Vitali set

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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}


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.


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.




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.


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.




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