TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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Assuming choice on the cosets of $Q$ in $R$ , a Vitali set in $[0, 1]$ exists

Statement

Assume the Axiom of Choice. Then there exists a Vitali set $V \subseteq [0, 1]$ in the sense of Vitali set on $[0, 1]$ .

Facts & Assumptions

Given: The Axiom of Choice.

[F1]

A Vitali set on $[0, 1]$ is a subset meeting every equivalence class of $x \sim y ⟺ x - y \in Q$ in exactly one point (Vitali set on $[0, 1]$ ).

[A1]

Every family of nonempty sets has a choice function (The Axiom of Choice).

[F2]

A choice function for a family $F$ is a function $g$ with domain $F$ such that $g (S) \in S$ for every $S \in F$ (Choice function).

Proof

technique · direct

1.1F1construct

Let $F$ be the family of all sets of the form $(x + Q) \cap [0, 1]$ with $x \in [0, 1]$ . Each member of $F$ is nonempty because it contains its defining point, and distinct members are exactly the equivalence classes of the relation in [F1].

2.1step 1.1A1F2construct

By [A1] and [F2] there is a choice function $g$ on $F$ . Put $V : = {g (S) : S \in F} \subseteq [0, 1]$ . Then $V$ meets every member of $F$ , and it meets each of them in exactly one point because $g$ assigns one value to each set in its domain.

3.1step 1.1step 2.1F1∎

Since the members of $F$ are exactly the equivalence classes of $x \sim y ⟺ x - y \in Q$ on $[0, 1]$ , step 2.1 says precisely that $V$ is a Vitali set on $[0, 1]$ .

Remarks

The proof uses one simultaneous selector on the family of $Q$ -cosets meeting $[0, 1]$ . Nothing in the proof reduces that family to a countable one.

Depends on

Used by

Every subset of ℝ of positive Lebesgue outer measure contains a nonmeasurable subset Corollary
A function can have measurable level sets without being measurable Counterexample
A nonmeasurable subset of a null line shows that the product of complete measures need not be complete Counterexample
A Vitali set shows that not every subset of ℝ is Lebesgue measurable Counterexample
An uncountable supremum of measurable indicators can be nonmeasurable Counterexample
Two disjoint nonmeasurable subsets of [0,1] can have the measurable union [0,1] Counterexample
FALSE: assuming the Axiom of Choice, every subset of ℝ is Lebesgue measurable False statement
FALSE: if every level set of a real-valued function is measurable, then the function is measurable False statement
FALSE: if the absolute value is measurable, then the function is measurable False statement
FALSE: the supremum of an arbitrary family of measurable functions is always measurable False statement
Hausdorff measure is countably additive on every subset False statement
What the Vitali set, Bernstein sets and free ultrafilters cost in choice Remark
Assuming the Axiom of Choice, no translation-invariant measure on P(ℝ) is both finite and nonzero on [0,1] Theorem

Dependency tree · two levels

8 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

Jacek Cichoń, Aleksander Kharazishvili, and Bogdan Węglorz, Subsets of the Real Line, Chapter 8 (standard reference, not scraped)
Vitali set (Wikipedia) (standard reference, not scraped)

Assuming choice on the cosets of QQ in RR, a Vitali set in [0,1][0,1] exists