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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.
Oh, I agree with your argument - and it's very neat. I'm just not yet convinced it invalidates mine. The alternative is that the operation is theoretically impossible. As PBE6 mentioned, we're relying on taking the limit of an infinite sequence to get round Zeno's paradox. So why can't the same approach be applied to the number of balls?
Originally posted by PBE6no balls are left in the bag"If you've done six impossible things this morning, why not round if off with breakfast at Millyways?" - Douglas Adams
Suppose you and your friend come into possession of a bag of infinite capacity, along with an infinite supply of numbered balls and a stopwatch with an infinite number of gradations. After watching the universe explode ...[text shortened]...
[b]Question: How many balls will be left in the bag when the stopwatch reads 0 seconds?[/b]
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Originally posted by mtthwSorry, I don't think I understand... Why isn't the proof that there can be no non-empty set without a smallest numbered ball enough?
Oh, I agree with your argument - and it's very neat. I'm just not yet convinced it invalidates mine. The alternative is that the operation is theoretically impossible. As PBE6 mentioned, we're relying on taking the limit of an infinite sequence to get round Zeno's paradox. So why can't the same approach be applied to the number of balls?
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Originally posted by PalynkaBecause I haven't seen any argument that suggests how a process that increases the number of balls 1 every step can result in a final number of zero.
Sorry, I don't think I understand... Why isn't the proof that there can be no non-empty set without a smallest numbered ball enough?
You've proven that it can't be any number other than zero. But you haven't shown how zero can be reached. I'm proposing that what we've actually done is shown the problem is ill-defined.
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Originally posted by mtthwI'm not sure how this problem can be ill-defined. Either the balls diverge or they do not, right?
Because I haven't seen any argument that suggests how a process that increases the number of balls 1 every step can result in a final number of zero.
You've proven that it can't be any number other than zero. But you haven't shown how zero can be reached. I'm proposing that what we've actually done is shown the problem is ill-defined.
To show zero is "reached", you can show that there is a bijection between the two sets, so the sets have the same cardinality. Since you're not skipping any balls, then there must be no balls left at the end.
It is indeed very counter intuitive that say, even numbers have the same cardinality as the natural numbers. But one can show by bijection that it is so. Despite that every step there seems to be double the amount of natural numbers, the nature of infinity is such that such "double" doesn't matter because we are going to infinity and double is still infinity.
Originally posted by PalynkaAnd so you've just proved that the sequence 1, 2, 3, 4, 5, .... converges to zero. That's why I think it's ill-defined. Unless that issue can be resolved.
To show zero is "reached", you can show that there is a bijection between the two sets, so the sets have the same cardinality. Since you're not skipping any balls, then there must be no balls left at the end.
Originally posted by PalynkaExactly!
What? I definitely did not prove that.
My argument shows that the number of balls increases by one with every step. If the final value is defined, this sequence ought to converge to it.
Your argument has ignored my argument, and relies on proving that if the final value is defined, it must be zero.
Unless someone can point out where my argument is flawed, I'm tempted to believe that the key to the problem is that "If".
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Originally posted by mtthwNo, the bijection argument shows that it is defined (I think).
Exactly!
My argument shows that the number of balls increases by one with every step. [b]If the final value is defined, this sequence ought to converge to it.
Your argument has ignored my argument, and relies on proving that if the final value is defined, it must be zero.
Unless someone can point out where my argument is flawed, I'm tempted to believe that the key to the problem is that "If".[/b]
But, ok, intermediate step. Do you agree that the cardinality of natural numbers and odd numbers is the same?
Originally posted by PalynkaYes, I do. I don't have any problem with that part of the argument.
No, the bijection argument shows that it is defined.
But, ok, intermediate step. Do you agree that the cardinality of natural numbers and odd numbers is the same?
For my intermediate question, I'm going to pose a related problem. Same time-steps. We start with one ball, labelled 1. At every step, the ball remains in place, but is relabelled, increasing the number by one. What is the number on the ball at time t = 0?
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Originally posted by mtthwInfinity is not a number, so there is no step number = infinity. So that problem has, indeed, no well defined solution as the limit is divergent.
For my intermediate question, I'm going to pose a related problem. Same time-steps. We start with one ball, labelled 1. At every step, the ball remains in place, but is relabelled, increasing the number by one. What is the number on the ball at time t = 0?
But the argument of bijection shows that there is a one-to-one correspondence so to each element of one side of the correspondence corresponds exactly one element on the other side. Hence all balls are removed. No divergence.
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But (I think) it's basically the same problem. The number of balls is divergent.
What we're essentially doing is solving the following:
lim(n -> infinity) {n+1, n + 2, n + 3, ..., 2n}
And I don't see how that converges to the empty set. It doesn't converge at all. My interpretation is that it would need to converge to make the original problem well-defined.
But fair enough, I could be wrong. 🙂
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Originally posted by mtthwMy head hurts. 🙂
But (I think) it's basically the same problem. The number of balls is divergent.
What we're essentially doing is solving the following:
lim(n -> infinity) {n+1, n + 2, n + 3, ..., 2n}
And I don't see how that converges to the empty set. It doesn't converge at all. My interpretation is that it would need to converge to make the original problem well-defined.
But fair enough, I could be wrong. 🙂
Maybe someone else can shed some light on this...
Edit - Maybe they are equivalent but the key thing is that you are pushing the lowest number further out with your series. If infinity was a number it's easy to see it goes to the empty set...since it's not then it's less intuitive. As you can see, what was clear to me before is less clear now. Thanks a lot, buddy!
Anyway, maybe PBE6 can explain it better than me...and make it clearer for me as well.