Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)
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.

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

Statement

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

B(x,r)=Bˉ(x,r),\overline{B(x,r)} = \bar 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 Rn\mathbb{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}[1,2]\{0\} \cup [1,2].

Facts & Assumptions

Given: The real line with its usual metric dR(u,v)=uvd_{\mathbb{R}}(u,v) = |u-v| (The absolute value makes R\mathbb{R} a metric space: d(x,y)=xyd(x,y) = |x-y| is a metric, its open balls are the intervals (xr,x+r)(x-r, x+r), and it is unbounded); the subset X:={0}[1,2]RX := \{0\} \cup [1,2] \subseteq \mathbb{R} with the subspace metric d:=dR(X×X)d := d_{\mathbb{R}} \restriction (X \times X) (Isometry, isometric embedding, and the subspace metric on a subset, Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length); an arbitrary metric space (X0,d0)(X_0, d_0) with xX0x \in X_0 and a real r>0r > 0.

[L2]
[L3]
[L4]

Absolute value and order: u=u|u| = u when u0u \ge 0, u=u|-u| = |u|, and 0<10 < 1; and by trichotomy u1u \ge 1 rules out u<1u < 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: UU is open when every point of UU has a ball around it inside UU; 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: Bˉ(x,r)\bar B(x,r) is closed and contains B(x,r)B(x,r), so the smallest closed superset of B(x,r)B(x,r) satisfies B(x,r)Bˉ(x,r)\overline{B(x,r)} \subseteq \bar B(x,r).

L1L2
1.2

In the witness X={0}[1,2]X = \{0\} \cup [1,2], every y[1,2]y \in [1,2] satisfies y1y \ge 1, hence d(0,y)=0y=y=y1d(0,y) = |0 - y| = |{-y}| = y \ge 1; and d(0,0)=0<1d(0,0) = 0 < 1.

givenL3L4
2.1

Therefore BX(0,1)={yX:d(0,y)<1}={0}B_X(0,1) = \{\, y \in X : d(0,y) < 1 \,\} = \{0\} and BˉX(0,1)={yX:d(0,y)1}={0,1}\bar B_X(0,1) = \{\, y \in X : d(0,y) \le 1 \,\} = \{0, 1\}, since 1[1,2]X1 \in [1,2] \subseteq X has d(0,1)=1d(0,1) = 1 while every other y[1,2]y \in [1,2] has d(0,y)=y>1d(0,y) = y > 1 or y=1y = 1.

step 1.2L3L4
2.2

The set [1,2]=X{0}[1,2] = X \setminus \{0\} is open in XX: for y[1,2]y \in [1,2] the ball BX(y,1)B_X(y,1) omits 00, because d(0,y)1d(0,y) \ge 1 by step 1.2, so BX(y,1)X{0}=[1,2]B_X(y,1) \subseteq X \setminus \{0\} = [1,2]. Hence {0}\{0\} is closed in XX.

step 1.2L3L5
3.1

Since {0}\{0\} is closed it equals its own closure, so BX(0,1)={0}={0}\overline{B_X(0,1)} = \overline{\{0\}} = \{0\} by step 2.1, while BˉX(0,1)={0,1}\bar B_X(0,1) = \{0,1\}; and {0}{0,1}\{0\} \ne \{0,1\} because 101 \ne 0.

step 2.1step 2.2L2L4
4.1

The witness (X,d)(X, d) with x=0x = 0 and r=1r = 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 Rn\mathbb{R}^n with the Euclidean metric (Rn\mathbb{R}^n as the set of functions nRn \to \mathbb{R}, and d1d_1, d2d_2, dd_\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 XX lies strictly between 00 and 11, so the point 11 of the closed ball is not approached from inside B(0,1)={0}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 B(p,1)={p}\overline{B(p,1)} = \{p\} and Bˉ(p,1)\bar 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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 47 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources