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I set up the background for the question(s) linking the cardinality of the Vitali set to the Continuum Hypothesis.

BACKGROUND The Vitali set is proven--based on arguments of Cantor--to be not Lebesgue measurable. The Axiom of Choice selects distinct real numbers from the 'Vitali' equivalence classes. By the Archimedean property, a rational number exists between any two distinct real numbers.

By construction, the members of one Vitali equivalence class are the rational numbers on the given interval of finite and positive Lebesgue measure. The rationals can be represented by countably infinite sets of countably infinite elements. The membership of each remaining Vitali equivalence class are countably infinite irrational numbers. Therefore, a Vitali set has only one rational number as a member.

The Archimedean Property, applied to the Vitali set, bounds from 'above' the cardinality of the Vitali set. This bound is puzzling when accepting Cantor's diagonal argument that a bijection exists between the natural numbers and the elements of countably infinite sets that each have countably infinite members. The problem is if the rationals are countable, then the Vitali set must be uncountable if not Lebesgue measurable, but that conflicts with the constraints implied by the Archimedean property.

Cantor's diagonal weaves its way through countably infinite sets of countably infinite members, assuming finite cardinality of elements on each diagonal, but that there are an infinite number of these 'finite' diagonals. This breaks down when realizing that more than one diagonal will be countably infinite. Think of a n x n matrix, then extend the matrix to infinity in both directions. At the nth row, Cantor's diagonals would have essentially accounted for half of the elements in the matrix. As the number of rows goes to infinity, that diagonal becomes infinite and one can't finish enumerating the rest of the matrix. This is not a problem if there are a finite number of sets of countably infinite members, as the sets can be arranged so that infinity is only in 'one direction'. Then a bijection with the natural numbers follows intuitively and formally, as there k first elements, k second elements, etc. This is not the case with countably infinite sets, each with countably infinite members.

That there would be more than one Cantor diagonal of countably infinite elements implies that there is no bijection between the natural numbers and the set of all rational numbers on any interval of finite and positive Lebesgue measure. This is both intuitive and consistent with the implications of the Archimedean Property.

However, treating the cardinality of rationals on such an interval as Aleph One implies that there are Aleph One elements in each Vitali equivalence class.

From the aforementioned, three cases result with implications for the Continuum Hypothesis. The question I pose, or perhaps more correctly three cases of questions, are detailed below.

QUESTION(S) ABOUT CARDINALITY OF VITALI SET AND CONTINUUM HYPOTHESIS

Case 1. A Vitali set has a finite number of elements. It is then of measure zero, since there are Aleph One sets of measure zero, so it's possible the 'sum' of Aleph sets of measure zero is the same as the Lebesgue measure of the interval that was partitioned.

Continuum Hypothesis? The cardinality of the reals on any interval of finite and positive Lebesgue measure would exceed that of the rationals on the same interval if and only if two or more finite number of sets with Aleph One members represents a higher cardinality.

Case 2. A Vitali set has countably infinite members. It is of measure zero.

Continuum Hypothesis? There are countably infinite more equivalence classes of irrational numbers with Aleph One members there are rational numbers on any given interval of finite and Lebesgue measure. If this represents only one higher number in cardinality from Aleph One, then the Continuum Hypothesis would be true. That's because there are countably infinite more rationals (reals) on the real line than there are rationals (reals) in any given interval of finite and positive Lebesgue measure, so comparison between the cardinality of the reals and the rationals for the purposes of the Continuum Hypothesis can be made on an interval of finite and positive Lebesgue measure.

Case 3. A Vitali set has at least Aleph One members. It might not be of measure zero, which could conflict with the Lebesgue measure of the interval which was partitioned. In Case 3, it might be that the cardinality of the Vitali set is Aleph One and of measure zero.

Continuum Hypothesis? It seems to me that the cardinality of the reals would uncountably higher than that of the rationals, so that the hypothesis would be false. Unless the definition of Aleph Two was Aleph One ^ (Aleph One), then the hypothesis would be true.

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closed as not a real question by Carl Mummert, Zhen Lin, tomasz, t.b., William Aug 23 '12 at 16:26

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

6  
I haven't actually read all that, but all Vitali sets have the cardinality of the continuum. CH doesn't seem to be relevant. –  Chris Eagle Mar 24 '12 at 17:34
    
Also, trying to introduce entire background of a question is not a very good idea. A wall of text is a very effective deterrent for potential readers, and those who are likely to be able to help you will probably be aware of the background anyway (or, if they're not, but are particularly interested, will look it up themselves). And in any case, you should always make sure to make it clear what is your question, exactly. I'm not sure of it (though admittedly, I did only a take a quick glance at what you've written). –  tomasz Aug 18 '12 at 2:16
    
@tomasz, and also those voting to close, I did sit down to read this question in its entire length. The OP misunderstood a concept in mathematics, and suggested a possible use of this concept which resulted in a long and confusing text. I encourage people not to close this question. –  Asaf Karagila Aug 18 '12 at 7:53
    
@AsafKaragila: I figured as much, but I think this perfectly qualifies as „not a real question”, since, well, I can't see what's the question here. –  tomasz Aug 18 '12 at 11:54
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1 Answer 1

Assuming the axiom of choice if we take a partition of $\mathbb R$ into countable pieces then there are $2^{\aleph_0}$ many parts. If we chose one from each part then we have exactly $2^{\aleph_0}$ many elements.

The Vitali set, therefore must have the same cardinality of the continuum regardless to the cardinality of the continuum.

The Vitali set is never measurable as well, so it cannot have measure zero.

(If we don't assume the axiom of choice, and are in a universe where - for example - every set is Lebesgue measurable then the Vitali set does not exist at all.)

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