|
|
| (One intermediate revision by one other user not shown) |
| Line 1: |
Line 1: |
| In [[mathematics]], a '''Vitali set''' is an elementary example of a set of [[real number]]s that is not [[Lebesgue measure|Lebesgue measurable]], found by {{harvs|txt|authorlink=Giuseppe Vitali|first=Giuseppe |last=Vitali|year=1905}}. The '''Vitali theorem''' is the [[existence theorem]] that there are such sets. There are [[uncountably many]] Vitali sets, and their existence is proven on the assumption of the [[axiom of choice]].
| | Greetings. The [http://www.Britannica.com/search?query=author%27s author's] name is Dalton but it's not the almost all masucline name out now there are. To drive is one of some of the things he loves most people. His wife and him chose to reside in South Carolina and as well as his family loves them. Auditing is where his primary income comes from. He is running and maintaining one specific blog here: http://circuspartypanama.com/<br><br>my web site ... [http://circuspartypanama.com/ clash of clans hack tool v8.8] |
| | |
| == Measurable sets ==
| |
| Certain sets have a definite 'length' or 'mass'. For instance, the [[interval (mathematics)|interval]] [0, 1] is deemed to have length 1; more generally, an interval [''a'', ''b''], ''a'' ≤ ''b'', is deemed to have length ''b''−''a''. If we think of such intervals as metal rods with uniform density, they likewise have well-defined masses. The set [0, 1] ∪ [2, 3] is composed of two intervals of length one, so we take its total length to be 2. In terms of mass, we have two rods of mass 1, so the total mass is 2.
| |
| | |
| There is a natural question here: if E is an arbitrary subset of the real line, does it have a 'mass' or 'total length'? As an example, we might ask what is the mass of the set of [[rational number]]s, given that the mass of the interval [0, 1] is 1. The rationals are [[Dense_set|dense]] in the reals, so any non negative value may appear reasonable.
| |
| | |
| However the closest generalization to mass is [[sigma additivity]], which gives rise to the [[Lebesgue measure]]. It assigns a measure of ''b'' − ''a'' to the interval [''a'', ''b''], but will assign a measure of 0 to the set of rational numbers because it is [[countable]]. Any set which has a well-defined Lebesgue measure is said to be "measurable", but the construction of the Lebesgue measure (for instance using [[Carathéodory's extension theorem]]) does not make it obvious whether there exist non-measurable sets. The answer to that question involves the [[axiom of choice]].
| |
| | |
| == Construction and proof ==
| |
| A Vitali set is a subset <math>V</math> of the [[interval (mathematics)|interval]] <nowiki>[0, 1]</nowiki> of [[real number]]s which for each real number ''r'' contains exactly one number <math>v \in V</math> such that ''v''−''r'' is a [[rational number]]. Vitali sets exist because the rational numbers '''Q''' form a [[subgroup]] of the real numbers '''R''' under [[addition]], and this allows the construction of the additive [[quotient group]] '''R'''/'''Q''' of these two groups which is the group formed by the [[coset]]s of the rational numbers as a subgroup of the real numbers under addition. This group '''R'''/'''Q''' consists of [[disjoint sets|disjoint]] "shifted copies" of the rational numbers in the sense that each element of this quotient group is a set of the form {{nowrap|'''Q''' + ''r''}} for some ''r'' in '''R'''. The [[uncountable set|uncountably many]] elements of '''R'''/'''Q''' [[partition of a set|partition]] '''R''', and each element is [[dense set|dense]] in '''R'''. Each element of '''R'''/'''Q''' intersects [0, 1], and the [[axiom of choice]] guarantees the existence of a subset of [0, 1] containing exactly one representative out of each element of '''R'''/'''Q'''. A set formed this way is called a Vitali set.
| |
| | |
| Every Vitali set ''V'' is uncountable, and ''v''−''u'' is irrational for any <math>u,v \in V, u \neq v</math>.
| |
| | |
| A Vitali set is non-measurable. To show this, we assume that ''V'' is measurable and we derive a contradiction. Let ''q''<sub>1</sub>, ''q''<sub>2</sub>, ... be an enumeration of the rational numbers in [−1, 1] (recall that the rational numbers are [[countable]]). From the construction of ''V'', note that the translated sets <math>V_k=V+q_k=\{v+q_k : v \in V\}</math>, ''k'' = 1, 2, ... are pairwise disjoint, and further note that <math>[0,1]\subseteq\biguplus_k V_k\subseteq[-1,2]</math>. (To see the first inclusion, consider any real number ''r'' in [0, 1] and let ''v'' be the representative in ''V'' for the equivalence class [''r'']; then ''r''−''v'' = ''q'' for some rational number ''q'' in [-1, 1].)
| |
| | |
| 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 λ(''V'') yields either zero or infinity, according to whether the constant is zero or positive. In neither case is the sum in [1, 3]. So ''V'' cannot have been measurable after all, i.e., the Lebesgue measure λ must not define any value for λ(''V'').
| |
| | |
| ==See also==
| |
| *[[Non-measurable set]]
| |
| *[[Banach–Tarski paradox]]
| |
| | |
| ==References==
| |
| * {{cite book|last=Herrlich|first=Horst|title=Axiom of Choice|page=120|publisher=Springer|year=2006}}
| |
| * {{cite journal|last=Vitali|first=Giuseppe|authorlink=Giuseppe Vitali|year=1905|title= Sul problema della misura dei gruppi di punti di una retta|journal=Bologna, Tip. Gamberini e Parmeggiani}}
| |
| | |
| [[Category:Sets of real numbers]]
| |
| [[Category:Measure theory]]
| |
| [[Category:Articles containing proofs]]
| |
Greetings. The author's name is Dalton but it's not the almost all masucline name out now there are. To drive is one of some of the things he loves most people. His wife and him chose to reside in South Carolina and as well as his family loves them. Auditing is where his primary income comes from. He is running and maintaining one specific blog here: http://circuspartypanama.com/
my web site ... clash of clans hack tool v8.8