05 Oct '10 12:31>1 edit
Originally posted by mtthwThis is why the numbering is crucial.
But here's another argument. Each time we add two balls, and remove one. The net effect is to add one.
If we start at time T, then at time T(1/2)^n, there are n + 2 balls in the bag.
Let n -> infinity
Have we just proved that the operation is impossible, then?
By Removing #1,#2,#3,... you ensure all balls are removed as n goes to infinity. But if you remove #1, #3,#5,...etc then your argument kicks in and there are infinite balls in the bag and you can count them: #2,#4,#6, etc.
I think the best way to think about it is to prove that since positive integers are bounded from below then any non-empty set of balls left must have a lowest numbered ball. So anything that leads to a contradiction in that must be false.