TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-27

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Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences

Statement

Assume the Axiom of Countable Choice for assertion 5 and the consequent existence claim (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $C$ be the set of all Cauchy sequences in $X$ (Cauchy sequence in a metric space). Then:

For all $x = (x_{n})$ and $y = (y_{n})$ in $C$ the real sequence $(d (x_{n}, y_{n}))_{n}$ converges, so $ρ (x, y) : = \lim_{n} d (x_{n}, y_{n})$ is a single well-determined real (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges, A sequence has at most one limit).
The relation $x \sim y : ⟺ ρ (x, y) = 0$ is an equivalence relation on $C$ . Write $\hat{X} : = C / ⁣ \sim$ for the set of its classes and $[x]$ for the class of $x$ .
$\hat{d} ([x], [y]) : = ρ (x, y)$ does not depend on the chosen representatives, and $\hat{d}$ is a metric on $\hat{X}$ .
The map $ι : X \to \hat{X}$ sending $p$ to the class of the constant sequence at $p$ is an isometric embedding with dense image (Isometry, isometric embedding, and the subspace metric on a subset, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).
$(\hat{X}, \hat{d})$ is complete.

Consequently $((\hat{X}, \hat{d}), ι)$ is a completion of $(X, d)$ (A completion of a metric space: a complete metric space together with an isometric embedding onto a dense subspace), and every metric space has a completion.

The notation is kept honest. A Cauchy sequence in $X$ need not converge in $X$ , so no symbol $\lim_{n} x_{n}$ appears anywhere below; the only limits taken are limits of real sequences, and each is written only after its existence has been proved. The equivalence relation is defined and verified here rather than cited, as was done for The integers as equivalence classes of pairs of naturals, so that the construction is self-contained and its transitivity argument is visible at the point of use.

Facts & Assumptions

Given: For assertion 5, the Axiom of Countable Choice; a metric space $(X, d)$ ; the set $C$ of Cauchy sequences in $X$ ; elements $x = (x_{n})$ , $y = (y_{n})$ , $z = (z_{n})$ of $C$ ; a real $ε > 0$ .

[A1]

Cauchyness: for every real $ε > 0$ there is $K$ with $d (x_{n}, x_{m}) < ε$ for all $n, m \geq K$ (Cauchy sequence in a metric space, The rationals embed densely in the reals).

[L1]

Reverse triangle inequality: $∣ d (u, w) - d (v, w) ∣ \leq d (u, v)$ (The reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space); with the triangle inequality for the absolute value (Basic properties of the absolute value) this gives the quadrilateral estimate $∣ d (u, v) - d (u^{'}, v^{'}) ∣ \leq d (u, u^{'}) + d (v, v^{'})$ .

[L2]

Every Cauchy sequence of reals converges, and the limit of a real sequence is unique, which licenses the notation $\lim_{n} a_{n}$ for a sequence already known to converge (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges, A sequence has at most one limit, Limits and Cauchy sequences of reals).

[L3]

Limits of reals preserve non-strict inequalities holding eventually, and behave additively (Limits preserve non-strict inequalities, Algebra of limits: sums, scalar multiples, products and quotients); a constant sequence converges to that constant.

[L4]

The metric axioms (M1), (M2), (M3) and nonnegativity (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Nonnegativity of a metric is a consequence of the other axioms, not an axiom).

[L5]

Density and convergence are tested with balls and with real $ε > 0$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, Open ball, closed ball and sphere in a metric space, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , The rationals embed densely in the reals).

[L6]

Countable choice: a family $(A_{k})_{k \in N}$ of nonempty sets admits $k \mapsto a_{k}$ with $a_{k} \in A_{k}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L7]

For every real $η > 0$ there is a natural $K$ with $1 / (K + 1) < η$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L8]

Two naturals have a maximum (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

Proof

technique · constructive

1.1

The quadrilateral estimate of [L1] gives $∣ d (x_{n}, y_{n}) - d (x_{m}, y_{m}) ∣ \leq d (x_{n}, x_{m}) + d (y_{n}, y_{m})$ for all $n, m$ .

2.1

Given a real $ε > 0$ , [A1] supplies $K_{x}$ and $K_{y}$ with $d (x_{n}, x_{m}) < ε / 2$ and $d (y_{n}, y_{m}) < ε / 2$ for indices beyond them; with $K : = \max {K_{x}, K_{y}}$ the sequence $(d (x_{n}, y_{n}))_{n}$ satisfies $∣ d (x_{n}, y_{n}) - d (x_{m}, y_{m}) ∣ < ε$ for $n, m \geq K$ , so it is a Cauchy sequence of reals.

step 1.1A1L8

3.1

By [L2] that sequence converges and its limit is unique, so $ρ (x, y) : = \lim_{n} d (x_{n}, y_{n})$ is a single well-determined real: claim 1.

step 2.1L2construct

4.1

$ρ$ is nonnegative and symmetric, and satisfies $ρ (x, z) \leq ρ (x, y) + ρ (y, z)$ : the terms $d (x_{n}, y_{n})$ are nonnegative, $d (x_{n}, y_{n}) = d (y_{n}, x_{n})$ , and $d (x_{n}, z_{n}) \leq d (x_{n}, y_{n}) + d (y_{n}, z_{n})$ for every $n$ , and all three pass to the limit.

step 3.1L3L4

5.1

Claim 2: $\sim$ is reflexive since $d (x_{n}, x_{n}) = 0$ for every $n$ ; symmetric since $ρ$ is; and transitive, since $ρ (x, y) = ρ (y, z) = 0$ gives $0 \leq ρ (x, z) \leq 0 + 0 = 0$ . So $\sim$ is an equivalence relation and $\hat{X} = C / ⁣ \sim$ is defined.

step 4.1L4construct

6.1

Claim 3, well-definedness: if $x \sim x^{'}$ and $y \sim y^{'}$ then, by [L1] applied termwise, $∣ d (x_{n}, y_{n}) - d (x_{n}^{'}, y_{n}^{'}) ∣ \leq d (x_{n}, x_{n}^{'}) + d (y_{n}, y_{n}^{'})$ ; passing to the limit gives $∣ ρ (x, y) - ρ (x^{'}, y^{'}) ∣ \leq ρ (x, x^{'}) + ρ (y, y^{'}) = 0$ , so $ρ (x, y) = ρ (x^{'}, y^{'})$ . Hence $\hat{d} ([x], [y]) : = ρ (x, y)$ is a well-defined function on $\hat{X} \times \hat{X}$ .

step 3.1step 5.1L1L3

7.1

$\hat{d}$ is a metric: symmetry and the triangle inequality are step 4.1 read on classes, and $\hat{d} ([x], [y]) = 0$ says $ρ (x, y) = 0$ , which says $x \sim y$ , which says $[x] = [y]$ . This completes claim 3.

step 4.1step 5.1step 6.1L4

8.1

Claim 4: for $p \in X$ the constant sequence at $p$ is Cauchy, so $ι (p) : = [(p)_{n}]$ is defined, and $\hat{d} (ι (p), ι (q)) = \lim_{n} d (p, q) = d (p, q)$ , a constant sequence; so $ι$ is an isometric embedding.

step 7.1L3construct

9.1

Density: let $[x] \in \hat{X}$ and let $ε > 0$ be real. By [A1] there is $N$ with $d (x_{n}, x_{m}) < ε / 2$ for all $n, m \geq N$ ; in particular $d (x_{n}, x_{N}) \leq ε / 2$ for all $n \geq N$ , so $\hat{d} ([x], ι (x_{N})) = \lim_{n} d (x_{n}, x_{N}) \leq ε / 2 < ε$ . Hence every ball around $[x]$ meets $ι [X]$ , that is $ι [X]$ is dense, completing claim 4.

step 8.1A1L3L5

10.1

Claim 5: let $(ξ^{(k)})$ be a Cauchy sequence in $(\hat{X}, \hat{d})$ . For each $k$ the set $A_{k} : = {p \in X : \hat{d} (ξ^{(k)}, ι (p)) < 1 / (k + 1)}$ is nonempty by step 9.1, so [L6] supplies $a_{k} \in A_{k}$ for every $k$ , that is a sequence $(a_{k})$ in $X$ with $\hat{d} (ξ^{(k)}, ι (a_{k})) < 1 / (k + 1)$ .

step 9.1L6L7choose

11.1

$(a_{k})$ is Cauchy in $X$ : since $ι$ is isometric, $d (a_{k}, a_{l}) = \hat{d} (ι (a_{k}), ι (a_{l})) \leq 1 / (k + 1) + \hat{d} (ξ^{(k)}, ξ^{(l)}) + 1 / (l + 1)$ ; given a real $ε > 0$ , choose $K$ so large that $1 / (K + 1) < ε / 3$ and $\hat{d} (ξ^{(k)}, ξ^{(l)}) < ε / 3$ for all $k, l \geq K$ , and then $d (a_{k}, a_{l}) < ε$ for all $k, l \geq K$ . So $a : = (a_{k}) \in C$ and $[a] \in \hat{X}$ .

step 8.1step 10.1L4L7L8

12.1

$ξ^{(k)} \to [a]$ : given a real $ε > 0$ take $K$ as in step 11.1 for $ε / 3$ , so that $d (a_{k}, a_{n}) < ε / 3$ for all $k, n \geq K$ , enlarged if necessary so that also $1 / (K + 1) < ε / 3$ . For $k \geq K$ we have $d (a_{k}, a_{n}) \leq ε / 3$ for all $n \geq K$ , hence $\hat{d} (ι (a_{k}), [a]) = \lim_{n} d (a_{k}, a_{n}) \leq ε / 3$ , and therefore $\hat{d} (ξ^{(k)}, [a]) \leq \hat{d} (ξ^{(k)}, ι (a_{k})) + \hat{d} (ι (a_{k}), [a]) < ε / 3 + ε / 3 < ε$ .

step 10.1step 11.1L3L4L5

13.1

So every Cauchy sequence in $(\hat{X}, \hat{d})$ converges in it, which is claim 5; with claims 1 to 4 this exhibits $((\hat{X}, \hat{d}), ι)$ as a completion of the arbitrary metric space $(X, d)$ .

step 3.1step 5.1step 7.1step 9.1step 12.1discharge-construct∎

Remarks

What the new points are. A point of $\hat{X}$ is a class of Cauchy sequences of $X$ , two sequences being identified exactly when the distance between their $n$ -th terms tends to $0$ . The old points reappear as the classes of constant sequences, and the new ones are the classes of Cauchy sequences that had nowhere to go. This is the same move that builds $R$ out of $Q$ , and the resulting completion of $Q$ really is $R$ (The completion of $Q$ under the usual metric is $R$ ↗).
Where choice is spent, and where it is not. Only at step 10.1, which selects one point of $X$ per natural number: that is exactly ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Claims 1 to 4 are choice free. The selection could not be avoided by a uniqueness argument, because a point of $X$ within $1 / (k + 1)$ of $ξ^{(k)}$ is in general far from unique.
The quadrilateral estimate does most of the work. Step 1.1 is used twice: at step 2.1, to show $(d (x_{n}, y_{n}))_{n}$ is Cauchy, and at step 6.1, to show $ρ$ is independent of representatives. It is the reverse triangle inequality (The reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space) applied twice and added. The triangle inequality for $ρ$ , and hence for $\hat{d}$ , does not use it: step 4.1 gets that from the triangle inequality of $d$ applied termwise and passed to the limit.
Nothing here needs $X$ to be nonempty. If $X = \emptyset$ then $C = \emptyset$ , $\hat{X} = \emptyset$ , and the empty metric space is complete, vacuously. The construction degenerates correctly rather than requiring a hypothesis.

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Sources

Complete metric space (Wikipedia) (standard reference, not scraped)
Cauchy sequence (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)