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why the Vitali set has size $2^{\aleph_0}$ and therefore there is a bijection between $\mathbb R$ and the Vitali set? thanks

Asaf Karagila
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2 Answers2

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Note that there are $2^{\aleph_0}$ equivalence classes in $\Bbb{R/Q}$. Since a choice function from these classes is injective, this shows that there are exactly $2^{\aleph_0}$ elements in a Vitali set, and certainly there are not more.

To see that there are $2^{\aleph_0}$ equivalence classes, note that the vector spaces $\Bbb R$ and $\Bbb{R/Q}$ are isomorphic as vector spaces over $\Bbb Q$. Therefore they have the same dimension and cardinality.

Asaf Karagila
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Hint. What is $\bigcup \limits_ {q \in \mathbb Q} V+q$?

Karolis Juodelė
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  • Interestingly both argument use the axiom of choice in a very fundamental way (as they should, of course). You are using the fact that $|A|\aleph_0=\max{|A|,\aleph_0}$, and I use the fact that every vector space has a basis and that taking a quotient by a finite dimensional space doesn't change the dimension of an infinite dimensional space. – Asaf Karagila Nov 18 '13 at 16:17
  • @AsafKaragila The axiom of choice is needed here only to get the existence of a Vitali set (meaning a set containing exactly one member from each coset in $\mathbb R/\mathbb Q$). It's not hard to explicitly find $2^{\aleph_0}$ cosets, so a Cantor-Schröder-Bernstein argument shows that any Vitali set has cardinality $2^{\aleph_0}$. – Andreas Blass Nov 18 '13 at 16:50
  • @Andreas: Yes, I am aware of that. But both the argument presented also use the axiom of choice in a different way. This answer uses cardinal arithmetic, and mine uses vector spaces. – Asaf Karagila Nov 18 '13 at 16:53