First take a sphere S2 in 3D space. Next consider the free group G generated by 2 elements, i.e. <a,b> which consists of all possible combinations of a, b, a-1, b-1. Each element of G can be written in a unique way. G can be split into 5 disjoint sets: G = {e} U S(a) U S(a-1) U S(b) U S(b-1) where S(x) represents the set of elements in G whose 'string' starts with x.
It is clear that aS(a-1) is equal to {e} U S(a-1) U S(b) U S(b-1) (the a 'cancels' the a-1 at the start of each element in S(a-1) leaving all elements starting with either a-1, b or b-1. Similarly bS(b-1) is equal to {e} U S(b-1) U S(a) U S(a-1).
So aS(a-1) U S(a) = G. Same for bS(b-1) U S(b).
Now consider a and b to be rotations of the sphere about orthogonal axes by some irrational number of degrees - say 1 radian. Every element of G is a combination of rotations about these 2 axes, and is thus a rotation. Each element of G (except e) will therefore have 2 fixed points on the sphere. Let the set of all these fixed points be Q, which is countable (as the elements of G are countable, being...equinumerous with N?...Wait! This is wrong!
I thought I understood this, but it doesn't work. The elements of G aren't countable, are they? Perhaps they are. Back to the drawing board. Rest of the proof involves the action of G on the set of points in the sphere, writing the set of points as a disjoint set of the orbits of G acting on S-Q, invoking the axiom of choice to select one point from each orbit which are put into a set M. This set is then composed with the disjoint bits of G to decompose the point-set of the sphere into 4 bits, two of which are translated across (say MS(a) and MS(a-1)) and then MS(a-1) is rotated by rotation a, and hey presto you've got your original sphere back. Do the same with the non-translated bits and you've got 2 spheres.
Few technical details with regard to the set Q - you have to rotate it by something not a member of G or something like that, and to extend the transformation to a solid ball rather than a hollow sphere you have to include the centre as a separate point and use the projections of each point to the centre, but it's largely the same.
It's all nonsense anyway - axiom of choice is clearly false, certainly when it comes to maths that could be applied to the real world. At least that's my opinion.
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"Now consider a and b to be rotations of the sphere about orthogonal axes by some irrational number of degrees - say 1 radian. Every element of G is a combination of rotations about these 2 axes, and is thus a rotation. Each element of G (except e) will therefore have 2 fixed points on the sphere. Let the set of all these fixed points be Q, which is countable (as the elements of G are countable, being...equinumerous with N?...Wait! This is wrong!"
The free group generated by two elements is countable. I think the reason for this is that the strings of a's and b's are isomorphic to Z X Z which is countable. The reason for this is that it does not matter whether one does the generating rotations on each axis in one go or interleave them so the strings are of the form ambn. This is different from a set of strings where the position of the a's and b's matters and is uncountable.
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I don't think that's right. A free group generated by a and b (with inverses) is surely equivalent to a binary string, which is uncountable as you can apply a diagonal argument.
Consider infinite strings of a's and b's, each of which has an exponent. As the free group is non-Abelian these elements are unique. List them, choose the first a and b from the first element etc, add one to their exponents... Hey presto you've got an element of the free group not in your list, so isn't it uncountable?
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Agreed : you are correct about the group being non-Abelian.
Looking at various references on the net e.g. wikipedia Banach-Tarski paradox article, the elements for the free group are the finite strings of a, b, a-1, and b-1 - this is a countable set. This is reasonable to me as I don't think it makes sense for the strings to be infinite in length (what rotation would be specified by an infinite string?).
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The strings, or at least some of them, have to be infinitely long, otherwise you have not just a countable group but a finite one. If you include infinite strings you end up with an uncountable set - I think - just like the set of reals expressed as binary strings.New comment
A set of all finite strings within an alphabet is countably infinite - an example is the set of finite strings comprising characters from alphabet { 0, 1,2,3,4,5,6,7,8,9 } maps to N (or Z to include the strings beginning with 0).
I checked my copy of Groups and Symmetry by Armstrong and that defines a free group's (non-identity) elements as comprising the finite strings of generators of infinite order. To use the symmetries, it does not make sense for them to be infinite string e.g. where would such an infinite combination of rotations map a particular point?
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OK I think I see. The rotations are all finite length strings, which are infinite in number - whichever finite length you choose there will always be strings one longer. Therefore countable as you can set up a bijection with N.
This is why I've given up on this infinity stuff and have moved on to knot theory and differential geometry.
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"OK I think I see. The rotations are all finite length strings, which are infinite in number - whichever finite length you choose there will always be strings one longer. Therefore countable as you can set up a bijection with N."
Yes, that's my understanding.
"This is why I've given up on this infinity stuff and have moved on to knot theory and differential geometry."
Taking the easy route then?
Good luck in your future studies.