Completeness Axiom

The completeness axiom is the defining property of the real numbers. It states that $\mathbb{R}$ has "no gaps" — every nonempty set of reals that is bounded above has a least upper bound. This axiom is the foundation for virtually all results in real analysis, from the convergence of sequences to the intermediate value theorem.

The least upper bound property

Definition1.1Completeness Axiom (Least Upper Bound Property)

Let $S \subseteq \mathbb{R}$ be nonempty and bounded above. Then $S$ has a least upper bound (supremum) in $\mathbb{R}$ . That is, there exists $\alpha \in \mathbb{R}$ such that:

$x \leq \alpha$ for all $x \in S$ (upper bound).
If $\beta$ is any upper bound of $S$ , then $\alpha \leq \beta$ (least).

We write $\alpha = \sup S$ .

RemarkUniqueness

The supremum, if it exists, is unique. If $\alpha$ and $\alpha'$ are both least upper bounds, then $\alpha \leq \alpha'$ and $\alpha' \leq \alpha$ , so $\alpha = \alpha'$ .

RemarkInfimum

By symmetry, every nonempty set $S$ that is bounded below has a greatest lower bound (infimum), denoted $\inf S$ . Indeed, $\inf S = -\sup(-S)$ , where $-S = \{-x \mid x \in S\}$ .

ExampleSupremum of a bounded set

Let $S = \{1 - 1/n \mid n \in \mathbb{N}\} = \{0, 1/2, 2/3, 3/4, \ldots\}$ . Then $S$ is bounded above (by $1$ , for instance). The least upper bound is $\sup S = 1$ . Note that $1 \notin S$ , so the supremum need not be an element of $S$ .

ExampleSupremum when no maximum exists

Let $S = \{x \in \mathbb{R} \mid x^2 < 2\}$ . Then $S$ is bounded above (e.g., by $2$ ). The supremum is $\sup S = \sqrt{2}$ , which is not in $S$ (since $(\sqrt{2})^2 = 2$ , not $< 2$ ). The set $S$ has no maximum element, but it has a least upper bound.

ExampleWhen sup is attained

Let $S = [0, 1] = \{x \in \mathbb{R} \mid 0 \leq x \leq 1\}$ . Then $\sup S = 1$ , and $1 \in S$ . In this case, the supremum is also the maximum: $\max S = 1$ . A set has a maximum if and only if $\sup S \in S$ .

ExampleUnbounded sets

Let $S = \mathbb{N} = \{1, 2, 3, \ldots\}$ . Then $S$ is not bounded above, so the completeness axiom does not apply. We write $\sup S = +\infty$ by convention, but this is not a real number.

Equivalent formulations

The completeness axiom has several equivalent formulations, each useful in different contexts.

Theorem1.1Equivalent forms of completeness

The following are equivalent for an ordered field $F$ :

(Least upper bound property) Every nonempty subset of $F$ that is bounded above has a supremum in $F$ .
(Greatest lower bound property) Every nonempty subset of $F$ that is bounded below has an infimum in $F$ .
(Monotone convergence) Every bounded monotone sequence in $F$ converges.
(Nested intervals) If $\{I_n\}$ is a sequence of closed bounded intervals with $I_{n+1} \subseteq I_n$ for all $n$ , then $\bigcap_{n=1}^\infty I_n \neq \varnothing$ .
(Cauchy completeness) Every Cauchy sequence in $F$ converges.

RemarkProof strategy

These are proved in Monotone Convergence Theorem and Cauchy Criterion. The key idea is that the least upper bound property implies monotone convergence, which implies nested intervals, which implies Cauchy completeness, which in turn implies the least upper bound property (by taking the limit of Cauchy sequences of upper bounds).

ExampleNested intervals property

Let $I_n = [0, 1/n]$ for $n \geq 1$ . Then $I_{n+1} \subseteq I_n$ for all $n$ , and $\bigcap_{n=1}^\infty I_n = \{0\}$ (nonempty, as required). The completeness axiom ensures this intersection is always nonempty when the intervals are nested and closed.

ExampleOpen intervals can have empty intersection

Let $J_n = (0, 1/n)$ (open intervals). Then $J_{n+1} \subseteq J_n$ for all $n$ , but $\bigcap_{n=1}^\infty J_n = \varnothing$ . The nested interval property requires closed intervals.

Applications of completeness

The completeness axiom is the engine behind virtually every major theorem in real analysis.

ExampleIntermediate Value Theorem

The Intermediate Value Theorem (IVT) states: if $f : [a, b] \to \mathbb{R}$ is continuous and $f(a) < 0 < f(b)$ , then there exists $c \in (a, b)$ with $f(c) = 0$ .

Proof sketch: Define $S = \{x \in [a, b] \mid f(x) < 0\}$ . By completeness, $c = \sup S$ exists. Using continuity of $f$ , one shows $f(c) = 0$ . Without completeness (e.g., over $\mathbb{Q}$ ), the IVT fails: $f(x) = x^2 - 2$ is continuous on $\mathbb{Q}$ , but has no rational root.

ExampleExtreme Value Theorem

The Extreme Value Theorem (EVT) states: if $f : [a, b] \to \mathbb{R}$ is continuous, then $f$ attains its maximum and minimum on $[a, b]$ .

Proof sketch: Let $M = \sup\{f(x) \mid x \in [a, b]\}$ (exists by completeness). One can find a sequence $x_n \in [a, b]$ with $f(x_n) \to M$ . By Bolzano-Weierstrass (which relies on completeness), $\{x_n\}$ has a convergent subsequence $x_{n_k} \to c \in [a, b]$ . By continuity, $f(c) = M$ .

ExampleBolzano-Weierstrass Theorem

Every bounded sequence in $\mathbb{R}$ has a convergent subsequence. This is a consequence of completeness via the nested intervals property (bisect intervals repeatedly to trap infinitely many sequence terms in arbitrarily small intervals).

ExampleHeine-Borel Theorem

A subset $K \subseteq \mathbb{R}$ is compact (every open cover has a finite subcover) if and only if $K$ is closed and bounded. The proof crucially uses completeness (via nested intervals or sequential compactness).

Archimedean property and density

Theorem1.2Archimedean Property

For all $x, y \in \mathbb{R}$ with $x > 0$ , there exists $n \in \mathbb{N}$ such that $nx > y$ .

Proof

Suppose not: $nx \leq y$ for all $n \in \mathbb{N}$ . Then the set $S = \{nx \mid n \in \mathbb{N}\}$ is bounded above by $y$ . By completeness, $\alpha = \sup S$ exists. Since $(n+1)x \in S$ , we have $(n+1)x \leq \alpha$ for all $n$ , hence $nx \leq \alpha - x$ . Thus $\alpha - x$ is also an upper bound of $S$ , contradicting that $\alpha$ is the least upper bound (since $x > 0$ ).

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RemarkNo infinitesimals in R

The Archimedean property says there are no "infinitely large" or "infinitely small" real numbers. Every positive real can be exceeded by some integer multiple of any other positive real.

Theorem1.3Density of Q in R

Between any two real numbers, there exists a rational number. That is, if $x < y$ in $\mathbb{R}$ , there exists $q \in \mathbb{Q}$ with $x < q < y$ .

Proof

By the Archimedean property, choose $n \in \mathbb{N}$ such that $n(y - x) > 1$ . Consider the set $S = \{m \in \mathbb{Z} \mid m > nx\}$ . By the Archimedean property, $S$ is nonempty. Let $m_0$ be the least element of $S$ (by well-ordering of $\mathbb{Z}$ ). Then $m_0 - 1 \leq nx < m_0$ , so $nx < m_0 \leq nx + 1 < nx + n(y - x) = ny$ . Thus $q = m_0/n$ satisfies $x < q < y$ .

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ExampleApproximating irrationals

Since $\mathbb{Q}$ is dense in $\mathbb{R}$ , every real number is the limit of a sequence of rationals. For instance, $\pi = \lim_{n \to \infty} r_n$ , where $r_n$ are the decimal truncations $3, 3.1, 3.14, 3.141, \ldots$ This is the basis for decimal expansions.

The gap between Q and R

RemarkQ is incomplete

The rationals $\mathbb{Q}$ do not satisfy the completeness axiom. For example, $S = \{q \in \mathbb{Q} \mid q^2 < 2\}$ is bounded above in $\mathbb{Q}$ , but has no supremum in $\mathbb{Q}$ (the supremum would be $\sqrt{2}$ , which is irrational). This "gap" is precisely what $\mathbb{R}$ fills.

ExampleFailure of completeness in Q

Let $a_n = (1 + 1/n)^n$ for $n \in \mathbb{N}$ . Then $\{a_n\}$ is a Cauchy sequence in $\mathbb{Q}$ (with rational terms), but its limit is $e \approx 2.71828\ldots$ , which is irrational. In $\mathbb{Q}$ , this Cauchy sequence does not converge. Completeness fails.

RemarkR is the completion of Q

$\mathbb{R}$ is the completion of $\mathbb{Q}$ with respect to the usual metric $|x - y|$ . Every Cauchy sequence of rationals converges to a (unique) real number. This is why we say $\mathbb{R}$ "completes" $\mathbb{Q}$ .

Summary

The completeness axiom is the heart of real analysis:

Every nonempty bounded set has a supremum (and infimum).
Equivalent formulations include monotone convergence, nested intervals, and Cauchy completeness.
Completeness implies the Archimedean property and density of $\mathbb{Q}$ in $\mathbb{R}$ .
Major theorems (IVT, EVT, Bolzano-Weierstrass, Heine-Borel) all rely on completeness.

Without completeness, analysis over $\mathbb{Q}$ would fail catastrophically. The next chapters develop the theory of sequences and limits, building directly on the completeness axiom. See Monotone Convergence Theorem for the first major application.