choice set for equivalence classes

Question

I'm trying to understand the proof of the Paradox of Banach-Tarski. I was reading Wagon's work about the paradox and tried to understand the proof of Theorem 1.5, only I haven't warmed up to group theory yet. Maybe someone can explain to me what exactly Wagon means by:

Let M be a choice set for equivalence classes of the relation on $S^1$ given by calling two points equivalent if one is obtainable from the other by a rotation about the origin through a (positive or negative) rational multiple $2\pi$ radians.

I would appreciate it if someone could explain the statements so that I can continue to work on the proof and to understand it better. Thanks for your help!

Answer 1

So you have a point $A=(x,y) = (\cos \theta, \sin \theta)$ and a point $B=(x',y') = (\cos \theta', \sin\theta')$. We say the two points are equivalent if $\theta' = \theta + 2r\pi$ for some $r \in \mathbb Q$.

(That is to say, if you can get from point $A$ to point $B$ by rotating the circle by rational proportion of a full turn they are equivalent. If, however, you can only get to $B$ by rotating an irrational proportion they are not equivalent.)

(Or if it's any easier you can get from any point $A$ to any point $B$ by rotating the circle. If the amount you rotate be is rational multiple of a full turn the points are equivalent and consider to be essentially the same. If the amount you rotate by is an irrational proportion of a full turn they are not equivalent.)

As there are uncountably many points and only countably many rationals there will be an uncountable number of equivalence classes.

Are you familiar with the concept of equivalence relations and equivalence classes?

Answer 2

Equivalence classes are easy. Let's say all the even numbers are equivalent to each other and all the odd numbers are equivalent to each other. This partitions the integers into two sets:

$\{\ldots,-1,1,3,5,\ldots\}$ and $\{\ldots,-2,0,2,4,\ldots\}$

We say that all of the members of the same set are equivalent to each other. So $-1\sim1\sim3\ldots$ and so on.

You should be able to see that every equivalence relation $\sim$ on a set partitions it.

If we map the real numbers onto the circle via the morphism $x\mapsto e^{2i\pi x}$ you should see that $0$ maps to the same point on the circle as $1$ and as $2,3,4,\ldots$. They all go to the same place on the circle. You can think of this as a projection in that it loses some information because if you were to pick a point on the circle and send that point back to where it came from in $\Bbb R$ you wouldn't know which point to send it back to. You would have an infinite set $\{x+i:i\in\Bbb Z\}$. All of those points could be thought of as equivalent.

Technically an equivalence relation is reflexive, symmetric and transitive. This means:

• every element is equivalent to itself, and
• if $a\sim b$ then $b\sim a$, and
• if $a\sim b$ and $b\sim c$ then $a\sim c$

But intuitively what really captures these three properties is the fact that an equivalence relation partitions a set. A partition of a set is a disjoint family of sets that covers the original set.