# Vitali set

The Vitali set was one of the first examples of a non-measurable set, and is often used to demonstrate the sometimes strange consequences of relying on the axiom of choice in a proof.

The strange property of the Vitali set is that the union of a countable number of Vitali sets has finite non-zero length. So if the Vitali set can't possibly have length 0, because then the length of a countable union of Vitali sets would have zero length, and it can't possibly have length l greater than 0, because then the length of a countable union of Vitali sets would have infinite length.

## Construction of a Vitali set

Imagine the unit interval $[0,1]\;$. We introduce an equivalence relation ~ on this interval by:

$p \sim q \Leftrightarrow |p-q| \in \mathbf{Q}$

So, to numbers between 0 and 1 are called equivalent if they differ by a rational distance.

And now we look at the equivalence classes: $\mathbf{Q} \cap [0,1]$ is one of these, as the difference of two rational numbers is rational again. Another one is $\sqrt{2} + \mathbf{Q} \mod 1$, i.e., the fractional part of all numbers which differ from $\sqrt{2}$ only by a rational number.

As usual for equivalence classes, two such classes are either identical or don't intersect at all.

The cardinality of our classes is equivalent to the cardinality of $\mathbf{Q} \cap [0,1]$, so the classes have countably infinite elements.

How many of these classes are there? As each only has countably infinite elements, but each element of $[0,1]\;$ lies in one class, there are uncountably infinite classes.

Now, take one element out of each class and put it in the set V... That's easily said, but here is no way to build an algorithm to extract a single element out of each class, and it's impossible to enumerate them, as there are too many!

In fact, only if you postulate the Axiom of Choice, you can state that the set V exists!

And this set V has a property: Let's have a look on it's length.

First a few musings: if you add a rational value $q \in \mathbf{Q} \cap ]0,1[$ to each element of V, and then take only the fractional part, you get a new set Vq. And $V \cap V_q = \empty$, in fact, for two rational $p,q \in \mathbf{Q} \cap ]0,1[$ with $p \neq q$, we have $V_p \cap V_q = \empty$. But each element of $[0,1]\;$ lies in one Vq, so $[0,1]\;$ is the union of a countable number of disjoint sets:

$[0,1]\ = \bigcup_{q \in \mathbf{Q} \cap ]0,1[} V_q$

Now, here comes the rub: the measure of $[0,1]\$ is obviously the length of the interval, i.e., 1.

The measure of a countable union of disjoint sets is the sum of the measures of the sets. So, back to the question: what's the measure of Vq?

First, all Vq are of equal length: you get a Vp out of a Vq by shifting it by |p-q|, cutting the bit which sticks out of our interval $[0,1]\;$ and pasting it on the other side... These operations don't change the length, they are so called isometrics.

Now, if the length of one - and therefore all - Vq would be zero, the length of their union would be zero again: but we know that it is 1.

And if each Vq had a length of ε (however small), the sum of infinite of such ε would be an infinite number, again, and not one!

So, Vq - and therefore V - has not length: our set V doesn't behave as we are used to expect from sets, we can't apply a meaningful measure to it.

There are (at least) two ways to cope with this dilemma:

• don't accept the Axiom of Choice, as it leads to such strange sets. Basically, this is the approach of intuitionism.
• accept that there are sets for which it doesn't make sense to speak of a measure: this is the usual approach of modern measure theorists.