Vitali Sets

Set Theory · Axiom Academy

LESSON Vitali Sets and Non-Measurability Constructing sets that cannot be assigned a meaningful volume We start by defining an equivalence relation on the interval [0,1]. Two real numbers are equivalent if their difference is rational: This is indeed an equivalence relation: Reflexive: x ~ x because x - x = 0 ∈ ℚ Symmetric: If x ~ y then x - y ∈ ℚ, so y - x = -(x - y) ∈ ℚ, thus y ~ x Transitive: If x ~ y and y ~ z, then (x - y) + (y - z) = x - z ∈ ℚ, so x ~ z This equivalence relation partitions [0,1] into equivalence classes. Each class contains all numbers that differ by rationals. Each equivalence class is countably infinite (it contains x + q for every rational q such that x + q ∈ [0,1]) There are uncountably many equivalence classes The classes are disjoint and cover all of [0,1] Here's where the Axiom of Choice becomes essential. We need to select exactly one representative from each equivalence class. This construction is inherently non-constructive: We cannot describe which elements are in V explicitly There is no algorithm to determine if a given number is in V Without AC, we cannot prove such a set exists The choice is made simultaneously for uncountably many classes Now we prove that the Vitali set V cannot be Lebesgue measurable. The proof uses a clever counting argument involving translations. Let r₁, r₂, r₃, ... enumerate the rationals in [-1,1] For each rational rₙ, form the translate Vₙ = V + rₙ (mod 1)

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