I don’t understand what you mean. The upgraded individuals are better off than the non-upgraded individuals, with everything else staying the same, so it is an application of Pareto.
Now, I can understand the intuition that (a) and (b) aren’t directly comparable due to identity of individuals. That’s what I mean with the caveat “(Unless we add an arbitrary ordering relation on the utilities or some other kind of structure.)”
Okay the pareto thing applies but the formal contradiction has a problem in the (b) prong. Consider 1N which is 1,2,3,4,5,6,7… if you took each other out from that yu would get 1,3,5,7… There is no 2 in there so there is no copy of 1N remaining in there. Sure if you have 0N which is 0,0,0,0,0… you can have it as a multiset but multisets track amounts. It is not sufficient that the members are of the same object the amount needs to match too. And in that dropping the amounts (atleast ought to) change. So 0N2 is not the same as 0N. So you get 2N∪0N≺N∪0N2 which is not an exact mirror of the (a) prong.
The number of elements in 0N won’t change when removing every other element from it. The cardinality of 0N is countable. And when you remove every other element, it is still countable, and indistinguishable from 0N. If you’re unconvinced, ask yourself how many elements 0N with every other element removed contains. The set is certainly not larger than N, so it’s at most countable. But it’s certainly not finite either. Thus you’re dealing with a set of countably many 0s. As there is only one such multiset, 0N equals 0N with every other element removed.
That there is only one such multiset follows from the definition of a multiset, a set of pairs (a,c), where a is an element and c is its cardinality. It would also be true if we define multisets using sets containing all the pairs (a,1),(a,2),… -- provided we ignore the identity of each pair. I believe this is where our disagreement lies. I ignore identities, working only with sets. I think you want to keep the identities intact. If we keep the identities, the set {(0,1),(0,2),(0,3),…} is not equal to {(0,1),(0,3),(0,5),(0,7),…}, and my argument (as it stands) fails.
To my mind the reduced set has ω2 elements which is less than ω. But yeah its part of a bigger pattern where I don’t think cardinality is a very exhaustive concept when it comes to infinite set sizes. But I don’t have that much knowledge to have a good alternative working conception around “ordinalities”.
Pareto explicitly says that you have to keep identities intact, because the definition stipulates that w1 and w2 “contain the same people.” If you don’t preserve identities, you can’t verify that that condition is met, in which case Pareto isn’t applicable.
I don’t understand what you mean. The upgraded individuals are better off than the non-upgraded individuals, with everything else staying the same, so it is an application of Pareto.
Now, I can understand the intuition that (a) and (b) aren’t directly comparable due to identity of individuals. That’s what I mean with the caveat “(Unless we add an arbitrary ordering relation on the utilities or some other kind of structure.)”
Okay the pareto thing applies but the formal contradiction has a problem in the (b) prong. Consider 1N which is 1,2,3,4,5,6,7… if you took each other out from that yu would get 1,3,5,7… There is no 2 in there so there is no copy of 1N remaining in there. Sure if you have 0N which is 0,0,0,0,0… you can have it as a multiset but multisets track amounts. It is not sufficient that the members are of the same object the amount needs to match too. And in that dropping the amounts (atleast ought to) change. So 0N2 is not the same as 0N. So you get 2N∪0N≺N∪0N2 which is not an exact mirror of the (a) prong.
The number of elements in 0N won’t change when removing every other element from it. The cardinality of 0N is countable. And when you remove every other element, it is still countable, and indistinguishable from 0N. If you’re unconvinced, ask yourself how many elements 0N with every other element removed contains. The set is certainly not larger than N, so it’s at most countable. But it’s certainly not finite either. Thus you’re dealing with a set of countably many 0s. As there is only one such multiset, 0N equals 0N with every other element removed.
That there is only one such multiset follows from the definition of a multiset, a set of pairs (a,c), where a is an element and c is its cardinality. It would also be true if we define multisets using sets containing all the pairs (a,1),(a,2),… -- provided we ignore the identity of each pair. I believe this is where our disagreement lies. I ignore identities, working only with sets. I think you want to keep the identities intact. If we keep the identities, the set {(0,1),(0,2),(0,3),…} is not equal to {(0,1),(0,3),(0,5),(0,7),…}, and my argument (as it stands) fails.
To my mind the reduced set has ω2 elements which is less than ω. But yeah its part of a bigger pattern where I don’t think cardinality is a very exhaustive concept when it comes to infinite set sizes. But I don’t have that much knowledge to have a good alternative working conception around “ordinalities”.
Pareto explicitly says that you have to keep identities intact, because the definition stipulates that w1 and w2 “contain the same people.” If you don’t preserve identities, you can’t verify that that condition is met, in which case Pareto isn’t applicable.