This Banach-Tarski explanation is nice at a very beginner level, but worse than useless above that. Here is a very important related fact: The Banach-Tarski paradox is simply NOT TRUE on the line and the plane. You can not do such a rearrangement with a circle to get two circles of the same size.
The difference between 2D and 3D that causes this change is very interesting. The isomorphism group has a much more complex structure in 3D. In particular, the group of 3D rotations contains a free group. This means that there exist 3D rotations a and b and their respective inverses A and B such that a list of successively applied such rotations is the identity if and only if it is the identity for formal reasons. (For example, aaBbAA is the identity for formal reasons.) How does this lead to the paradox? There are two main ideas involved:
This free subgroup has a paradoxical behavior. (We now treat it as a subset of the unit sphere.) The elements of this set are defined by rotations such as AbAbABaAABBaBbb. This set can be divided into four disjoint subsets depending on their first letter. These four subsets are isomorphic to each other, but each is also isomorphic to their union.
We can use the axiom of choice to pick representative elements from the cosets of our free group. (Basically, to get a maximal subset of the sphere such that the above free rotations never move one element into another.) Such a set will behave very similarly to a single element of the free group, but it has the advantage that its rotated versions together give the whole sphere, not just a sparse subset of it.
These were the main ideas. EDIT: One minor idea that I originally forgot is nicely explained by earthwormchuck163 in a reply to this comment. The complication is that rotations have fixed points, and the relief is that there are only countably many of them.
One very minor idea is that if you have a paradox for the sphere using rotations, you can get a paradox for the ball. This is a nice exercise.
The paradoxical decomposition of F2 only gives a decomposition for a dense subset of the sphere, because you have to throw away the (countably many) fixed points of all the rotations involved to make the correspondence between F2 and the orbits of various points. To go the rest of the way and you need to use something other than rotations about the origin, ie something more than just the action of F2. But it’s certainly fair to say that Banach-Tarski works because of the structure of F2.
We seem to be talking about different things, I’m talking about doubling the surface of the sphere. You’re talking about how to get the center once you’ve doubled the surface.
This Banach-Tarski explanation is nice at a very beginner level, but worse than useless above that. Here is a very important related fact: The Banach-Traski paradox is simply NOT TRUE on the line and the plane. You can not do such a rearrangement with a circle to get two equally sized circles.
I seem to recall reading about a way to divide the interval [0, 1] in subsets, translating some of them, and getting [0, 2] (involving the Vitali set or something like that), but maybe my memory fails me.
These were the main ideas. One very minor idea is that if you have a paradox for the sphere using rotations, you can get a paradox for the ball. This is a nice homework.
Whfg pbafvqre gur onyy nf orvat znqr hc ol enqvv. Or am I missing something?
I seem to recall reading about a way to divide the interval [0, 1] in subsets, translating some of them, and getting [0, 2] (involving the Vitali set or something like that), but maybe my memory fails me.
This is possible if you use infinitely many subsets. With an uncountably infinite number of pieces it is true by definition, with a countably infinite number of pieces it can be proven using the Vitali set, and with a finite number of pieces it is not true.
Or am I missing something?
What Oscar_Cunningham said, but basically, no, you are not.
Okay. In the original formulation of the paradox, the task is to cut a ball into pieces, and assemble two balls from the pieces. If I am not mistaken, you have solved a slightly easier task: cut a ball into pieces, and covered two balls with the pieces (with overlaps). A part of the “nice exercise” is to bridge this gap.
This Banach-Tarski explanation is nice at a very beginner level, but worse than useless above that. Here is a very important related fact: The Banach-Tarski paradox is simply NOT TRUE on the line and the plane. You can not do such a rearrangement with a circle to get two circles of the same size.
The difference between 2D and 3D that causes this change is very interesting. The isomorphism group has a much more complex structure in 3D. In particular, the group of 3D rotations contains a free group. This means that there exist 3D rotations a and b and their respective inverses A and B such that a list of successively applied such rotations is the identity if and only if it is the identity for formal reasons. (For example, aaBbAA is the identity for formal reasons.) How does this lead to the paradox? There are two main ideas involved:
This free subgroup has a paradoxical behavior. (We now treat it as a subset of the unit sphere.) The elements of this set are defined by rotations such as AbAbABaAABBaBbb. This set can be divided into four disjoint subsets depending on their first letter. These four subsets are isomorphic to each other, but each is also isomorphic to their union.
We can use the axiom of choice to pick representative elements from the cosets of our free group. (Basically, to get a maximal subset of the sphere such that the above free rotations never move one element into another.) Such a set will behave very similarly to a single element of the free group, but it has the advantage that its rotated versions together give the whole sphere, not just a sparse subset of it.
These were the main ideas. EDIT: One minor idea that I originally forgot is nicely explained by earthwormchuck163 in a reply to this comment. The complication is that rotations have fixed points, and the relief is that there are only countably many of them.
One very minor idea is that if you have a paradox for the sphere using rotations, you can get a paradox for the ball. This is a nice exercise.
The paradoxical decomposition of F2 only gives a decomposition for a dense subset of the sphere, because you have to throw away the (countably many) fixed points of all the rotations involved to make the correspondence between F2 and the orbits of various points. To go the rest of the way and you need to use something other than rotations about the origin, ie something more than just the action of F2. But it’s certainly fair to say that Banach-Tarski works because of the structure of F2.
To go the rest of the way you still only use rotations, just not the rotations in F2.
The way I always did it was to use rotations about some fixed line that doesn’t go through 0.
We seem to be talking about different things, I’m talking about doubling the surface of the sphere. You’re talking about how to get the center once you’ve doubled the surface.
Ahh yes, you’re right.
Indeed. I worked from memory, and forgot about this complication. Edited the text to cite your comment.
I seem to recall reading about a way to divide the interval [0, 1] in subsets, translating some of them, and getting [0, 2] (involving the Vitali set or something like that), but maybe my memory fails me.
Whfg pbafvqre gur onyy nf orvat znqr hc ol enqvv. Or am I missing something?
This is possible if you use infinitely many subsets. With an uncountably infinite number of pieces it is true by definition, with a countably infinite number of pieces it can be proven using the Vitali set, and with a finite number of pieces it is not true.
What Oscar_Cunningham said, but basically, no, you are not.
I was expecting something harder given that you called it “a nice exercise”, so I pretty much assumed that mine was not the right solution...
Okay. In the original formulation of the paradox, the task is to cut a ball into pieces, and assemble two balls from the pieces. If I am not mistaken, you have solved a slightly easier task: cut a ball into pieces, and covered two balls with the pieces (with overlaps). A part of the “nice exercise” is to bridge this gap.
Jurer qbrf gur prager tb?
Jvgu gur abegu cbyr (be nal bgure neovgenel qverpgvba). Vf gurer nal ceboyrz jvgu gung V pnaabg frr?