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.

Borel hierarchy exhaustion and preservation by continuous pullback

Statement

In ZFC, for every topological space $X$ ,

$B (X) = ⋃_{1 \leq α < ω_{1}} Σ_{α}^{0} (X) .$

Continuous inverse images preserve $Σ_{α}^{0}$ , $Π_{α}^{0}$ and $Δ_{α}^{0}$ at every positive countable rank. If $Y \subseteq X$ has the subspace topology, its $Σ_{α}^{0}$ and $Π_{α}^{0}$ sets are exactly the traces of the corresponding classes on $X$ . No trace assertion for $Δ$ is made. The inverse-image proof uses no choice beyond the supplied representations.

Facts & Assumptions

[F1]

The countable Borel hierarchy and its limit convention defines the positive ranks and the least Borel sigma-algebra.

[F2]

Under countable choice, countable subsets of $ω_{1}$ have countable suprema: Assuming countable choice: every at most countable subset of $ω_{1}$ is bounded below $ω_{1}$ , so no at most countable subset of $ω_{1}$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable.

[F3]

Continuity of a map of topological spaces at a point and globally gives the open-neighbourhood criterion.

[F4]

Opens of a subspace are traces: Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace.

[F5]

Transfinite induction permits induction over the positive countable ordinals.

[A1]

Assume The Axiom of Choice, including its restriction to countable families.

Proof

Given: The indicated spaces, ranks and axiom assumptions.

1.1

By F5, every rank lies in $B (X)$ : rank one consists of opens; the progressive step takes complements and countable unions of earlier Borel sets. Write $H$ for the union in the statement. If $A \in Σ_{α}^{0}$ , then $X ∖ A \in Π_{α}^{0} \subseteq Σ_{α + 1}^{0}$ , the latter inclusion using a constant sequence. Thus $H$ is closed under complements.

F1F5

1.2

Let $f : X \to Z$ be continuous. For open $U \subseteq Z$ , every point of $f^{- 1} [U]$ has an open neighbourhood inside that preimage by F3; their union proves it open. Induct on the rank by F5. For a represented union $A = ⋃_{n} B_{n}$ , $f^{- 1} [A] = ⋃_{n} f^{- 1} [B_{n}]$ , and the induction hypotheses place each preimage in its original lower $Π$ rank. Also $f^{- 1} [Z ∖ A] = X ∖ f^{- 1} [A]$ . These identities prove both classes at the next rank. Membership in both gives the $Δ$ assertion, without selecting any representations simultaneously.

F1F3F5

2.1

Given $A_{n} \in H$ , let $α_{n}$ be its least positive $Σ$ rank. The preceding complement argument applied twice gives $A_{n} \in Π_{α_{n} + 1}^{0}$ . By F2 and A1, $δ = \sup_{n} (α_{n} + 2) < ω_{1}$ , and each $α_{n} + 1 < δ$ . Hence $⋃_{n} A_{n} \in Σ_{δ}^{0}$ . The empty union is $\emptyset \in Σ_{1}^{0}$ . Thus $H$ is a sigma-algebra containing the opens and so contains $B (X)$ by its leastness; step 1.1 gives equality.

F1F2A1step 1.1

3.1

For the inclusion $i : Y \to X$ , $i^{- 1} [U] = Y \cap U$ is open by F4, so $i$ is continuous and step 1.2 gives one trace inclusion. Conversely induct by F5. Opens lift by F4. If $B = ⋃_{n} B_{n} \in Σ_{α}^{0} (Y)$ with lower- $Π$ constituents, each $B_{n}$ has an ambient lift in its own rank by induction. Their sets of lifts are nonempty subsets of $P (X)$ ; A1 chooses lifts $C_{n}$ . Then $⋃_{n} C_{n} \in Σ_{α}^{0} (X)$ has trace $B$ . If $B = Y ∖ D$ is $Π_{α}^{0}$ , lift $D$ to $C$ and use $X ∖ C$ , whose trace is $B$ . This proves the reverse inclusion in both classes. For $Y = \emptyset$ the same identities apply, with $\emptyset$ an available lift throughout. QED.

F1F4F5A1step 1.2

Depends on

Used by

Every set of reals is Borel False statement
Borel payoffs admit unraveling covers Theorem
Universal Borel sets and strictness on Cantor space Theorem

Dependency tree · two levels

27 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 2.5(i–ii), Lemma 2.6(iv), Exercise 2.8(a), printed pp14–15; local exhaustion argument fills the abbreviated sigma-algebra step (standard reference, not scraped)