The Real Number System

Attempt these from memory before reading.

• State the completeness axiom (least upper bound property).
• What is a supremum? How does it differ from a maximum?
• State the Archimedean property.
• Give a Cauchy sequence in $\mathbb{Q}$ that does not converge in $\mathbb{Q}$.
• Name three statements equivalent to completeness.

Check your answers against the text. Prompt 4 is the one that shows completeness is a property $\mathbb{Q}$ does not have, which is what makes it worth stating.

The rationals are a field. They are ordered. They are dense: between any two rationals there is another. And yet calculus does not work on them.

The reason is a single missing property, and it is visible in one construction. Let $a_n = (1 + 1/n)^n$. This sequence is increasing and bounded above by 3, so it ought to converge. Its limit is $e \approx 2.71828$, which is irrational. Working entirely inside $\mathbb{Q}$, the sequence is increasing, bounded, and converges to nothing.

The completeness axiom is the statement that this cannot happen in $\mathbb{R}$: every non-empty set bounded above has a least upper bound. That single axiom is what separates $\mathbb{R}$ from $\mathbb{Q}$, and it is the engine behind every existence theorem on this site — the intermediate value theorem, the extreme value theorem, the monotone convergence theorem, the convergence of Newton’s method, and the Riemann integral.

This page states the axiom, derives the properties that follow, and constructs $\mathbb{R}$ from $\mathbb{Q}$ by Dedekind cuts so that the axiom is not merely asserted.

Exercise. Show that $\sqrt{2} \notin \mathbb{Q}$ by contradiction: if $\sqrt 2 = p/q$ in lowest terms, then $p$ and $q$ are both even. Deduce that the set $\{x \in \mathbb{Q} : x^2 < 2\}$ has no least upper bound in $\mathbb{Q}$.

This chapter assumes the reader is comfortable with:

• Proof techniques: direct proof, proof by contradiction, mathematical induction.
• Basic set theory: sets, subsets, unions, intersections, the power set.
• Functions: domain, range, injectivity, surjectivity, composition.
• Informal calculus: limits, continuity, differentiation, and integration as studied at A-Level or IB Mathematics.

The axioms stated below are taken as given; the purpose of this chapter is to derive their consequences, not to justify the axioms themselves. Readers without a proof-writing background should consult an introductory discrete mathematics text first.

The real numbers $\mathbb{R}$ form a complete ordered field. The field axioms guarantee closure under addition, subtraction, multiplication, and division (by non-zero elements), together with the Usual commutative, associative, and distributive laws.

The order relation $\leq$ on $\mathbb{R}$ satisfies:

1. Reflexivity: $a \leq a$
2. Antisymmetry: $a \leq b$ and $b \leq a$ implies $a = b$
3. Transitivity: $a \leq b$ and $b \leq c$ implies $a \leq c$
4. Totality: for all $a, b$ Either $a \leq b$ or $b \leq a$
5. Compatibility with addition: $a \leq b$ implies $a + c \leq b + c$
6. Compatibility with multiplication: $a \leq b$ and $0 \leq c$ implies $ac \leq bc$

The completeness axiom (also called the least upper bound property) is what distinguishes $\mathbb{R}$ from $\mathbb{Q}$:

Axiom (Completeness). Every non-empty subset of $\mathbb{R}$ that is bounded above has a least Upper bound (supremum) in $\mathbb{R}$.

Let $S \subseteq \mathbb{R}$ be a non-empty set that is bounded above.

Definition. The supremum (or least upper bound) of $S$ Denoted $\sup(S)$ Is the real number $u$ satisfying:

1. $u$ is an upper bound: $s \leq u$ for all $s \in S$.
2. $u$ is the least upper bound: if $v$ is any upper bound of $S$ Then $u \leq v$.

Similarly, the infimum (or greatest lower bound), $\inf(S)$ Is the greatest number $l$ such that $l \leq s$ for all $s \in S$.

Proposition 1.1. $\sup(S)$ exists if and only if $S$ is non-empty and bounded above.

Proposition 1.2 (Approximation Property). If $u = \sup(S)$ Then for every $\varepsilon > 0$ There Exists $s \in S$ such that $u - \varepsilon \lt s \leq u$.

Proof. If no such $s$ existed, then $u - \varepsilon$ would be an upper bound of $S$ strictly less than $u$ Contradicting the definition of $\sup(S)$. $\blacksquare$

Example. Let $S = \{x \in \mathbb{R} : x^2 \lt 2\}$. Then $\sup(S) = \sqrt{2}$. Note that $\sqrt{2} \notin \mathbb{Q}$, so $\mathbb{Q}$ does not satisfy the completeness axiom.

Theorem 1.1 (Archimedean Property). For every $x \in \mathbb{R}$ There exists $n \in \mathbb{N}$ Such that $n \gt x$.

Proof. Suppose, for contradiction, that $\mathbb{N}$ is bounded above. By the completeness axiom, $s = \sup(\mathbb{N})$ exists in $\mathbb{R}$. Then $s - 1$ is not an upper bound for $\mathbb{N}$ so there exists $n \in \mathbb{N}$ with $n \gt s - 1$ I.e., $n + 1 \gt s$. But $n + 1 \in \mathbb{N}$ Contradicting that $s$ is an upper bound. $\blacksquare$

Corollary 1.2. For every $\varepsilon > 0$ There exists $n \in \mathbb{N}$ such that $1/n \lt \varepsilon$.

Proof. By the Archimedean property, choose $n \in \mathbb{N}$ with $n \gt 1/\varepsilon$. Then $1/n \lt \varepsilon$. $\blacksquare$

Corollary 1.3 (Density of $\mathbb{Q}$). Between any two distinct real numbers $a \lt b$ There Exists a rational number $q \in \mathbb{Q}$ with $a \lt q \lt b$.

Proof. Since $b - a > 0$ By Corollary 1.2 there exists $n \in \mathbb{N}$ with $1/n \lt b - a$ So $1 \lt n(b - a) = nb - na$. Let $m = \lfloor na \rfloor + 1 \in \mathbb{Z}$. Then $m - 1 \leq na \lt m$ Giving $m \leq na + 1 \lt na + n(b - a) = nb$. Hence $a \lt m/n \lt b$ And $m/n \in \mathbb{Q}$. $\blacksquare$

Proposition 1.4. If $A$ and $B$ are non-empty bounded subsets of $\mathbb{R}$ Then $\sup(A + B) = \sup(A) + \sup(B)$ Where $A + B = \{a + b : a \in A, b \in B\}$.

Proof. For all $a \in A$ and $b \in B$: $a \leq \sup(A)$ and $b \leq \sup(B)$ So $a + b \leq \sup(A) + \sup(B)$. Thus $\sup(A) + \sup(B)$ is an upper bound for $A + B$ So $\sup(A + B) \leq \sup(A) + \sup(B)$.

For the reverse inequality, let $\varepsilon > 0$. By the approximation property, there exist $a \in A$ And $b \in B$ with $a > \sup(A) - \varepsilon/2$ and $b > \sup(B) - \varepsilon/2$. Then $a + b > \sup(A) + \sup(B) - \varepsilon$ So $\sup(A + B) \geq \sup(A) + \sup(B) - \varepsilon$. Since $\varepsilon > 0$ is arbitrary, $\sup(A + B) \geq \sup(A) + \sup(B)$. $\blacksquare$

Proposition 1.5. For any non-empty bounded set $S \subseteq \mathbb{R}$, $\inf(S) = -\sup(-S)$ Where $-S = \{-s : s \in S\}$.

Proof. Let $u = \sup(-S)$. Then $-s \leq u$ for all $s \in S$ So $s \geq -u$ for all $s \in S$ Meaning $-u$ is a lower bound for $S$. If $v$ is any lower bound for $S$ Then $-v$ is an upper bound for $-S$ So $u \leq -v$ I.e., $-u \geq v$. Hence $-u = \inf(S)$. $\blacksquare$

Worked Example: Find $\sup$ and $\inf$ of $S = \{(-1)^n + 1/n : n \in \mathbb{N}\}$

Solution. The first few terms are $0, 3/2, -2/3, 5/4, -4/5, 7/6, \ldots$.

For even $n = 2k$: $(-1)^{2k} + 1/(2k) = 1 + 1/(2k)$ Which decreases toward $1$ from above. For odd $n = 2k-1$: $(-1)^{2k-1} + 1/(2k-1) = -1 + 1/(2k-1)$ Which increases toward $-1$ from below.

The even terms form the sequence $3/2, 5/4, 7/6, \ldots$ with limit $1$ So $\sup(S) = 3/2$ (the first even term). The odd terms form $0, -2/3, -4/5, \ldots$ with limit $-1$ And since $0$ Is an odd-indexed term, $\inf(S) = -1$ (approached but not attained). $\blacksquare$

Remark. The following outline shows how $\mathbb{R}$ can be constructed from $\mathbb{Q}$ Making the completeness axiom a theorem rather than an axiom.

Definition (Dedekind Cut). A Dedekind cut is a subset $\alpha \subseteq \mathbb{Q}$ satisfying:

1. $\alpha \neq \emptyset$ and $\alpha \neq \mathbb{Q}$
2. If $p \in \alpha$ and $q \lt p$ (with $q \in \mathbb{Q}$), then $q \in \alpha$ (downward closure)
3. $\alpha$ has no greatest element: for every $p \in \alpha$ There exists $q \in \alpha$ with $p \lt q$

Definition. The set of real numbers $\mathbb{R}$ is defined as the set of all Dedekind cuts.

The order, addition, and multiplication are defined as follows:

• Order: $\alpha \lt \beta$ if and only if $\alpha \subsetneq \beta$
• Addition: $\alpha + \beta = \{p + q : p \in \alpha, q \in \beta\}$
• Multiplication: For $\alpha, \beta \geq 0^*$: $\alpha \cdot \beta = \{p \cdot q : p \in \alpha, q \in \beta, p \geq 0, q \geq 0\} \cup \{r \in \mathbb{Q} : r \lt 0\}$

Here $0^* = \{q \in \mathbb{Q} : q \lt 0\}$ represents the real number $0$.

Theorem. With these definitions, $\mathbb{R}$ is a complete ordered field, and $\mathbb{Q}$ embeds Into $\mathbb{R}$ via $q \mapsto \{r \in \mathbb{Q} : r \lt q\}$.

Proof (sketch). Verifying the field axioms and order axioms is lengthy but straightforward. The key Step is the completeness axiom: if $\mathcal{A}$ is a non-empty set of Dedekind cuts bounded above, Then $\alpha = \bigcup_{\beta \in \mathcal{A}} \beta$ is itself a Dedekind cut and $\alpha = \sup(\mathcal{A})$. $\blacksquare$

The completeness axiom can be formulated in several equivalent ways. Each implies the others:

1. Least Upper Bound Property: Every non-empty set bounded above has a supremum.
2. Monotone Convergence Theorem: Every bounded monotone sequence converges.
3. Nested Interval Property: Every nested sequence of closed intervals $I_1 \supseteq I_2 \supseteq \cdots$ with $\mathrm{length}(I_n) \to 0$ has exactly one point in $\bigcap I_n$.
4. Bolzano-Weierstrass Property: Every bounded sequence has a convergent subsequence.
5. Cauchy Completeness: Every Cauchy sequence converges.

Proposition 1.6. In any ordered field, (1) $\iff$ (2) $\iff$ (3) $\iff$ (4) $\iff$ (5).

Proof (outline). We have shown $(1) \Rightarrow (2)$ (MCT in Section 2.2), $(2) \Rightarrow (4)$ (via the bisection argument in Bolzano-Weierstrass), $(4) \Rightarrow (5)$ (Cauchy completeness …/1-number-and-algebra/3_proof-and-logic In Section 2.3), and $(5) \Rightarrow (1)$ can be shown by constructing a Cauchy sequence converging To $\sup S$ from the approximation property. The equivalence $(1) \Rightarrow (3)$ follows from the Nested interval argument, and $(3) \Rightarrow (1)$ follows by constructing nested intervals that Shrink to $\sup S$. $\blacksquare$

Remark. The field $\mathbb{Q}$ satisfies none of these properties, which is why it must be Extended to $\mathbb{R}$ for analysis.

The completeness axiom is the single property that separates $\mathbb{R}$ from $\mathbb{Q}$ and makes calculus possible. Without completeness, limits of Cauchy sequences might not exist, the intermediate value theorem fails, and the Bolzano-Weierstrass theorem is false.

A concrete failure in $\mathbb{Q}$. Consider the sequence $(x_n)$ in $\mathbb{Q}$ defined by $x_1 = 1$ and $x_{n+1} = \frac{x_n}{2} + \frac{1}{x_n}$. This is Newton’s method for solving $x^2 = 2$. The sequence is Cauchy (the terms get arbitrarily close to each other) and every term is rational. But the limit is $\sqrt{2} \notin \mathbb{Q}$. In $\mathbb{Q}$, this Cauchy sequence does not converge. In $\mathbb{R}$, completeness guarantees that it does.

Why the other axioms are not enough. The rational numbers $\mathbb{Q}$ form an ordered field: they satisfy all the field axioms and the order axioms. But $\mathbb{Q}$ has “gaps” --- the irrational numbers are missing. The completeness axiom fills these gaps by requiring that every “gap” (every set bounded above) has a supremum in $\mathbb{R}$. This is why $\mathbb{R}$ is the smallest complete ordered field containing $\mathbb{Q}$.

Connection to physics. Physical measurements are inherently finite, but the mathematical models we use to describe nature (differential equations, probability theory, quantum mechanics) assume the continuum. The completeness of $\mathbb{R}$ is what makes these models well-defined: it guarantees that the solutions to differential equations exist, that probabilities sum to 1, and that infinite processes (like series and integrals) converge when they “should.”

Problem. Find $\sup(S)$ and $\inf(S)$ where $S = \{x \in \mathbb{R} : x^2 - 3x + 2 < 0\}$.

Solution

Factor: $x^2 - 3x + 2 = (x - 1)(x - 2)$. The inequality $(x - 1)(x - 2) < 0$ holds when $1 < x < 2$. Therefore $S = (1, 2)$.

$\sup(S) = 2$ (the least upper bound; $2 \notin S$ but every element of $S$ is less than 2). $\inf(S) = 1$ (the greatest lower bound; $1 \notin S$ but every element of $S$ is greater than 1).

Note that neither the supremum nor the infimum belongs to $S$, since $S$ is an open interval. This illustrates that the completeness axiom guarantees the existence of $\sup$ and $\inf$ in $\mathbb{R}$, but they need not be elements of the set.

Approximation property check: For any $\varepsilon > 0$, there exists $x \in S$ with $2 - \varepsilon < x \leq 2$ (take $x = 2 - \varepsilon/2$ for small enough $\varepsilon$, as long as $x > 1$). This confirms that $\sup(S) = 2$. $\blacksquare$

The real numbers are the unique complete ordered field. The completeness axiom says that every non-empty set bounded above has a least upper bound, which is the property that fills in all the “gaps” in the rationals. The rational numbers have holes: for example, the set of rationals less than $\sqrt{2}$ is bounded above but has no supremum in $\mathbb{Q}$ because $\sqrt{2}$ is irrational. Completeness fills these holes, ensuring that limits of Cauchy sequences always converge.

The practical consequence is that calculus works. Without completeness, sequences that “should” converge might not. The sequence $1, 1.4, 1.41, 1.414, \ldots$ (successive approximations to $\sqrt{2}$) is Cauchy in $\mathbb{Q}$ but has no limit there. In $\mathbb{R}$, completeness guarantees convergence. The supremum and infimum are the workhorses of analysis: they allow you to define limits, continuity, and integrals without assuming the limit point exists. The supremum is the “best possible upper bound,” which is weaker than requiring a maximum (an element of the set that achieves the bound). This distinction is crucial: $\sup(0,1) = 1$ but $\max(0,1)$ does not exist.

The completeness axiom is often misstated as "every bounded set has a supremum." The set must be non-empty. Also, completeness does not say every set has a maximum;

$\sup(S)$ need not belong to $S$. For example, $\sup\{1/n : n \in \mathbb{N}\} = 1$ which belongs to the set, but $\sup(0, 1) = 1$ which does not belong to $(0, 1)$.

• Do not confuse supremum with maximum. The maximum of a set must be an element of the set; the supremum need not be. If $\sup(S) \in S$, then $\sup(S) = \max(S)$.
• The completeness axiom is specific to $\mathbb{R}$. In $\mathbb{Q}$, the set $\{x \in \mathbb{Q} : x^2 < 2\}$ is bounded above but has no supremum in $\mathbb{Q}$ (since $\sqrt{2} \notin \mathbb{Q}$).

The density row is the one that surprises people: $\mathbb{Q}$ is dense, so “no gaps you can see” is not what completeness means. The gaps are not visible as missing points between others; they are visible only as limits of Cauchy sequences that fail to arrive. That is why the completeness axiom has to be stated in terms of suprema rather than density.

The equivalences of completeness. The following are equivalent in an ordered field: (a) every non-empty bounded set has a supremum; (b) every Cauchy sequence converges; (c) every bounded monotone sequence converges; (d) the intermediate value theorem holds for every continuous function. Proving the equivalences is the cleanest way to see that completeness is one idea with four faces, and Sequences and Limits uses (a) and (b) while Continuity uses (d).

Mix these with material from Sequences and Limits, Series and Continuity.

1. (Suprema.) Find $\sup\{x \in \mathbb{Q} : x^2 < 2\}$ in $\mathbb{R}$, and prove it is the supremum using $x^2 < 2 \implies x < \sqrt 2$.
2. (Archimedean.) Show that for every real $x$ there is an integer $n$ with $n

   x$, and deduce that$1/n \to 0$.

3. (Completeness.) Show that the intersection of the nested intervals $[n, \infty)$ is empty, and explain why this does not contradict the nested interval property (which requires bounded intervals).
4. (Dedekind cuts.) Show that the cut $\{x \in \mathbb{Q} : x^2 < 2\}$ is a valid Dedekind cut, and that it corresponds to an irrational.
5. (Equivalences.) Show that the monotone convergence theorem implies the existence of suprema, giving one direction of the equivalence.

• Site Home: Main landing page for Mathematics notes.

• Linear Algebra: Vector spaces, matrices, and linear transformations.

• Real Analysis: Rigorous treatment of real numbers and calculus.

• Practice: Practice problems for revision.

• Quantum Mechanics

• Graph Theory

• Classical Mechanics

• Electromagnetism

• Statistical Learning

• Statistical Mechanics

• Confusing supremum with maximum: The maximum must be an element of the set; the supremum need not be. $\sup(0,1) = 1$ but $\max(0,1)$ does not exist. Always check whether $\sup(S) \in S$ before claiming it is a maximum.
• Assuming every bounded set has a supremum in $\mathbb{Q}$: The completeness axiom holds only in $\mathbb{R}$. In $\mathbb{Q}$, sets like $\{x \in \mathbb{Q} : x^2 < 2\}$ are bounded above but have no supremum in $\mathbb{Q}$ because $\sqrt{2} \notin \mathbb{Q}$.
• Omitting “non-empty” when stating completeness: The completeness axiom requires the set to be non-empty and bounded above. An empty set is bounded above but has no supremum (or, by convention, $\sup \varnothing = -\infty$, which is not a real number).
• Treating density and completeness as the same property: The rationals are dense in $\mathbb{R}$ (between any two reals there is a rational), but they are not complete. Density means no gaps between elements; completeness means no missing limit points.

• Continuity
• Differentiability
• Problem Set
• Riemann Integration