> $\mathbb{R}$. If it were, $b = \sup \mathbb{N}$ would exist; but $b - 1$ is not an upper bound,
> so some $m \in \mathbb{N}$ has $m > b - 1$, whence $m + 1 > b$, contradicting that $b$ bounds
> $\mathbb{N}$. $\blacksquare$

Part (1) says that stacking copies of any fixed $x > 0$ eventually passes any $y$.

$$
% caption: The Archimedean property: repeatedly adding a fixed positive step,
% however small, eventually carries past any prescribed target on the line.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\definecolor{red}{HTML}{C0392B}
\draw[->, black] (-0.3,0) -- (10.4,0);
% the target y
\draw[red, thick] (8.4,0.5) -- (8.4,-0.3);
\node[text=red, anchor=south] at (8.4,0.52) {target $y$};
% unit steps of size x
\foreach \k in {0,1,2,3,4,5,6,7} {
  \fill[acc] (\k*1.15,0) circle (1.6pt);
}
\foreach \k in {0,1,2,3,4,5,6} {
  \draw[acc, ->] (\k*1.15,0.22) -- ({(\k+1)*1.15},0.22);
}
\node[text=acc, anchor=south] at (0.58,0.24) {$x$};
\node[anchor=north, black] at (0,-0.08) {$0$};
\node[anchor=north, text=acc] at (8.05,-0.08) {$nx$};
\node[anchor=west, black] at (3.2,-0.6) {every step adds the same $x$; step $n$ clears $y$};
\end{tikzpicture}
$$

> **Proof of (2).** Take $x \ge 0$ first. By (1) pick $n$ with $n(y - x) > 1$, so
> $y - x > 1/n$. By (1) again the set $\{k \in \mathbb{N} : k > nx\}$ is nonempty; let $m$ be its
> least element (well-ordering). Then $m > nx$, and $m - 1 \le nx$ because $m$ is least.
> Combining $m \le nx + 1 < ny$ gives $x < m/n < y$, so $r = m/n$ works. The case $x < 0$
> reduces to this one. $\blacksquare$

> **Worked example.** $\inf\{1/n : n \in \mathbb{N}\} = 0$. Zero is a lower bound, since
> $1/n > 0$ for every $n$. For any $a > 0$, the Archimedean property gives an $n$ with $na > 1$,
> that is $a > 1/n$, so $a$ is not a lower bound. No positive number bounds the set below, and
> the greatest lower bound is $0$.

$$
% caption: The points 1/n crowd toward 0 from the right; every positive number is
% eventually overtaken, so the infimum is 0 even though 0 is never attained.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\definecolor{red}{HTML}{C0392B}
\draw[->, black] (-0.3,0) -- (10.4,0);
% the infimum at 0
\fill[red] (0.2,0) circle (2.4pt);
\node[text=red, anchor=north] at (0.2,-0.12) {$0$};
\node[text=red, anchor=south] at (0.35,0.16) {$\inf$};
% the points 1/n at positions scaled so 1 -> far right
\foreach \n/\xp/\lab in {1/9.4/1, 2/4.8/{\frac{1}{2}}, 3/3.2/{\frac{1}{3}}, 4/2.45/{\frac{1}{4}}, 5/1.98/{}, 6/1.68/{}, 7/1.46/{}, 8/1.3/{}} {
  \fill[acc] (\xp,0) circle (1.7pt);
}
\node[text=acc, anchor=south] at (9.4,0.1) {$1$};
\node[text=acc, anchor=south] at (4.8,0.1) {$\frac{1}{2}$};
\node[text=acc, anchor=south] at (3.2,0.1) {$\frac{1}{3}$};
\node[text=acc, anchor=south] at (2.45,0.12) {$\frac{1}{4}$};
\node[anchor=north, black] at (3.4,-0.6) {crowding toward zero};
\end{tikzpicture}
$$

The sequence $1/n$ therefore gets arbitrarily small, the basic fact behind the theory of
[limits](/real-analysis/sequences-series/sequences-limits).

Density means every open interval of reals, however short, contains a rational (and, since
the irrationals are dense too, an irrational).

> **Worked example.** A rational between $\sqrt2$ and $\sqrt2 + \tfrac{1}{100}$. Follow the
> density proof: with $y - x = \tfrac{1}{100}$, any $n > 100$ satisfies $n(y - x) > 1$; take
> $n = 101$. Then $nx = 101\sqrt2 \approx 142.83$, and the least integer exceeding it is
> $m = 143$, giving $r = \tfrac{143}{101} \approx 1.41584$. Since
> $\sqrt2 \approx 1.41421 < 1.41584 < 1.42421 \approx \sqrt2 + \tfrac{1}{100}$, the rational
> $r$ lies in the interval.

## Suprema under arithmetic

Suprema and infima interact predictably with shifting and scaling a set. For $A \subset
\mathbb{R}$ and $c \in \mathbb{R}$, write $c + A = \{c + a : a \in A\}$ and $cA = \{ca : a \in
A\}$.[^lebl-12]

| Operation | Supremum | Infimum |
| --- | --- | --- |
| shift by $c$ | $\sup(c + A) = c + \sup A$ | $\inf(c + A) = c + \inf A$ |
| scale by $c > 0$ | $\sup(cA) = c \sup A$ | $\inf(cA) = c \inf A$ |
| scale by $c < 0$ | $\sup(cA) = c \inf A$ | $\inf(cA) = c \sup A$ |

Multiplying by a negative number swaps the roles of $\sup$ and $\inf$, which is the same sign
reversal that flips inequalities. A second rule is used constantly when comparing two sets.

> **Proposition (Separated sets).** If $A, B \subset \mathbb{R}$ are nonempty and $x \le y$
> whenever $x \in A$ and $y \in B$, then $\sup A \le \inf B$.

The proof is immediate: every $x \in A$ is a lower bound for $B$, so $x \le \inf B$; hence
$\inf B$ is an upper bound for $A$, so $\sup A \le \inf B$. The strict version fails: take
$A = \{0\}$ and $B = \{1/n : n \in \mathbb{N}\}$; then $x < y$ for
every pair, yet $\sup A = 0 = \inf B$. Strict inequalities are not preserved by passing to
suprema. This same pair of separated sets reappears in the nested-interval proof of the
uncountability of $\mathbb{R}$.

[^lebl-11]: **Lebl**, _Basic Analysis I_, §1.1 — Basic properties: ordered sets, upper and lower bounds, the supremum and infimum, the least-upper-bound property, the field and ordered-field axioms, and the sign rules for an ordered field.
[^lebl-12]: **Lebl**, _Basic Analysis I_, §1.2 — The set of real numbers: the existence and uniqueness of $\mathbb{R}$ as a complete ordered field, the existence of $\sqrt2$ via a supremum, the Archimedean property and density of $\mathbb{Q}$, and the behavior of suprema and infima under translation, scaling, and set separation.