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Vitali set
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== Construction and proof == A Vitali set is a subset <math>V</math> of the [[interval (mathematics)|interval]] <math>[0,1]</math> of [[real number]]s such that, for each real number <math>r</math>, there is exactly one number <math>v \in V</math> such that <math>v-r</math> is a [[rational number]]. Vitali sets exist because the rational numbers <math>\mathbb{Q}</math> form a [[normal subgroup]] of the real numbers <math>\mathbb{R}</math> under [[addition]], and this allows the construction of the additive [[quotient group]] <math>\mathbb{R}/\mathbb{Q}</math> of these two groups which is the group formed by the [[coset]]s <math>r+\mathbb{Q}</math> of the rational numbers as a subgroup of the real numbers under addition. This group <math>\mathbb{R}/\mathbb{Q}</math> consists of [[disjoint sets|disjoint]] "shifted copies" of <math>\mathbb{Q}</math> in the sense that each element of this quotient group is a set of the form <math>r+\mathbb{Q}</math> for some <math>r</math> in <math>\mathbb{R}</math>. The [[uncountable set|uncountably many]] elements of <math>\mathbb{R}/\mathbb{Q}</math> [[partition of a set|partition]] <math>\mathbb{R}</math> into disjoint sets, and each element is [[dense set|dense]] in <math>\mathbb{R}</math>. Each element of <math>\mathbb{R}/\mathbb{Q}</math> intersects <math>[0,1]</math>, and the [[axiom of choice]] guarantees the existence of a subset of <math>[0,1]</math> containing exactly one [[representative (mathematics)|representative]] out of each element of <math>\mathbb{R}/\mathbb{Q}</math>. A set formed this way is called a Vitali set. Every Vitali set <math>V</math> is uncountable, and <math>v-u</math> is irrational for any <math>u,v \in V, u \neq v</math>. ===Non-measurability=== [[File:diagonal argument.svg|thumb|A possible enumeration of the positive rational numbers]] A Vitali set is non-measurable. To show this, we assume that <math>V</math> is measurable and we derive a contradiction. Let <math>q_1,q_2,\dots</math> be an enumeration of the rational numbers in <math>[-1,1]</math> (recall that the rational numbers are [[countable]]). From the construction of <math>V</math>, we can show that the translated sets <math>V_k=V+q_k=\{v+q_k : v \in V\}</math>, <math>k=1,2,\dots</math> are pairwise disjoint. (If not, then there exists distinct <math>v,u \in V</math> and <math>k,\ell \in \mathbb{N}</math> such that <math>v+q_k = u + q_{\ell} \implies v-u = q_{\ell}-q_k \in \mathbb{Q}</math>, a contradiction.) Next, note that :<math>[0,1]\subseteq\bigcup_k V_k\subseteq[-1,2].</math> To see the first inclusion, consider any real number <math>r</math> in <math>[0,1]</math> and let <math>v</math> be the representative in <math>V</math> for the equivalence class <math>[r]</math>; then <math>r-v=q_i</math> for some rational number <math>q_i</math> in <math>[-1,1]</math> which implies that <math>r</math> is in <math>V_i</math>. Apply the Lebesgue measure to these inclusions using [[sigma additivity]]: :<math>1 \leq \sum_{k=1}^\infty \lambda(V_k) \leq 3.</math> Because the Lebesgue measure is translation invariant, <math>\lambda(V_k) = \lambda(V)</math> and therefore :<math>1 \leq \sum_{k=1}^\infty \lambda(V) \leq 3.</math> But this is impossible. Summing infinitely many copies of the constant <math>\lambda(V)</math> yields either zero or infinity, according to whether the constant is zero or positive. In neither case is the sum in <math>[1,3]</math>. So <math>V</math> cannot have been measurable after all, i.e., the Lebesgue measure <math>\lambda</math> must not define any value for <math>\lambda(V)</math>.
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