True, that is the subtle bit -- but I think most people misunderstand (I'm not saying you are!), and don't realise you can always split an infinity into two -- this is just about splitting a sphere into some 'point clouds' such that you can cleverly stitch them back together into the same original space. In particular, the 'cutting' really makes no real-world sense (at least, as far as I can understand).
I agree that the "cutting" makes no real world sense. In a way, that's one of the points of the exercise.
And I agree that most people don't (initially) understand that an infinite set can be divided into two infinite sets that kinda "look the same", such as dividing Z (or N) into the evens and odds.
But BT is more than that. What follows isn't really for you, but is for anyone following the conversation.
Let's take a set A. It's a subset of the unit sphere, and it's a carefully chosen, special set, not just any random set. It's complicated to define, and requires the Axiom of Choice to do so, but that's what the BT theorem does ... it shows us how to define the set A.
One of the properties of A is that we can rotate it into a new position, r(A), where none of the points of r(A) are in the same position as any points of the original position, A. So the sets r(A) and A have a zero intersection. For the set A there are lots of possible choices of r ... we pick a specific one that has some special properties. Again, the BT theorem is all about showing us how to do this.
Now we take the union: B = A u r(A)
The bizarre thing is this. If we've chosen A and r (and therefore by implication, B) carefully enough, it ends up that there's another rotation, call it s, such that s(B)=A, the set we started with.
So whatever the volume of A, the volume of A u r(A) must be twice that, but that's B, and B can be rotated to give A back to us. So B must have the same volume as A. So 2 times V(A) must equal V(A), so A must have zero volume.
Well, we can kinda cope with that.
But if we've chosen A carefully enough, we find that a small, finite number of them, carefully chosen and rotated appropriately, together make up effectively the entire sphere (we miss out countably many points, but they have zero total volume, and we can fix that up later). So if finitely many copies of A make up a solid sphere, they can't have zero volume.
And that's the "paradox".
The conclusion is that we can't assign a concept of "volume" to the set A, and this is explained a little more in a blog post I've submitted here before:
There's a lot more going on than just the "I can split infinite sets into multiple pieces that kinda look the same as the original", although that is certainly part of it, and lots of people already find that hard to take.
To any who has got this far, I hope that's useful.
What do you mean "spread out"? Aren't there the same amount of even numbers as natural numbers?, because they both are countable sets. https://en.wikipedia.org/wiki/Countable_set
>>> ... if you take the natural numbers, and split them into odd and even, you get two copies of the natural numbers ...
>> ... the "two copies of the natural numbers" is sorta fine, except that they're more "spread out" ...
> What do you mean "spread out"? Aren't there the same amount of even numbers as natural numbers?
Yes, there are the same number, but when you look at just the even numbers, they are each distance 2 from their neighbours, whereas the natural numbers are all distance 1 from their neighbours. So people are less surprised, because the even numbers are "spread out", they are less dense in any given area. To map the even numbers back onto the natural numbers you have to "compress" them.
But this is not the case with the Banach-Tarski Theorem. There is a set, A, and another set B, which is just A rotated around, and they are disjoint. So they have a union, C=AuB. But when you rotate C, you can get an exact copy of A. There's no squashing or spreading needed.
So we have A and B, with B=r(A), and A intersect B is empty. Then we have C=AuB. No problem here.
The challenge comes that there is a rotation, s, such that s(C)=A.
So even though C is made up of two copies of A, it's actually identical to A. So start with C, divide it into A and B, then rotate B back to become a copy of A, and then rotate each of those to become copies of C. So you start with C, do some "cutting" and rotations, and you get two copies of C.
Finally, when you take a few of these and put them together, you get a full sphere, so you can't say they have zero volume.
Does that make sense? Does that answer your question?
The surprising thing about BT is that the "pieces" are "moved around" ... there's no expansion or contraction.
Yes, the natural number thing helps to understand that simply counting things doesn't help, but the "rigid motion" aspect of BT takes it further.