CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09

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Finite sets cannot be replaced by arbitrary countable sets

Statement refuted

Every $ω_{1}$ -sized family of countable sets has an uncountable delta subsystem. This would replace finite sets by countable sets in the finite delta-system lemma.

Facts & Assumptions

Given: The family $F = {α : ω \leq α < ω_{1}}$ of ordinals, each regarded as the set of its predecessors.

[F1]

A delta system requires a single pairwise intersection for all distinct members. Delta systems and roots

[F2]

The valid theorem requires finite members and a regular uncountable cardinal. The finite delta-system lemma at a regular uncountable cardinal

Counterexample

1.1

Every member of $F$ is countable by $α < ω_{1}$ , and infinite because $ω \subseteq α$ . There are $ω_{1}$ many such ordinals: the family is a subset of $ω_{1}$ , and if it were countable its union with the countable initial segment $ω$ would make $ω_{1}$ countable. Thus this family satisfies the proposed countability hypothesis but not F2's finiteness hypothesis.

F2given

2.1

For $α < β < γ$ in $F$ , ordinal inclusion gives $α \cap β = α$ and $β \cap γ = β$ . These intersections differ since $α \neq β$ . Therefore no three members form a delta system, and in particular no uncountable subfamily does. This refutes the asserted strengthening.

F1given∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

Monk, Set theory following Jech (2024), Theorem 9.20, printed pp77–78; explicit pairwise-intersection instance (standard reference, not scraped)