LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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.

Closed subsets of Baire space are tree bodies

Statement

In ZF, $F \subseteq N = N^{N}$ is closed if and only if $F = [T]$ for a tree $T$ on $N$ . When $F$ is closed, its prefix tree

$T_{F} = {s \in N^{< ω} : (\exists y \in F) s \subseteq y}$

has body $F$ . If $F$ is nonempty, $T_{F}$ is nonempty and pruned; if $F$ is empty, $T_{F}$ is empty.

Facts & Assumptions

[F1]

Baire cylinders $N_{s}$ form a basis, including $N_{\emptyset} = N$ ; see Baire sequence space $N^{N}$ and its cylinder topology.

[F2]

A tree is prefix closed, and $x \in [T]$ means every finite prefix of $x$ lies in $T$ ; see Trees and their bodies.

Proof

Given: A subset $F \subseteq N$ and the definitions above.

1.1F1F2

For any tree $T$ and $x \notin [T]$ , F2 gives $n$ with $x ↾ n \notin T$ . If $y \in N_{x ↾ n}$ , it has that same excluded prefix, so $y \notin [T]$ . Thus each point of the complement has a basic neighbourhood in the complement, which proves $[T]$ closed. This includes $n = 0$ and $T = \emptyset$ .

1.2givenF2

Suppose $F$ is closed. If $s \in T_{F}$ , one witness $y \in F$ extending $s$ also extends every restriction of $s$ . Thus $T_{F}$ is a tree. Each $y \in F$ has all its prefixes in $T_{F}$ , so $F \subseteq [T_{F}]$ .

2.1givenF1step 1.2

Let $x \in [T_{F}]$ . If $x \notin F$ , closedness and F1 give a cylinder $N_{x ↾ n}$ disjoint from $F$ . But $x ↾ n \in T_{F}$ has an extending witness $y \in F$ , a contradiction to this disjointness. Therefore $[T_{F}] \subseteq F$ , and equality follows.

3.1F2step 1.1step 2.1

If $F = \emptyset$ , no prefix has a witness and $T_{F} = \emptyset$ . If $F \neq \emptyset$ , its empty prefix belongs to $T_{F}$ . For each individual $s \in T_{F}$ , a witness $y$ extends it to $y ↾ (∣ s ∣ + 1) \in T_{F}$ , proving pruning. These are separate existential deductions at each node, not a simultaneous choice of witnesses. Together with the closed-body implication this establishes both directions and all additional claims. QED.

Depends on

Used by

Dependency tree · two levels

5 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

Lemma 1.14(ii), Definition 1.15 (standard reference, not scraped)