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.)”

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.