MrMind comments on Open thread, Oct. 12 - Oct. 18, 2015 - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (250)
Yes. It's not the Choice axiom which is problematic, but the infinity itself. So it doesn't mater if ZF or ZFC.
Why do I believe this? It's known for some time now, that you can't have an uniform probability distribution over the set of all naturals. That would be an express road to paradoxes.
The problem is, that even if you have a probability distribution where P(0)=0.5, P(1)=0.25, P(2)=0.125 and so on ... you can then invite a super-task of swapping two random naturals (using this distribution) at the time 0. Then the next swapping at 0.5. Then the next swapping at 0.75 ... and so on.
The question is, what is the probability that 0 will remain in its place? It can't be more than 0, after the completion of the super-task after just a second. On the other hand, for every other number, that probability of being on the leftmost position is also zero.
We apparently can construct an uniform distribution over the naturals. Which is bad.
I doubt that any proof in FAI will use infinitary methods.
I'm not sure why you think that. This may depend strongly on what you mean by an in infinitary method. Is induction infinitary? Is transfinite induction infinitary?