LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06

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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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An uncountable almost-disjoint family of subsets of the naturals

Statement

There is an uncountable family $A$ of infinite subsets of $N$ such that $A \cap B$ is finite whenever $A, B \in A$ are distinct.

Facts & Assumptions

Given: A fixed enumeration $(q_{n})_{n \in N}$ of $Q$ and the set $I$ of irrational real numbers.

[F1]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

[F2]

Every nonempty interval in $R$ contains a rational, and $I$ is uncountable (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable, The irrationals are uncountable).

Proof

technique · direct

1.1F1F2givenconstruct

For $x \in I$ , recursively let $n_{k} (x)$ be the least unused index $n$ with $q_{n} \in (x - 1 / k, x + 1 / k)$ , and put $A_{x} = {n_{k} (x) : k \geq 1}$ . Such an index exists by [F2] because every interval contains infinitely many rationals; [F1] makes the choice deterministic.

2.1step 1.1given

Each $A_{x}$ is infinite since its indices are chosen unused. If $x \neq y$ , then for sufficiently large $k, ℓ$ the intervals $(x - 1 / k, x + 1 / k)$ and $(y - 1 / ℓ, y + 1 / ℓ)$ are disjoint; any common selected index must therefore arise among finitely many early choices. Thus $A_{x} \cap A_{y}$ is finite.

3.1step 2.1F2∎

The map $x \mapsto A_{x}$ is injective by the finite-intersection conclusion, so ${A_{x} : x \in I}$ is uncountable by [F2] and has the required property.

Depends on

Used by

The quotient ell-infinity/c₀ has no countable separating family Lemma

Dependency tree · two levels

39 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

Piotr Hajlasz, Functional Analysis, Lemma 10.20 (standard reference, not scraped)