So I guess the sum of the cardamum is that proving division by n in set theory without the axiom of choice gets you in a bit of a pickle because the structure isn't necessarily preserved, meaning that the socks analogy of unordered sets won't necessarily divide by n since to do so requires a bijection meaning 1-to-1 and onto which is difficult to prove if multiplication of sets isn't defined as a repeated sum. With finite sets we are used to multiplication being a repeated sum but with infinite sets that doesn't necessarily hold.
I mean, the comment is badly wrong and a substantial parts of it are basically nonsense. Ideally one should write a reply explaining why, but this takes time and people aren't always going to be willing to do that. Like right now, I need to go to sleep I'm afraid!
All their comments are auto-dead going back 11 months except for ones people have vouched for. Vouched comments can still have a zero or negative score and so can remain grayed out.
I think this paper is refuting Conway (and others') proof of the claim that a set can be divided into 3+ parts without relying on the Axiom of Choice. It does so by observing that the proof only applies to sets that have an ordering, which is to say that there is an algorithm for picking an element from the set, which am extra constraint beyond having only a test for determining whether a known object is an element is in the set. It then goes on to prove that it is not possible to evenly divide a set that lacks such a selection algorithm.
> I think this paper is refuting Conway (and others') proof of the claim that a set can be divided into 3+ parts without relying on the Axiom of Choice.
This paper is not refuting Conway's, and Conway's paper does not prove the claim that a set can be divided in 3+ parts without relying on AC.
What the Conway's paper proves is that, without assuming AC, if there is a bijection between A×n and B×n for some finite n, then there is a bijection between A and B. Axn can be equivalently written as the union of a×n, as a ranges over the elements of A, similarly B×n can be written as the union over b×n. This paper shows that if instead you take the union over a×N_a, where the sets N_a are pairwise disjoint and have n elements, and similarly instead of considering B×n you consider the union of b×N_b, where the sets N_b are pairwise disjoint and have n elements, then the existence of a bijection between those two unions is not sufficient to construct a bijection between A and B if we're not assuming AC. The main point here is that without choice we cannot order all of the N_a's and N_b's at the same time, while in Conway's paper, since N_a=N_b=n={0,1,...,n-1}, they are already uniformly ordered and no such issue arises.
