Monday, August 28, 2017

Open Problem: Shearing Vitali Sets

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.

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