CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Assuming choice, a Lindelöf space whose square is not Lindelöf: the lower-limit line

Statement refuted

Assuming countable choice, Lindelöfness is productive.

Facts & Assumptions

Given: The lower-limit line $S$ and its square $S^2$ .

[L1]

The lower-limit line has Lindelöf degree $\aleph_0$ , hence is Lindelöf (For the lower-limit line, $\chi=d=L=c=\aleph_0$ and $w=2^{\aleph_0}$ under choice).

[L2]

The lower-limit plane $S^2$ is not Lindelöf (Assuming choice, the lower-limit plane is first countable, separable, and ccc, but not second countable or Lindelöf).

[L3]

The false statement being exhibited asserts that products of Lindelöf spaces are Lindelöf (Assuming countable choice, refuted: Lindelöfness is productive).

Counterexample

technique · direct

1.1

The factor $S$ is Lindelöf by [L1].

1.2

Its square $S^2$ is not Lindelöf by [L2].

2.1

Hence the product $S\times S$ fails the conclusion while each displayed factor satisfies the hypothesis, refuting the assertion in [L3].

step 1.1step 1.2L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 111 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

UCR General Topology Notes (standard reference, not scraped)
Sorgenfrey plane (Wikipedia) (standard reference, not scraped)
Lower limit topology (Wikipedia) (standard reference, not scraped)