Understanding Vitali Sets

1. Feb 11, 2013

Artusartos

I'm not sure if I understood Vitali Sets correctly, so I just want to write what I understood (because I don't know if it's right):

We have an equivalence relation where $x \sim y \iff x-y \in Q$. So if we look at the interval [0,1], each irrational number will have its own equivalence class...and we will have one equivalence class for all rational numbers, right? Now, using the axiom of choice, we take one element from each equivalence class as a representative and form the set A. And then we form a new collection of sets $A_q = \{q+a | a \in A\}$. We know that this collection has a countable number of sets, because each set corresponds to one rational number between 0 and 1...and the rational numbers are countable. We also know that the sets are disjoint. Then, when we take the union of these sets, we just need to add their measure. But we know that each set has the same measure, since measure is translation invariant. But we also know that there are infinte number of rational numbers between 0 and 1, so there are infinite amound of sets...so the measure must be infinity. But that can't be true since [0,1] has measure 1. So that's a contradiction, and the vitali set is not measureable.

Do you think my understanding is correct? If not can you please correct me?

Last edited: Feb 11, 2013
2. Feb 11, 2013

joeblow

I think you got it.....

3. Feb 12, 2013

Stephen Tashi

It isn't clear what you mean by that.

Yes, if you mean to say that all rational numbers are in the same equivalence class.

4. Feb 12, 2013

jbunniii

Pretty close, but I would state it as follows. As you said, if $A$ is measurable, then each $A_q$ is measurable and has the same measure, due to translation invariance. Also, the $A_q$ form a countable partition of $[0,1]$, so we must have
$$\sum_{q\in \mathbb{Q}} m(A_q) = 1$$
But all of the $m(A_q)$ are equal to $m(A)$, so the sum on the left is either zero or infinity, depending on whether $m(A) = 0$ or $m(A) > 0$. In either case we have a contradiction.

5. Feb 12, 2013

jbunniii

No, a single equivalence class is of the form $\{x + q \textrm{ }|\textrm{ } q \in \mathbb{Q}\}$, so every equivalence class contains a countably infinite number of elements. There is one equivalence class containing all of the rationals (and no irrationals). Every other equivalence class contains a countably infinite number of irrationals (and no rationals).

There are of course uncountably many equivalence classes. $A$ contains one element from each equivalence class, by construction. The same is true of each $A_q$.

Last edited: Feb 12, 2013
6. Feb 15, 2013

Thanks