TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)

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The reals are complete

Statement

Every Cauchy sequence of real numbers (Limits and Cauchy sequences of reals) converges to a real number. Together with The reals form a totally ordered field, this completes the construction: $R$ is a complete totally ordered field. The proof uses no form of the axiom of choice.

Facts & Assumptions

Given: A Cauchy sequence $(x_{k})_{k \geq 1}$ of reals.

[L1]

Rational approximation: for any real $z$ and rational $η > 0$ there is $q$ with $∣ z - \hat{q} ∣ < \hat{η}$ (The rationals embed densely in the reals).

[L2]

Archimedean property: for rational $ε > 0$ there is $k$ with $1 / k < ε$ (The rationals are Archimedean).

[L3]

Cauchy definitions in $Q$ and $R$ (Cauchy sequence of rationals, Limits and Cauchy sequences of reals).

[L4]

The embedding preserves and reflects order and arithmetic; triangle inequality in $R$ (The rationals embed densely in the reals, The reals form a totally ordered field, Order on the reals).

[L5]

Reals are classes of rational Cauchy sequences (The real numbers).

[L6]

Every rational has a positive-denominator integer representative; the nonnegative integers are the embedded naturals, with compatible arithmetic and order (Every rational has a positive-denominator representative, The naturals embed in the integers, The integers form a totally ordered ring).

[L7]

Every nonempty subset of $N$ has a unique least element (The well-ordering principle). Cartesian products are sets (The Cartesian product $A \times B : = {z \in P (P (A \cup B)) : \exists a \in A \exists b \in B z = (a, b)}$ ), and a uniquely specified subset of an existing set is formed by Separation (The Axiom Schema of Separation: for each formula $φ$ , $\forall \overset{ˉ}{p} \forall x \exists y \forall z (z \in y \leftrightarrow (z \in x \land φ (z, \overset{ˉ}{p})))$ ).

Proof

technique · direct

1.1

For a fixed $k \geq 1$ , call a triple $(h, b, j)$ of naturals admissible when $h \geq 1$ , $1 \leq b \leq h$ , $0 \leq j \leq 2 h$ , and $∣ x_{k} - \hat{(j - h) / b} ∣ < \hat{1 / k}$ . Here $j - h$ is an integer. Such a triple exists: [L1] supplies one approximating rational $a / b$ ; [L6] makes $b$ a positive natural, and either $a$ or $- a$ is a nonnegative integer. Thus some natural $h$ satisfies $h \geq b$ and $- h \leq a \leq h$ . Then $j = a + h$ is a natural with $j \leq 2 h$ and $(j - h) / b = a / b$ . This proves nonemptiness separately for each $k$ ; it does not choose a family of witnesses.

L1L6construct

2.1

Let $h_{k}$ be the least first coordinate of an admissible triple; with $h_{k}$ fixed, let $b_{k}$ be the least admissible second coordinate; with both fixed, let $j_{k}$ be the least admissible third coordinate. Each minimum exists and is unique by [L7]. Define $q_{k} = (j_{k} - h_{k}) / b_{k}$ . This unique rule defines the graph of $(q_{k})$ as a subset of $N_{\geq 1} \times Q$ by Separation. Consequently $∣ x_{k} - {\hat{q}}_{k} ∣ < \hat{1 / k}$ for every $k$ , without choosing representatives of all the $x_{k}$ or invoking any choice axiom.

step 1.1L6L7construct

3.1

$(q_{k})$ is Cauchy in $Q$ : given rational $ε > 0$ , pick $k_{0}$ with $1 / k_{0} < ε / 3$ and $K$ with $∣ x_{k} - x_{l} ∣ < \hat{ε / 3}$ for $k, l \geq K$ ; then for $k, l \geq \max (k_{0}, K)$ , $\hat{∣ q_{k} - q_{l} ∣} \leq ∣ {\hat{q}}_{k} - x_{k} ∣ + ∣ x_{k} - x_{l} ∣ + ∣ x_{l} - {\hat{q}}_{l} ∣ < \hat{1 / k} + \hat{ε / 3} + \hat{1 / l} \leq 3 \hat{ε / 3} = \hat{ε}$ , and the embedding reflects order, so $∣ q_{k} - q_{l} ∣ < ε$ .

step 2.1L2L3L4

4.1

Set $x : = [(q_{k})] \in R$ , the class of this rational Cauchy sequence.

step 3.1L5

5.1

$x_{k} \to x$ : given rational $ε > 0$ , pick $k_{1}$ with $1 / k_{1} < ε / 3$ and $K_{2}$ with $∣ q_{k} - q_{l} ∣ < ε / 3$ for $k, l \geq K_{2}$ ; for $k \geq \max (k_{1}, K_{2})$ , the difference ${\hat{q}}_{k} - x$ has representative $(q_{k} - q_{l})_{l}$ , whose absolute values $∣ q_{k} - q_{l} ∣$ are eventually below $ε / 3$ , so $∣ {\hat{q}}_{k} - x ∣ \leq \hat{ε / 3}$ , and $∣ x_{k} - x ∣ \leq ∣ x_{k} - {\hat{q}}_{k} ∣ + ∣ {\hat{q}}_{k} - x ∣ < \hat{1 / k} + \hat{ε / 3} \leq 2 \hat{ε / 3} < \hat{ε}$ .

step 2.1step 3.1step 4.1L4

6.1

Every Cauchy sequence of reals converges in $R$ : the reals are complete.

step 5.1∎

Depends on

Used by

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Sources

T. Tao, Analysis I, 3rd ed., §6.4 (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)
L. S. Krapp, Constructions of the real numbers: a set theoretical approach (Oxford, 2014) (standard reference, not scraped)
W. Aitken, Constructing the Real Numbers, Section 8, Lemma 30 and footnote 3 (standard reference, not scraped)