Mathematics.

sequences and convergence

Completeness of the Real Numbers

Real Analysis35 minDifficulty4 out of 10

You should know: cauchy sequences, real numbers

Overview

Completeness is the property distinguishing the real numbers ℝ from the rationals ℚ: informally, ℝ has 'no gaps.' It can be stated in several logically equivalent ways — the least-upper-bound (supremum) property, the Cauchy completeness property, the Monotone Convergence property, and the Nested Interval property — all of which hold in ℝ but fail in ℚ. The most common formulation, due to Dedekind, is the least-upper-bound property: every nonempty subset of ℝ that is bounded above has a least upper bound (supremum) in ℝ. Completeness is the single axiom (beyond the ordered field axioms shared with ℚ) that makes calculus work: it underlies the Intermediate Value Theorem, the Extreme Value Theorem, and the existence of limits of Cauchy sequences.

Intuition

Think of ℝ as ℚ with every 'missing point' filled in. The set {q ∈ ℚ : q² < 2} is bounded above in ℚ (by 2, for instance) but has no least upper bound within ℚ — the natural candidate √2 simply isn't a rational number, so the supremum is 'missing.' Completeness asserts that in ℝ this can never happen: bounded-above sets always have a genuine least upper bound sitting in ℝ itself. Equivalently, if you build a sequence of rational approximations that keep tightening in on some target (a Cauchy sequence), ℝ guarantees that target actually exists as a real number, whereas ℚ offers no such guarantee.

Formal Definition

Definition

Several equivalent formulations of completeness for an ordered field such as ℝ:

∀S⊆R, S≠∅, S bounded above  ⟹  sup⁡S exists in R\forall S \subseteq \mathbb{R},\ S \neq \emptyset,\ S \text{ bounded above} \implies \sup S \text{ exists in } \mathbb{R}
Least-upper-bound (supremum) property
(an) is Cauchy  ⟺  (an) converges in R(a_n) \text{ is Cauchy} \iff (a_n) \text{ converges in } \mathbb{R}
Cauchy completeness
Every bounded monotone sequence in R converges.\text{Every bounded monotone sequence in } \mathbb{R} \text{ converges.}
Monotone Convergence property
I1⊇I2⊇I3⊇⋯ ,  ∣In∣→0  ⟹  ⋂n=1∞In≠∅ (exactly one point)I_1 \supseteq I_2 \supseteq I_3 \supseteq \cdots,\ \ |I_n| \to 0 \implies \bigcap_{n=1}^{\infty} I_n \neq \emptyset \text{ (exactly one point)}
Nested Interval property (for closed bounded intervals Iₙ)

Worked Examples

  1. S is bounded above in ℚ (e.g. by 2), so if ℚ were complete, sup S would be a rational number r with r² = 2.

    r2=2r^2 = 2

  2. But √2 is irrational (classical proof by contradiction using unique factorization), so no rational r satisfies r² = 2.

    2∉Q\sqrt{2} \notin \mathbb{Q}

  3. Hence S has an upper bound in ℚ but no least upper bound in ℚ — the supremum 'exists' only once we pass to ℝ.

    sup⁡S=2∈R∖Q\sup S = \sqrt{2} \in \mathbb{R} \setminus \mathbb{Q}

Answer: S is bounded above in ℚ but sup S = √2 ∉ ℚ, so ℚ fails the least-upper-bound property.

Practice Problems

Difficulty 4/10

Give the supremum and infimum of S = {1 − 1/n : n ∈ ℕ, n ≥ 1} in ℝ.

Difficulty 3/10

Which of the following is NOT equivalent to completeness of an ordered field?

Difficulty 6/10

Assuming the least-upper-bound property, prove that every Cauchy sequence of real numbers converges.

Quiz

The least-upper-bound property states that:
Why does {q ∈ ℚ : q² < 2} fail to have a supremum in ℚ?

Summary

  • Completeness of ℝ can be stated as the least-upper-bound property: every nonempty, bounded-above subset of ℝ has a supremum in ℝ.
  • It is equivalent to Cauchy completeness, the Monotone Convergence property, and the Nested Interval property.
  • ℚ fails completeness — e.g. {q∈ℚ : q²<2} is bounded above but has no rational least upper bound — which is exactly why ℝ is needed for calculus.

References

  1. WebsiteWikipedia — Completeness of the real numbers