CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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

Statement refuted

Refuted claim: in every metric space, the closure of the open ball $B (x, r)$ is the closed ball $\overset{ˉ}{B} (x, r)$ (FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$ ).

The witness is the metric subspace

$X : = {0} \cup [1, 2] \subseteq R$

of the real line with its usual metric (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, Isometry, isometric embedding, and the subspace metric on a subset, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), with $x = 0$ and $r = 1$ . In it

$B_{X} (0, 1) = {0}, \overline{B_{X} (0, 1)} = {0}, {\overset{ˉ}{B}}_{X} (0, 1) = {0, 1},$

so the closure of the open ball is a proper subset of the closed ball of the same centre and radius. The inclusion $\overline{B (x, r)} \subseteq \overset{ˉ}{B} (x, r)$ that does hold in general is proved in FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$ and is not repeated.

Facts & Assumptions

Given: The real line with $d_{R} (u, v) = ∣ u - v ∣$ , and $X : = {0} \cup [1, 2]$ with the subspace metric $d : = d_{R} ↾ (X \times X)$ .

[L1]

The subspace metric makes $X$ a metric space, and its balls are traces of the balls of $R$ (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).

[L2]

Balls: $B_{X} (a, s) = {y \in X : d (a, y) < s}$ and ${\overset{ˉ}{B}}_{X} (a, s) = {y \in X : d (a, y) \leq s}$ (Open ball, closed ball and sphere in a metric space).

[L3]

Open and closed sets of a metric space, and the closure as the set of adherent points (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, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

[L4]

A closed set equals its own closure, the closure being the smallest closed superset (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).

[L5]

Absolute value and order: $∣ t ∣ = t$ for $t \geq 0$ , $∣ - t ∣ = ∣ t ∣$ , and $0 < 1$ ; by trichotomy $t \geq 1$ excludes $t < 1$ (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)).

[L6]

Intervals: $[1, 2] = {t \in R : 1 \leq t \leq 2}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Counterexample

technique · direct

1.1

Every $y \in [1, 2]$ has $y \geq 1$ , so $d (0, y) = ∣ 0 - y ∣ = ∣ - y ∣ = y \geq 1$ ; and $d (0, 0) = 0 < 1$ .

L1L5L6

2.1

Hence $B_{X} (0, 1) = {y \in X : d (0, y) < 1} = {0}$ , since no $y \in [1, 2]$ satisfies $d (0, y) < 1$ ; and ${\overset{ˉ}{B}}_{X} (0, 1) = {y \in X : d (0, y) \leq 1} = {0, 1}$ , since among the points of $[1, 2]$ exactly $y = 1$ satisfies $y \leq 1$ .

step 1.1L2L5L6

2.2

The set $[1, 2] = X ∖ {0}$ is open in $X$ : for $y \in [1, 2]$ the ball $B_{X} (y, 1)$ cannot contain $0$ , because $d (0, y) \geq 1$ by step 1.1, so $B_{X} (y, 1) \subseteq [1, 2]$ . Therefore ${0}$ is closed in $X$ .

step 1.1L2L3

3.1

Since ${0}$ is closed it equals its own closure, so $\overline{B_{X} (0, 1)} = \overline{{0}} = {0}$ , whereas ${\overset{ˉ}{B}}_{X} (0, 1) = {0, 1}$ contains the point $1 \neq 0$ .

step 2.1step 2.2L4L5

4.1

The two sets are therefore different, and $(X, d)$ with $x = 0$ , $r = 1$ refutes the claim: the closure of an open ball can be a proper subset of the closed ball of the same centre and radius.

step 3.1∎

Remarks

Nothing pathological is used. $X$ is an unremarkable bounded subset of the real line and the metric is the one inherited from $R$ ; the only feature exploited is that $X$ has a gap, so that the point $1$ of the closed ball cannot be approached from inside $B_{X} (0, 1)$ .
A starker version lives in the discrete metric (The discrete metric induces the discrete topology, in which every subset is clopen) on a set with at least two points, where $\overline{B (p, 1)} = {p}$ while $\overset{ˉ}{B} (p, 1)$ is the entire space; the two witnesses refute the claim in the same way, at different scales.
The equality does hold in $R^{n}$ with any of $d_{1}$ , $d_{2}$ , $d_{\infty}$ , which is where the false intuition comes from. This library does not prove it, and neither $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it nor The metrics $d_{1}$ , $d_{2}$ and $d_{\infty}$ on $R^{n}$ are metrics and are Lipschitz equivalent, with explicit constants contains it; the usual argument runs along the segment from the centre to the point in question, and no such segment need exist in a general metric space.

Depends on

Used by

Nothing in the library uses this result yet.

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)

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