General Paradoxes

Let \( G \) be a group acting on a set \( X \) and suppose \( E \subseteq X \) is a nonempty subset of \( X \). Then \( E \) is \(G\)-paradoxical (or paradoxical with respect to \( G \)) if, for some positive integers \( m, n \), there are pairwise disjoint subsets \( A_1, \ldots, A_n, B_1, \ldots, B_m \) of \( E \) and \( g_1, \ldots, g_n, h_1, \ldots, h_m \in G \) such that \[ E = \bigcup_{i=1}^n g_i(A_i) \quad \text{and} \quad E = \bigcup_{j=1}^m h_j(B_j). \]

Consider the simplest example of a paradox: the generating set \( F \) of a free non-abelian group of rank 2. Let us denote by \( W(f) \) the set of all elements of \( F \) starting with \( f \) and perform certain transformations.

Well, we took 4 disjoint sets and from each pair of sets obtained the entire set. And this is called a paradox!

Another important fact in this section is the equivalence of the following three statements:

1. \( |X| = 2|X| \).
2. \( X \) is paradoxical with respect to the group of all permutations of \( X \), that is, all bijections from \( X \) to \( X \).
3. \( X \) is infinite or empty.

And while the implication (b) \(\rightarrow\) (c) is quite obvious, and (c) \(\rightarrow\) (a) is difficult to prove and not very relevant to our topic, the implication (a) \(\rightarrow\) (b) has a clear and interesting proof.

The equality \( |X| = 2|X| \) actually means that there exist bijections from \( X \) to \( A \) and from \( X \) to \( B \), where \( A \) and \( B \) are disjoint and their union gives \( X \).

Let’s define h as the composition of these bijections and consider the sets \(A,h(A),h(h(A)),...\)

Now, let’s apply the bijection \(g ^{−1}\) to \(B\), separately considering \(h(C)\) and \(D =X∖C\).

Then let’s choose a bijection from \(X\) to \(X\) corresponding to this given bijection.

Now, let’s perform similar actions for \(h\) equal to \(f\circ f\) and \(B\), obtaining another bijection.

And a bit of magic and obvious analogies. (It works because \(f_1^{-1}\) works on \(f(A)\) the same as on \(f(C)\).)

Here comes the paradox! It’s easy to see that \(f(A),f(B),g(A),g(B)\) are non-intersecting sets in X.

And the final paradox, an example of a countably paradoxical set (the sizes of sets A and B in the definition can be countable, not necessarily finite): consider a circle under rotations about itself.

First, let's identify what's known as the Vitali set on the circle (dividing the set into equivalence classes based on the principle 'one point maps to another under rotation by a rational angle in degrees' and choosing one point from each class).

Now, let's rotate our set by various rational values ( \(\rho \) is the set of rotations by rational angles).

Here we have obtained the representation of the circle through rotations using only even and odd rotations of the Vitali set separately — just what we wanted to achieve!