LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

Dense constructible subsets contain an open

Statement

If a constructible subset $C \subseteq X$ has nonempty irreducible closure $Z$ , then $C$ contains a nonempty open subset of $Z$ .

Work over a fixed algebraically closed field $k$ , with the Axiom of Choice. Classical varieties are separated and admit finite affine covers; they may be reducible or empty unless irreducibility is specified. Irreducible means nonempty. All fibres and points below are classical closed-point fibres and points.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[F1]

A subset $S$ of a classical variety $X$ is locally closed if $S = U \cap Z$ for some open $U \subseteq X$ and closed $Z \subseteq X$ . A subset is constructible if it is a finite union of locally closed subsets. The empty union is allowed, so $\emptyset$ is constructible. Work over a fixed algebraically closed field $k$ , with the Axiom of Choice. Classical varieties are separated and admit finite affine covers; they may be reducible or empty unless irreducibility is specified. Irreducible means nonempty. All fibres and points below are classical closed-point fibres and points. (Locally closed and constructible subsets).

Proof

1.1

Write $C = ⋃_{i = 1}^{m} (U_{i} \cap Z_{i})$ with $U_{i}$ open and $Z_{i}$ closed, discarding empty pieces. The family is nonempty because $Z$ is nonempty. As $Z = \overline{C} \subseteq ⋃_{i} Z_{i}$ , irreducibility implies $Z \subseteq Z_{j}$ for some $j$ .

2.1

Then $Z \cap U_{j} \subseteq Z_{j} \cap U_{j} \subseteq C$ . This open subset of $Z$ is nonempty because the retained piece $U_{j} \cap Z_{j}$ is nonempty and contained in $C \subseteq Z$ . Thus it is the required open.

F1step 1.1∎

Depends on

Locally closed and constructible subsets

Used by

A dominant image contains a dense open Corollary
Chevalley: images of constructible sets are constructible Theorem

Dependency tree · two levels

2 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

Milne Proposition 9.6 (standard reference, not scraped)