30 Nov
2008
30 Nov
'08
5:13 p.m.
Luke Palmer wrote:
The other nice one problem is allowing the argument itself to be infinite (you have to require all of the lists to be nonempty).
I think the requirement has to be a lot stronger for that to work. If every sublist has two elements, the answer is 2^infinity lists which is uncountable. In order for the answer to be countable, you have to require that only a finite number of sublists contain more than one element, at which point you can use your omega monad again. Martijn.