LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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 $T = ⋃_{n \in N} T_{n}$ with every level $T_{n}$ finite and nonempty, and nonempty compact sets $(K_{s})_{s \in T}$ , such that $T_{0}$ has one root $r$ with $K_{r} = K$ , every node has a finite nonempty set of children whose sets cover its set, every child set is contained in its parent set, and $diam (K_{s}) \leq 2^{- n}$ for $s \in T_{n}$ 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 $T = ⋃_{n \in N} T_{n}$ 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 $F \subseteq X$ be closed in $X$ (def-metric-topology). Then $F$ is a compact subset of $X$ : the metric subspace $(F, d_{F})$ 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 $A, B \subseteq X$ . $A$ is bounded if $A = \emptyset$ or there are $x_{0} \in X$ and a real $r > 0$ with $A \subseteq B (x_{0}, r)$ ; and for nonempty bounded $A$ the diameter $diam (A)$ is the supremum of $D (A) = {d (a, b) : a, b \in A}$ , 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 $R \subseteq X \times X$ be a relation, called entire on $X$ when for every $x \in X$ there is $y \in X$ with $x R y$ . The Axiom of Dependent Choice is the statement: for every nonempty set $X$ , every relation $R$ entire on $X$ , and every $a \in X$ , there is a sequence $x : N \to X$ with $x_{0} = a$ and $x_{n} R x_{n + 1}$ for every $n \in N$ (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Proof

technique · direct

1.1

At stage zero take the single root $r$ with $K_{r} = K$ , a nonempty compact set, so $T_{0}$ is finite and nonempty.

givenF2F1F3

2.1

Let $L$ be a nonempty compact set and $ε > 0$ . The open balls $B (x, ε / 3)$ for $x \in L$ form an open cover of $L$ , so by the compactness clause of [F1] finitely many of them, say about $x_{0}, \dots, x_{m}$ , already cover $L$ . The sets $L \cap \overline{B (x_{i}, ε / 3)}$ 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 $2 ε / 3 < ε$ 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 $ε$ .

step 1.1F2F1F3

3.1

Apply step 2.1 with $ε = 2^{- (n + 1)}$ to every node of level $n$ to obtain that node's children, and let $T_{n + 1}$ 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.

step 2.1F1F4

4.1

The preceding construction and implications establish the assertion.

step 3.1∎

Depends on

Used by

Under Dependent Choice, every nonempty compact metric space is a continuous image of Cantor space Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 67 results over 13 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

David Marker, Descriptive Set Theory, §§1–2 (standard reference, not scraped)
Michael Kunzinger, General Topology, §§11.3–11.4 (standard reference, not scraped)
MFF General Topology course summary, §4.3 (standard reference, not scraped)
Jesse Peterson, Real Analysis, §§3.6–3.7 (standard reference, not scraped)