Thanks! But is that correct? I notice that your argument seems to work for finite sequences as well (or even single rational numbers), but clearly we can order the rational numbers.
I think the issue here is whether you’re comparing functions (which allow self-reference in a way that can break ordering) or numbers (which don’t); for ordering arbitrary computable sequences, you need to have a way to avoid the sort of diagonalization that makes your choices affect the sequences you have to sort.
FYI, I was still confused about this so I posted on math .se. Someone responded that the above proof is incorrect, but they gave their own proof that there is no computable ordering over Z∞ which respects Pareto.
Thanks! But is that correct? I notice that your argument seems to work for finite sequences as well (or even single rational numbers), but clearly we can order the rational numbers.
I think the issue here is whether you’re comparing functions (which allow self-reference in a way that can break ordering) or numbers (which don’t); for ordering arbitrary computable sequences, you need to have a way to avoid the sort of diagonalization that makes your choices affect the sequences you have to sort.
FYI, I was still confused about this so I posted on math .se. Someone responded that the above proof is incorrect, but they gave their own proof that there is no computable ordering over Z∞ which respects Pareto.