**Prerequisites:**

As everybody knows, the rational numbers are a countable subset of the reals. The reals are uncountable; almost all real numbers are not rational numbers. It just so happens that the difference between any two rational numbers is also a rational number.

If you add to each rational number a certain irrational number such as sqrt(2), you arrive at another countable subset of the real numbers, but a disjoint one from the original rational numbers. Again, the difference between any two elements of the set Q + sqrt(2) is a rational number, because the sqrt(2)'s cancel out when they are subtracted.

For example, above the dotted line is the set Q of rational numbers (some points are omitted). Below the dotted line is the set Q + sqrt(2) of rational numbers increased by sqrt(2). The set below the line looks very much like the set above the line - except all of the points are shifted to the right.

Consider the following equivalence relation ~ on the real numbers.

We say x ~ y (x is equivalent to y) if there exists a rational number q such that x = y + q.

Under this relation, all of the elements of Q (the points above the dotted line) are equivalent to one another; and all of the elements of Q + sqrt(2) (the points below the dotted line) are equivalent to one another.

A vitali set is a set with exactly one representative from each equivalence class.

In other words a vitali set is a set V such that for each real number r, there is exactly one v in V such that r - v is rational.

**Problem:**

Consider a vitali set V.

Let F by the set of bijections from V to [0, 1]. There are uncountably infinitely many of these bijections.

For each f in F, consider the set W(f) = {v + f(v) : v in V}. (That is, take each element v of the vitali set V and add to it f(v), its index from [0, 1].) For how many functions f in F is W(f) a vitali set? Almost all? Almost none?

I don't know the answer to this question; it seems rather interesting.