False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)

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FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$

Statement

False claim: for every metric space $(X, d)$ , every $x \in X$ and every real $r > 0$ ,

$\overline{B (x, r)} = \overset{ˉ}{B} (x, r),$

that is, the closure of the open ball (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space) is the closed ball of the same centre and radius (Open ball, closed ball and sphere in a metric space).

One inclusion is a theorem and the other is false. The names open ball and closed ball do not by themselves license the equality, and the intuition behind it comes from $R^{n}$ with a Euclidean metric, where it happens to be true; it fails already in a subspace of the real line with a gap, and the witness used below is ${0} \cup [1, 2]$ .

Facts & Assumptions

Given: The real line with its usual metric $d_{R} (u, v) = ∣ u - v ∣$ (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded); the subset $X : = {0} \cup [1, 2] \subseteq R$ with the subspace metric $d : = d_{R} ↾ (X \times X)$ (Isometry, isometric embedding, and the subspace metric on a subset, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length); an arbitrary metric space $(X_{0}, d_{0})$ with $x \in X_{0}$ and a real $r > 0$ .

[L1]

The closed ball $\overset{ˉ}{B} (x, r)$ is a closed set, and it contains $B (x, r)$ (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, Open ball, closed ball and sphere in a metric space).

[L2]

The closure of a set is the smallest closed superset of it, and a closed set equals its own closure (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

[L3]

The subspace metric makes $X$ a metric space and its balls are traces: $B_{X} (a, s) = B_{R} (a, s) \cap X$ (Isometry, isometric embedding, and the subspace metric on a subset, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Open ball, closed ball and sphere in a metric space).

[L4]

Absolute value and order: $∣ u ∣ = u$ when $u \geq 0$ , $∣ - u ∣ = ∣ u ∣$ , and $0 < 1$ ; and by trichotomy $u \geq 1$ rules out $u < 1$ (Absolute value in an ordered field, Basic properties of the absolute value, The multiplicative identity is positive, Ordered field, Complete ordered field (least-upper-bound property)).

[L5]

Open sets of a metric space: $U$ is open when every point of $U$ has a ball around it inside $U$ ; a set is closed when its complement is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Refutation

technique · direct

1.1

The inclusion that does hold, in every metric space: $\overset{ˉ}{B} (x, r)$ is closed and contains $B (x, r)$ , so the smallest closed superset of $B (x, r)$ satisfies $\overline{B (x, r)} \subseteq \overset{ˉ}{B} (x, r)$ .

L1L2

1.2

In the witness $X = {0} \cup [1, 2]$ , every $y \in [1, 2]$ satisfies $y \geq 1$ , hence $d (0, y) = ∣ 0 - y ∣ = ∣ - y ∣ = y \geq 1$ ; and $d (0, 0) = 0 < 1$ .

givenL3L4

2.1

Therefore $B_{X} (0, 1) = {y \in X : d (0, y) < 1} = {0}$ and ${\overset{ˉ}{B}}_{X} (0, 1) = {y \in X : d (0, y) \leq 1} = {0, 1}$ , since $1 \in [1, 2] \subseteq X$ has $d (0, 1) = 1$ while every other $y \in [1, 2]$ has $d (0, y) = y > 1$ or $y = 1$ .

step 1.2L3L4

2.2

The set $[1, 2] = X ∖ {0}$ is open in $X$ : for $y \in [1, 2]$ the ball $B_{X} (y, 1)$ omits $0$ , because $d (0, y) \geq 1$ by step 1.2, so $B_{X} (y, 1) \subseteq X ∖ {0} = [1, 2]$ . Hence ${0}$ is closed in $X$ .

step 1.2L3L5

3.1

Since ${0}$ is closed it equals its own closure, so $\overline{B_{X} (0, 1)} = \overline{{0}} = {0}$ by step 2.1, while ${\overset{ˉ}{B}}_{X} (0, 1) = {0, 1}$ ; and ${0} \neq {0, 1}$ because $1 \neq 0$ .

step 2.1step 2.2L2L4

4.1

The witness $(X, d)$ with $x = 0$ and $r = 1$ therefore refutes the claim; all that survives in general is the inclusion of step 1.1, and it can be strict.

step 1.1step 3.1∎

Remarks

Where the intuition comes from and why it does not transfer. In $R^{n}$ with the Euclidean metric ( $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it) the segment from the centre to a point of the closed ball lies in the space, and running along it approaches that point from inside the open ball; that is the usual route to the equality there, and this library does not prove it. A metric space need not contain any such segment: in the witness above, nothing of $X$ lies strictly between $0$ and $1$ , so the point $1$ of the closed ball is not approached from inside $B (0, 1) = {0}$ at all.
The failure is not exotic. A discrete metric on a set with at least two points produces the same phenomenon in a starker form, with $\overline{B (p, 1)} = {p}$ and $\overset{ˉ}{B} (p, 1)$ the whole space; the companion page carries both witnesses.
The sphere is not the boundary of the ball either, and that failure is recorded separately on the companion page.

Depends on

Used by

In {0} ∪ [1,2] with the metric of ℝ, the closure of B(0,1) = {0} is {0} while the closed ball is {0,1} Counterexample

Dependency tree · two levels

34 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ball (mathematics) (Wikipedia) (standard reference, not scraped)
Closure (topology) (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)
Isolated point (Wikipedia) (standard reference, not scraped)

FALSE: in every metric space the closure of B(x,r)B(x,r) is the closed ball of radius rr