ucim wrote:Xias wrote:lim_{x→∞}(x-x) = 0.

Well, that's not infinity minus infinity. If you subtract an actual infinity from another actual infinity, you can get duck soup if you want to.

I was just illustrating that trying to solve the problem using arithmetic on infinity does not work.

Jose

We agree on all points. I just wanted to make it clear that what is happening in the balls and jugs problem is not the same as performing arithmetic on infinity as though it were a number. You can take two infinite things and take the difference, and if the things are defined correctly.

lim

_{x→∞}(x-x) = 0 is not infinity minus infinity, just like |

N \

N| is not infinity minus infinity, even though |

N| - |

N| is.

I wanted to make that distinction because many people in the thread have dismissed the "zero balls" solution as us just using arithmetic on infinity to get the answer we want, and that's not what we're doing at all.

kryptonaut wrote:How do you say that, in the procedure described above (appending '0' to lowest ball) that you can end up with balls with 'infinite' numbers on them, whereas in the case of removing the lowest and adding 10 new ones you can't get balls with infinite numbers on? In both cases there is no defined step where 'infinity' appears on a ball, so the only way to justify it is by saying it happens at step 'infinity' when midnight chimes - but I don't see how the two scenarios really differ at step 'infinity'.

Consider the following examples (with credit to mward), where balls and step times are all defined as in the original problem:

1. You have an empty jug. At each step, you add the ten lowest numbered balls. At midnight, how many balls are outside of the jug? How many are inside?

2. You have an empty jug. Before the first step, you place all of the balls in it. At each step, you remove the lowest numbered ball. At midnight, how many balls are outside of the jug? How many are inside?

3. You have two empty jugs. Before the first step, place all of the balls in jug A. At each step, remove the lowest numbered ball from jug A and place it in jug B. At midnight, how many balls are outside of the jugs? How many are in jug A, and how many are in jug B?

I think we have to agree on the answers to all of those problems to really get to an agreement with the original problem and any variant. Do you agree that the answers are:

1: Zero balls outside, infinite balls inside.

2. Infinite balls outside, zero balls inside.

3. Zero balls outside, zero balls inside A, infinite balls inside B.

Do you agree with all of those? If not, why?

If you do agree, then consider this: In (1), at every step, you have a finite number of balls in the jug, but at midnight you have an infinite number. Yet, there is no infinitieth step. It just happens after an infinite number of steps in which you've done a finite thing. In (2), at every step, you have an infinite number of balls left inside at every step, yet a finite number at midnight. Yet, there is no infinitieth step. And the same goes for (3).

Now, if we are at agreement there, then consider this:

4. You have a ball with the number "1" on it. At each step, you add a 0 to the end of it. At midnight, what is the number you have?

Hopefully we agree that you have a 1 followed by infinite zeros. Obviously, if there were a finite number of zeros at the end, then I could tell you which step I added one more. How did we get from having a natural number at each step to having a non-natural number at the end? The same way we ended up with infinite numbers in variants (1), (2), and (3) above. As it was in your blocks variant, the natural/non-natural state on the ball is defined by what is on the ball, not by the procedure. The procedure doesn't state, as a rule, that we have to end up with a natural number. We just happen to end up with a natural number when we append a 0 to a natural number. When we take the procedure to infinity, though, there is a discontinuity, so in order to find the natural/non-natural, or finite/non-finite, status of the number on the ball, we have to examine what the procedure tells us is on the ball at that point.

So let's examine the set in my "take the lowest ball and append a 0" variant. The set looks the same after every step, right? So the set of balls in the jug at the end of step 1 looks like:

{2, 3, 4, 5, 6, 7, 8, 9, 10}

which is the same as the set at the end of step 1 in the original. However, the element 10 in the set is not defined as "the 10th natural number," as it is in the original, but rather "The natural number 1 with a 0 appended onto it," which is 10. So what it really is at the end of step 1 is:

{2, 3, 4, 5, 6, 7, 8, 9, 1*10^1}

And after step 2, it looks like {3, 4, ... , 19, 20} but really it is

{3, 4, 5, 6, 7, 8, 9, 1*10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2*10}

which happens to look the same, but is defined differently. In the end, you end up with a set that looks like {k(10^∞): all k not divisible by 10}.

kryptonaut wrote:Why should we use the set-of-natural-numbers argument in the case where balls are removed, but not in the case where numbers are removed/ changed but balls left behind (with a different number on them)? It seems that in both cases all the numbers get removed.

Indeed, all of the natural numbers are removed. But the procedure created new non-natural, non-finite numbers that now exist on the balls. The original procedure is not defined in a way that creates those numbers.