Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedaudited 2026-09-12
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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.

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The Zariski topology on the affine line over an infinite field is not Hausdorff

Statement refuted

Over an infinite field, the Zariski topology on Ak1 is Hausdorff.

Example

Let ab be two points of Ak1. By On the affine line, the classical Zariski topology is cofinite, every nonempty open subset of Ak1 is the complement of a finite set. If U and V are nonempty open neighborhoods of a and b, then UV=Ak1(FG) for finite sets F and G. Because k is infinite, FG is still a proper subset of k, so UV.

Thus no two distinct points admit disjoint neighborhoods. The refuted statement is therefore false: the classical Zariski topology on Ak1 is not Hausdorff.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources