The Real Number System

Prerequisites

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.

1.1 Field Axioms

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.

1.2 Order and the Completeness Axiom

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 $\mathbb{R}$.

1.3 Supremum and Infimum

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.

1.4 Archimedean Property

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 $m/n \in \mathbb{Q}$. $\blacksquare$

1.5 Properties of Supremum and Infimum

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 $-u = \inf(S)$. $\blacksquare$

1.6 Construction of $\mathbb{R}$ via Dedekind Cuts

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 $\alpha = \sup(\mathcal{A})$. $\blacksquare$

1.7 Equivalences of Completeness

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.

:::caution Common Pitfall 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 $(0, 1)$.

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