1

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!

  • This is called a quotient. Given an equivalence relation $\sim$ on a set $S$, there is a quotient set $S/\sim$ essentially got by pretending any two elements related by $\sim$ are actually the same element. A common example is modular arithmetic - two numbers are considered "the same" mod $n$ if they differ by a multiple of $n$ (which is an equivalence relation). Quotient sets are quotient spaces if there is topology involved, or quotient groups if there is group structure involved. These are keywords you can google. – anon Mar 12 '21 at 20:35

2 Answers2

0

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?

fleablood
  • 124,253
  • I never got equivalence classes explained really good, that's why it's hard for me to understand it well, your explanation is very good and understandable, thank you! –  Mar 12 '21 at 20:45
  • so M is countable because it consists of all points which are equivalent by rational proportion and because we know that the set of rationals is countable? And because there are uncountably many points, there are uncountably many choice sets like M? does this make any sense or have I still not quite understood it? –  Mar 12 '21 at 20:59
  • No... M is uncountable because its the set of classes. Each element of $M$ is the set of all points that are equivalent to a certain point. For point $A$ there are countably many points equivalent to it. all those points are in the same set... call it oh... I'll use $\alpha_A$. That set is a single element of $M$. You have an $\alpha_K$ for every point of $S^1$. (Although if $A$ rational turn from $B$ then $\alpha_A=\alpha_B$). ANd $M$ is the collection of all the sets $\alpha_K$. There are uncountably many such sets. (as each set has only countably many elements.) – fleablood Mar 12 '21 at 21:52
  • Let $A=(\cos \theta, \sin\theta)$. Let $\alpha_A ={(\cos r\theta, \sin r\theta)|r\in \mathbb Q}$. (Note that if $B\in \alpha_A$ then $\alpha_B$ and $\alpha_A$ are the exact same set). Then $M = {\alpha_K| K\in S^1}$. – fleablood Mar 12 '21 at 21:55
  • Oh I have finally understood it, thank you very much!! –  Mar 12 '21 at 22:53
0

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.

Pock Suppet
  • 126