## Infinite Balls and Jugs [solution]

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Demki
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### Re: Infinite Balls and Jugs [solution]

The ordinal omega is not a natural number, where did a ball with label omega come from in the S T D game? Did it pop out of thin air into the jug?

You can't say it was in S because you agreed S had only natural numbered balls.

Cauchy
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### Re: Infinite Balls and Jugs [solution]

Xias wrote:So, I think that we can resolve the paradox in this way: We have a sequence of sets of balls that is different from the sequence of sets of balls in the first game; and a sequence of sets of labels which is the same as the sequence of sets of labels in the original game. At midnight, we have an infinite set of balls, which each corresponds with a value in A, that were never removed. We also have a sequence of sets of labels which converges to the null set, so we could say that there are no labels. Or, we could examine instead the sequence of values that each label took, and find that each of those sequences diverge, and upon close inspection to the rules of the game, diverge because they each take a value equal to n with an infinite number of zeroes. I'm partial to the latter, which i think is okay to do since the labels are not constrained to natural numbers in the same way that balls with unchanging labels are. However, I'm not opposed to accepting that there are no labels either, if there is such a constraint. What matters is, the sets of balls aren't actually the same at each step n, and the paradox is resolved.

I can't see the balls as anything but interchangeable. That is, if I take the balls with labels 2 and 3, and swap their labels, I haven't changed anything in the jug.

This also fails the lights test I described, which I firmly believe is a valid test. If the lights go out between each step so that all you can see is the final result of each step, then you can't get a different answer or say "I don't know anymore", because the limit only cares about the discrete sequence of sets that are the results of each step. If you want to say "I don't know", then that has to be your answer no matter what the process is.

Regarding your "essence of balls" argument, I can set up the same family of bijections in the "remove balls" scenario. I can say that balls 1, 10, 100, etc. are mystically linked, because just as one enters, the previous one leaves. As such, one of that family of balls is always in the jug at each step, so there should be one member of that family of balls present in the limit. Except, that's not true, and there's no reason it should be true in the "append-0" scenario.

I don't see why balls being numbered out of existence is that much harder to swallow than f(n) = n catching up to f(n) = 10n at infinity.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.

kryptonaut
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### Re: Infinite Balls and Jugs [solution]

Demki wrote:The ordinal omega is not a natural number, where did a ball with label omega come from in the S T D game? Did it pop out of thin air into the jug?

You can't say it was in S because you agreed S had only natural numbered balls.

Omega is a label for a number greater than any natural number, which is what the supertask inevitably leads to in this case. The numbers in the set of balls are always higher than the step number, which takes on every number in N

Just like the irrational e is defined as an infinite sum of rationals - no irrational numbers go into the recipe but one pops out at the end of the infinite sum.

Omega may be an 'impossible number' in reality but that doesn't mean it isn't a useful concept that can lead to sensible and consistent results, like complex numbers or infinitely precise irrationals do. This whole puzzle is an impossible scenario, trying to imagine what happens if an infinite task actually finishes.

ucim
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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:Omega is a label for a number greater than any natural number, which is what the supertask inevitably leads to in this case.
No, it does not lead to that. The only numbers involved are natural numbers. But there are a lot of them.
kryptonaut wrote:The numbers in the set of balls are always higher than the step number, which takes on every number in N
... and for every number in N, there is always a bigger number, also finite, and in N. Always. And they all make it from S to T to D, even though there's no caboose.

kryptonaut wrote:Omega may be an 'impossible number' in reality but that doesn't mean it isn't a useful concept that can lead to sensible and consistent results
It's not an impossible number. It's just not in N, and therefore is never in S or T or D. Neither are emoji or ham sandwiches.

kryptonaut wrote:This whole puzzle is an impossible scenario, trying to imagine what happens if an infinite task actually finishes.
Spoiler:
"It couldn't be done", so they said, but he went right to it.
He tackled that task that couldn't be done...
...and he couldn't do it.
Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

Xias
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### Re: Infinite Balls and Jugs [solution]

Cauchy wrote:I can't see the balls as anything but interchangeable. That is, if I take the balls with labels 2 and 3, and swap their labels, I haven't changed anything in the jug.

I disagree. If you look at a single ball in the first game, you see it enter at a step, and leave at a step, and those steps are defined. If you look at a single ball in the second game, you see it enter at a step, but there is no step for which it leaves. At the end, I think we agree that "there can be no ball with a naturally numbered label on it." But I see no reason to believe that if looking at the balls independently from the labels, that sequence also converges to the null set. Applying the same reasoning to the original game, the sequence of balls does converge to the null set, regardless of labels.

I posted a variant of Hilbert's Hotel some posts back: Let's say that on Friday night you run into the original HH problem, where you have a guest in each room n and a countably infinite number of new guests arrive. You can set up a supertask to fill the rooms, where at each step n, you take the guest in room (2n-1) and move them to room 2(2n-1), the guest in room 2(2n-1) and move them to 4(2n-1), the guest in room 2k(2n-1) to room 2k+1(2n-1), etc. This makes room (2n-1) available for incoming guest n. As you iterate through this supertask, every original guest moves exactly one time. If I ask you where any guest is based on where they started, you can give me an answer. Incoming guest n is in room 2n-1. Original guest m is in room 2m.

You hear that another countably infinite group is coming the next night, but you have Saturdays off. So you phone up the Saturday manager and tell him the strategy you used. However, he misunderstands. So Saturday comes around and the new guests arrive, and he does it like so: At each step n, he takes the guest in room 2kn and moves them to room 2k+1n to make room n available for incoming guest n.

By your lights test, at every step n the rooms are all full. If you couldn't tell the difference between any of the guests, and you didn't know the process by which a room was made available, the hotel at each step would appear indistinguishable from each step on Friday. However, at the end of the supertask in the second case, the original guests cannot occupy any natural-numbered room.

Cauchy
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### Re: Infinite Balls and Jugs [solution]

Xias wrote:
Cauchy wrote:I can't see the balls as anything but interchangeable. That is, if I take the balls with labels 2 and 3, and swap their labels, I haven't changed anything in the jug.

I disagree. If you look at a single ball in the first game, you see it enter at a step, and leave at a step, and those steps are defined. If you look at a single ball in the second game, you see it enter at a step, but there is no step for which it leaves. At the end, I think we agree that "there can be no ball with a naturally numbered label on it." But I see no reason to believe that if looking at the balls independently from the labels, that sequence also converges to the null set. Applying the same reasoning to the original game, the sequence of balls does converge to the null set, regardless of labels.

So you want to apply a second label to each ball that doesn't change when we change the first label. All I think you're showing is the weakness of supertasks as thought experiments.

What if we do it the other way? We modify the original supertask slightly: now, after removing ball n, we take ball 10n and erase the trailing zeroes in its label. In this variant, each label that doesn't end with a 0 will stay in the jug forever, since whenever we go to remove a ball, we change another ball to have its same label. So now at the end of the supertask, there are labels left in the jug, but no balls?

By your lights test, at every step n the rooms are all full. If you couldn't tell the difference between any of the guests, and you didn't know the process by which a room was made available, the hotel at each step would appear indistinguishable from each step on Friday. However, at the end of the supertask in the second case, the original guests cannot occupy any natural-numbered room.

If you can't tell the difference between guests, then "original guests" has no meaning. If you can pinpoint the original guests, then it's clear that the tasks are different after step 2.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.

Xias
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### Re: Infinite Balls and Jugs [solution]

Cauchy wrote:So you want to apply a second label to each ball that doesn't change when we change the first label. All I think you're showing is the weakness of supertasks as thought experiments.

We don't have to apply a second label. The ball has the property of having originally been indexed. Nothing about the game actually has to change. I'm talking about a more accurate mathematical representation of what is happening; one that captures all of the qualities of the game. The only reason why the sequences of sets even applies to the original game is because the sets behave in the exact same way as the balls. In the append-0 game, the exact same sets do not represent the balls anymore, because the sets don't offer us information about the memory of different elements.

Cauchy wrote:What if we do it the other way? We modify the original supertask slightly: now, after removing ball n, we take ball 10n and erase the trailing zeroes in its label. In this variant, each label that doesn't end with a 0 will stay in the jug forever, since whenever we go to remove a ball, we change another ball to have its same label. So now at the end of the supertask, there are labels left in the jug, but no balls?

This new game isn't clearly defined to me. Which ball are we removing? At step 1, we change the label on ball 10 to say 1, right? So at step 10, do we remove the 10th ball to enter, which now has a label 1? or do we remove no ball since no ball has a 10 on it? Or do we remove the ball labeled 11 for some reason?

There are different ways that you can define which ball you remove, that results in different answers for what's in the jug at midnight; I can't actually think of one that ends with the result you're describing. Maybe if you go into more detail?

Cauchy wrote:If you can't tell the difference between guests, then "original guests" has no meaning. If you can pinpoint the original guests, then it's clear that the tasks are different after step 2.

I see no reason to say that we can't tell the difference between guests, nor to say that we can't tell the difference between balls. As the supertask is defined, at any step n I can tell you what label is on each ball, and also when each ball was placed in the jug based on that label, and also which labels each ball has had before, and at which points it had them. The ball with a 10 on it at the end of step 1 in the append-0 game has a different history than the ball with a 10 on it at the end of step 1 in the remove-the-lowest game, which is different as well from the ball with a 10 on it at the end of step 1 in the remove-the-highest game. When we construct mathematical objects to model these different situations, why should we ignore those differences?

And then, since we can tell the difference between different guests, and can therefore pinpoint the original guests, then you're right: "it's clear that the tasks are different after step 2." They are different, despite resulting in the same "labels" at every n! And the append-0 game is different despite the same labels at every n.

Also, to amend my response to this for clarity:

Cauchy wrote:I can't see the balls as anything but interchangeable. That is, if I take the balls with labels 2 and 3, and swap their labels, I haven't changed anything in the jug.

I do agree that the contents of the jug are the same in the sense that it has the same balls in it and it has the same labels in it. But if I then tell you to "remove the lowest numbered ball," the ball you will remove will be different than if you hadn't swapped them. If the first ball placed in the jug was the one labeled 2 before the swap, then if I tell you to "remove the ball that has been in the jug the longest" then the ball you remove will be the same, but the label on it will be different than if you hadn't swapped them.

I think that something like {(2,2), (3,3)} becoming {(2,3), (3,2)} is a more accurate mathematical representation of what the swap did, since it captures those qualities better than merely {2, 3} becoming the same set.

kryptonaut
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### Re: Infinite Balls and Jugs [solution]

Xias wrote:You hear that another countably infinite group is coming the next night, but you have Saturdays off. So you phone up the Saturday manager and tell him the strategy you used. However, he misunderstands. So Saturday comes around and the new guests arrive, and he does it like so: At each step n, he takes the guest in room 2kn and moves them to room 2k+1n to make room n available for incoming guest n.

By your lights test, at every step n the rooms are all full. If you couldn't tell the difference between any of the guests, and you didn't know the process by which a room was made available, the hotel at each step would appear indistinguishable from each step on Friday. However, at the end of the supertask in the second case, the original guests cannot occupy any natural-numbered room.

This variant of the Hibert's Hotel paradox is the key (haha).

All the original guests end up in rooms whose numbers are strictly higher than those of all the new guests. The set of original guests and the set of new guests each has ordinal ω, but since we have combined them in such a way that one set is completely 'after' the other we end up with a set with ordinal ω·2. If the original guests had blue key fobs and the new guests were given red ones then despite there being infinite red ones, the blue ones would all be in higher room numbers.

In the 1000-ball puzzle we similarly keep pushing the lower bound of the set of balls in the jug up by 1 at every step, for every natural number, so that by the end of the supertask the numbers on the balls are strictly after every natural number.

Let's imagine painting a dot on ball n at step n instead of removing it from the jug. Now at the end of the supertask, call it phase 1, we have a set of dotted balls which correspond to every natural number, and a set of 1000 undotted balls which sort strictly after the dotted ones. The ordinal of this combined set is ω+1000. In a separate supertask, phase 2, we can remove the dotted balls, which have a 1:1 mapping to N, leaving us with the 1000 balls in an ordered set with ordinal 1000. The numbers on these balls are different, sequential, and higher than any number you can imagine.

A similar argument applies to the 'remove the lowest' puzzle - if we split it into two phases putting dots in phase 1 on the balls to be removed in phase 2, then we can see that after phase 1 we have a countably infinite set of dotted balls, followed by a countably infinite set of undotted balls. The combined set has ordinal ω·2. In phase 2 we remove the dotted ones, leaving us with a countably infinite ordered set of undotted balls with ordinal ω. The numbers on these balls are also higher than any number you can imagine.

This is the same process by which we allow an infinite set to be removed from N and still leave an infinite set behind - for example we can partition N into odd and even subsets and then define a sorting order so that we rank any odd number as less than any even number, using normal sorting to compare numbers with the same 'oddness'. So in our new ordering we have {1,3,5,7,... 2,4,6,8,...} - this set has ordinal ω·2. Having defined this ordering we can now map each number in the first part of the set to a number in N and remove them in a supertask, leaving us with the set {2,4,6,8,...} having ordinal ω

Xias
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### Re: Infinite Balls and Jugs [solution]

Kryptonaut, why is the end result of the supertask made different by adding balls in sequence, than it would be if we added all of the balls at once before we started?

If we start with a jug with natural numbered balls for all n, and then remove the lowest one at a time, we end up with an empty jug. If we instead put dots on the balls, we end up with every ball having a dot on it. Yet somehow, in your mind, the act of adding the balls sequentially while doing this leads to additional balls being put in the jug.

You haven't given any good reason why this is so. You're trying to get around the fact that every ball that is added is eventually removed by defining into existence balls that are added and not removed. But to be added and not removed, they have to be added. There is no mathematical basis for such balls to be added in the first place, much less exist at all. You're only saying that the balls exist and are added but not removed because you say so.

Tell me, where do you see the problem with the jug being empty? Can you articulate it without inventing another variant? Because in my mind:

Spoiler:
A. The jug being empty is consistent with the fact that, for any ball added, there is a defined step for which it is removed;

B. The jug being empty is consistent with the limit of the sequence of sets of natural numbers represented by the balls;

C. The jug being empty is consistent with the isolation of both sub-supertasks (that is, if we do all of the adding first, and then do all of the removing, we still end up with an empty jug);

Meanwhile, the only contradictions you've offered are

Spoiler:
1. The lowest/highest numbered ball is always increasing for every finite step;

2. The difference between the lowest/highest numbered ball is always increasing for every finite step;

3. The number of balls in the jug is always increasing for every finite step;

But in my mind, these are not actually contradictions at all, since something that is true for every finite step is not necessarily true for the limit. See my discussion on the sets, sequences, and series related to the sum to e.

In order for them to be contradictions, the following implications must be true:

Spoiler:
If (1), then the jug at midnight (JAM) has a lowest/highest numbered ball which is higher than any natural number.

If (2), then the difference between the lowest and highest numbered ball in the JAM is infinite.

If (3), then the number of balls in the JAM is infinite.

All three of these statements are false. I proved it with set theory, by showing that (1), (2), and (3) can all be true of a sequence of sets that converges to the null set. If (A and B) is true, then (A implies ~B) is not true. You can't just offer the implications as granted; you have to prove them. Can you?

kryptonaut
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### Re: Infinite Balls and Jugs [solution]

Xias wrote:If we start with a jug with natural numbered balls for all n, and then remove the lowest one at a time, we end up with an empty jug. If we instead put dots on the balls, we end up with every ball having a dot on it. Yet somehow, in your mind, the act of adding the balls sequentially while doing this leads to additional balls being put in the jug.

I'm assuming we're talking about the 'remove lowest number' problem. How do you see this ending up with every ball having a dot on it? None of the balls higher than n at any stage has a dot.

Can you not see that it is very similar to the second example of the Hilbert Hotel where the existing guests are shuffled up infinite times to make room for the new ones? You end up with an infinite set having red keyfobs in the first infinite number of rooms, and a second infinite set with blue keyfobs in the second infinite number of rooms. Each set can be independently mapped to N. It is the fact that there is a set which is defined explicitly to lie at the far end of the natural numbers which causes this configuration.

In the puzzle we are growing two countably infinite sets simultaneously, with the condition that one is explicitly defined to lie after the other. The first set is the set of numbers on the removed balls, call it X, the second set is the set of numbers on the balls in the jug, call it J. Both sets grow to become countably infinite sets, each with ordinal ω. The union of these two sets is not N, it is {X1, X2, X3, ..., J1, J2, J3....} with ordinal ω·2

The error in your set-theory proof is that you don't make this distinction, assuming the union of X and J is N. It has the same cardinality, but not the same ordinal.

So no, not every ball that is added is removed, only those in X are removed. This is a countably infinite set, but with a lesser ordinal than the entire set of balls we are dealing with in the problem. When X is removed from this union we are left with J, another countably infinite set.

With this explanation, everything that is true for each finite step is also true at the end, we just have to accept that we can't assign a number to J1 - it is something that is bigger than any natural number you choose to compare it with. Infinity, if you will, or ω, or a positive and unboundedly large number, or some such phrase. Note that J is ordered so J1 is less than all other elements of J

Xias wrote:Meanwhile, the only contradictions you've offered are

Spoiler:
Show
1. The lowest/highest numbered ball is always increasing for every finite step;

2. The difference between the lowest/highest numbered ball is always increasing for every finite step;

3. The number of balls in the jug is always increasing for every finite step;

But in my mind, these are not actually contradictions at all, since something that is true for every finite step is not necessarily true for the limit. See my discussion on the sets, sequences, and series related to the sum to e.

In order for them to be contradictions, the following implications must be true:

Spoiler:
Show
If (1), then the jug at midnight (JAM) has a lowest/highest numbered ball which is higher than any natural number.

If (2), then the difference between the lowest and highest numbered ball in the JAM is infinite.

If (3), then the number of balls in the JAM is infinite.

All three of these statements are false. I proved it with set theory, by showing that (1), (2), and (3) can all be true of a sequence of sets that converges to the null set. If (A and B) is true, then (A implies ~B) is not true. You can't just offer the implications as granted; you have to prove them. Can you?

From the above proof, all three statements are not only true, but also follow on from the finite step invariants, and as a bonus they agree with intuition. Which is nice.

Xias
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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:
Xias wrote:If we start with a jug with natural numbered balls for all n, and then remove the lowest one at a time, we end up with an empty jug. If we instead put dots on the balls, we end up with every ball having a dot on it. Yet somehow, in your mind, the act of adding the balls sequentially while doing this leads to additional balls being put in the jug.

I'm assuming we're talking about the 'remove lowest number' problem. How do you see this ending up with every ball having a dot on it? None of the balls higher than n at any stage has a dot.

- The balls are indexed by the natural numbers; therefore, there is a bijection between the balls and the natural numbers.

- For every n in the natural numbers, there is a defined step before midnight.

- Therefore, if at each step we put a dot on a ball, chosen in such a way that matches the bijection, then at midnight every single ball has a dot on it.

If you disagree with this, then feel free to read on, but know that any disagreements that follow can't be resolved without resolving this one first.

kryptonaut wrote:Can you not see that it is very similar to the second example of the Hilbert Hotel where the existing guests are shuffled up infinite times to make room for the new ones? You end up with an infinite set having red keyfobs in the first infinite number of rooms, and a second infinite set with blue keyfobs in the second infinite number of rooms. Each set can be independently mapped to N. It is the fact that there is a set which is defined explicitly to lie at the far end of the natural numbers which causes this configuration.

In the puzzle we are growing two countably infinite sets simultaneously, with the condition that one is explicitly defined to lie after the other. The first set is the set of numbers on the removed balls, call it X, the second set is the set of numbers on the balls in the jug, call it J. Both sets grow to become countably infinite sets, each with ordinal ω. The union of these two sets is not N, it is {X1, X2, X3, ..., J1, J2, J3....} with ordinal ω·2

The error in your set-theory proof is that you don't make this distinction, assuming the union of X and J is N. It has the same cardinality, but not the same ordinal.

I don't even agree that the original Hotel guests end up being associated with a nonempty set of room numbers, but even if it did, that example was defined in such a way as to create it. The original balls and jugs game is not.

With the balls and jugs, J is the null set. By definition, J is the difference between the balls added and the balls removed. The limit of that set is the null set. Even if we don't define it as B \ A, the sequence of sets {n+1, n+2, ... , 10n} (which corresponds to the set of balls left in the jug at step n) for all n does not converge to a set of infinitely large numbers. The limit is the null set. By the limit of sequences of sets: Take any element x in J. Then x is in {n+1, n+2, ... , 10n} for all n greater than some arbitrarily value. Then there exists n* such that x is in {n*+1, n*+2, ... , 10n*}. Then x is not in {10n*+1, 10n*+2, ... 100n*}. Since 10n*>n*, there is a contradiction, and there exist no x in J. So J is the null set. Also, the union of the set of balls removed with J will be the set of balls that were added to the jug. The set of balls added to the jug is the union of the sets {10n-9, 10n-8, ... , 10n} for all n. Which is N. So N U J = N. So J is the null set.

In fact, J being the null set also means that J1 cannot be defined, and that it is ordered in such a way that Jn<Jn+1 for all n where Jn exists, since Jn does not exist for any n. So even all of those conditions you've created hold for J when J is the null set.

By every way that we can measure J, it is the null set. Except for your method, which is to say that it has infinitely large numbers because...? The Hotel game really doesn't help you, since it's a totally different supertask. Even if I agreed that the original guests end up in infinitely large numbered rooms (I don't), that doesn't have any influence over the balls and jugs. If you want to say that the same numbers in the hotel game can be used in the balls and jug game, you have to justify that.

kryptonaut wrote:From the above proof, all three statements are not only true, but also follow on from the finite step invariants, and as a bonus they agree with intuition. Which is nice.

Except that J is the null set, so the statements were not proven true.

kryptonaut
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### Re: Infinite Balls and Jugs [solution]

Xias wrote:I don't even agree that the original Hotel guests end up being associated with a nonempty set of room numbers, but even if it did, that example was defined in such a way as to create it. The original balls and jugs game is not.

I think it is.

There is a fundamental difference between the ordered sets {a,b,c,1,2,3,4,5,6,7,...} and {1,2,3,4,5,6,7,... a,b,c}. The first can be mapped 1:1 to the set of natural numbers by mapping a->1, b->2, c->3, 1->4, etc. The second set cannot, since the final members a,b,c are beyond the limit ... so the set has an ordinal ω+3
See https://en.wikipedia.org/wiki/Ordinal_arithmetic for more information.

The puzzle is designed exactly to create a situation like this, except the sequence a,b,c is also infinite and we end up dealing with {1,2,3,... a,b,c,...} with ordinal ω+ω = ω·2.

Subset {a,b,c,...} represents the numbers left in the jug. The subset {1,2,3,...} is the balls that are removed.

Crucially, the set {1,2,3,... a,b,c,...} cannot be mapped to N

ucim
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### Re: Infinite Balls and Jugs [solution]

If you count to infinity, what number do you reach? Yes, it's a silly question, but it seems to be isomorphic to the supertask questions being bandied about.

Hilbert hotel: There is one guest in room one. At every supertask step she is moved to the next room over, in order to make room for a new guest. At the end of the supertask, what room is she in? Is she still even in the hotel? Remember that the Hilbert Hotel does not have a room numbered omega, or infinity, or aleph-anything.

Balls and jugs: At the end of the supertask, how can the jug be empty? (Pretty much any variant - I'll use the T jug in my S-T-D version)

Algebraic functions and limits: Consider the graph of y=1/x. If you "ride the curve" (an algebraic supertask) towards zero, what's the value of y when you reach zero?

If you are comfortable giving an answer to either of these questions, you should be comfortable with a similar answer for the rest of them.

Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

Xias
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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:There is a fundamental difference between the ordered sets {a,b,c,1,2,3,4,5,6,7,...} and {1,2,3,4,5,6,7,... a,b,c}.

That is correct.

kryptonaut wrote:The puzzle is designed exactly to create a situation like this, except the sequence a,b,c is also infinite and we end up dealing with {1,2,3,... a,b,c,...} with ordinal ω+ω = ω·2.

I'm tired of telling you that you are wrong over and over, but really, I've gone to great lengths to explain why this isn't true. You haven't done anything other than simply say that it is.

kryptonaut wrote:Subset {a,b,c,...} represents the numbers left in the jug.

One thing you can do is attempt to prove that {a,b,c,...} is not the null set, in this specific game. Just telling me that the elements are infinite is not proof. Imagining other supertasks that we can't agree on isn't proof. Prove that in this game, the set is nonempty, and then we can start talking about what elements are in it. I've offered multiple proofs that it is the null set.

kryptonaut wrote:The subset {1,2,3,...} is the balls that are removed.

I have no idea how you can determine that without taking the union of all sets of balls removed at each step; that is, the union of {1} with {2} and {3} and so on. And if that is true, then the set of balls added can only be the same set.

This is why we can't agree. You seem to think that somehow the act of removing balls also adds balls. If removing any set of balls, finite or infinite, leaves any nonempty set of balls, also finite or infinite, then the latter set must be a subset of the balls added. When we remove no balls, it's obvious that the balls added are equivalent to N. You're saying that removing balls somehow adds additional balls not in N. That's far more contradictory and hard to accept than anything that I've suggested, and you've done nothing to justify it.

kryptonaut wrote:Crucially, the set {1,2,3,... a,b,c,...} cannot be mapped to N

Well, it can, as illustrated by the Hotel: map {1,a,2,b,3,c,...} to {1,2,3,4,5,6,...}. But again, since there is no a, b, c in the balls and jugs game, it's not relevant.

kryptonaut
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### Re: Infinite Balls and Jugs [solution]

ucim wrote:If you count to infinity, what number do you reach? Yes, it's a silly question, but it seems to be isomorphic to the supertask questions being bandied about.

Hilbert hotel: There is one guest in room one. At every supertask step she is moved to the next room over, in order to make room for a new guest. At the end of the supertask, what room is she in? Is she still even in the hotel? Remember that the Hilbert Hotel does not have a room numbered omega, or infinity, or aleph-anything.

Balls and jugs: At the end of the supertask, how can the jug be empty? (Pretty much any variant - I'll use the T jug in my S-T-D version)

Algebraic functions and limits: Consider the graph of y=1/x. If you "ride the curve" (an algebraic supertask) towards zero, what's the value of y when you reach zero?

If you are comfortable giving an answer to either of these questions, you should be comfortable with a similar answer for the rest of them.

Counting to infinity: you reach some arbitrarily large number with the property that it's bigger than any finite number you can compare it with.

Hotel: She is in the 'last' room, with a countable infinity of rooms below her room number. Her room number is higher than any finite number you compare it with. This room can't be accessed directly because its actual number is unknowable. The ordinal of the set of rooms below her is ω. The ordinal of the set of rooms including hers is ω+1. The cardinality of the whole set is Aleph_null. If the hotel is not constructed to handle this then she's out in the car park.

Balls&jugs: The jug T isn't empty, it contains 10 balls with sequential numbers higher than any finite number you compare them with. The ordinal of the set of balls in T is 10.

Algebraic function: I don't quite see how this is a comparable situation. If you can explain how it is, then I can comment further.

ucim
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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:Algebraic function: I don't quite see how this is a comparable situation. If you can explain how it is, then I can comment further.
Rephrase it as a supertask. Consider the function y=1/x. Starting at time t=(-1) and continuing through t=0 (stopping at 0), you evaluate y where x=t. When you stop, you look at your paper. What is the last number on it? (You can make this countable rather than uncountable by checking only when t=1/(2^n) as n goes from 1 and increases without bound.) Since you do this increasingly fast, you will reach t=0.

In all these cases, there is no last number. You never "reach" infinity, because there's no such number. You can't get there from here.

There is a discontinuity "at" infinity. The algebra one is a comfortable example... the graph of y=1/x jumps at x=0, but your head doesn't explode because you don't view it as a "task". You don't "reach" zero. It's just there, and the function is undefined at that point. Turning it into a carnival ride (or supertask) throws this discontinuity up and makes it sound like a contradiction, but it isn't. It's isomorphic to the original curve, which your head was fine with.

If you can follow this, then apply the same kind of analogy to the set version of supertask solutions. Sure the jug has 10 balls in it for all finite steps, but "at" the "infinith" step, the jug is empty. This can't happen unless there is a discontinuity of some sort, so you think it can't happen.

But it does.

Therefore, there must be a discontinuity "at" infinity. And that's what happens. You can't "get to" infinity even when you turn it into a supertask. Nonetheless, you do an "infinite number of steps".

The jug never "gets emptied". But the jug ends up empty.

1/x never "goes from" negative to positive while passing through zero. But it does get there.

kryptonaut wrote:The ordinal of the set of rooms below her is ω.
That word.... it doesn't mean what I think you think it means.

Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

kryptonaut
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### Re: Infinite Balls and Jugs [solution]

ucim wrote:Rephrase it as a supertask. Consider the function y=1/x. Starting at time t=(-1) and continuing through t=0 (stopping at 0), you evaluate y where x=t. When you stop, you look at your paper. What is the last number on it? (You can make this countable rather than uncountable by checking only when t=1/(2^n) as n goes from 1 and increases without bound.) Since you do this increasingly fast, you will reach t=0.

As I am constantly being told, you won't reach t=0 but you will come closer to it than any interval you can name. The last number is more negative than any number you can name. 1/0 is never evaluated.

ucim wrote:
kryptonaut wrote:The ordinal of the set of rooms below her is ω.

That word.... it doesn't mean what I think you think it means.

The ordered set of rooms below hers is isomorphic to N. The ordered set of rooms including hers is not.
If we call her room number G then G is by definition higher than any other room number, since we inserted all of N before her. So we've ended up with {1,2,3,... G} which has an upper bound and so cannot be isomorphic to N

Xias
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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:
ucim wrote:
kryptonaut wrote:The ordinal of the set of rooms below her is ω.

That word.... it doesn't mean what I think you think it means.

The ordered set of rooms below hers is isomorphic to N. The ordered set of rooms including hers is not.
If we call her room number G then G is by definition higher than any other room number, since we inserted all of N before her. So we've ended up with {1,2,3,... G} which has an upper bound and so cannot be isomorphic to N

If room numbers are only natural numbers and G is a room number, then G cannot exist (since that would create a contradiction) and therefore her room number does not exist and she is not in a room. This answer is consistent with the parameters of the puzzle. Your answer has to account for creating a room that didn't exist before.

From the first line of the description of the paradox on wikipedia: "Consider a hypothetical hotel with a countably infinite number of rooms, all of which are occupied." If the set of hotel rooms was the ordinal omega edit: omega + 1, then it would not be a countably infinite number of rooms.
Last edited by Xias on Thu Nov 24, 2016 2:02 pm UTC, edited 1 time in total.

ucim
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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:As I am constantly being told, you won't reach t=0...
And Zeno's arrow never reaches the target. The whole point of a supertask is that you do reach t=0. This generates what looks like a contradiction, but is actually a discontinuity. It's also why real supertasks are not possible in the first place.

But if you manage to do that impossible thing, the jug will be empty, despite never being emptied. There is a discontinuity "at" infinity in the cardinality of the set of balls in the jug. This is not mindbending so long as infinity is infinitely far away. A supertask "compresses" infinity so that mentally we can picture going beyond infinity, even though there's no such thing (despite surreal numbers and the different sizes of infinity). Surreals aren't really "beyond" infinity; you can't get there from here. Rather, it might be better to think of them as a completely separate set, disjoint from the reals. It's a "once you're there, this is what you can do", with no actual method of "getting there". A supertask makes it seem like getting there is seamless, but it's actually impossible to "get there" (from here), even though you can "be there".

kryptonaut wrote:If we call her room number G then G is by definition higher than any other room number, since we inserted all of N before her. So we've ended up with {1,2,3,... G} which has an upper bound and so cannot be isomorphic to N.
Right. And the Hilbert Hotel only has rooms isomorphic to N. So, she can't be anywhere, because the task of moving her cannot finish, despite it being phrased as a supertask.

Supertasks cannot finish, despite being phrased with decreasing time intervals.

You can't count to ω, even if you count increasingly fast. You can never "reach" ω. To get past this, you have to make a (please forgive me, physicists) quantum leap, and when you do that, some things that were true before (the cardinality of T being 10) are no longer true. However, the things that are true are still defined by the problem. Everything in the jug had to get there somehow. If there never were surreals, emoji or ham sandwiches to put in the jug, there won't be surreals, emoji or ham sandwiches in the jug.

Xias wrote:If the set of hotel rooms was the ordinal omega
Do you mean "If the set of hotel rooms contained the ordinal omega+1"? (or "contained a room labeled by the ordinal omega+1")? (or "the cardinality of the set...was...")?

Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

Xias
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### Re: Infinite Balls and Jugs [solution]

ucim wrote:
Xias wrote:If the set of hotel rooms was the ordinal omega
Do you mean "If the set of hotel rooms contained the ordinal omega+1"? (or "contained a room labeled by the ordinal omega+1")? (or "the cardinality of the set...was...")?

Jose

By "is the ordinal omega + 1" I mean "is the well-ordered set containing all natural numbers and omega" but I can see why that might not be the best usage for this discussion.

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### Re: Infinite Balls and Jugs [solution]

@xias, omega+1 has the same cardinality as N, that is it is countably infinite, but yes, there is no order preserving bijection between omega+1 and omega.

But yes, HH is usually defined to be isomorphic to omega

kryptonaut
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### Re: Infinite Balls and Jugs [solution]

Xias wrote:If room numbers are only natural numbers and G is a room number, then G cannot exist (since that would create a contradiction) and therefore her room number does not exist and she is not in a room. This answer is consistent with the parameters of the puzzle. Your answer has to account for creating a room that didn't exist before.

G is a number, but it is defined relative to the end of the set N - that was how it was generated. She has a room number higher than any number you can think of. As I said before, if the hotel is not constructed to handle this then she's out in the car park. If you ask me "What happens if X occurs in a system where X isn't allowed to occur" then I can't give you an answer.

ucim wrote:You can't count to ω, even if you count increasingly fast. You can never "reach" ω.

ω is just a definition, representing the unknowable count of elements in N. If I define a sorting order for the naturals so that 1 is to be treated as larger than any number you compare it with, then the ordered set {2,3,4,... 1} has ordinal ω+1. In this ordering it would be impossible to count downwards from 1, although any other number would have a predecessor. But you can get to ω just by defining or arranging for your position to be greater than any other natural number. Which is what happens to the lowest member of the set of balls left in the jug, or the poorly-handled guest in the Hilbert hotel, or a number that is incremented for every member of N in a supertask. In the jug game, the lowest number in the set left behind is always higher than the set removed, which can eventually be mapped to N.

From Wikipedia on well-ordered sets https://en.wikipedia.org/wiki/Well-order
wikipedia wrote:The standard ordering ≤ of the natural numbers is a well ordering and has the additional property that every non-zero natural number has a unique predecessor.

Another well ordering of the natural numbers is given by defining that all even numbers are less than all odd numbers, and the usual ordering applies within the evens and the odds:

0 2 4 6 8 ... 1 3 5 7 9 ...
This is a well-ordered set of order type ω + ω. Every element has a successor (there is no largest element). Two elements lack a predecessor: 0 and 1.

Xias
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### Re: Infinite Balls and Jugs [solution]

Demki wrote:@xias, omega+1 has the same cardinality as N, that is it is countably infinite, but yes, there is no order preserving bijection between omega+1 and omega.

But yes, HH is usually defined to be isomorphic to omega

Ah, I think I was conflating the two ideas. Obviously they're not the same thing or you wouldn't be able to map omega + omega to N. For Omega + 1, you can just map omega to 1 and go from there and you've counted the set. My mistake.

So the last paragraph in that post can be disregarded. The paragraph before it still applies though, right?

kryptonaut wrote:
Xias wrote:If room numbers are only natural numbers and G is a room number, then G cannot exist (since that would create a contradiction) and therefore her room number does not exist and she is not in a room. This answer is consistent with the parameters of the puzzle. Your answer has to account for creating a room that didn't exist before.

G is a number, but it is defined relative to the end of the set N - that was how it was generated. She has a room number higher than any number you can think of. As I said before, if the hotel is not constructed to handle this then she's out in the car park. If you ask me "What happens if X occurs in a system where X isn't allowed to occur" then I can't give you an answer.

You can give me an answer: "She's out in the car park." Or, more generally, "There is no room for her to be in." If you want to change the parameters so that a room for her to be in exists, that's one thing, but then you're filling up a different hotel. The parameters as they are do not allow for it.

The same goes for the balls in jugs. If you want to change the game to include balls and steps with numbers greater than any natural number, you can do that, but it's a different game. The game as it's defined leads to the jug being empty, because a ball with properties that allows it to remain in the jug does not exist.

kryptonaut
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### Re: Infinite Balls and Jugs [solution]

Xias wrote:The same goes for the balls in jugs. If you want to change the game to include balls and steps with numbers greater than any natural number, you can do that, but it's a different game. The game as it's defined leads to the jug being empty, because a ball with properties that allows it to remain in the jug does not exist.

There is nothing about the rules of the game that prevents us from ending up with numbers of arbitrarily large value.

Let me ask something.
Define a set B to be all the numbers {1,2,3,... n} for some number n where n=10m.
Define a set D to be a set consisting of the lowest 10% of the members of B {1,2,3,... n/10}.
Then:
What can you say about D as n tends to infinity and B tends to N?
What are the properties of a set consisting of the first 10% of N?
What is in B-D, the upper 90% of B?
What are the properties of a set consisting of the upper 90% of N?
Can you describe a 1:1 mapping between the elements of D and B?

Xias
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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:
Xias wrote:The same goes for the balls in jugs. If you want to change the game to include balls and steps with numbers greater than any natural number, you can do that, but it's a different game. The game as it's defined leads to the jug being empty, because a ball with properties that allows it to remain in the jug does not exist.

There is nothing about the rules of the game that prevents us from ending up with numbers of arbitrarily large value.

Arbitrarily large? Or infinitely large? Those two words mean different things - "arbitrarily large n" is still in N.

Define a set B to be all the numbers {1,2,3,... n} for some number n where n=10m.
Define a set D to be a set consisting of the lowest 10% of the members of B {1,2,3,... n/10}.
Then:
What can you say about D as n tends to infinity and B tends to N?
D tends to N. An element d is in D if and only if there exists b in B such that b is greater than or equal to 10d. If we replace B with N, then for any natural number d, d is in D. So D tends to N.
What are the properties of a set consisting of the first 10% of N?
There is no set that fits that definition.
What is in B-D, the upper 90% of B?
An element is in B - D if and only if it is in B but not in D. As n tends to infinity, there is no such element, so B - D becomes the null set.
What are the properties of a set consisting of the upper 90% of N?
There is no set that fits that definition.
Can you describe a 1:1 mapping between the elements of D and B?
Not for any n.

ucim
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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:ω is just a definition, representing the unknowable count of elements in N.
The count of elements in N is knowable. It is aleph-0, which is the cardinality of the set N. There is no element of N whose ordinality... ("there is no ordinal in set N which"...) ... is equal to the cardinality of set N.

However, the set {1,2,3,4...k} (which I'll call set K) does not have this property. The cardinality of the set is k, and the highest ordinal is k. This is true as k increases without bound, but it is not true of N ("where k "has already increased without bound")

quoting wikipedia, kryptonaut wrote:0 2 4 6 8 ... 1 3 5 7 9 ...
This is a well-ordered set of order type ω + ω. Every element has a successor (there is no largest element). Two elements lack a predecessor: 0 and 1.
You can never reach 1 by starting from 0 and using the successor function. If you design a single-decreasing-interval supertask that puts balls in the jug in the given order, no odd numbers will get into the jug. Do you agree?

I suppose you could design a double-decreasing-interval supertask that would put 1 in the jug in the half-interval after the first (half of the double) supertask is done, and 3 would go in in the half-interval after that... but that's a different game, and it depends on the odd numbers being in the set in the first place. Emoji won't go into the jug. Neither will ham sandwiches. But in any case, even though 1 is a member of the set, it will not be in the jug immediately after the first (half of the double) supertask finishes.

Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

kryptonaut
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### Re: Infinite Balls and Jugs [solution]

ucim wrote:
quoting wikipedia, kryptonaut wrote:0 2 4 6 8 ... 1 3 5 7 9 ...
This is a well-ordered set of order type ω + ω. Every element has a successor (there is no largest element). Two elements lack a predecessor: 0 and 1.
You can never reach 1 by starting from 0 and using the successor function. If you design a single-decreasing-interval supertask that puts balls in the jug in the given order, no odd numbers will get into the jug. Do you agree?

Yes, I agree. By defining this sorting order we have partitioned N into two parts, each individually isomorphic to N but one of them (the odd numbers) 'unreachable' by virtue of the definition that all its elements are 'greater than' all elements in the other (the even numbers). That's how we manage to cram another infinite number of guests into the hotel, or check out an infinite number but still leave the hotel full.

ucim wrote:I suppose you could design a double-decreasing-interval supertask that would put 1 in the jug in the half-interval after the first (half of the double) supertask is done, and 3 would go in in the half-interval after that... but that's a different game, and it depends on the odd numbers being in the set in the first place. Emoji won't go into the jug. Neither will ham sandwiches. But in any case, even though 1 is a member of the set, it will not be in the jug immediately after the first (half of the double) supertask finishes.

But the ball and jug game produces an ordering that is isomorphic with this even-then-odd ordering. We have a set (the 'discard' set) starting at 1 and going up to some arbitrarily large value, followed by another 'unreachable' set (the balls in the jug) carrying on forever. The 'jug' set are defined to be higher than the 'discard' set. Both sets can be mapped to N, you just can't say where the set in the jug starts - any more than you can say how many even balls you have to count before you reach the first odd one.

What you can say is that if you remove all the even balls then you are left with the odd ones. And if you remove all the 'discard' balls you are left with the 'jug' set, which is isomorphic with N - it contains numbers which are ordered, even though they are all arbitrarily large.

What you can't say is "I started with a set that mapped to N, then removed a set that mapped to N, therefore there's nothing left."

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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:
ucim wrote:[given {0,2,4,6...1,3,5,7...}, you ] can never reach 1 by starting from 0 and using the successor function. If you design a single-decreasing-interval supertask that puts balls in the jug in the given order, no odd numbers will get into the jug. Do you agree?

Yes, I agree. By defining this sorting order we have partitioned N into two parts, each individually isomorphic to N but one of them (the odd numbers) 'unreachable' by virtue of the definition that all its elements are 'greater than' all elements in the other (the even numbers).
Similarly, you can never reach omega in the jug scenario. The jug will never contain a ball labeled omega. It is unreachable, even if you consider the set {0,1,2,3,4...omega, omega+1...}{/i] as your source. And in fact, omega and its successors are [i]not even in the source (N). There will never be a ball labeled omega in the jug. And since all natural numbers are removed from the jug, the jug ends up empty even if there is no "moment" where it gets emptied. (That mythical "moment" would correspond to the greatest number less than 1, which doesn't exist even though 1 exists and all numbers less than one exist).

kryptonaut wrote:What you can't say is "I started with a set that mapped to N, then removed a set that mapped to N, therefore there's nothing left."
True. But that's not what's being said. What is being said is "I added all the natural numbers one at a time, and I took away all the natural numbers one at a time, so there are none left". It happens that I added numbers faster than I took them away, but after an infinite number of steps, that doesn't matter.

Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

kryptonaut
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### Re: Infinite Balls and Jugs [solution]

ucim wrote:
kryptonaut wrote:
ucim wrote:[given {0,2,4,6...1,3,5,7...}, you ] can never reach 1 by starting from 0 and using the successor function. If you design a single-decreasing-interval supertask that puts balls in the jug in the given order, no odd numbers will get into the jug. Do you agree?

Yes, I agree. By defining this sorting order we have partitioned N into two parts, each individually isomorphic to N but one of them (the odd numbers) 'unreachable' by virtue of the definition that all its elements are 'greater than' all elements in the other (the even numbers).
Similarly, you can never reach omega in the jug scenario. The jug will never contain a ball labeled omega. It is unreachable, even if you consider the set {0,1,2,3,4...omega, omega+1...}{/i] as your source. And in fact, omega and its successors are [i]not even in the source (N). There will never be a ball labeled omega in the jug. And since all natural numbers are removed from the jug, the jug ends up empty even if there is no "moment" where it gets emptied. (That mythical "moment" would correspond to the greatest number less than 1, which doesn't exist even though 1 exists and all numbers less than one exist).

ω is not a number, it is an ordinal for an element indicating it's the start of an 'unreachable' section. An element acquires this ordinal when it becomes unreachable at the end of the supertask, or when a sorting order has been defined that splits N into one arbitrarily long ordered set followed by another ordered set. In the ordering {1,3,5,... 2,4,6,...} the number 2 has the ordinal ω because you can't count up to it in this ordering. But 2 exists and was in the set N to start with, it's just been moved to an unreachable place. There is no element whose successor is 2, but I can always say in this ordering that 2 comes later than 1, or 36, or any other odd number. And 6 comes after 4, which comes after 2.

In the jug game we have {1,2,3,... j,j+1,j+2,...} where j is a number bigger than any finite number you care to compare it with, but smaller than j+1. You can't count to j, just like you can't count to the room number of the guest who got pushed out of the Hilbert hotel. There is no number x such that the next number after x is j. Just as in the odd-then-even sorting where there is no odd number that immediately precedes '2'.

No, clearly ω is not in the source N because the ordering of N does not contain this discontinuity. It's the infinite process of the game that creates the discontinuity, ending up with the set described where j is unreachable from below.

Spoiler:
If the instructions had been 'feed the balls in to the jug using the odd ones first, then the even ones', then there would have been a ball with ordinal ω (ball 2) in the original set, but we would never have reached it.

ucim wrote:
kryptonaut wrote:What you can't say is "I started with a set that mapped to N, then removed a set that mapped to N, therefore there's nothing left."
True. But that's not what's being said. What is being said is "I added all the natural numbers one at a time, and I took away all the natural numbers one at a time, so there are none left". It happens that I added numbers faster than I took them away, but after an infinite number of steps, that doesn't matter.

We add the natural numbers one at a time but we are simultaneously sorting them into two sets, one for discarding and one for keeping. The 'keepers' are defined to have higher numbers than the 'discards'. Both sets become infinitely big. That is the situation the puzzle sets up, it's why it's completely different from just putting all the balls in the jug and then pulling them all out again. This is crucial.

If we split the balls into two jugs, odd and even, you are quite happy that we can set up a supertask that maps n to each ball 2n-1 in the odd jug, discards it, and leaves the odd jug empty and the even jug full. The numbers on the balls once they have been separated don't matter, it's the ordering and the fact that you can select one jug and map its contents 1:1 to the natural numbers that allows you to discard all of them one at a time.

So let's relabel the balls in the odd jug o1,o2,o3,... We can label the even ones e1,e2,e3,... or to make things more fun we could call them aardvark, banana, crocodile, ... (assuming an infinite dictionary) - it doesn't matter, so long as there is a well defined sorting order. Now we can pour the two jugs into one, and provided we stipulate that aardvark is greater than any of the balls in {o1,o2,o3,...} - we can run a supertask discarding the n'th ball which is clearly defined as on. We'll be left with {aardvark, banana, crocodile,...}. Their former numeric values are irrelevant, it's the ordering that's important.

In the puzzle, we have a task which is carefully designed to create two infinite sets - the discards and the keepers - the keepers by definition to be sorted after the discards. We discard the lower infinite set and are left with the infinite set of keepers.

You appear to still be saying that 'infinite balls went in, infinite balls came out, none are left.' I agree that you added natural numbers one at a time. I agree that you removed balls with natural numbers one at a time. But you did not remove the balls numbered j,j+1,j+2,... which the puzzle carefully contrived to have numbers bigger than any number you choose to compare them with.

ucim
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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:ω is not a number, it is an ordinal for
Ordinals are numbers. ω is just not a natural number, and thus not in the original jug.
kryptonaut wrote:An element acquires this ordinal when it becomes unreachable at the end of the supertask,
I'm not sure what you mean here. If we keep track of the order in which numbers are moved around (labeling a ball as it is moved), then ok, balls "acquire" ordinals. If we move balls that are already labeled (in black), and move them starting with 1 using the successor function in supertask time and labeling them in red as we do this, then every ball will "acquire" a red label (an ordinal) and it will match the black number. You are ordering the set as you move the balls. No ball will acquire the red ω label or ordinal because the ωth ball is never reached. It's not even in the source set (N).

With the odds...evens ordered set, or the "move the odds first, then the evens", you'll never reach the evens.

kryptonaut wrote:In the jug game we have {1,2,3,... j,j+1,j+2,...} where j is a number bigger than any finite number you care to compare it with, but smaller than j+1. You can't count to j...
Which jug game? In the S T D game, where S contains N and only N, you don't have j in the jug. If we do a modified STD game where S contains {1,2,3,... j,j+1,j+2,...}, j never makes it into T. The end is S has {j,j+1,j+2,...}, T is empty and D has {1,2,3,...}.

kryptonaut wrote:If the instructions had been 'feed the balls in to the jug using the odd ones first, then the even ones', then there would have been a ball with ordinal ω (ball 2) in the original set, but we would never have reached it.
... and it wouldn't end up in the jug.

kryptonaut wrote:We add the natural numbers one at a time but we are simultaneously sorting them into two sets, one for discarding and one for keeping.
No, because every ball gets discarded at some point. None are sorted into a "keep forever" set. The "forever" part is the kicker. It's subtly (but importantly) different that we don't discard the lower balls before we discard the upper balls; we discard the lower balls while we discard the balls that were once upper balls but no longer are. Every upper ball becomes a lower ball at some finite time. This is not true of the odd-even partition you proposed.

kryptonaut wrote:The numbers on the balls once they have been separated don't matter, it's the ordering...
The number on the balls determine the ordering. They matter.

Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

Xias
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### Re: Infinite Balls and Jugs [solution]

Kryptonaut: The "keepers" in the original game converges to the null set. Then everything that you say is consistent with the jug being empty.

A major relevant difference between a game in which you sequentially remove every naturally numbered ball (because there is a naturally numbered step assigned to removing every naturally numbered ball) and one in which you sequentially remove every 2nd ball or every 10th ball or whatever is that in the former case the set of balls remaining is not a nondecreasing sequence of sets.

Let's call, in any game, the "keepers" set Kn for all n and K at midnight, and K is defined as the limit of the sequence of Kn.

You've gone out of your way to carefully contrive, as you put it, a variant of the game in which Kn ⊂ Kn+1 for all n. This is called a nondecreasing sequence of sets, since once an element is in some Kn, it never leaves. Every set is bigger than the set before it, not just by measure of cardinality but by the fact that you're continually adding elements without taking them away. The sets of balls added form a nondecreasing sequence. The sets of balls removed create a nondecreasing sequence, and if you remove every 2nd ball or every 10th ball or whatever then the sets of "keepers" form a nondecreasing sequence.

When you have a nondecreasing sequence, the limit is easy to calculate: If x is in Kn for any n, then x is in K. This is one way that we know that the sets of balls added converges to N, as well as the sets of balls removed. It is also one way we know that the sets of keepers if we remove every odd ball converges to the set of even balls.

We know that the limit can be found this way because if x is in Kn, then it is in Kn+1; then it is in Kn+2; etc. for all n. So we can easily come up with a value M where x is in Kn for all n>M. And this is how we prove that, say, 6 is in K in the "remove only odd numbered balls" game; because 6 never gets removed.

When it comes to the original problem, the sequence of Kn is not a nondecreasing subsequence. K1 is not a subset of K2. In fact, K1 is distinct from all Kn where n>9.

There is actually a pattern here, where for any element in Kn, it is not in K10n or any K thereafter. And it follows that there is no element that is in infinitely many sets in the sequence. The limit, then, is the null set.

And one very very important point: the union of {all of the odd numbers} and {all of the even numbers} is N, which is the set of balls added. In any game of any definition, the union of {all of the discards} and {all of the keepers} will be {all of the added balls}. If the discards are N and the keepers are K, then all of the added balls must be A = N U K. So if K is nonempty, you have to account for how you added the set A through the supertask. Can you account for A being added to the jug under the parameters of the supertask?

kryptonaut
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### Re: Infinite Balls and Jugs [solution]

ucim wrote:Ordinals are numbers. ω is just not a natural number, and thus not in the original jug.

Ordinals relate to the position of an element in an ordered set. They start counting from 0, so in the set {0,1,2,3,...} every element is the same as its ordinal. But if the set is ordered {1,3,5,7,0,2,4,6} the number 0 has ordinal 4 - if you step through the set counting 'zero, first, second, third,...' then when you get the the number '0' you just said 'fourth' . ω is defined as the first ordinal coming after all natural ordinals. You can't count to it, but if you define the set of all odd numbers followed by all even numbers, then the first even number has ordinal ω. Like any ordinal, ω isn't a particular number that's in a set and can be moved into another set, it describes the position of an element within an ordered set, like the words 'first, second, third'. You might informally think of ω as meaning 'infinityeth', although it's not quite so straightforward.

ucim wrote:No, because every ball gets discarded at some point. None are sorted into a "keep forever" set. The "forever" part is the kicker. It's subtly (but importantly) different that we don't discard the lower balls before we discard the upper balls; we discard the lower balls while we discard the balls that were once upper balls but no longer are. Every upper ball becomes a lower ball at some finite time. This is not true of the odd-even partition you proposed.

Firstly, the 'keeper' set can be defined as 'keep all balls whose ordinal is higher than that of the discard set'.
Secondly, if you argue that every upper ball becomes a lower ball, I can argue that every discarded ball creates 10 new upper balls so the set never decreases in size.
The lowest number in the set of 'keepers' increases without bound, just as the highest number in the set of 'discards'. By the definition of ω, when the lowest number has increased without bound (as happens in the supertask) it has ordinal ω in the set of (discards + keepers) - you can't count up to it, it's always higher than any number you can think of. It is still a number. But if you step through the discards one at a time, matching each one to a number in N, you won't get to the first number in the keepers. That's just how it works out.

I suggest the following line of reasoning - please justify your argument where you disagree:

Q1) The limit of 1/x (towards zero from a positive direction) is an arbitrarily large number, bigger than anything you can compare it with. Agreed?
A1) This is the case.

Q2) If I have a supertask which starts with a ball numbered 1 and increments the number at each step, what can you say about the number at the end of the supertask?
A2) The number on the ball is bigger than any number you can compare it with, we can't evaluate it so let's call it j.

Q3) If I start the supertask with a set {1,2,3} and increment each of them at each step, what can you say about the set at the end of the supertask?
A3) The numbers on the balls are all bigger than any number they can be compared with, but there is still 1 between each. Let's call them {j,j+1,j+2}

Q4) If I start with a set {1,2,3,...} and increment all of them at each step, what do I end up with?
A4) The ordered set {j,j+1,j+2,...}

Q5) If I start with a set labelled {1,2,3,...} in red, then at each step n I increment the red-labelled balls and introduce a blue-labelled n, what do I end up with? In particular, what's the first number on a red-labelled ball?
A5) The ordered set {1,2,3,... j,j+1,j+2,...} The first red ball has a number higher than any natural number you can compare it with.

Q6) If I start with a set {1} labelled in red, then at step n I increment all red-labelled balls and add a new red one with number 2n+1, and a blue one number n, what do I get?
A6) The ordered set {1,2,3,... j,j+1,j+2,...}, as in A5.

The set we end up with in A5 and A6 is the same as the result of the 'remove lowest' jug game. The element labelled j has ordinal ω - we can't count up to it because every number we think of is less than j. But we can discard the blue balls in the supertask since they are countable, and and we will then be left with the red ones, a countable infinite set of them, bearing numbers higher than any we care to compare them with.

Xias wrote:The "keepers" in the original game converges to the null set. Then everything that you say is consistent with the jug being empty.

No they don't, see reasoning above. Your argument that they do is based on an assumption that the union of discards and keepers can be mapped to N, which it cannot.

Xias wrote:A major relevant difference between a game in which you sequentially remove every naturally numbered ball (because there is a naturally numbered step assigned to removing every naturally numbered ball) and one in which you sequentially remove every 2nd ball or every 10th ball or whatever is that in the former case the set of balls remaining is not a nondecreasing sequence of sets.

The set of balls remaining is a non-decreasing set relative to its lowest element. At no point does the set get smaller. If element (first+k) is in the set at step n then element (first+k) is in the set at step n+1, although (first) will have changed. By the end of the supertask, (first) is an element with ordinal ω in the set of (discards + keepers).

Xias wrote:And one very very important point: the union of {all of the odd numbers} and {all of the even numbers} is N, which is the set of balls added. In any game of any definition, the union of {all of the discards} and {all of the keepers} will be {all of the added balls}. If the discards are N and the keepers are K, then all of the added balls must be A = N U K. So if K is nonempty, you have to account for how you added the set A through the supertask. Can you account for A being added to the jug under the parameters of the supertask?

The Q&A session above shows how the set A is added to the jug.

You are making the mistake of identifying numbers with ordinals. They are not the same, as I pointed out above. This is a tricky point, particularly with infinite sets, but it is the crux of the matter.

In your assertion that the keepers ends up empty, you need to show that's the case without assuming it first. Clearly at any finite step we have a set D of discarded balls followed by a set K of keepers with strictly higher numbers than anything in D. Without first assuming K becomes empty, you need to show a 1:1 correspondence between the n'th element of D and the n'th element of (D followed by K). Since we have redefined the sets you cannot just assume that D followed by K is 1:1 mappable to the original set N.

Just like the unfortunate guest who started with a natural number but got pushed out of the Hilbert hotel by being bumped an infinite number of times, so has the set of keepers been bumped until it is a set of numbers bigger than any natural number you choose to compare them with.

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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:No they don't, see reasoning above. Your argument that they do is based on an assumption that the union of discards and keepers can be mapped to N, which it cannot.

They have to map to the balls actually added to the jug in order for K to be in the jug. You have yet to show how the balls added by the parameters outlined in the supertask we are discussing includes K in the first place. The union of D and K must be equivalent - not merely "map to" - the set of balls added. If the set of balls added is equivalent to N (which I've shown multiple times), and the set D is also equivalent N (which I've also shown multiple times), then K must be the null set. This isn't about mere "mapping," it's about the sets being equivalent. Equivalence means they have the same elements.

kryptonaut wrote:
Xias wrote:A major relevant difference between a game in which you sequentially remove every naturally numbered ball (because there is a naturally numbered step assigned to removing every naturally numbered ball) and one in which you sequentially remove every 2nd ball or every 10th ball or whatever is that in the former case the set of balls remaining is not a nondecreasing sequence of sets.

The set of balls remaining is a non-decreasing set relative to its lowest element.

That's not what non-decreasing means. It has nothing to do with the lowest element in the set. A non-decreasing sequence of sets is one for which every set in the sequence is a subset of the next set in the sequence. The set {1} is not a subset of {2}, so the sets {1}, {2}, {3} is not non-decreasing (nor is it increasing). On the other hand, the sets {1}, {1, 2}, {1, 2, 3}, {1, 2, 3, 4} are non-decreasing (and more specifically, increasing, but non-decreasing is good enough.)

A "non-decreasing sequence of sets" has this particular definition, that allows you to draw conclusions about it. I discussed what those conclusions are. The sequence of keeper sets does not fit this definition and thus we cannot use the same reasoning to draw the same conclusions about it.

kryptonaut wrote:The Q&A session above shows how the set A is added to the jug.

I'm sorry, but at best the Q&A session above shows how the set A gets added to the jug in totally different supertasks. And you've only even shown that much if we agree on those supertasks. What you need to do is show how K is a subset of the balls added by the parameters outlined in the particular supertask we are discussing.

kryptonaut wrote:In your assertion that the keepers ends up empty, you need to show that's the case without assuming it first. Clearly at any finite step we have a set D of discarded balls followed by a set K of keepers with strictly higher numbers than anything in D. Without first assuming K becomes empty, you need to show a 1:1 correspondence between the n'th element of D and the n'th element of (D followed by K). Since we have redefined the sets you cannot just assume that D followed by K is 1:1 mappable to the original set N.

Lol. I didn't assume that K became empty. I proved it. Multiple times. By every single method of finding the elements in K, the solution is "none." You still haven't provided such a proof of the opposite. Here, I will restate them, along with a couple extras (though really, they are all equivalent, since they all imply each other).

1. Kn is equivalent to Bn \ An, where Bn is "All balls that have been added once step n is complete" and An is "all balls that have been removed once step n is complete." Then K is equivalent to B \ A.

2. Kn is equivalent to {n+1, n+2, n+3, ... , 10n} for all n. Then x is in K iff for arbitrarily large M, x is in Kn for all n>M. No such x exists.

3. The union of the set D "discards" and the set K "keepers" must be the set A "added." The limit of the sequence of sets that define A and D are both N. Then D U K = N U K = A = N, and K is the null set.

4. The limit of a sequence of sets can also be found by taking the limit supremum and the limit infimum of the set and seeing if they are equivalent. limsup Kn = liminf Kn = {}.

5. For any x, define a sequence of 0s and 1s (tn) such that the nth term of the sequence is 1 if x is in Kn and 0 otherwise. Then x is in K iff (tn) converges to 1. No such x exists.

6*. Let Jn be the null set for all n. Let Ln be {n+1, n+2, ...} for all n. Then the limit of both Jn and Ln is the null set. Since Jn ⊂ Kn ⊂ Ln for all n, then Kn must also converge to the null set, by the same reasoning as the squeeze theorem in sequences of real numbers.

None of those take K = {} as given. Yet they all have the same conclusion: There exists no element x in K. Where is your proof that K is nonempty? Can you prove that some element x is in the set without simply defining x to be some element you claim is in K? If anyone's argument here is circular, it's yours: K has infinitely large elements in it because there are numbers greater than any natural number in the set. "Why are there numbers greater than any natural number in the set?" Because the infinitely large elements in the set are bigger than any natural number, of course!

*If the jug starts empty, and at every step we add ball n and remove ball n, obviously the jug ends up empty. Let's call the "keepers" in that set Jn (which happens to be the null set for all n). If the jug starts full of N, and at every step we remove ball n, obviously the jug ends up empty. Let's call the "keepers" in that set Ln (which is the set {n+1, n+2, ...} for all n). Don't believe me about the limit of Ln? Well, you'll have to take it up with set theory then because the sequence of Lns is nonincreasing, and thus the limit is the intersection of every set, which is the null set.

ucim
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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:You can't count to it, but if you define the set of all odd numbers followed by all even numbers, then the first even number has ordinal ω.
Sure, if you define an ordered set that way. (That doesn't make ordinals not numbers though, which was my point. Perhaps one of us was just a bit sloppy using the word at one point.)

kryptonaut wrote:I suggest the following line of reasoning - please justify your argument where you disagree:

Q1) The limit of 1/x (towards zero from a positive direction) is an arbitrarily large number, bigger than anything you can compare it with. Agreed?
A1) This is the case.
Right there. The limit "is infinity", which is a way of saying "there is no limit", but using the form "the limit is {this}". Infinity is not a number. More specifically, it is not ω, not because "ω is not a number" (it is a number), but because limits are cardinal numbers, not ordinal numbers, and ω is an ordinal number. The cardinal number that corresponds to it (being the first transfinite number of the given type) is aleph null.

We speak of infinite limits as if we can reach them, but we cannot reach them. It's just a figure of speech, which means "there is no limit". And in the jug version, if we cannot reach it, we cannot put that ball in the jug.

kryptonaut wrote:Q2) If I have a supertask which starts with a ball numbered 1 and increments the number at each step, what can you say about the number at the end of the supertask?
A2) The number on the ball is bigger than any number you can compare it with, we can't evaluate it so let's call it j.
Here too. What I can say about it is there is no such number. There is no such ball. There is no last step. And there is no number that is "closest to zero", even though we must cross "such a number" while traveling to zero, which (unlike infinity) is clearly reachable.

And a jug which only contains a ball which cannot exist must be empty, no matter how it got that way.

Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

kryptonaut
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### Re: Infinite Balls and Jugs [solution]

ucim wrote:Sure, if you define an ordered set that way. (That doesn't make ordinals not numbers though, which was my point. Perhaps one of us was just a bit sloppy using the word at one point.)

And the puzzle is a way of defining an infinite set (the discards) followed by another infinite set (the keepers) in just this way. I don't understand why you can't see that. Everything is set up so that the limits work that way - the discards are {1,2,3...n} followed by the keepers {n+1,n+2,...10n} which converges to N followed by another set isomorphic to N but higher than any number in N. How would you go about constructing such a set in a supertask, if you were asked to?

ucim wrote:We speak of infinite limits as if we can reach them, but we cannot reach them. It's just a figure of speech, which means "there is no limit". And in the jug version, if we cannot reach it, we cannot put that ball in the jug.

"There is no limit" to me means "arbitrarily high, bigger than anything you compare it with". Do you see some kind of distinction?

Do you accept any of the stuff written here: https://en.wikipedia.org/wiki/Ordinal_number ?
It's a rich vein of mathematics which you seem determined to avoid exploring.

ucim wrote:Here too. What I can say about it is there is no such number. There is no such ball. There is no last step. And there is no number that is "closest to zero", even though we must cross "such a number" while traveling to zero, which (unlike infinity) is clearly reachable.

Either there is a last step and the answer is 'infinity, game over' or there is no last step and the answer is 'an arbitrarily large number'. Which do you want?

I don't know what an imaginary number of anything looks like, but I'm happy to work with complex numbers and they produce useful and interesting results. For the same reason it's worth exploring sets of numbers with ordinal ω or higher to see what happens. It turns out that if you allow them then this puzzle produces consistent and intuitive results, instead of making balls magically vanish.

Xias wrote:They have to map to the balls actually added to the jug in order for K to be in the jug. You have yet to show how the balls added by the parameters outlined in the supertask we are discussing includes K in the first place. The union of D and K must be equivalent - not merely "map to" - the set of balls added. If the set of balls added is equivalent to N (which I've shown multiple times), and the set D is also equivalent N (which I've also shown multiple times), then K must be the null set. This isn't about mere "mapping," it's about the sets being equivalent. Equivalence means they have the same elements.

Except 'the same elements' is a slippery concept when dealing with an infinite set. The initial set were taken from N, but by partitioning them 'partway up' so to speak, we create two distinct sets that both map to N. Just like the odds and evens only instead of being nice easy-to-imagine numbers one set has arbitrarily big numbers.

Xias wrote:That's not what non-decreasing means.

Well if you don't accept that there is anything relevant in the fact that the set only ever increases in size, then mentioning non-decreasing sets doesn't prove or disprove anything. But the set of keepers is increasing in size for every n, and as has been stated many times there is no last step where something special happens, so I fail to see how you justify a special step where an infinite number of balls disappear into thin air.

Xias wrote:I'm sorry, but at best the Q&A session above shows how the set A gets added to the jug in totally different supertasks. And you've only even shown that much if we agree on those supertasks. What you need to do is show how K is a subset of the balls added by the parameters outlined in the particular supertask we are discussing.

Q6 was exactly like a version of the jug game where you add 2 and remove the lowest. The blue balls correspond to the discarded ones, the red to the keepers.

At every stage of the original puzzle the boundary between the discards and the keepers is 10% of the way through the number of balls introduced. At the end of the puzzle this leads to a splitting of N into two portions, one strictly above the other. That's the whole purpose of the puzzle, to set up this strange situation and explore what happens. A logical consequence of this is that the boundary is related to the size of N, is unreachable, arbitrarily high (or infinite if you like), so the bottom set maps to N. The top set's start position is the same unreachable value which has no defined predecessor. The top set is also countable and infinite. If you don't believe in the possibility of such numbers or the utility of imagining them, then we'll get nowhere. But there are many references to them such as the one I gave earlier in this post.

xias wrote:1. Kn is equivalent to Bn \ An, where Bn is "All balls that have been added once step n is complete" and An is "all balls that have been removed once step n is complete." Then K is equivalent to B \ A.

Bn is All balls that have been added once step n is complete - in the configuration in which they are now presented, as a partitioned sequence with a defined internal boundary. The set is {1,2,...n,n+1...10n} This may be the same as {1,2...10n} for finite n but is distinct if n is allowed to become arbitrarily large. If you don't see this then we'll get nowhere.

Step back a bit and re-examine the problem logically.
If I just put 10 unmarked balls in the jug and take one out each step, nobody has any qualms about ending up with an infinite number of them. What if you are watching me do this, and only at the end do I reveal that actually I had secretly numbered the balls and was using some system to remove them? Does the universe unwind and make the balls vanish?

If we add nine and renumber the lowest ball, does some cosmic overseer notice our sleight of hand and allow us to end up with infinite balls instead of erasing them?

Or does the puzzle behave exactly the same in all scenarios, adding 9 balls at a time, ending up with an infinite set each time, but with different numbers on them depending on which numbers we chose to remove?

ucim
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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:I don't understand why you can't see that.[...]It's a rich vein of mathematics which you seem determined to avoid exploring.
This is bordering on ad hominum. It is unworthy of someone exploring mathematical puzzles and (apparent?) paradoxes.

kryptonaut wrote:And the puzzle is a way of defining an infinite set (the discards) followed by another infinite set (the keepers) in just this way.
I'm losing track of which of these supertasks we're talking about.

kryptonaut wrote:"There is no limit" to me means "arbitrarily high, bigger than anything you compare it with". Do you see some kind of distinction?
There is no limit to the natural numbers, but there is also no natural number that is bigger than anything I can compare it with. You can call the number that is "bigger than anything you can compare it with" j if you like, and j exists in the surreals, but it does not exist in the set of natural numbers. If you are selecting balls from the jug of natural numbers, you will never select j. It's not in there.

If you like, you can define the set of supernatural numbers, which is the union of the natural numbers and the surreal integers, but then it's a different game, with a different result.

kryptonaut wrote:Either there is a last step and the answer is 'infinity, game over' or there is no last step and the answer is 'an arbitrarily large number'. Which do you want?
If there is no last step, the answer is not "an arbitrarily large number". It is "there is no such number".

kryptonaut wrote:I don't know what an imaginary number of anything looks like, but I'm happy to work with complex numbers...
yes, and on the surface omega and/or aleph null looks like it "fixes things", but I'm not convinced it does.

Suppose we take a partition of the natural numbers into two sets{
L={1,2,3,4...n}
R={n+1, n+2, n+3...}

The cardinality of L is finite (equal to n), and the cardinality of R is infinite (equal to aleph null)

At every supertask step, we toss the lowest element of R into L and increment n.

At every step of this supertask, the cardinality of L is finite (equal to n), and the cardinality of R is infinite (equal to aleph null)

At the end of the supertask, the cardinality of L is infinite (equal to aleph null), and the cardinality of R is finite (equal to zero). Any other conclusion leads to a contradiction. This happens, but there is no step at which it happens, because there are an infinite number of steps.

Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

Xias
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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:Everything is set up so that the limits work that way - the discards are {1,2,3...n} followed by the keepers {n+1,n+2,...10n} which converges to N followed by another set isomorphic to N but higher than any number in N.

If the set of keepers converges to the null set, then, magically, the whole thing converges to [b]N. And I have shown multiple times that that is the case, and I have shown that there is no contradiction.

kryptonaut wrote:"There is no limit" to me means "arbitrarily high, bigger than anything you compare it with". Do you see some kind of distinction?

What if the limit were negative infinity? Or what if we had an oscillating sequence that oscillated between pi at its lower bound and 17 at its upper bound? There are many types of limits that don't exist for different reasons. "The limit is infinity" is a way to describe a particular type of "there is no limit," that provides an extra detail about the nature of the set. It does not mean "arbitrarily high, bigger than anything you compare it with." In fact, you're not even using "arbitrarily" correctly.

kryptonaut wrote:I don't know what an imaginary number of anything looks like, but I'm happy to work with complex numbers and they produce useful and interesting results. For the same reason it's worth exploring sets of numbers with ordinal ω or higher to see what happens. It turns out that if you allow them then this puzzle produces consistent and intuitive results, instead of making balls magically vanish.

You seem very resistant to exploring the nuances of set theory, though, which produces the interesting result that the jug is empty.

kryptonaut wrote:Except 'the same elements' is a slippery concept when dealing with an infinite set. The initial set were taken from N, but by partitioning them 'partway up' so to speak, we create two distinct sets that both map to N. Just like the odds and evens only instead of being nice easy-to-imagine numbers one set has arbitrarily big numbers.

Sets are defined by the elements they are in. Two sets are identical if they have the same elements. You can order the set differently all you want; in fact, the odds and evens case is an interesting result of being able to take the same set and reorder it to have ordinal omega plus omega. However, the "all of N followed by the keepers" set is not reordered - it is exactly the same set, in exactly the same order.

kryptonaut wrote:Well if you don't accept that there is anything relevant in the fact that the set only ever increases in size, then mentioning non-decreasing sets doesn't prove or disprove anything.

Did you bother to read and understand the paragraphs that followed my first mention of non-decreasing sequences of sets?

kryptonaut wrote:But the set of keepers is increasing in size for every n, and as has been stated many times there is no last step where something special happens, so I fail to see how you justify a special step where an infinite number of balls disappear into thin air.

The whole purpose of my set-based proof was to show how the cardinality of the set at any point in the sequence does not matter. It does this with no relation to balls and jugs at all - just purely set theory, showing that "the cardinality increases monotonically" and "the sequence converges to the null set" are not incompatible or contradictory statements.

kryptonaut wrote:Q6 was exactly like a version of the jug game where you add 2 and remove the lowest. The blue balls correspond to the discarded ones, the red to the keepers.

No, it's not the same, but I'm unwilling to discuss your Qs 1-6 until I'm confident that you can approach my responses with the same respect and nuance that I would put into them.

kryptonaut wrote:That's the whole purpose of the puzzle, to set up this strange situation and explore what happens.

Yes, and the end result is the very interesting and paradoxical "the jug is empty" (as in, HH paradoxical, not "this statement is a lie" paradoxical). The purpose of the original puzzle is to prime someone to then consider variants that are further paradoxical. The "take the lowest ball" is a particular variant of the "ross-littlewood paradox" which has the end result of the jug being empty, which is quite peculiar and fun and why it's a paradox in the first place.

kryptonaut wrote:Bn is All balls that have been added once step n is complete - in the configuration in which they are now presented, as a partitioned sequence with a defined internal boundary. The set is {1,2,...n,n+1...10n} This may be the same as {1,2...10n} for finite n but is distinct if n is allowed to become arbitrarily large. If you don't see this then we'll get nowhere.

As I said before, claiming that the act of removing balls somehow changes the balls that were added is far more contradictory than anything I've said, and I welcome your attempt at a proof.

In particular, "bit is distinct if n is allowed to become arbitrarily large" is not a true statement in any realm of mathematics I've ever heard of. In fact, I had allowed myself to briefly commit to that reasoning with regard to the append-0 game, and Cauchy's responses caused me to reflect and realize how wrong it is. Sets are sets; the "keepers" set is a collection of the same elements whether you describe it as B - A = K or B = A + K or K = {f(n)} or anything. And in this case, the sequence of "keepers" converges to 0 no matter how you define it.

kryptonaut wrote:Step back a bit and re-examine the problem logically.
If I just put 10 unmarked balls in the jug and take one out each step, nobody has any qualms about ending up with an infinite number of them. What if you are watching me do this, and only at the end do I reveal that actually I had secretly numbered the balls and was using some system to remove them? Does the universe unwind and make the balls vanish?

If you perform the supertask in a way that makes the balls vanish, then the balls will vanish regardless of my being aware of the way you are doing it. If it turns out that the whole time you had been sequentially removing the balls, then I would watch you do it and the jug would end up empty. If I watch you do it and the jug is full of infinite balls, then that means you had not done it in that way.

kryptonaut wrote:If we add nine and renumber the lowest ball, does some cosmic overseer notice our sleight of hand and allow us to end up with infinite balls instead of erasing them? Or does the puzzle behave exactly the same in all scenarios, adding 9 balls at a time, ending up with an infinite set each time, but with different numbers on them depending on which numbers we chose to remove?

If you would like my take on this, go look at my most recent response to Cauchy. I'm not willing to put more time into discussing it with you when we don't even agree on the limit of the balls in the current game.

kryptonaut
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### Re: Infinite Balls and Jugs [solution]

ucim wrote:This is bordering on ad hominum. It is unworthy of someone exploring mathematical puzzles and (apparent?) paradoxes.

Yes, I apologise. I was getting frustrated. Thank you for responding gracefully. ucim wrote:There is no limit to the natural numbers, but there is also no natural number that is bigger than anything I can compare it with. You can call the number that is "bigger than anything you can compare it with" j if you like, and j exists in the surreals, but it does not exist in the set of natural numbers. If you are selecting balls from the jug of natural numbers, you will never select j. It's not in there.

If you like, you can define the set of supernatural numbers, which is the union of the natural numbers and the surreal integers, but then it's a different game, with a different result.

In the ordering {1,3,5,... 2,4,6...} 2 is bigger than any odd number you compare it with, and has ordinal omega. In the ordering {1,2,3,... 0} 0 is bigger than any number you compare it with and has ordinal omega. It's all a matter of how you define the sorting order. I guess that's where the 'ord' in 'odinal' comes from. In the case of the 'remove the lowest' ball game I think the boundary between the two sets (discards and keepers) is just the same. If you call it infinity then it's the kind of infinity that has degrees of infiniteness, insofar as there is an ordered set of infinite values. They are infinite relative to N but not relative to each other - does that make sense?

If that doesn't fit in with the natural numbers then I guess I'm talking supernatural numbers, but I maintain that this is analogous to what's going on when partitioning the Hilbert Hotel into odd and even rooms, so it's not really so esoteric.

ucim wrote:Suppose we take a partition of the natural numbers into two sets{
L={1,2,3,4...n}
R={n+1, n+2, n+3...}

The cardinality of L is finite (equal to n), and the cardinality of R is infinite (equal to aleph null)

At every supertask step, we toss the lowest element of R into L and increment n.

At every step of this supertask, the cardinality of L is finite (equal to n), and the cardinality of R is infinite (equal to aleph null)

At the end of the supertask, the cardinality of L is infinite (equal to aleph null), and the cardinality of R is finite (equal to zero). Any other conclusion leads to a contradiction. This happens, but there is no step at which it happens, because there are an infinite number of steps.

Ok fair point.

In attacking this puzzle (the 'remove lowest' one) are you prepared to consider a set whose structure is defined as {1,2,3,... j,j+1,j+2,...} ? Because that is essential to my view of it, and if you don't believe such a thing can arise then I don't think we'll ever come to a conclusion. But I would ask how you would represent the room number of the guest who got shunted out of the hotel by being bumped up by an infinite number of new arrivals.

If you do accept such a set, how would you refer to the number j? Would you call it infinite?, transfinite? omega? In this sorting order it comes after all the natural numbers, but it still has the property of being less than j+1 etc.

Demki
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### Re: Infinite Balls and Jugs [solution]

3 things I'd like to address about kryptonaut's posts(should you wish, you may ignore this post, and I don't really plan on writing here in the near future):
1. The natural numbers are already ordered as the ordinal omega, so when we say 'at step n, where n is a natural number, do this_n', we mean that the set of steps is also ordered as the ordinal omega, and thus there is no step that corresponds to a 'number' greater than all natural numbers. Should we wish to have such a step we'd explicitly state the intended order.
2. 'Arbitrarilly large number' could you give a concrete definition of what you mean by this? Because I don't think it means what you think it means, and in particular, by what I think you mean by it, there is no such number in the natural numbers, yet you claim such number appears in set T in the S-T-D scenario even though every number in this scenario is a member of the natural numbers.
3. This is about limits of numbers
Let f(n) be a function from N to R(ie a sequence of real numbers)
lim[n->inf] f(n) = inf doesn't mean that the sequence f(n) is convergent. It means that it is divergent, aka doesn't have a limit. This is just a classification of divergent sequences, it is simply a statement that f(n) increases without bound as n increases, that is for any real number we want M, we can find a natural number N such that for all n>N, f(n)>M.
This is usually where we use 'arbitrarily large', when we refer to M, meaning we can choose any M we want, even large M, but M will still be a real number and will still have real numbers larger than it.
It doesn't mean f(inf) = inf, or f(aleph_0) = aleph_0, because neither of inf or aleph_0 are in the domain of f, nor in the codomain of f, and as a reminder f(x)=k requires that x is in the domain of f and k is in the codomain of f.
If you knew that already, I am sorry for bringing it up, but from your post you seem to have a wrong idea when it comes to what 'lim[n->inf] f(n) = inf' means.

ucim
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### Re: Infinite Balls and Jugs [solution]

kryptonaut wrote:
ucim wrote: Suppose we take a partition of the natural numbers into two sets
L={1,2,3,4...n}
R={n+1, n+2, n+3...}
Spoiler:
The cardinality of L is finite (equal to n), and the cardinality of R is infinite (equal to aleph null)

At every supertask step, we toss the lowest element of R into L and increment n.

At every step of this supertask, the cardinality of L is finite (equal to n), and the cardinality of R is infinite (equal to aleph null)
At the end of the supertask, the cardinality of L is infinite (equal to aleph null), and the cardinality of R is finite (equal to zero). Any other conclusion leads to a contradiction. This happens, but there is no step at which it happens, because there are an infinite number of steps.
Ok fair point.
Good. Before addressing the rest of your post, let me propose a minor elaboration. Partition the natural numbers into three sets:
L={1,2,3,4...n-1}
C={n}
R={n+1, n+2, n+3...}
with the supertask of moving the lowest element of R into C, the lowest element of C into L, and incrementing n.

Consider now the cardinalities of the sets. Specifically, I claim that at the end,
the cardinality of L is infinite (equal to aleph null),
the cardinality of R is finite (equal to zero), and
the cardinality of C is also finite (equal to zero). Anything else leads to a contradiction, for the same reasons.

Do you agree with this?

Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

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