Ordered Fields and the Completeness Axiom

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 $\sqrt{2}$ 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.¹

Ordered sets, bounds, and suprema

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.

The supremum, when it exists, is unique: two least upper bounds $b,b^{\prime }$ each satisfy $b\le b^{\prime }$ and $b^{\prime }\le b$, so $b=b^{\prime }$. 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.

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

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

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

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.

From these two compatibility rules the familiar sign laws follow, the ones used in nearly every later inequality:¹

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.

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 $ϵ>0$ silently means $x,ϵ\in \mathbb{R}$.

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 $\inf A=-\sup B$.

Read together with its mirror image, this says: to prove $x=0$, show $|x|\le ϵ$ for every $ϵ>0$; to prove $x\le y$, show $x\le y+ϵ$ for every $ϵ>0$. 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.

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 $\mathbb{R}$ 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.

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

The sequence $1/n$ therefore gets arbitrarily small, the basic fact behind the theory of limits.

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

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 $cA=\{ca:a\in A\}$.²

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.

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 $\mathbb{R}$.