Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedverified 2026-09-26 (gpt-6-sol)
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.

Under Dependent Choice, a compact metric space carries a finitely branching refining tree of covers of arbitrarily small diameter

Statement

Assume Dependent Choice. If K is a nonempty compact metric space, there is a rooted levelled tree T=⋃n∈NTn with every level Tn finite and nonempty, and nonempty compact sets (Ks)s∈T, such that T0 has one root r with Kr=K, every node has a finite nonempty set of children whose sets cover its set, every child set is contained in its parent set, and diam⁡(Ks)≤2−n for s∈Tn after a harmless rescaling of the metric.

The tree is finitely branching with finite levels; it is not itself a finite set. Since every node has at least one child, induction from the root puts a node at every level, so T=⋃n∈NTn is infinite.

Facts & Assumptions

Given: The objects, hypotheses, and choice principles stated above.

[F1]

Let (X,d) be a metric space (def-metric-space), with open sets as in def-metric-topology and balls as in def-metric-ball. An open cover of (X,d) is a family U of open subsets of X with X=⋃U; a subcover is a subfamily that is itself an open cover; and (X,d) is compact when every open cover of it has a finite subcover. (Open cover, subcover, compact metric space, and compact subset of a metric space).

[F2]

Let (X,d) be a compact metric space (def-metric-compactness, def-metric-space) and let F⊆X be closed in X (def-metric-topology). Then F is a compact subset of X: the metric subspace (F,dF) is a compact metric space (def-isometry-and-metric-embedding). No choice principle is used. (A closed subset of a compact metric space is compact).

[F3]

Let (X,d) be a metric space (def-metric-space) and let A,B⊆X. A is bounded if A=∅ or there are x0∈X and a real r>0 with A⊆B(x0,r); and for nonempty bounded A the diameter diam⁡(A) is the supremum of D(A)={d(a,b):a,b∈A}, diameters being written in this library for nonempty bounded sets only. (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[F4]

Let X be a set and let R⊆X×X be a relation, called entire on X when for every x∈X there is y∈X with xRy. The Axiom of Dependent Choice is the statement: for every nonempty set X, every relation R entire on X, and every a∈X, there is a sequence x:N→X with x0=a and xnRxn+1 for every n∈N (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain).

Proof

technique · direct
1.1givenF1F3

First the balls B(x,1), x∈K, cover K. Compactness gives finitely many centers covering it, so the triangle inequality bounds all distances in K by one finite real number. Hence D=diam⁡d(K) exists. Replace d by d′=d/max⁡{1,D}; this positive constant rescaling preserves the compact topology and gives diam⁡d′(K)≤1. Use d′ for every ball and diameter below. At stage zero take the single root r with Kr=K, so T0 is finite and nonempty and its diameter bound holds.

2.1step 1.1F2F1F3

Let L be a nonempty compact set and ε>0. The open balls B(x,ε/3) for x∈L form an open cover of L, so by the compactness clause of [F1] finitely many of them, say about x0,…,xm, already cover L. The sets L∩B(xi,ε/3)‾ are then finitely many nonempty-or-discardable subsets of L covering L; each is closed in L and hence compact by [F2], and each has diameter at most 2ε/3<ε by the triangle inequality and the diameter clause of [F3]. Discarding the empty ones leaves a finite nonempty family of nonempty compact subsets of L, covering L, each of diameter below ε.

3.1step 2.1F1F4

Apply step 2.1 with ε=2−(n+1) to every node of level n to obtain that node's children, and let Tn+1 be the resulting finite set of children. Each level is finite because level n is finite and each of its nodes gets finitely many children, and each level is nonempty because every node has at least one child. The passage from one level to the next makes a selection — step 2.1 supplies at least one admissible finite family per node but names none canonically — and the family available at level n+1 is not known until level n is fixed, so the recursion is licensed by Dependent Choice, applied via [F4] to the relation "is an admissible next level for" on finite levelled labellings, taking the root labelling of step 1.1 as the prescribed starting point. This is the Statement's hypothesis and the only place it is used.

4.1step 3.1∎

The preceding construction and implications establish the assertion.

Depends on

Used by

Dependency tree · two levels

25 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