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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

In the discrete metric the boundary of $B (p, 1)$ is empty while the sphere of radius $1$ is everything but $p$

Statement refuted

Refuted claim: in every metric space, the boundary of the open ball is the sphere of the same centre and radius,

$\partial B (x, r) = S (x, r)$

(Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, Open ball, closed ball and sphere in a metric space).

The witness is any set $X$ with at least two points, carrying the discrete metric $δ$ (The discrete metric induces the discrete topology, in which every subset is clopen), together with $x = p \in X$ and $r = 1$ . There

$B (p, 1) = {p}, \partial B (p, 1) = \emptyset, S (p, 1) = X ∖ {p} \neq \emptyset .$

Facts & Assumptions

Given: A set $X$ with at least two points, the discrete metric $δ$ on it, a point $p \in X$ and a point $q \in X$ with $q \neq p$ .

[L1]

The discrete metric: $δ$ is a metric, $B (p, 1) = {p}$ , $S (p, 1) = X ∖ {p}$ , and every subset of $X$ is both open and closed (The discrete metric induces the discrete topology, in which every subset is clopen, 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, 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).

[L2]

Interior, closure, boundary: $int (A)$ is the largest open subset of $A$ , $\overline{A}$ the smallest closed superset of $A$ , and $\partial A = \overline{A} ∖ int (A)$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, 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).

Counterexample

technique · direct

1.1givenL1

In $(X, δ)$ the open ball is $B (p, 1) = {p}$ and the sphere is $S (p, 1) = X ∖ {p}$ , which is nonempty because $q \neq p$ lies in it.

1.2L1

The set ${p}$ is open and closed in $(X, δ)$ , every subset of a discrete metric space being clopen.

2.1step 1.2L2

Hence $int ({p}) = {p}$ , since ${p}$ is an open subset of itself and the interior is the largest one; and $\overline{{p}} = {p}$ , since ${p}$ is a closed superset of itself and the closure is the smallest one.

3.1step 1.1step 2.1L2

Therefore $\partial B (p, 1) = \overline{{p}} ∖ int ({p}) = {p} ∖ {p} = \emptyset$ , while $S (p, 1)$ contains $q$ and is not empty.

4.1step 3.1∎

The two sets differ, so the discrete metric on any set with at least two points refutes the claim.

Remarks

One inclusion does survive. In any metric space $\partial B (x, r) \subseteq S (x, r)$ : the ball $B (x, r)$ is open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed), so it is its own interior and $\partial B (x, r) = \overline{B (x, r)} ∖ B (x, r)$ , which sits inside $\overset{ˉ}{B} (x, r) ∖ B (x, r) = S (x, r)$ because $\overline{B (x, r)} \subseteq \overset{ˉ}{B} (x, r)$ (FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$ , Open ball, closed ball and sphere in a metric space). What the witness above shows is that the inclusion can be strict, and as strict as possible: empty on the left, everything but the centre on the right.
It is the same defect as FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$ : the names open ball, closed ball and sphere are labels for three sets defined by three inequalities (Open ball, closed ball and sphere in a metric space), and none of the topological relations suggested by the words is automatic.
Every point of a discrete space is isolated, so no ball has any boundary at all; the post-office metric (The post-office metric $d (x, y) = ∥ x ∥ + ∥ y ∥$ for $x \neq y$ on $R^{n}$ , and its isolated points) shows the intermediate case, where all but one point is isolated.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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

Discrete space (Wikipedia) (standard reference, not scraped)
Boundary (topology) (Wikipedia) (standard reference, not scraped)
Ball (mathematics) (Wikipedia) (standard reference, not scraped)

In the discrete metric the boundary of B(p,1)B(p,1) is empty while the sphere of radius 11 is everything but pp