Axioms of the Real Numbers, Supremum and Infimum

Every theorem of real analysis — every limit, every continuity argument, every integral — ultimately rests on a single fact about $\mathbb{R}$ that the rationals $\mathbb{Q}$ do not share: completeness. This chapter makes that fact precise. We treat $\mathbb{R}$ axiomatically, as Rudin does, rather than constructing it from $\mathbb{Q}$ by Dedekind cuts or Cauchy sequences; what matters for everything that follows is not how $\mathbb{R}$ is built but which properties it has, and that those properties suffice to prove the existence of suprema and infima.

Have to hand: the field axioms (associativity, commutativity, distributivity, existence of $0,1$ and of additive/multiplicative inverses) and the notion of an order relation. We assume these are familiar from school algebra and spend our effort on the one axiom that is new: completeness, stated as the existence of least upper bounds.

Why this matters. Without completeness, $\sup\{x \in \mathbb{Q} : x^2 < 2\}$ does not exist in $\mathbb{Q}$. Every later theorem — Bolzano–Weierstrass, the Intermediate Value Theorem, the existence of $\int_a^b f$ — is a disguised consequence of the least upper bound property. If you can only remember one definition from this course, make it this one.

1. Ordered Fields

A field $(F, +, \cdot)$ is a set with two operations satisfying the usual nine field axioms (associativity and commutativity of $+$ and $\cdot$, distributivity, the two identities with $0 \ne 1$, additive inverses, and multiplicative inverses for nonzero elements; closure is built into the notion of an operation). An ordered field is a field $F$ equipped with a relation $<$ satisfying, for all $x, y, z \in F$:

• (Trichotomy) exactly one of $x < y$, $x = y$, $y < x$ holds;
• (Transitivity) $x < y$ and $y < z$ imply $x < z$;
• (Translation invariance) $x < y$ implies $x + z < y + z$;
• (Compatibility with multiplication) $x < y$ and $0 < z$ imply $xz < yz$.

Both $\mathbb{Q}$ and $\mathbb{R}$ are ordered fields. What distinguishes them is the next axiom, which $\mathbb{Q}$ fails.

2. Upper Bounds, Supremum and the Completeness Axiom

Let $S \subseteq \mathbb{R}$ be nonempty.

Definition. $M \in \mathbb{R}$ is an upper bound of $S$ if $x \le M$ for every $x \in S$. $S$ is bounded above if it has an upper bound. A number $\alpha \in \mathbb{R}$ is the supremum (least upper bound) of $S$, written $\alpha = \sup S$, if

1. $\alpha$ is an upper bound of $S$, and
2. for every upper bound $M$ of $S$, $\alpha \le M$.

Lower bounds, "bounded below", and the infimum $\inf S$ (greatest lower bound) are defined symmetrically, reversing all inequalities.

Axiom (Completeness / Least Upper Bound Property). Every nonempty subset $S \subseteq \mathbb{R}$ that is bounded above has a supremum $\sup S \in \mathbb{R}$.

This is an axiom, not a theorem — it is exactly the property that separates $\mathbb{R}$ from $\mathbb{Q}$. Everything else in this chapter is a consequence of it. The dual statement — every nonempty set bounded below has an infimum — follows immediately: if $S$ is bounded below, apply the axiom to $-S = \{-x : x \in S\}$, which is bounded above, and check that $\sup(-S) = -\inf S$.

3. An Equivalent Characterisation of the Supremum

Checking condition 2 of the definition directly (against every upper bound) is rarely convenient. The following criterion is the one used in practice.

Proposition. Let $S$ be nonempty and bounded above, and let $\alpha$ be an upper bound of $S$. Then $\alpha = \sup S$ if and only if for every $\varepsilon > 0$ there exists $x \in S$ with $x > \alpha - \varepsilon$.

Proof. ($\Rightarrow$) Suppose $\alpha = \sup S$ and, for contradiction, that some $\varepsilon > 0$ has no $x \in S$ exceeding $\alpha - \varepsilon$; then $\alpha - \varepsilon$ is an upper bound of $S$ smaller than $\alpha$, contradicting minimality of $\alpha$ among upper bounds.

($\Leftarrow$) Suppose the $\varepsilon$-condition holds. If $\alpha$ were not the least upper bound, some upper bound $M < \alpha$ would exist; take $\varepsilon = \alpha - M > 0$. The condition gives $x \in S$ with $x > \alpha - \varepsilon = M$, contradicting that $M$ is an upper bound. $\blacksquare$

Worked example. Let $S = \left\{1 - \dfrac{1}{n} : n \in \mathbb{N}\right\}$. Every element satisfies $1 - \frac{1}{n} < 1$, so $1$ is an upper bound. Given $\varepsilon > 0$, choose (by the Archimedean property, proved in Section 4) $n$ with $\frac{1}{n} < \varepsilon$; then $1 - \frac{1}{n} > 1 - \varepsilon$. By the proposition, $\sup S = 1$, even though $1 \notin S$ — the supremum need not belong to the set.

4. The Archimedean Property

Theorem (Archimedean Property). For every $x \in \mathbb{R}$ there exists $n \in \mathbb{N}$ with $n > x$. Equivalently, for every $\varepsilon > 0$ there exists $n \in \mathbb{N}$ with $\dfrac{1}{n} < \varepsilon$.

Proof. Suppose not: some $x \in \mathbb{R}$ satisfies $n \le x$ for all $n \in \mathbb{N}$, so $\mathbb{N}$ is bounded above by $x$. Since $\mathbb{N}$ is nonempty, completeness gives $\alpha = \sup \mathbb{N} \in \mathbb{R}$. By the characterisation of Section 3 with $\varepsilon = 1$, there is $n \in \mathbb{N}$ with $n > \alpha - 1$, so $n + 1 > \alpha$. But $n + 1 \in \mathbb{N}$, contradicting that $\alpha$ is an upper bound of $\mathbb{N}$. $\blacksquare$

Key distinction. The Archimedean property is not a consequence of the ordered field axioms alone — there exist ordered fields (containing "infinitesimals") where it fails. It is proved here entirely from completeness. This is the first of many results in this course that look "obvious" but need a hypothesis beyond the ordered-field axioms; here completeness supplies it.

5. Density of $\mathbb{Q}$ in $\mathbb{R}$

Theorem. Between any two real numbers $a < b$ there exists a rational number $r$ with $a < r < b$.

Proof. Since $b - a > 0$, the Archimedean property gives $n \in \mathbb{N}$ with $n(b-a) > 1$, i.e. $nb - na > 1$. Consider the set of integers $m$ with $m > na$; this set is nonempty and bounded below (again by the Archimedean property applied to $-na$), so it has a least element $m_0$ (a bounded-below nonempty subset of $\mathbb{Z}$ has a minimum, by the well-ordering property of $\mathbb{N}$ applied after a shift). Minimality gives $m_0 - 1 \le na$, hence

Dividing by $n > 0$: $a < \dfrac{m_0}{n} < b$. Take $r = \dfrac{m_0}{n} \in \mathbb{Q}$. $\blacksquare$

A near-identical argument (applied to $a/\sqrt 2 < b/\sqrt2$, say, granting the existence of irrationals) shows the irrationals $\mathbb{R}\setminus\mathbb{Q}$ are also dense in $\mathbb{R}$. Density of $\mathbb{Q}$ is what makes it possible to approximate any real number by rationals to arbitrary precision — the starting point of every numerical scheme.

6. Existence of $\sqrt 2$: Completeness in Action

This is the paradigm example of why completeness is indispensable.

Proposition. There exists $\alpha \in \mathbb{R}$ with $\alpha^2 = 2$, but no such $\alpha$ exists in $\mathbb{Q}$.

Proof sketch. Let $S = \{x \in \mathbb{R} : x > 0,\ x^2 < 2\}$. $S$ is nonempty ($1 \in S$) and bounded above (by $2$), so completeness gives $\alpha = \sup S \in \mathbb{R}$. Trichotomy forces one of $\alpha^2 < 2$, $\alpha^2 = 2$, $\alpha^2 > 2$. If $\alpha^2 < 2$, a short computation shows that for small enough $h > 0$, $(\alpha+h)^2 = \alpha^2 + 2\alpha h + h^2 < 2$ (choose $h < \min\left(1, \dfrac{2-\alpha^2}{2\alpha+1}\right)$), so $\alpha + h \in S$, contradicting that $\alpha$ is an upper bound. If $\alpha^2 > 2$, an analogous computation produces $k > 0$ with $(\alpha - k)^2 > 2$ and $\alpha - k$ still an upper bound of $S$, contradicting minimality of $\alpha$. Hence $\alpha^2 = 2$. That no rational squares to $2$ is the classical parity argument: if $\alpha = p/q$ in lowest terms, $p^2 = 2q^2$ forces $p$ even, $p = 2p'$, then $2p'^2 = q^2$ forces $q$ even too, contradicting $\gcd(p,q)=1$. $\blacksquare$

This is exactly why $S = \{x \in \mathbb{Q} : x^2 < 2\}$ has rational upper bounds but no least rational upper bound: $\mathbb{Q}$ is an ordered field that is not complete.

7. The Nested Interval Property

Theorem. Let $I_n = [a_n, b_n]$ be a sequence of nonempty closed bounded intervals with $I_1 \supseteq I_2 \supseteq I_3 \supseteq \cdots$. Then $\bigcap_{n=1}^{\infty} I_n \ne \emptyset$. If, in addition, $b_n - a_n \to 0$, the intersection is a single point.

Proof. The nesting gives $a_1 \le a_2 \le \cdots \le b_n \le \cdots \le b_2 \le b_1$ for every $n$, so the set $\{a_n : n \in \mathbb{N}\}$ is bounded above (by any $b_m$). By completeness, $\alpha = \sup_n a_n$ exists. For each fixed $m$, every $b_m$ is an upper bound of $\{a_n\}$ (since $a_n \le b_n \le b_m$ for $n \ge m$, and $a_n \le a_m \le b_m$ for $n < m$), so $\alpha \le b_m$; also $a_m \le \alpha$ by definition of supremum. Hence $\alpha \in I_m$ for every $m$, so $\alpha \in \bigcap_n I_n$. If $b_n - a_n \to 0$, uniqueness follows because any two points of the intersection differ by less than $b_n - a_n$ for every $n$, forcing them equal. $\blacksquare$

This theorem is the real-line precursor to compactness (Chapter 3) and is used to construct Cauchy sequences with prescribed properties throughout analysis.

Common Exam Pitfalls

• Writing "$\max S$" when the problem asks for $\sup S$. The supremum need not belong to $S$ (Section 3's example); only ask for $\max S$ when you can exhibit an element of $S$ equal to it.
• Proving "$\alpha$ is an upper bound" and stopping. That only shows $\alpha \ge \sup S$; you must also show no smaller upper bound exists, via the $\varepsilon$-criterion or a direct minimality argument.
• Treating the Archimedean property as "obvious". It is a theorem that uses completeness; in an exam that asks you to derive it, quoting it without proof loses the marks.
• Confusing bounded above with bounded. "Bounded" means bounded above and below; a set can be bounded above with no lower bound at all (e.g. $(-\infty, 0]$).
• Sign errors when passing from sup to inf via $-S$. Remember $\inf S = -\sup(-S)$, not $\sup(-S)$; check with a one-element example if unsure.
• Assuming every bounded set has a supremum that is easy to compute. For sets defined by a condition like $x^2 < 2$, the supremum is characterised by the proof technique of Section 6, not by solving an equation directly.

Summary

• An ordered field satisfies trichotomy, transitivity, translation invariance and compatibility of $<$ with multiplication by positives; $\mathbb{Q}$ and $\mathbb{R}$ both qualify.
• Completeness axiom: every nonempty $S \subseteq \mathbb{R}$ bounded above has $\sup S \in \mathbb{R}$; this is the property $\mathbb{Q}$ lacks.
• $\alpha = \sup S$ iff $\alpha$ is an upper bound and $\forall \varepsilon>0\ \exists x \in S: x > \alpha - \varepsilon$.
• Archimedean property: $\forall x \in \mathbb{R}\ \exists n \in \mathbb{N}: n > x$ — proved from completeness, not assumed.
• Density of $\mathbb{Q}$ (and of $\mathbb{R}\setminus\mathbb{Q}$) in $\mathbb{R}$: between any $a<b$ there is a rational (and an irrational).
• $\sqrt 2 \in \mathbb{R} \setminus \mathbb{Q}$: constructed as $\sup\{x>0 : x^2<2\}$, the archetypal example of why completeness matters.
• Nested Interval Property: $I_1 \supseteq I_2 \supseteq \cdots$ closed bounded intervals have nonempty intersection, a single point if $b_n - a_n \to 0$.