False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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.

ZF proves that countable unions of countable sets are countable

Statement refuted

ZF proves that every countable union of countable sets is countable.

Assuming $Con (Z F)$ , this statement is false: it is not a theorem of ZF.

Facts & Assumptions

Given: $Con (Z F)$ . The conclusion is conditional syntactic nonprovability; it does not assert a transitive model from bare consistency.

[F1]

Relative consistency of the Feferman–Levy choice failures over ZF proves consistency of ZF with a sequence of countable real layers whose union is the whole real line.

[F2]

$R$ is uncountable (Cantor's nested intervals, 1874) proves in ZF, without Choice, that the real line is uncountable.

Proof

technique · contradiction with the consistent Feferman--Levy target theory

1.1

Let $T$ be the consistent theory supplied by F1. It contains ZF and asserts that a sequence $⟨ R_{m} : m < ω ⟩$ consists pointwise of countable sets and satisfies $R = ⋃_{m < ω} R_{m}$ . Since $T$ contains ZF, it also proves from F2 that $R$ is not countable.

F1F2

Boundary check. The witness is not the empty family or a one-set union: its domain is all of $ω$ , beginning with index $0$ , and its union contains the zero real and hence is nonempty. Repeated or empty individual layers would not affect the argument; only pointwise countability and the exact union equality are used. No enumeration is selected from the family, because the false principle is assumed only as a single theorem for contradiction.

2.1

Suppose for contradiction that ZF proved the statement refuted. Then $T$ would inherit that theorem. Applying it to the specific sequence in step 1.1 would make $R$ countable, contradicting the same step's ZF proof that $R$ is uncountable. Thus $T$ would be inconsistent, contrary to F1, and the claimed ZF theorem is not provable under the stated consistency hypothesis.

F1step 1.1assume-contradischarge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 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

Thomas Jech, The Axiom of Choice, Theorem 10.6, printed pp. 142–144 (standard reference, not scraped)