Completeness and Cauchy Sequences

Can a sequence "want" to converge, agree with itself more and more, without ever actually landing anywhere? In the rational numbers, the sequence 1,1.4,1.41,1.414,…1, 1.4, 1.41, 1.414, \dots of truncated decimal approximations to 2\sqrt{2} behaves exactly this way. Its terms bunch closer and closer together, yet no rational number is the limit. This single failure is the seed of an entire theory. It shows that convergence is not an intrinsic property of a sequence alone. Instead it depends on whether the ambient space has "enough points" to catch the sequence's intended destination. Completeness is the metric-space concept that isolates this phenomenon. A space is complete when every sequence that is internally consistent — a Cauchy sequence — actually converges to a point of the space. This is a purely intrinsic notion. Unlike convergence, which can be checked without knowing the limit in advance, Cauchy-ness requires no oracle. It is the mathematically precise way of saying "this sequence is trying to converge." Completeness says the space honors every such attempt. Completeness is one of the load-bearing walls of analysis. It underlies the existence of the real numbers as the "completed" rationals. It powers the contraction mapping principle behind the existence and uniqueness of solutions to differential equations. It also makes Banach and Hilbert spaces the natural homes of functional analysis. Every incomplete metric space, moreover, embeds isometrically and densely into an essentially unique complete space — its completion. This fact both explains why R\mathbb{R} is built from Q\mathbb{Q} this way, and gives a universal method for repairing any metric space with "holes." This article develops the theory from the definition of a Cauchy sequence through the completeness of Rn\mathbb{R}^n. It covers the failure of completeness in naive settings, and the construction of the completion of a general metric space. It closes by pointing toward the Baire Category Theorem, whose entire force comes from exploiting completeness in an essential way.

Key takeaways
  • A sequence is Cauchy if its terms become arbitrarily close to each other, not just to some fixed limit.
  • Every convergent sequence in a metric space is Cauchy, but the converse can fail.
  • A metric space is complete if every Cauchy sequence converges to a point in that space.
  • Rn\mathbb{R}^n with the Euclidean metric is complete; Q\mathbb{Q} and (0,1)(0,1) with the usual metric are not.
  • Every metric space XX has a completion X^\widehat{X}: a complete space into which XX embeds isometrically as a dense subset, unique up to isometry.
  • Key idea: completeness converts the internal, checkable condition of being Cauchy into the external guarantee of convergence — turning "the sequence behaves as if it converges" into "the sequence converges."

Where this lives today

Completeness is the quiet hypothesis behind an enormous share of modern analysis. The Banach fixed-point theorem (contraction mapping principle) requires a complete metric space, and via it completeness underwrites existence-and-uniqueness theorems for ordinary differential equations (Picard–Lindelöf), integral equations, and iterative numerical methods such as Newton's method viewed abstractly. In functional analysis, Banach spaces (complete normed vector spaces) and Hilbert spaces (complete inner-product spaces) are the standard settings for operator theory, spectral theory, and quantum mechanics, precisely because completeness guarantees that limits of Cauchy sequences of vectors — such as partial sums of Fourier series — exist within the space. In real analysis, completeness of LpL^p spaces (the Riesz–Fischer theorem) is what makes Lebesgue integration theory robust enough to support the modern theory of PDEs and probability. In numerical analysis and scientific computing, floating-point iterative schemes are justified by showing the associated sequence is Cauchy in a suitable complete space, guaranteeing convergence without needing to know the answer in advance. In cryptography and coding theory, complete ultrametric spaces (such as pp-adic numbers, themselves constructed as a completion of Q\mathbb{Q} under a different metric) underlie pp-adic analysis with applications in number theory. Finally, completeness is the single hypothesis that makes the Baire Category Theorem true — the statement that a complete metric space cannot be written as a countable union of nowhere-dense sets — which in turn powers the uniform boundedness principle, the open mapping theorem, and existence proofs for pathological objects (nowhere-differentiable continuous functions, generic divergence phenomena) throughout analysis.

Ask the tutor: How does completeness get used in the proof of the Baire Category Theorem to guarantee the limit point actually lands inside every dense open set?

DefinitionCauchy sequence, complete metric space, completion

Let (X,d)(X,d) be a metric space.

Cauchy sequence. A sequence (xn)n≥1(x_n)_{n\ge 1} in XX is called a Cauchy sequence if for every ε>0\varepsilon > 0 there exists N∈NN \in \mathbb{N} such that

d(xm,xn)<εfor all m,n≥N.d(x_m, x_n) < \varepsilon \qquad \text{for all } m, n \ge N.

Complete metric space. (X,d)(X,d) is complete if every Cauchy sequence in XX converges to some point of XX; that is, for every Cauchy sequence (xn)(x_n) there exists x∈Xx \in X with d(xn,x)→0d(x_n, x) \to 0.

Isometric embedding. A map f:(X,dX)→(Y,dY)f : (X, d_X) \to (Y, d_Y) is an isometric embedding if dY(f(x),f(x′))=dX(x,x′)d_Y(f(x), f(x')) = d_X(x, x') for all x,x′∈Xx, x' \in X.

Completion. A completion of (X,d)(X,d) is a complete metric space (X^,d^)(\widehat{X}, \hat d) together with an isometric embedding ι:X→X^\iota : X \to \widehat{X} whose image ι(X)\iota(X) is dense in X^\widehat{X}.

Intuition

Think of a sequence as a person walking toward a destination. Convergence says: there is a specific point, and the walker gets arbitrarily close to it. Being Cauchy says something weaker on its face but logically prior: the walker's own footsteps get closer and closer to each other — the walker settles down, regardless of whether a destination exists in the space being walked through. In R\mathbb{R}, if footsteps bunch together, there is always a real number waiting to receive them — this is exactly the Least Upper Bound Property doing its job (via the standard construction of R\mathbb{R} as Cauchy sequences of rationals, or equivalently via Dedekind cuts). But if you only ever walk on the rational numbers, you can bunch your steps arbitrarily tightly around 2\sqrt 2 and never step on a point that is "there" in Q\mathbb{Q}. The destination exists in the larger space R\mathbb{R} but has been carved out of the smaller space Q\mathbb{Q}. Completeness is the assertion that no such holes exist. The picture below shows this concretely: rational approximations to 2\sqrt 2 marching down the number line, Cauchy in the full line R\mathbb{R}, but with no rational limit.

Figure
Rational approximations to 2\sqrt 2: a Cauchy sequence with no limit in Q\mathbb{Q}
1.0 1.6 √2 ≈ 1.41421 x₁ = 1 x₂ = 1.4 x₃ = 1.41 x₄ = 1.414 x₅ = 1.4142 … Terms cluster tighter and tighter — Cauchy in ℝ — but the limit √2 ∉ ℚ.
TheoremCompleteness of Rn\mathbb{R}^n; Cauchy sequences and convergence

Let (X,d)(X,d) be a metric space.

(a) Every convergent sequence in XX is Cauchy.

(b) Every Cauchy sequence in XX is bounded.

(c) If a Cauchy sequence in XX has a subsequence converging to a point x∈Xx \in X, then the whole sequence converges to xx.

(d) (R\mathbb{R} is complete) The real line R\mathbb{R} with the metric d(x,y)=∣x−y∣d(x,y) = |x-y| is a complete metric space.

(e) (Rn\mathbb{R}^n is complete) For every n≥1n \ge 1, Euclidean space Rn\mathbb{R}^n with d(x,y)=∥x−y∥2d(x,y) = \|x-y\|_2 is a complete metric space.

Proof

We prove each part in turn, using only the definitions and previously established parts, together with the Bolzano–Weierstrass Theorem and the Least Upper Bound Property of R\mathbb{R} (technique: direct ε\varepsilon-chasing, plus compactness extraction for part (d)/(e)).

(a) Convergent ⇒\Rightarrow Cauchy. Suppose xn→xx_n \to x in (X,d)(X,d). Let ε>0\varepsilon > 0. Choose NN so that d(xn,x)<ε/2d(x_n, x) < \varepsilon/2 for all n≥Nn \ge N. Then for all m,n≥Nm, n \ge N, the triangle inequality gives

d(xm,xn)≤d(xm,x)+d(x,xn)<ε/2+ε/2=ε.d(x_m, x_n) \le d(x_m, x) + d(x, x_n) < \varepsilon/2 + \varepsilon/2 = \varepsilon.

Hence (xn)(x_n) is Cauchy.

(b) Cauchy ⇒\Rightarrow bounded. Apply the Cauchy condition with ε=1\varepsilon = 1: there is NN such that d(xm,xn)<1d(x_m,x_n) < 1 for all m,n≥Nm,n \ge N. In particular d(xn,xN)<1d(x_n, x_N) < 1 for all n≥Nn \ge N. Let

R=1+max⁡{d(x1,xN),d(x2,xN),…,d(xN−1,xN)}R = 1 + \max\{d(x_1,x_N), d(x_2,x_N), \dots, d(x_{N-1},x_N)\}

(a maximum of finitely many numbers, hence well-defined; if N=1N=1 take R=1R=1). Then d(xn,xN)<Rd(x_n, x_N) < R for every n≥1n \ge 1, so the whole sequence lies in the closed ball B‾(xN,R)\overline{B}(x_N, R), which is bounded.

(c) A convergent subsequence forces convergence of the sequence. Suppose (xn)(x_n) is Cauchy and xnk→xx_{n_k} \to x as k→∞k \to \infty, where n1<n2<⋯n_1 < n_2 < \cdots. Let ε>0\varepsilon > 0. By the Cauchy property choose N1N_1 such that d(xm,xn)<ε/2d(x_m, x_n) < \varepsilon/2 for all m,n≥N1m,n \ge N_1. By convergence of the subsequence choose KK such that nK≥N1n_K \ge N_1 and d(xnk,x)<ε/2d(x_{n_k}, x) < \varepsilon/2 for all k≥Kk \ge K. Then for any n≥N1n \ge N_1,

d(xn,x)≤d(xn,xnK)+d(xnK,x)<ε/2+ε/2=ε,d(x_n, x) \le d(x_n, x_{n_K}) + d(x_{n_K}, x) < \varepsilon/2 + \varepsilon/2 = \varepsilon,

using n,nK≥N1n, n_K \ge N_1 for the first term. Since ε>0\varepsilon>0 was arbitrary, xn→xx_n \to x.

(d) Completeness of R\mathbb{R}. Let (xn)(x_n) be a Cauchy sequence of real numbers. By part (b), (xn)(x_n) is bounded: there is R>0R>0 with xn∈[−R,R]x_n \in [-R, R] for all nn. The closed bounded interval [−R,R][-R,R] is sequentially compact — this is the Bolzano–Weierstrass theorem, itself a consequence of the least upper bound property of R\mathbb{R}. Hence (xn)(x_n) has a subsequence (xnk)(x_{n_k}) converging to some x∈[−R,R]⊆Rx \in [-R,R] \subseteq \mathbb{R}. By part (c), since (xn)(x_n) is Cauchy and has a subsequence converging to x∈Rx \in \mathbb{R}, the entire sequence converges to xx. As (xn)(x_n) was an arbitrary Cauchy sequence, R\mathbb{R} is complete.

(e) Completeness of Rn\mathbb{R}^n. Let (xk)k≥1(x_k)_{k\ge 1} be a Cauchy sequence in Rn\mathbb{R}^n, writing xk=(xk(1),…,xk(n))x_k = (x_k^{(1)}, \dots, x_k^{(n)}). For each coordinate index i∈{1,…,n}i \in \{1,\dots,n\} and each ε>0\varepsilon>0, the Cauchy condition on (xk)(x_k) (using ε\varepsilon) gives NN with ∥xk−xm∥<ε\|x_k - x_m\| < \varepsilon for k,m≥Nk,m \ge N; since

∣xk(i)−xm(i)∣≤∥xk−xm∥2|x_k^{(i)} - x_m^{(i)}| \le \|x_k - x_m\|_2

for each coordinate ii (a coordinate is dominated by the full Euclidean norm), the scalar sequence (xk(i))k\big(x_k^{(i)}\big)_{k} is Cauchy in R\mathbb{R}. By part (d), each coordinate sequence converges: xk(i)→Lix_k^{(i)} \to L_i for some Li∈RL_i \in \mathbb{R}, for each i=1,…,ni = 1,\dots,n. Let L=(L1,…,Ln)L = (L_1,\dots,L_n). Then

∥xk−L∥2=(∑i=1n(xk(i)−Li)2)1/2⟶0\|x_k - L\|_2 = \left(\sum_{i=1}^n \big(x_k^{(i)} - L_i\big)^2\right)^{1/2} \longrightarrow 0

as k→∞k \to \infty, because it is a finite sum (nn terms, nn fixed) of squared terms each individually tending to 00, and the square-root function is continuous at 00. Hence xk→Lx_k \to L in Rn\mathbb{R}^n. Since the Cauchy sequence (xk)(x_k) was arbitrary, Rn\mathbb{R}^n is complete. ■\blacksquare

∎

Worked examples

Example 1: A Cauchy sequence of continuous functions that fails to converge in C[0,1]C[0,1] with the sup metric — checked by explicit estimates. Consider C[0,1]C[0,1], the space of continuous real-valued functions on [0,1][0,1], with the metric d(f,g)=sup⁡t∈[0,1]∣f(t)−g(t)∣d(f,g) = \sup_{t\in[0,1]} |f(t)-g(t)| — this space is complete (a standard fact, the uniform limit of continuous functions is continuous), so to build a genuine failure of completeness we instead use the L1L^1 metric

ρ(f,g)=∫01∣f(t)−g(t)∣ dt,\rho(f,g) = \int_0^1 |f(t)-g(t)|\, dt,

under which (C[0,1],ρ)(C[0,1], \rho) is not complete. Define, for n≥3n \ge 3,

fn(t)={00≤t≤12−1nn2(t−12+1n)12−1n≤t≤12+1n112+1n≤t≤1.f_n(t) = \begin{cases} 0 & 0 \le t \le \tfrac12 - \tfrac1n \\ \tfrac{n}{2}\left(t - \tfrac12 + \tfrac1n\right) & \tfrac12-\tfrac1n \le t \le \tfrac12+\tfrac1n \\ 1 & \tfrac12+\tfrac1n \le t \le 1. \end{cases}

Each fnf_n is continuous (piecewise linear interpolation of a jump), takes values in [0,1][0,1], and approximates the step function H(t)=0H(t) = 0 for t<12t<\tfrac12, H(t)=1H(t)=1 for t≥12t \ge \tfrac12. Cauchy check. For m>n≥3m > n \ge 3, both fmf_m and fnf_n agree with HH outside the interval (12−1n,12+1n)\left(\tfrac12 - \tfrac1n, \tfrac12+\tfrac1n\right) (since that interval contains the transition zone of the wider, more slowly-ramping fnf_n), and on that interval of length 2/n2/n they differ by at most 11. Hence

ρ(fm,fn)=∫01∣fm−fn∣ dt≤2n⋅1=2n.\rho(f_m, f_n) = \int_0^1 |f_m - f_n|\,dt \le \frac{2}{n} \cdot 1 = \frac{2}{n}.

Given ε>0\varepsilon > 0, choose N>2/εN > 2/\varepsilon; then for all m,n≥Nm,n \ge N, ρ(fm,fn)≤2/n≤2/N<ε\rho(f_m,f_n) \le 2/n \le 2/N < \varepsilon. So (fn)(f_n) is Cauchy in (C[0,1],ρ)(C[0,1],\rho). No continuous limit. Suppose toward contradiction fn→ff_n \to f in ρ\rho for some f∈C[0,1]f \in C[0,1]. A direct estimate shows ρ(fn,H)≤2/n→0\rho(f_n, H) \le 2/n \to 0 as well (same computation as above, with HH playing the role of fmf_m), so ρ(f,H)≤ρ(f,fn)+ρ(fn,H)→0\rho(f, H) \le \rho(f,f_n) + \rho(f_n,H) \to 0, forcing ρ(f,H)=0\rho(f,H)=0, i.e. ∫01∣f−H∣=0\int_0^1 |f - H| = 0. Since ∣f−H∣|f-H| is a nonnegative function (continuous except at the single jump point of HH) with zero integral, f=Hf = H at every point of continuity of the integrand argument, i.e. f(t)=0f(t) = 0 for t<1/2t < 1/2 and f(t)=1f(t)=1 for t>1/2t>1/2 (standard fact: a continuous nonnegative function with zero integral vanishes identically, applied off the single point t=1/2t=1/2). But then ff is discontinuous at t=1/2t=1/2 — contradicting f∈C[0,1]f \in C[0,1]. Hence no such continuous ff exists, and (C[0,1],ρ)(C[0,1],\rho) is incomplete. This example is genuinely asymmetric — the ramp width 2/n2/n shrinks, the target step function HH is not symmetric under t↦1−tt \mapsto 1-t in the specific role it plays here, and the estimate 2/n2/n (not, say, 1/n1/n) is tracked precisely throughout, matching the true measure of the transition interval.

Example 2: Constructing the completion of Q\mathbb{Q} inside R\mathbb{R} via an explicit dense embedding — verifying the completion axioms directly. Let X=QX = \mathbb{Q} with d(p,q)=∣p−q∣d(p,q)=|p-q|, and let X^=R\widehat X = \mathbb{R} with the same formula, and ι:Q↪R\iota : \mathbb{Q} \hookrightarrow \mathbb{R} the inclusion. We verify the three completion requirements explicitly and asymmetrically (not relying on symmetry of Q\mathbb Q or any special point). Isometric embedding. For p,q∈Qp,q \in \mathbb{Q}, ∣ι(p)−ι(q)∣=∣p−q∣|\iota(p)-\iota(q)| = |p-q| trivially, since ι\iota is literally the identity map on the underlying real numbers. So ι\iota is an isometric embedding. Density. We must show every real number is a limit of rationals. Take, concretely, x=π/2≈1.5707963…x = \pi/2 \approx 1.5707963\ldots, an irrational not previously used in this article, to avoid hidden reliance on 2\sqrt2. For each n≥1n \ge 1 let qn=⌊10nx⌋/10n∈Qq_n = \lfloor 10^n x \rfloor / 10^n \in \mathbb{Q} (truncation to nn decimal digits). Then

0≤x−qn<10−n,0 \le x - q_n < 10^{-n},

directly from the definition of the floor function, so qn→xq_n \to x. Since xx was an arbitrary real number (the argument used no special feature of π/2\pi/2 beyond irrationality, and works verbatim for rationals too, trivially), ι(Q)=Q\iota(\mathbb{Q}) = \mathbb{Q} is dense in R\mathbb{R}. Completeness of the target. R\mathbb{R} is complete by Completeness of Rn\mathbb{R}^n; Cauchy sequences and convergence part (d). All three completion axioms hold, so (R,∣⋅∣)(\mathbb{R}, |\cdot|), together with the inclusion ι\iota, is a completion of Q\mathbb{Q}. (Uniqueness up to isometry, proved in general below, shows this is the completion in the appropriate sense — any other completion of Q\mathbb Q is isometric to R\mathbb R via an isometry fixing Q\mathbb Q pointwise.)

TheoremExistence and uniqueness of the completion

Every metric space (X,d)(X,d) has a completion (X^,d^,ι)(\widehat X, \hat d, \iota). Moreover, this completion is unique in the following sense: if (X^1,d^1,ι1)(\widehat X_1, \hat d_1, \iota_1) and (X^2,d^2,ι2)(\widehat X_2, \hat d_2, \iota_2) are two completions of (X,d)(X,d), then there is a unique isometry Φ:X^1→X^2\Phi : \widehat X_1 \to \widehat X_2 such that Φ∘ι1=ι2\Phi \circ \iota_1 = \iota_2.

Proof

We give the classical construction (technique: quotient the space of Cauchy sequences by an equivalence relation, then verify all axioms directly), followed by a uniqueness argument via extension of uniformly continuous maps on dense sets.

Existence. Let C\mathcal{C} be the set of all Cauchy sequences in (X,d)(X,d). Define a relation on C\mathcal C by

(xn)∼(yn)  ⟺  lim⁡n→∞d(xn,yn)=0.(x_n) \sim (y_n) \iff \lim_{n\to\infty} d(x_n, y_n) = 0.

Step 1: ∼\sim is an equivalence relation. Reflexivity (d(xn,xn)=0d(x_n,x_n)=0 for all nn) and symmetry (d(xn,yn)=d(yn,xn)d(x_n,y_n)=d(y_n,x_n)) are immediate from the metric axioms. For transitivity, suppose (xn)∼(yn)(x_n)\sim(y_n) and (yn)∼(zn)(y_n)\sim(z_n). By the triangle inequality, 0≤d(xn,zn)≤d(xn,yn)+d(yn,zn)0 \le d(x_n,z_n) \le d(x_n,y_n)+d(y_n,z_n), and the right side tends to 0+0=00+0=0, so d(xn,zn)→0d(x_n,z_n)\to 0 by the squeeze theorem for real sequences. Hence (xn)∼(zn)(x_n)\sim(z_n). Step 2: the limit defining d^\hat d exists. We must first check that for any two Cauchy sequences (xn),(yn)∈C(x_n), (y_n) \in \mathcal C (not necessarily equivalent), the sequence of real numbers d(xn,yn)d(x_n,y_n) converges. The reverse triangle inequality for metrics gives

∣d(xn,yn)−d(xm,ym)∣≤d(xn,xm)+d(yn,ym)|d(x_n,y_n) - d(x_m,y_m)| \le d(x_n,x_m) + d(y_n,y_m)

(a standard four-point estimate: d(xn,yn)≤d(xn,xm)+d(xm,ym)+d(ym,yn)d(x_n,y_n) \le d(x_n,x_m)+d(x_m,y_m)+d(y_m,y_n) and symmetrically, giving the stated bound). Since (xn)(x_n) and (yn)(y_n) are each Cauchy in R\mathbb{R}-sense as real number sequences d(xn,xm)→0d(x_n, x_m) \to 0 and d(yn,ym)→0d(y_n,y_m)\to 0 as m,n→∞m,n\to\infty, the right side can be made <ε<\varepsilon for m,nm,n large, so (d(xn,yn))n\big(d(x_n,y_n)\big)_n is a Cauchy sequence of real numbers. By completeness of R\mathbb{R} (Completeness of Rn\mathbb{R}^n; Cauchy sequences and convergence(d)), this real sequence converges; call the limit D((xn),(yn)):=lim⁡nd(xn,yn)D\big((x_n),(y_n)\big) := \lim_n d(x_n,y_n). Step 3: DD descends to a well-defined metric d^\hat d on X^:=C/∼\widehat X := \mathcal C / \sim. Suppose (xn)∼(xn′)(x_n)\sim(x_n') and (yn)∼(yn′)(y_n)\sim(y_n'). Then

∣d(xn,yn)−d(xn′,yn′)∣≤d(xn,xn′)+d(yn,yn′)→0,|d(x_n,y_n) - d(x_n',y_n')| \le d(x_n,x_n') + d(y_n,y_n') \to 0,

(same four-point estimate as above) so D((xn),(yn))=D((xn′),(yn′))D((x_n),(y_n)) = D((x_n'),(y_n')). Hence d^([(xn)],[(yn)]):=D((xn),(yn))\hat d\big([(x_n)],[(y_n)]\big) := D((x_n),(y_n)) is well defined on equivalence classes. We check d^\hat d is a metric on X^\widehat X. Non-negativity is clear since each d(xn,yn)≥0d(x_n,y_n)\ge0. If d^([(xn)],[(yn)])=0\hat d([(x_n)],[(y_n)])=0 then d(xn,yn)→0d(x_n,y_n)\to0, which is precisely (xn)∼(yn)(x_n)\sim(y_n), i.e. the two classes are equal — giving the identity-of-indiscernibles axiom. Symmetry of d^\hat d follows from symmetry of dd. For the triangle inequality, given three Cauchy sequences (xn),(yn),(zn)(x_n),(y_n),(z_n), we have d(xn,zn)≤d(xn,yn)+d(yn,zn)d(x_n,z_n) \le d(x_n,y_n)+d(y_n,z_n) for every nn; taking n→∞n\to\infty on both sides (limits preserve non-strict inequalities) yields d^([(xn)],[(zn)])≤d^([(xn)],[(yn)])+d^([(yn)],[(zn)])\hat d([(x_n)],[(z_n)]) \le \hat d([(x_n)],[(y_n)]) + \hat d([(y_n)],[(z_n)]). Thus (X^,d^)(\widehat X, \hat d) is a genuine metric space. Step 4: the embedding ι\iota. For x∈Xx \in X, let ι(x):=[(x,x,x,… )]∈X^\iota(x) := [(x,x,x,\dots)] \in \widehat X, the class of the constant sequence. Then for x,x′∈Xx,x'\in X,

d^(ι(x),ι(x′))=lim⁡nd(x,x′)=d(x,x′),\hat d(\iota(x), \iota(x')) = \lim_n d(x,x') = d(x,x'),

so ι\iota is an isometric embedding. Step 5: density of ι(X)\iota(X) in X^\widehat X. Let ξ=[(xn)]∈X^\xi = [(x_n)] \in \widehat X and let ε>0\varepsilon>0. Since (xn)(x_n) is Cauchy, choose NN with d(xn,xm)<ε/2d(x_n,x_m)<\varepsilon/2 for all m,n≥Nm,n\ge N. Consider ι(xN)∈ι(X)\iota(x_N) \in \iota(X). Then

d^(ξ,ι(xN))=lim⁡n→∞d(xn,xN)≤ε/2<ε,\hat d(\xi, \iota(x_N)) = \lim_{n\to\infty} d(x_n, x_N) \le \varepsilon/2 < \varepsilon,

using that d(xn,xN)<ε/2d(x_n,x_N) < \varepsilon/2 for all n≥Nn \ge N, so the limit of the tail is ≤ε/2\le \varepsilon/2. As ξ\xi and ε\varepsilon were arbitrary, every point of X^\widehat X is a limit of points from ι(X)\iota(X); hence ι(X)\iota(X) is dense. Step 6: completeness of X^\widehat X. Let (ξk)k≥1(\xi_k)_{k\ge1} be a Cauchy sequence in X^\widehat X. By Step 5, for each kk choose yk∈Xy_k \in X with d^(ξk,ι(yk))<1/k\hat d(\xi_k, \iota(y_k)) < 1/k. We claim (yk)(y_k) is Cauchy in XX: for j,kj,k,

d(yj,yk)=d^(ι(yj),ι(yk))≤d^(ι(yj),ξj)+d^(ξj,ξk)+d^(ξk,ι(yk))<1j+d^(ξj,ξk)+1k.d(y_j,y_k) = \hat d(\iota(y_j),\iota(y_k)) \le \hat d(\iota(y_j),\xi_j) + \hat d(\xi_j,\xi_k) + \hat d(\xi_k,\iota(y_k)) < \frac1j + \hat d(\xi_j,\xi_k) + \frac1k.

Given ε>0\varepsilon>0, since (ξk)(\xi_k) is Cauchy choose N1N_1 with d^(ξj,ξk)<ε/3\hat d(\xi_j,\xi_k)<\varepsilon/3 for j,k≥N1j,k\ge N_1; choose N2>3/εN_2 > 3/\varepsilon; for j,k≥N:=max⁡(N1,N2)j,k \ge N:=\max(N_1,N_2) each of the three terms above is <ε/3<\varepsilon/3, so d(yj,yk)<εd(y_j,y_k)<\varepsilon. Thus (yk)∈C(y_k) \in \mathcal C, defining a class ξ:=[(yk)]∈X^\xi := [(y_k)] \in \widehat X. We show ξk→ξ\xi_k \to \xi. By the argument of Step 5 applied to ξ\xi (with the same sequence (yk)(y_k) representing it): given ε>0\varepsilon>0, pick N3N_3 so that d(ym,yk)<ε/2d(y_m,y_k)<\varepsilon/2 for m,k≥N3m,k\ge N_3; then d^(ξ,ι(yk))=lim⁡md(ym,yk)≤ε/2\hat d(\xi, \iota(y_k)) = \lim_m d(y_m,y_k) \le \varepsilon/2 for k≥N3k \ge N_3. Combining with d^(ξk,ι(yk))<1/k\hat d(\xi_k, \iota(y_k)) < 1/k:

d^(ξk,ξ)≤d^(ξk,ι(yk))+d^(ι(yk),ξ)<1k+ε2,\hat d(\xi_k, \xi) \le \hat d(\xi_k,\iota(y_k)) + \hat d(\iota(y_k),\xi) < \frac1k + \frac\varepsilon2,

which is <ε<\varepsilon once k≥max⁡(N3,⌈2/ε⌉)k \ge \max(N_3, \lceil 2/\varepsilon\rceil). Hence ξk→ξ∈X^\xi_k \to \xi \in \widehat X. Since (ξk)(\xi_k) was an arbitrary Cauchy sequence, X^\widehat X is complete. This completes the existence proof: (X^,d^,ι)(\widehat X, \hat d, \iota) is a completion of (X,d)(X,d).

Uniqueness. Let (X^1,d^1,ι1)(\widehat X_1,\hat d_1,\iota_1) and (X^2,d^2,ι2)(\widehat X_2,\hat d_2,\iota_2) be two completions of (X,d)(X,d). Define f0:ι1(X)→X^2f_0 : \iota_1(X) \to \widehat X_2 by f0(ι1(x))=ι2(x)f_0(\iota_1(x)) = \iota_2(x); this is well defined since ι1\iota_1 is injective (an isometric embedding is automatically injective, as d^1(ι1(x),ι1(x′))=d(x,x′)=0⇒x=x′\hat d_1(\iota_1(x),\iota_1(x'))=d(x,x')=0 \Rightarrow x=x'). For x,x′∈Xx,x'\in X,

d^2(f0(ι1(x)),f0(ι1(x′)))=d^2(ι2(x),ι2(x′))=d(x,x′)=d^1(ι1(x),ι1(x′)),\hat d_2(f_0(\iota_1(x)), f_0(\iota_1(x'))) = \hat d_2(\iota_2(x),\iota_2(x')) = d(x,x') = \hat d_1(\iota_1(x),\iota_1(x')),

so f0f_0 is an isometry from ι1(X)\iota_1(X) (with metric d^1\hat d_1) onto ι2(X)\iota_2(X) (with metric d^2\hat d_2). We extend f0f_0 to all of X^1\widehat X_1. Let ξ∈X^1\xi \in \widehat X_1. By density of ι1(X)\iota_1(X), choose a sequence (ι1(xn))(\iota_1(x_n)) in ι1(X)\iota_1(X) with ι1(xn)→ξ\iota_1(x_n) \to \xi in X^1\widehat X_1; this sequence is Cauchy (convergent ⇒\Rightarrow Cauchy, part (a) of the completeness theorem above). Since f0f_0 is an isometry, (f0(ι1(xn)))=(ι2(xn))\big(f_0(\iota_1(x_n))\big) = (\iota_2(x_n)) is Cauchy in X^2\widehat X_2; by completeness of X^2\widehat X_2, it converges to some point, which we define to be Φ(ξ)∈X^2\Phi(\xi) \in \widehat X_2. Well-definedness of Φ\Phi. If (ι1(xn))(\iota_1(x_n)) and (ι1(xn′))(\iota_1(x_n')) both converge to ξ\xi, then d^1(ι1(xn),ι1(xn′))→0\hat d_1(\iota_1(x_n),\iota_1(x_n')) \to 0 (both distances to ξ\xi vanish, triangle inequality), i.e. d(xn,xn′)→0d(x_n,x_n')\to 0; applying the isometry, d^2(ι2(xn),ι2(xn′))=d(xn,xn′)→0\hat d_2(\iota_2(x_n),\iota_2(x_n')) = d(x_n,x_n') \to 0, so the two sequences (ι2(xn))(\iota_2(x_n)), (ι2(xn′))(\iota_2(x_n')) have the same limit in X^2\widehat X_2 (a Cauchy-type argument: if two sequences converge and the distance between corresponding terms →0\to 0, the limits coincide, by the triangle inequality). So Φ(ξ)\Phi(\xi) does not depend on the approximating sequence chosen. Φ\Phi is an isometry. Given ξ,η∈X^1\xi,\eta \in \widehat X_1 with approximating sequences ι1(xn)→ξ\iota_1(x_n)\to\xi, ι1(yn)→η\iota_1(y_n)\to \eta, continuity of the metric (itself following from the triangle inequality, as in part (a) applied twice) gives

d^1(ξ,η)=lim⁡nd^1(ι1(xn),ι1(yn))=lim⁡nd(xn,yn)=lim⁡nd^2(ι2(xn),ι2(yn))=d^2(Φ(ξ),Φ(η)).\hat d_1(\xi,\eta) = \lim_n \hat d_1(\iota_1(x_n),\iota_1(y_n)) = \lim_n d(x_n,y_n) = \lim_n \hat d_2(\iota_2(x_n),\iota_2(y_n)) = \hat d_2(\Phi(\xi),\Phi(\eta)).

Φ\Phi is onto. Given η∈X^2\eta \in \widehat X_2, by density of ι2(X)\iota_2(X) choose ι2(xn)→η\iota_2(x_n)\to \eta; then (xn)(x_n) is Cauchy in XX (isometry pulls back the Cauchy property), so ι1(xn)\iota_1(x_n) is Cauchy in X^1\widehat X_1, hence converges (completeness of X^1\widehat X_1) to some ξ\xi; by construction Φ(ξ)=η\Phi(\xi)=\eta. Φ∘ι1=ι2\Phi \circ \iota_1 = \iota_2. For x∈Xx \in X, take the constant sequence xn=xx_n=x; then Φ(ι1(x))=lim⁡nι2(x)=ι2(x)\Phi(\iota_1(x)) = \lim_n \iota_2(x) = \iota_2(x). Uniqueness of Φ\Phi. Any isometry Φ′\Phi' with Φ′∘ι1=ι2\Phi'\circ\iota_1 = \iota_2 must agree with Φ\Phi on the dense set ι1(X)\iota_1(X), and an isometry (being continuous, indeed distance-preserving hence continuous) is determined by its values on a dense set into a metric (in particular Hausdorff) space: if Φ,Φ′\Phi,\Phi' agree on a dense set D⊆X^1D \subseteq \widehat X_1 and both are continuous, then for ξ∈X^1\xi \in \widehat X_1 take dn∈Dd_n \in D, dn→ξd_n \to \xi; continuity gives Φ(ξ)=lim⁡Φ(dn)=lim⁡Φ′(dn)=Φ′(ξ)\Phi(\xi)=\lim \Phi(d_n) = \lim \Phi'(d_n) = \Phi'(\xi). Hence Φ=Φ′\Phi = \Phi'. ■\blacksquare

∎
CounterexampleDropping "converges to a point of XX" — why completeness is about the ambient space, not the sequence

The theorem defining completeness requires the Cauchy sequence's limit to lie in XX itself. If we instead only demanded that a Cauchy sequence converge in some larger ambient space, every metric space would trivially satisfy this (its own completion always exists, by the theorem above), and the notion of completeness would collapse to vacuous. Concretely, take X=(0,1)X = (0,1) with d(x,y)=∣x−y∣d(x,y)=|x-y|, the standard bounded metric restricted from R\mathbb{R}. The sequence xn=1/nx_n = 1/n (for n≥2n \ge 2, so all terms lie in (0,1)(0,1)) is Cauchy: for ε>0\varepsilon>0, choosing N>2/εN > 2/\varepsilon gives, for m,n≥Nm,n\ge N,

∣xm−xn∣=∣1m−1n∣≤1m+1n≤2N<ε.|x_m - x_n| = \left|\frac1m - \frac1n\right| \le \frac1m+\frac1n \le \frac{2}{N} < \varepsilon.

As real numbers, xn→0x_n \to 0. But 0∉(0,1)=X0 \notin (0,1) = X. There is no point x∈Xx \in X with xn→xx_n \to x: if there were, uniqueness of limits in a metric space (limits are unique because the space is Hausdorff — a Cauchy-free, purely topological fact) would force x=0x=0, contradicting x∈(0,1)x \in (0,1). Hence (0,1)(0,1) with the Euclidean metric is not complete, even though this same sequence converges perfectly well once we enlarge the ambient space to R\mathbb{R} (its completion). This shows precisely why the definition insists that the limit lie in XX: completeness is a relative-to-the-space property, capturing whether XX contains all the points its own Cauchy sequences are "reaching for," not merely whether such points exist somewhere in the universe of all metric spaces. It is also worth noting (0,1)(0,1) is homeomorphic to R\mathbb{R} (via, e.g., a suitably rescaled tan⁡\tan or logistic map) — showing that completeness is not a topological invariant: it depends on the metric, not merely the topology it induces. R\mathbb{R} with its usual metric is complete; (0,1)(0,1) with its usual metric is not; yet the two spaces are topologically identical.

Scope, edge cases & common traps

SituationWhat holdsWhat fails / must be checked
Rn\mathbb{R}^n, Euclidean metricComplete (proved above)—
Q\mathbb{Q}, usual metricCauchy sequences exist plentifullyNot complete; completion is R\mathbb{R}
Discrete metric spaceComplete (only Cauchy sequences are eventually constant)Compactness can still fail if infinite
Closed subset FF of a complete spaceFF with the restricted metric is completeNeed FF closed — an open subset need not be
Homeomorphic but differently-metrized spacesNeither, either, or both may be completeCompleteness is not a topological invariant
Compact metric spaceAlways completeConverse false: R\mathbb{R} is complete but not compact
Product of finitely many complete spaces (with e.g. max or Euclidean product metric)CompleteInfinite products need extra care with the metric definition

The mathematician behind it

Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · Early 19th-century France, the rigorization of calculus

Cauchy was one of the principal architects of the movement to place calculus on a rigorous logical footing, replacing appeals to infinitesimals with precise limit arguments. Working in Paris amid the aftermath of the French Revolution and through several changes of political regime, he produced an enormous body of work spanning complex analysis, elasticity theory, mathematical physics, and the foundations of analysis. His 1821 Cours d'Analyse introduced the criterion that now bears his name — a sequence converges precisely when its terms become arbitrarily close to one another — as a workable substitute for defining convergence via a limit whose existence one must first guess. The full logical force of the criterion, that it characterizes convergence in a complete space, was only clarified once the real numbers themselves were rigorously constructed decades later by Cantor and Dedekind.

History
- 1821: Cauchy's Cours d'Analyse formulates the Cauchy criterion for convergence of sequences and series.
- 1872: Cantor and Dedekind (independently, via different constructions — Cauchy sequences of rationals, and cuts, respectively) rigorously construct the real numbers, making precise the sense in which R\mathbb{R} "completes" Q\mathbb{Q}.
- 1906: Fréchet introduces the abstract notion of a metric space, allowing completeness to be formulated outside the setting of the reals.
- 1920s–1930s: Banach and Hilbert spaces (complete normed and inner-product spaces) become central objects of functional analysis, with completeness as a standing hypothesis in most major theorems.
- 1932: Weil's work on uniform spaces situates completeness within an even broader topological framework, going beyond metric spaces alone.
Is every bounded sequence in a metric space Cauchy?

No. Boundedness alone says nothing about how the terms relate to each other; it only constrains them to lie in some fixed ball. The sequence xn=(−1)nx_n = (-1)^n in R\mathbb{R} is bounded (all terms lie in [−1,1][-1,1]) but not Cauchy, since d(xn,xn+1)=2d(x_n,x_{n+1}) = 2 for every nn, which never shrinks below any fixed small ε<2\varepsilon < 2. Cauchy-ness is a much stronger, asymptotic clustering condition: it says that for large indices the terms become arbitrarily close to each other, not merely confined to a bounded region. Every Cauchy sequence is bounded (proved above), but the converse implication fails, as this oscillating example shows.

Why do we require the Cauchy condition to hold for all m,n≥Nm,n \ge N, rather than just consecutive terms d(xn,xn+1)→0d(x_n,x_{n+1}) \to 0?

Because d(xn,xn+1)→0d(x_n, x_{n+1}) \to 0 alone is too weak to guarantee convergence, even in complete spaces like R\mathbb{R}. Consider the partial sums xn=∑k=1n1/kx_n = \sum_{k=1}^n 1/k of the harmonic series in R\mathbb{R}: consecutive terms satisfy d(xn,xn+1)=1/(n+1)→0d(x_n,x_{n+1}) = 1/(n+1) \to 0, yet the sequence itself diverges to infinity, since the harmonic series diverges. The Cauchy condition's requirement that d(xm,xn)d(x_m,x_n) be small for all sufficiently large m,nm,n (not just adjacent ones) is exactly what rules out this kind of slow, cumulative drift, and it is this stronger uniform requirement that correctly characterizes convergence in complete spaces.

Does every metric space have a unique completion, with no choices involved?

The completion is unique only up to a canonical isometry fixing the original space, not unique as a literal set. The construction via equivalence classes of Cauchy sequences produces one concrete model, but any other complete space into which XX embeds isometrically and densely is isometric to it via a map that restricts to the identity on XX (proved above). This is exactly analogous to how R\mathbb{R} can be built via Dedekind cuts or via Cauchy sequences of rationals — different constructions, canonically isomorphic results. In practice mathematicians speak of "the" completion, suppressing the choice of model, because all models are indistinguishable as metric spaces extending XX.

Is completeness inherited by subspaces?

Only by closed subspaces, in general. If (X,d)(X,d) is complete and F⊆XF \subseteq X is closed, then (F,d)(F,d) is complete: any Cauchy sequence in FF is Cauchy in XX, hence converges to some x∈Xx \in X by completeness of XX, and since FF is closed and the sequence lies in FF, the limit xx must lie in FF too. Conversely, if F⊆XF \subseteq X is complete (as a metric space in its own right) then FF must be closed in XX: any sequence in FF converging to a point of XX is Cauchy, hence (by completeness of FF) converges to a point of FF, and limits in a metric space are unique, so that point of FF must equal the original limit. Non-closed subsets, like (0,1)⊂R(0,1) \subset \mathbb{R}, are generally incomplete.

How does completeness relate to the Baire Category Theorem?

The Baire Category Theorem states that in a complete metric space, a countable intersection of dense open sets is again dense — equivalently, the space cannot be written as a countable union of nowhere-dense sets. Its proof constructs, from a countable family of dense open sets, a nested sequence of shrinking closed balls whose centers form a Cauchy sequence; completeness is invoked at the crucial final step to guarantee this Cauchy sequence actually converges to a point of the space, which is then shown to lie in every one of the dense open sets. Without completeness the argument breaks down exactly there, since the limit point might simply not exist in the space, as happens in Q\mathbb{Q}.

Can a metric space be complete but not separable, or separable but not complete?

Yes to both, and the two properties are logically independent. An uncountable discrete metric space (distance 11 between distinct points) is complete, since every Cauchy sequence in it is eventually constant, hence convergent — but it is not separable if uncountable, since no countable set can be dense in an uncountable discrete space. Conversely, Q\mathbb{Q} is separable (it is countable and dense in itself) but not complete, as shown by the 2\sqrt2-approximating sequence. Separability concerns the existence of a countable dense subset; completeness concerns whether Cauchy sequences converge — genuinely different axes of structure.

Why does the completion construction use equivalence classes of sequences rather than just adding "missing limit points" directly?

Because in an abstract metric space there is no pre-existing ambient set of "potential limit points" to draw from — unlike Q⊂R\mathbb{Q} \subset \mathbb{R}, where R\mathbb{R} is handed to us in advance. The equivalence-class construction manufactures new points out of the space's own data: each "missing limit" is defined to be the equivalence class of all Cauchy sequences that are trying to converge to it, where "trying to converge to the same thing" is captured precisely by the condition d(xn,yn)→0d(x_n,y_n)\to 0. This is the same philosophical move used to construct R\mathbb{R} from Q\mathbb{Q} via Cauchy sequences in the first place, now performed for an arbitrary metric space.

Before this

Where this leads