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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

1.1F1F2F3givenalgebra

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 $(B, d)$ .

2.1F1F2F3step 1.1givenalgebra∎

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.

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)