ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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.

Canonical least-ball selection removes choice

Example

A coded least-pair rule is an alternative to the lexicographic rule in Separable complete metric spaces are Baire in ZF. Given a nonempty open $G$ and $ρ > 0$ , call $(k, m)$ admissible when $m \geq 1$ , $1 / m \leq ρ$ and $\overset{ˉ}{B} (s (k), 1 / m) \subseteq G$ . Choose the admissible pair of least code under $P (k, m) = 2^{k} (2 m + 1) - 1$ . The following three-stage calculation permits repetitions in the dense enumeration and spends no choice.

Facts & Assumptions

Given: Work in ZF. Let $(X, d)$ be a nonempty separable complete metric space, let $s : N \to X$ have dense range, let $(U_{n})_{n \in N}$ be a specified sequence of dense open subsets of $X$ , and let $W$ be nonempty open. Set $G_{0} = W \cap U_{0}$ and use radius bound $2^{- (n + 2)}$ at stage $n$ .

[F1]

Positive-radius balls contain their centers; open balls are open and finite intersections of open sets are open (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, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed).

[F2]

Density of the range of $s$ means it meets every nonempty open set (Separability: the existence of an at most countable dense subset). The metric triangle inequality and symmetry hold (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric); for every $ε > 0$ some integer $m \geq 1$ has $1 / m < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[F3]

The explicit $P = σ^{- 1} \circ J$ in $N \times N \approx N$ is a bijection $N^{2} \to N$ . Every nonempty subset of $N$ has a least element (The well-ordering principle), so the least element of $P [A]$ decodes to a unique member of every nonempty admissible set $A$ .

[F4]

A specified self-map of a set and an initial state determine a unique natural-number sequence by The recursion theorem.

Verification

1.1

For nonempty open $G$ and $ρ > 0$ , fix $y \in G$ and $r > 0$ with $B (y, r) \subseteq G$ . By [F2] take $m \geq 1$ with $1 / m < \min (r / 3, ρ)$ and $k$ with $s (k) \in B (y, 1 / m)$ . For $z \in \overset{ˉ}{B} (s (k), 1 / m)$ the triangle inequality gives $d (z, y) \leq d (z, s (k)) + d (s (k), y) < 2 / m < r$ . Thus $\overset{ˉ}{B} (s (k), 1 / m) \subseteq G$ , proving the admissible set nonempty. This is one finite existence argument for arbitrary $G, ρ$ , not a simultaneous choice of witnesses.

givenF1F2

2.1

Density of $U_{0}$ makes $G_{0}$ nonempty, and it is open. Thus step 1.1 with bound $1 / 4$ supplies an admissible pair.

givenF1step 1.1

3.1

Decode the least code to $(k_{0}, m_{0})$ and put $x_{0} = s (k_{0})$ , $r_{0} = 1 / m_{0}$ . Then $\overset{ˉ}{B} (x_{0}, r_{0}) \subseteq G_{0}$ . The set $G_{1} = B (x_{0}, r_{0}) \cap U_{1}$ is nonempty by density of $U_{1}$ , because the open ball contains $x_{0}$ ; it is open by finite intersection.

step 2.1F1F3given

4.1

Apply the same rule to $G_{1}$ with bound $1 / 8$ , and then to $G_{2} = B (x_{1}, r_{1}) \cap U_{2}$ with bound $1 / 16$ . At each stage density of the specified next $U_{n}$ keeps $G_{n}$ nonempty open, and step 1.1 and [F3] supply the unique next pair. Repeated values of $s$ do not affect uniqueness of the index-radius pair.

step 1.1step 3.1F1F3given

5.1

For a concrete example take $X = {p}$ , $d (p, p) = 0$ , $s (k) = p$ for every $k$ , and $W = U_{n} = X$ for every $n$ . The metric axioms hold, every sequence converges to $p$ , and the range of $s$ is dense, so the given hypotheses hold. All positive-radius open and closed balls equal $X$ . At stages $n = 0, 1, 2$ admissibility is exactly $m \geq 2^{n + 2}$ . Since $P (k, m) \geq 2 m$ , with equality exactly when $k = 0$ , the least pairs are $(0, 4), (0, 8), (0, 16)$ , with codes $8, 16, 32$ and radii $1 / 4, 1 / 8, 1 / 16$ . All centers are the same point $p$ , as required for an enumeration with repetitions.

step 4.1F1F3construct

6.1

For the general recursion use the set of states $(n, G)$ with $G$ nonempty open, together with a default state. The unique least-code pair defines the successor state $(n + 1, B (s (k), 1 / m) \cap U_{n + 1})$ on each such state; let the default state map to itself. The preceding nonemptiness argument makes this a total self-map. Apply [F4] from $(0, G_{0})$ and take the uniquely defined centers and radii. This is set recursion, not a choice of points from an arbitrary family; it gives the same closed-ball inclusions and radius bounds needed in the cited theorem, though its pairs need not equal the lexicographically least pairs there.

step 1.1step 2.1step 3.1step 4.1F3F4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Marianne Morillon, Synthese (standard reference, not scraped)