ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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An open Euclidean unit ball is metrically incomplete

Example

For every $n \geq 1$ , the open Euclidean unit ball $B = {x \in R^{n} : ∥ x ∥ < 1}$ with the restricted Euclidean distance is not a complete metric space. In dimension zero the ball is a singleton and is complete, so the positive-dimensional hypothesis matters.

Facts & Assumptions

Given: $n \geq 1$ and the restricted distance $d (x, y) = ∥ x - y ∥_{2}$ on $B$ .

[F1]

Cauchy sequence in a metric space gives the epsilon-tail definition of a Cauchy sequence, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ gives the epsilon-tail definition of convergence, and Complete metric space: every Cauchy sequence converges in the space says that a metric space is complete precisely when every Cauchy sequence in it converges to a point of that space. $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it makes $d_{2} (x, y) = ∥ x - y ∥_{2}$ a metric on all of $R^{n}$ for $n \geq 1$ , with the separation and triangle axioms of Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric; the displayed $d$ is its restriction to $B$ .

[F2]

For $n \geq 1$ , The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ supplies the coordinate vector $e_{0} \in R^{n}$ , while The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ gives $∥ e_{0} ∥_{2} = 1$ , $∥ (a - b) e_{0} ∥_{2} = ∣ a - b ∣$ , and the singleton zero space $R^{0}$ .

[F3]

For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ gives, for every real $ϵ > 0$ , a natural $r \geq 1$ with $1 / r < ϵ$ ; Inverses of positives are positive, and reciprocation reverses order makes reciprocation reverse inequalities between positive reals.

Verification

1.1

For $k \geq 0$ put $x_{k} = (1 - 1 / (k + 2)) e_{0}$ . By [F2], $∥ x_{k} ∥_{2} = 1 - 1 / (k + 2)$ , which lies strictly between $0$ and $1$ , so every $x_{k}$ lies in $B$ . If $m, k \geq N$ , then [F2] and [F3], after interchanging $m, k$ if necessary, give $d (x_{m}, x_{k}) = ∣ \frac{1}{m + 2} - \frac{1}{k + 2} ∣ \leq \frac{1}{N + 2} .$ Given $ϵ > 0$ , use [F3] to take $r \geq 1$ with $1 / r < ϵ$ and put $N = r$ ; then $1 / (N + 2) < 1 / r < ϵ$ . Thus [F1] makes $(x_{k})$ Cauchy in $(B, d)$ .

F1F2F3givenalgebra

2.1

Suppose $x_{k}$ converged in $B$ to some $z$ , and put $δ = d_{2} (z, e_{0})$ in the ambient Euclidean space. If $δ > 0$ , [F3] supplies a natural $r \geq 1$ with $1 / r < δ / 2$ , while convergence in [F1] supplies $N_{1}$ such that $d_{2} (z, x_{k}) = d (z, x_{k}) < δ / 2$ for $k \geq N_{1}$ . Put $k = \max {r, N_{1}}$ . Then [F2], [F3], and the ambient triangle inequality in [F1] give $δ = d_{2} (z, e_{0}) \leq d_{2} (z, x_{k}) + d_{2} (x_{k}, e_{0}) < \frac{δ}{2} + \frac{1}{k + 2} < δ,$ a contradiction. Hence $δ = 0$ , so ambient metric separation gives $z = e_{0}$ . But $∥ e_{0} ∥_{2} = 1$ and $e_{0} \notin B$ , another contradiction. Thus the Cauchy sequence has no limit in $B$ , and [F1] makes $B$ incomplete. If $n = 0$ , [F2] gives $B = {0}$ and every sequence is constant, so the ball is complete. All witnesses are prescribed by formulas, and no choice principle is used.

F1F2F3step 1.1givenalgebra∎

Source locator

Andrews, §11.5, Theorem 11.5.1 and its proof, printed pp.106--108 (PDF pp.6--8), discuss metric completeness in the context of geodesics. The radial Cauchy witness and the dimension-zero qualification above are local calculations, not attributed to that text.

Depends on

Used by

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Sources

Ben Andrews, Geodesics and Completeness, §11.5 (standard reference, not scraped)