LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-12

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.

Irreducibility is equivalent to every pair of nonempty open sets meeting

Statement

Let $k$ be an algebraically closed field, let $n \geq 0$ , and let $X \subseteq A_{k}^{n}$ be a nonempty subset with the subspace Zariski topology. Then the following are equivalent:

$X$ is irreducible.
Every pair of nonempty open subsets of $X$ has nonempty intersection.
Every nonempty open subset of $X$ is dense in $X$ .

Facts & Assumptions

Given: An algebraically closed field $k$ , an integer $n \geq 0$ , and a nonempty subset $X \subseteq A_{k}^{n}$ with the subspace Zariski topology.

Proof

technique · direct

1.1given

Assume $X$ is irreducible. If $U, V \subseteq X$ are nonempty open subsets with $U \cap V = \emptyset$ , then $X = (X ∖ U) \cup (X ∖ V)$ is a union of two proper closed subsets, contrary to irreducibility. Therefore every two nonempty open subsets of $X$ meet.

1.2givenalgebra

Conversely, assume every two nonempty open subsets of $X$ meet. If $X = A \cup B$ with $A$ and $B$ proper closed subsets, then $X ∖ A$ and $X ∖ B$ are nonempty disjoint open subsets, a contradiction. Hence $X$ is irreducible.

1.3givenalgebra

If every nonempty open subset of $X$ is dense and $U, V$ are nonempty open subsets, then $\overline{U} = X$ , so $U$ meets the nonempty open set $V$ . Thus every pair of nonempty open subsets meets.

2.1step 1.2given

Assume every two nonempty open subsets of $X$ meet, and let $U \subseteq X$ be nonempty open. If $\overline{U} \neq X$ , then $X ∖ \overline{U}$ is a nonempty open set disjoint from $U$ , contradiction. So $U$ is dense.

3.1step 1.1step 1.2step 2.1step 1.3∎

Steps 1.1 and 1.2 show that irreducibility is equivalent to pairwise intersection of nonempty opens, and steps 2.1 and 1.3 show that this is also equivalent to density of every nonempty open subset.

Irreducibility is equivalent to every pair of nonempty open sets meeting

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources