Cauchy wrote:It was argued that since balls have special memories of their past digits and Hilbert's Hotel has an incompetent assistant manager, the limits should actually be different.

You sound unconvinced.

1. In the original, remove-the-lowest-ball game, there exists at some point a ball with the label "100" on it. When was it put in the jug? 2. In the append-0 game, there exists at some point a ball with the label "100" on it. When was it put in the jug?

1. At step 10. You can tell because there was no ball labeled 100 before. 2. At step 10. You can tell because there was no ball labeled 100 before.

phlip wrote:They're not the same function, though. In order to view them as the same, you have to view the "append a 0" action as changing the ball's identity. But it's not, it's the same ball, it's just got a different label.

That is... we have a countably-infinite set of balls B, all identical except as parameters to the functions we're about to describe. At each step, each ball has a position p_{i}:B->{in, out} and a label l_{i}:B->N.

They're all identical except that they're clearly distinguishable as arguments of the function, is what you're saying. So they're not identical at all.

If balls have magical intrinsic immutable labels, then sure, the limits are whatever. I argue that they don't, and that MIILs are a crutch that you all are using because you don't want to accept some facet of pointwise limits. Two balls without labels are actually identical and indistinguishable, and I get the same result if I write 1 and 2 on two of them, or if I write 2 and 1 and then swap their positions.

At the very least, I think it's clear that this is a matter of opinion now, not math. You can clearly take the limits of the p and f functions philip outlined, and they have the obvious limits. Whether you think p or f describes the supertask better is not something that can be deduced, really.

(∫|p|^{2})(∫|q|^{2}) ≥ (∫|pq|)^{2} Thanks, skeptical scientist, for knowing symbols and giving them to me.

Cauchy wrote:It was argued that since balls have special memories of their past digits and Hilbert's Hotel has an incompetent assistant manager, the limits should actually be different.

You sound unconvinced.

1. In the original, remove-the-lowest-ball game, there exists at some point a ball with the label "100" on it. When was it put in the jug? 2. In the append-0 game, there exists at some point a ball with the label "100" on it. When was it put in the jug?

1. At step 10. You can tell because there was no ball labeled 100 before. 2. At step 10. You can tell because there was no ball labeled 100 before.

Ok. Then we are not justified in saying that no ball ever leaves the jug, and there is no paradox.

ucim wrote:Let's skip the labels and the balls and all the distractions.

The labels are important, I don't think they should just be dismissed.

ucim wrote:Now consider the three-partition L-C-R scenario. At the end of the supertask, all the natural numbers are in L. It doesn't matter how it gets there, it just matters that it gets there. Now, we'll call set U the union of C and R. Set U contains all the natural numbers that are not in L.

Do you agree that set U is empty? If so, this implies that set C must also be empty. Do you agree here?

Set C contains a single infinite (in the sense that if you start counting you'll never get there) element. Set L can be mapped to N. A defining property of an infinite set is that it can be mapped to a subset of itself. In this case, N maps to L whilst also leaving an element in C. In the same way you could remove element '1' and still map {2,3,...} to N

phlip wrote:It should be pretty clear that f∞ is 1 everywhere, while g∞ is 0 everywhere.

Are you sure that g∞ is 0 everywhere? If g_{i} is only defined for finite i then for very large i g has a very large number of 1's. If g is defined for infinite i then I'd argue that g∞ has an infinite number of 1's (for n=∞,∞+1,∞+2,...)

xias wrote:1. In the original, remove-the-lowest-ball game, there exists at some point a ball with the label "100" on it. When was it put in the jug? 2. In the append-0 game, there exists at some point a ball with the label "100" on it. When was it put in the jug?

1. At step 10 2. The ball, at step 1. The label first appeared at step 10.

At the end of the append-0 game, with an infinite number of balls in the jug, what does the set of labels look like? What does the set of numbers corresponding to those labels look like? Why is that set of numbers different from the set of numbers on the balls in the 'remove-lowest' game?

Going back to the start - correct me if I'm wrong, but there seem to be two aspects to the argument that the jug ends up empty in the 'remove-the-lowest' game: A) For any number n I can point to a time t before midnight when ball n was removed from the jug, or: B) Every number that enters the jug also leaves it at some point.

Point (A) should be clarified to read "For any finite number n...". Whilst it is true that only balls with finite numbers exist at any time before midnight, there are an infinite number of balls with infinite numbers (in the sense that you can never count that high) present at the end of the supertask - in the same way that infinitely long labels are generated in the 'append a zero' variant. These balls do not have a step when they are removed, they correspond to the final state of the infinite number of steps in the supertask.

Point (B) relates to the observation that the input to the process can be mapped to N, and - if we model the discarded balls as a set D {1,2,3,...n} and the balls in the jug as a set J {n+1,n+2,...10n} - that the set D also converges to N. The deduction from this observation is that J must therefore be empty. However as I have pointed out, an infinite set can be mapped 1:1 to a subset of itself - and that is what happens here. N can be mapped to (D+J) and to D without implying that J is empty.

This is why I believe that neither argument is a valid proof that the jug is empty at the end of the supertask. In fact both statements, when modified as shown, are compatible with the intuitive answer that there are an infinite number of balls bearing infinite numbers at the end of the supertask.

ucim wrote:Now consider the three-partition L-C-R scenario. At the end of the supertask, all the natural numbers are in L. It doesn't matter how it gets there, it just matters that it gets there. Now, we'll call set U the union of C and R. Set U contains all the natural numbers that are not in L.

Do you agree that set U is empty? If so, this implies that set C must also be empty. Do you agree here?

Set C contains a single infinite (in the sense that if you start counting you'll never get there) element. Set L can be mapped to N. A defining property of an infinite set is that it can be mapped to a subset of itself. In this case, N maps to L whilst also leaving an element in C. In the same way you could remove element '1' and still map {2,3,...} to N

If L were merely infinite, then your argument might have some merit. However, L is not merely infinite - it contains ALL the natural numbers. There isn't any mapping going on here.

kryptonaut wrote:

phlip wrote:It should be pretty clear that f∞ is 1 everywhere, while g∞ is 0 everywhere.

Are you sure that g∞ is 0 everywhere? If g_{i} is only defined for finite i then for very large i g has a very large number of 1's. If g is defined for infinite i then I'd argue that g∞ has an infinite number of 1's (for n=∞,∞+1,∞+2,...)

Why should g be defined for infinite? There are no infinite natural numbers.

kryptonaut wrote:

xias wrote:1. In the original, remove-the-lowest-ball game, there exists at some point a ball with the label "100" on it. When was it put in the jug? 2. In the append-0 game, there exists at some point a ball with the label "100" on it. When was it put in the jug?

1. At step 10 2. The ball, at step 1. The label first appeared at step 10.

At the end of the append-0 game, with an infinite number of balls in the jug, what does the set of labels look like? What does the set of numbers corresponding to those labels look like? Why is that set of numbers different from the set of numbers on the balls in the 'remove-lowest' game?

Look at your answer to these two questions. Can you not see that the paths the balls take are different? The ball in the original game entered the jug at step 10, and will leave at step 100. The ball in the append-0 game entered the jug at step 1, and will never leave. Doesn't that suggest that the end results will be different as well?

kryptonaut wrote:Point (A) should be clarified to read "For any finite number n...". Whilst it is true that only balls with finite numbers exist at any time before midnight, there are an infinite number of balls with infinite numbers (in the sense that you can never count that high) present at the end of the supertask - in the same way that infinitely long labels are generated in the 'append a zero' variant. These balls do not have a step when they are removed, they correspond to the final state of the infinite number of steps in the supertask.

Infinitely long labels exist in the 'append a zero' variant because the labels get modified. As the game goes, the labels are changed to get longer and longer. No ball corresponds to any one label, because they continually get changed. In the original game, no such changes occur. The balls correspond exactly to the finite labels, because there are no infinite natural numbers.

kryptonaut wrote:Point (B) relates to the observation that the input to the process can be mapped to N, and - if we model the discarded balls as a set D {1,2,3,...n} and the balls in the jug as a set J {n+1,n+2,...10n} - that the set D also converges to N. The deduction from this observation is that J must therefore be empty. However as I have pointed out, an infinite set can be mapped 1:1 to a subset of itself - and that is what happens here. N can be mapped to (D+J) and to D without implying that J is empty.

An infinite set can indeed be mapped 1:1 to a subset of itself - but D doesn't converge to a subset of N, it converges to N itself.

kryptonaut wrote:Point (A) should be clarified to read "For any finite number n...". Whilst it is true that only balls with finite numbers exist at any time before midnight, there are an infinite number of balls with infinite numbers (in the sense that you can never count that high) present at the end of the supertask - in the same way that infinitely long labels are generated in the 'append a zero' variant. These balls do not have a step when they are removed, they correspond to the final state of the infinite number of steps in the supertask.

There could be an infinite number of balls with infinite numbers, but if you want to convince anyone of that you have to prove it. We both agree that the jug is void of all natural-numbered balls. You've just encumbered yourself with a claim that there are some other balls in the jug, and you haven't proven it yet. In a way, you've described a handful of necessary qualities that such balls would have to have, but you've also ignored every other complication that you've introduced.

One such complication: We can split the supertask into two supertasks, one that involves only adding balls and one that involves only removing balls. If we perform the adding supertask first, and then the removing supertask, then the results should be the same. Under your argument, they are different. Under your argument, the act of removing balls somehow changes which balls are added, even though the balls added are not dependent on the balls removed. There is no justification for this, especially given the following:

kryptonaut wrote:Point (B) relates to the observation that the input to the process can be mapped to N, and - if we model the discarded balls as a set D {1,2,3,...n} and the balls in the jug as a set J {n+1,n+2,...10n} - that the set D also converges to N. The deduction from this observation is that J must therefore be empty. However as I have pointed out, an infinite set can be mapped 1:1 to a subset of itself - and that is what happens here. N can be mapped to (D+J) and to D without implying that J is empty.

The best you've done is show that we can reorder N by taking a partition of N, order it a particular way, and end up with infinite ordinals. The example that you used is {1, 3, 5, ... 2, 4, 6, ...}. You're not wrong. You can absolutely do that. The reason you can do that is because every element in that set is also in N. What this does not allow you to do is then partition N into N and a nonempty set.

The odd-even ordering can be constructed like so: Start with the the set N. Take the subset of odd numbers, O_{1}, O_{2}, etc, and the subset of even numbers E_{1}, E_{2}, etc. Then define a well-ordered set such that O_{n+1}>O_{n}, E_{m+1}>E_{m}, and E_{m}>O_{n} for all m and n. Then you end up with {1, 3, 5, ... 2, 4, 6, ...}. Note that every number in this set must also be in N.

We could do it instead by sorting into prime and composite numbers. Start with the set N. Take the subset of prime numbers, P_{1}, P_{2}, etc, and the subset of composite numbers C_{1}, C_{2}, etc. Then define a well-ordered set such that P_{n+1}>P_{n}, C_{m+1}>C_{m}, and C_{m}>P_{n} for all m and n. Then you end up with {2, 3, 5, 7, 11, ... 1, 4, 6, 8, 9, ...}. Note that every number in this set must also be in N.

We could even do this with sets that are not N. Start with the set O. Take the subset of prime numbers, P_{1}, P_{2}, etc, and the subset of composite numbers C_{1}, C_{2}, etc. Then define a well-ordered set such that P_{n+1}>P_{n}, C_{m+1}>C_{m}, and C_{m}>P_{n} for all m and n. Then you end up with {3, 5, 7, 11, ... 1, 9, 15, 21, 25, ...}. Note that every number in this set must also be in O.

We can then do the same thing with the balls and jugs. Start with the set A, which is the set of all balls added. Take the subset of discarded balls, D_{1}, D_{2}, etc, and the subset of balls that remain J_{1}, J_{2}, etc. Then define a well-ordered set such that D_{n+1}>D_{n}, J_{m+1}>J_{m}, and J_{m}>D_{n} for all m and n. Then you end up with {D_{1}, D_{2}, ... J_{1}, J_{2}, ...}. We seem to agree that at midnight, D is the set of all natural numbers. So then our final well-ordered set is {1, 2, 3, ... J_{1}, J_{2}, ...}. Like before, every element in this set must also be in the set we started with, A.

Now, your claim is that J is nonempty and that A is a proper superset of N. However, the only proof you seem to have of the latter is a claim that the former is true; and your only proof of the former seems to be a claim that the latter is true. You have not proven either statement in a noncircular way.

What I and others have been arguing is that A is N and J is 0. There have been multiple methods that we have shown. For example, A is the infinite union of all sets of balls added at each individual step: {1, ..., 10} U {11, ..., 20} U ... etc. This is equal to N in the same way that D is equal to N. Second, to take the point that I set aside earlier, "the limit of the sequence of sets ({n+1,n+2,...10n}) has been mathematically shown to be the null set."

kryptonaut wrote:In fact both statements, when modified as shown, are compatible with the intuitive answer that there are an infinite number of balls bearing infinite numbers at the end of the supertask.

I think that's the problem, kryptonaut. You are looking for something compatible with your intuition.

kryptonaut wrote:Point (B) relates to the observation that the input to the process can be mapped to N, and - if we model the discarded balls as a set D {1,2,3,...n} and the balls in the jug as a set J {n+1,n+2,...10n} - that the set D also converges to N. The deduction from this observation is that J must therefore be empty. However as I have pointed out, an infinite set can be mapped 1:1 to a subset of itself - and that is what happens here. N can be mapped to (D+J) and to D without implying that J is empty.

Talking about maping this puzzle be described as mapping every number in N to a range of numbers. We know when a ball enters and when it leaves. So we get for every number x (when the balls entered) to the numbers 10x-9 to 10x (turn the balls added at step x are removed). If you ignore the puzzle and only look at it mathematically would you consider there to be any number in the (turn removed) set that isn't in the (turn added) set? If not I don't think the subset argument works because with subsets you deliberately living stuff out. Here you don't map N to a subset of N, you map N to N just that you map one element of N to multiple elements of itself.

(Don't worry I have not changed my mind to demonstrate here another similarly pointless puzzle: "Two tireless immortal race each other on an endless way which has no goal because it is an endless way. Immortal A is far faster than B so A promises to wait when reaching the goal so that they can chat for an eternity or two. What happens once the race is over?")

Poker wrote:If L were merely infinite, then your argument might have some merit. However, L is not merely infinite - it contains ALL the natural numbers. There isn't any mapping going on here.

ALL the natural numbers from 1 to where? The only way to compare them is to set up a 1:1 mapping.

Poker wrote:Why should g be defined for infinite? There are no infinite natural numbers.

I didn't define g, ask Phlip. I was covering both bases.

Poker wrote:Look at your answer to these two questions. Can you not see that the paths the balls take are different? The ball in the original game entered the jug at step 10, and will leave at step 100. The ball in the append-0 game entered the jug at step 1, and will never leave. Doesn't that suggest that the end results will be different as well?

As I said, the ball entered at step 1. The fact that it doesn't leave makes it clear that there will be infinite balls at the end. The fact that it is being modeled using sets of numbers, and that exactly the same set of numbers as are present on the labels also models the 'remove lowest' case, should make it clear that both models end up the same, with an infinite set of infinite numbers.

Poker wrote:Infinitely long labels exist in the 'append a zero' variant because the labels get modified. As the game goes, the labels are changed to get longer and longer. No ball corresponds to any one label, because they continually get changed. In the original game, no such changes occur. The balls correspond exactly to the finite labels, because there are no infinite natural numbers.

Is it not apparent that every zero appended to a label also represents a number multiplied by ten? Is it not apparent that an infinitely long label also represents an infinitely big number? I can't understand how it's possible to say that labels get infinitely long but numbers don't get infinitely big. At any finite stage of the task the labels have finite length and the numbers have finite magnitude. At the end, they don't.

Poker wrote:An infinite set can indeed be mapped 1:1 to a subset of itself - but D doesn't converge to a subset of N, it converges to N itself.

Again, you can only say that D maps to N. Just as (D+J) does. That does not justify the claim that J is empty.

Xias wrote:One such complication: We can split the supertask into two supertasks

No we can't. For the same reason that you can't say 10+infinity-infinity=10. The whole paradox revolves around the fact that two supertasks run in parallel but at different rates, which fact is lost if you run them sequentially.

I studied this idea, and I find it hard to understand how the limits infimum and supremum can be defined for a divergent sequence such as this. But by mapping the set {n+1,n+2,...10n} to the range [1-2^{-(n+1)},1-2^{-10n}] as far as I can see the limit is [1,1] which maps back to infinite values.

Xias wrote:What this does not allow you to do is then partition N into N and a nonempty set.

Can you explain why not? The number of elements is not defined, all that matters is the 1:1 mapping. If I arrange to extract some numbers from the top end of the set (which is what the puzzle does) then the top of the remainder is still infinite, so the remainder still maps to 1:1 to N even though I have some numbers removed. Just as I can remove numbers from the bottom of N and map the remainder to N. The only difference is conceptual, that it's not possible to name the number where one subset ends and the other begins.

Xias wrote:What I and others have been arguing is that A is N and J is 0. There have been multiple methods that we have shown. For example, A is the infinite union of all sets of balls added at each individual step: {1, ..., 10} U {11, ..., 20} U ... etc. This is equal to N in the same way that D is equal to N. Second, to take the point that I set aside earlier, "the limit of the sequence of sets ({n+1,n+2,...10n}) has been mathematically shown to be the null set."

As I have shown, the fact that A is partitioned into two sets in no way implies that J is empty. I don't accept that the limit has been mathematically shown to be empty.

Xias wrote:I think that's the problem, kryptonaut. You are looking for something compatible with your intuition.

If I see a model and some reasoning claiming that fundamentally different things happen depending only on whether we choose to write numbers on things or not, then I suspect that the model or reasoning are at fault.

PeteP wrote: If not I don't think the subset argument works because with subsets you deliberately living stuff out. Here you don't map N to a subset of N, you map N to N just that you map one element of N to multiple elements of itself.

The set D {1,2,...n} is clearly a subset of the total balls added {1,2,...10n} And yet both map to N.

Neither maps to N as long as n is a number, and if you put it as 1 to infinity and 1 to 10 infinity there is nothing clear about one being the subset of the other.

kryptonaut wrote:The labels are important, I don't think they should just be dismissed.

Maybe, but if we can't agree on the pure version, we'll never agree on the farded version.

kryptonaut wrote:

ucim wrote:Now consider the three-partition L-C-R scenario. At the end of the supertask, all the natural numbers are in L. It doesn't matter how it gets there, it just matters that it gets there. Now, we'll call set U the union of C and R. Set U contains all the natural numbers that are not in L.

Do you agree that set U is empty? If so, this implies that set C must also be empty. Do you agree here?

Set C contains a single infinite (in the sense that if you start counting you'll never get there) element.

You're jumping the gun here.

Is set U empty? --> If set U is empty, then is set C empty? ----> If set C is not empty, but set U is, wat?

--> If set U is not empty, then is set R (in the two-partition game) empty? -----> If so, wat?

And it doesn't matter if one set can be "mapped to" another set. The question involves whether the intersection is empty - i.e. if they are the same set.

responding to xias, kryptonaut wrote:Point (A) should be clarified to read "For any finite number n...". Whilst it is true that only balls with finite numbers exist at any time before midnight, there are an infinite number of balls with infinite numbers (in the sense that you can never count that high) present at the end of the supertask - in the same way that infinitely long labels are generated in the 'append a zero' variant. These balls do not have a step when they are removed, they correspond to the final state of the infinite number of steps in the supertask.

These balls with infinite numbers... they do not have a step when they are added. So, they don't need a step where they are removed. They don't exist.

Here's another thing for your intuition to consider - for how much time does any given ball remain in the jug? As n goes to infinity, the amount of time that ball n stays in the (transfer) jug goes to zero. "At" infinity, the "infinityeth" ball spends zero time in the jug.

responding to Poker, kryptonaut wrote:The only way to compare them is to set up a 1:1 mapping.

Mapping, schmapping. No, you can also take the intersection of the two sets. This is what is relevant. Focusing on mapping rather than intersection is what's mustarding you up.

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 wrote:One such complication: We can split the supertask into two supertasks

No we can't. For the same reason that you can't say 10+infinity-infinity=10. The whole paradox revolves around the fact that two supertasks run in parallel but at different rates, which fact is lost if you run them sequentially.

Again, when you say that you are saying that removing balls somehow changes the balls that are added. I have yet to see any mathematical or even philosophical justification for that.

I studied this idea, and I find it hard to understand how the limits infimum and supremum can be defined for a divergent sequence such as this. [/quote]

Do you understand the limits infimum and supremum to begin with?

kryptonaut wrote:But by mapping the set {n+1,n+2,...10n} to the range [1-2^{-(n+1)},1-2^{-10n}] as far as I can see the limit is [1,1] which maps back to infinite values.

What is the set of numbers between 1 and 1? The null set. For what value is 2^{-x} equal to zero? "DNE."

kryptonaut wrote:

Xias wrote:What this does not allow you to do is then partition N into N and a nonempty set.

Can you explain why not?

You want me to explain what partitioning means?

kryptonaut wrote:The number of elements is not defined, all that matters is the 1:1 mapping. If I arrange to extract some numbers from the top end of the set (which is what the puzzle does) then the top of the remainder is still infinite, so the remainder still maps to 1:1 to N even though I have some numbers removed. Just as I can remove numbers from the bottom of N and map the remainder to N. The only difference is conceptual, that it's not possible to name the number where one subset ends and the other begins.

What is this mapping nonsense? When you remove the numbers "from the bottom of N" and those numbers are exactly all of the natural numbers then you have removed all of N and there is no remainder. It has nothing to do with mapping. There is no "top end" of the set. The numbers you are talking about don't exist in N for you to partition out in the first place.

kryptonaut wrote:

Xias wrote:What I and others have been arguing is that A is N and J is 0. There have been multiple methods that we have shown. For example, A is the infinite union of all sets of balls added at each individual step: {1, ..., 10} U {11, ..., 20} U ... etc. This is equal to N in the same way that D is equal to N. Second, to take the point that I set aside earlier, "the limit of the sequence of sets ({n+1,n+2,...10n}) has been mathematically shown to be the null set."

As I have shown, the fact that A is partitioned into two sets in no way implies that J is empty.

You haven't shown that.

If I partition A into two sets, and one of those sets is A, then the other is the null set. This isn't my opinion. This is how partitioning works. If both sets are nonempty, then A is the union of both of those sets.

If A is N, then when you partition A into N and another set, the second set must be the null set. If you are claiming to partition A into two nonempty sets, which is what you are claiming to do, then A has to be a superset of both N and J to begin with. I gave you three examples of this, Kryptonaut.

kryptonaut wrote:I don't accept that the limit has been mathematically shown to be empty.

I'm not sure what to do about that. It's true, regardless of whether you accept it or not. I'd be happy to explain it to you if it's a matter of you not understanding the language used, but if you're outright denying set theory in the defense of your intuition then there's nothing I can do.

kryptonaut wrote:

Xias wrote:I think that's the problem, kryptonaut. You are looking for something compatible with your intuition.

If I see a model and some reasoning claiming that fundamentally different things happen depending only on whether we choose to write numbers on things or not, then I suspect that the model or reasoning are at fault.

We could do the supertask with no labels, and as long as we tracked the balls correctly, the jug would be empty. The argument has nothing to do with the labels. The labels just help us find b_{n} to remove during step n.

This is why we offer a different answer to "remove the lowest" and "remove the highest." We aren't saying "well, we added infinity balls and removed infinity balls so there must be zero left." We are saying that we systematically removed the exact same set of balls that we added. If we systematically removed every odd ball, we would end up with only even balls. If we systematically removed the lowest ball that wasn't b_{1}, b_{2}, or b_{3}, then we would end up with exactly three balls left in the jug.

kryptonaut wrote:

PeteP wrote: If not I don't think the subset argument works because with subsets you deliberately living stuff out. Here you don't map N to a subset of N, you map N to N just that you map one element of N to multiple elements of itself.

The set D {1,2,...n} is clearly a subset of the total balls added {1,2,...10n} And yet both map to N.

Both of them are N. Not just mapping. And the set {1,2,...n} becomes not just a subset of, but the same set as {1, 2, ... 10n} at the limit, since both are N at the limit.

Last edited by Xias on Tue Dec 06, 2016 1:23 pm UTC, edited 1 time in total.

Poker wrote:If L were merely infinite, then your argument might have some merit. However, L is not merely infinite - it contains ALL the natural numbers. There isn't any mapping going on here.

ALL the natural numbers from 1 to where? The only way to compare them is to set up a 1:1 mapping.

"ALL the natural numbers from 1 to where?" That's the whole point of saying all the natural numbers - there is no "where" at which we stop! And as others on here have said in their responses, there are other ways to compare sets that don't rely on mapping.

kryptonaut wrote:

Poker wrote:Why should g be defined for infinite? There are no infinite natural numbers.

I didn't define g, ask Phlip. I was covering both bases.

Phlip didn't see a need to cover a base that doesn't exist.

kryptonaut wrote:

Poker wrote:Look at your answer to these two questions. Can you not see that the paths the balls take are different? The ball in the original game entered the jug at step 10, and will leave at step 100. The ball in the append-0 game entered the jug at step 1, and will never leave. Doesn't that suggest that the end results will be different as well?

As I said, the ball entered at step 1. The fact that it doesn't leave makes it clear that there will be infinite balls at the end. The fact that it is being modeled using sets of numbers, and that exactly the same set of numbers as are present on the labels also models the 'remove lowest' case, should make it clear that both models end up the same, with an infinite set of infinite numbers.

Poker wrote:Infinitely long labels exist in the 'append a zero' variant because the labels get modified. As the game goes, the labels are changed to get longer and longer. No ball corresponds to any one label, because they continually get changed. In the original game, no such changes occur. The balls correspond exactly to the finite labels, because there are no infinite natural numbers.

Is it not apparent that every zero appended to a label also represents a number multiplied by ten? Is it not apparent that an infinitely long label also represents an infinitely big number? I can't understand how it's possible to say that labels get infinitely long but numbers don't get infinitely big. At any finite stage of the task the labels have finite length and the numbers have finite magnitude. At the end, they don't.

You seem to have completely ignored what I said about the original game. So what if the numbers look the same at particular snapshots? The paths the balls themselves take are completely different.

Let me give you an example of how different paths can change answers: Let's say three identical cars are driving around a circular track. Their drivers are named Moe, Larry, and Curly. You do not pay attention to what the cars do when they are not near you, but every time the cars pass you on the track, you snap a picture. You discover that each time, Moe is on the outside, Larry in the middle, and Curly on the inside. Now, say you went halfway around the track and snapped a picture at that point. The obvious thing to guess is that Moe would still be on the outside, Larry would be in the middle, and Curly would be on the inside. That would be the obvious guess, but it's far from certain. You weren't watching what the cars did for the rest of the track. For all you know, when they're out of view, the cars could be weaving in and out of each other, and at the point halfway around the circle they could be completely reversed - or, indeed, in any order! To know what point the cars are at there, you need to know how they travel. It's the same principle with this problem - to know where the balls end up, we need to know how they travel.

kryptonaut wrote:

Poker wrote:An infinite set can indeed be mapped 1:1 to a subset of itself - but D doesn't converge to a subset of N, it converges to N itself.

Again, you can only say that D maps to N. Just as (D+J) does. That does not justify the claim that J is empty.

What are you talking about? By its very definition, D converges to N! There is no mapping involved - at the limit, D, in every single sense of the word, quite literally IS N!

kryptonaut wrote:

Xias wrote:What this does not allow you to do is then partition N into N and a nonempty set.

Can you explain why not? The number of elements is not defined, all that matters is the 1:1 mapping. If I arrange to extract some numbers from the top end of the set (which is what the puzzle does) then the top of the remainder is still infinite, so the remainder still maps to 1:1 to N even though I have some numbers removed. Just as I can remove numbers from the bottom of N and map the remainder to N. The only difference is conceptual, that it's not possible to name the number where one subset ends and the other begins.

You can certainly remove some numbers from N leaving some set (for example, removing the numbers 1-10 from N - for the sake of convenience, I will call this set K), and can claim (rightfully so) that there is a 1:1 mapping between K and N. But - and this is the critical point - K IS NOT N.

When you partition N into two sets (let's call them A and B, and since at least one of them has to be infinite let's have A be infinite), every element in N is in one and only one of A and B (and, incidentally, every element in either A and B is an element of N). So whatever elements are in B, you have to remove them from N to get A. And whenever you remove an element from N - any element - you are left with a set which has the same cardinality as N, is the same size as N, has a 1:1 mapping with N - and yet despite all that, because some element of N is missing, A IS NOT N. The only way for A to be N is if no element was removed in the first place - i.e. if B was the empty set.

phlip wrote:It should be pretty clear that f∞ is 1 everywhere, while g∞ is 0 everywhere.

Are you sure that g∞ is 0 everywhere? If g_{i} is only defined for finite i then for very large i g has a very large number of 1's. If g is defined for infinite i then I'd argue that g∞ has an infinite number of 1's (for n=∞,∞+1,∞+2,...)

The fact that g_{i} has a large number of 1s doesn't matter though, because we're doing a pointwise limit, we're only looking at one point at a time.

So, for instance, to find g_{∞}(5), we find the limit of: g_{0}(5), g_{1}(5), g_{2}(5), g_{3}(5), g_{4}(5), g_{5}(5), g_{6}(5), ... That is, we find the limit of: 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, ... which is clearly 0, since it just ends in an infinite sequence of 0s. So g_{∞}(5) = 0.

And we can do the same at any other point. For any natural number n (which, to reiterate, are all finite, and constitute the entirety of the domain of g_{∞}), the sequence of g_{i}(n) values starts at 0, then after a while becomes 1, then a while later becomes 0 again, and then stays 0 forever. The limit of that sequence is 0. The only exception is g_{∞}(0), as g_{i}(0) = 0 for every i, so it's even more obvious that the limit is 0.

This is what a pointwise limit means.

Xias wrote:↶ You want me to explain what partitioning means?

I think I will anyway, for clarity.

Two sets A and B are "partitions" of a set S if:

The intersection of A and B is empty (ie A and B are disjoint, do not overlap)

The union of A and B is S (note: not "maps 1:1 to S" or "is the same size as S" but "is S" ie contains exactly the same elements of S, ie every element of S is in A∪B and every element of A∪B is in S)

An alternative way to phrase this is: every element of S is in exactly one of A and B (and nothing else is in either A or B). Or another is: B is all the elements of S that aren't in A (symbolically, B=S\A).

So if S = A, then... well, every element of S is in exactly one of A and B, and it's already in A, so it can't be in B. None of the elements of S can be in B, so B must be empty. This is true regardless of whether S is finite or infinite, it doesn't matter. Because we're not doing any fancy mapping stuff here, we're just talking about set equality.

[edit] Forgot to mention: you can partition into more than 2 sets, and it means the same thing... each element of the overall set is in exactly one partition. This is just talking about partitioning into 2 sets, since that's what were discussing right now.

Xias wrote:↶ You want me to explain what partitioning means?

I think I will anyway, for clarity.

Two sets A and B are "partitions" of a set S if:

The intersection of A and B is empty (ie A and B are disjoint, do not overlap)

The union of A and B is S (note: not "maps 1:1 to S" or "is the same size as S" but "is S" ie contains exactly the same elements of S, ie every element of S is in A∪B and every element of A∪B is in S)

An alternative way to phrase this is: every element of S is in exactly one of A and B (and nothing else is in either A or B). Or another is: B is all the elements of S that aren't in A (symbolically, B=S\A).

So if S = A, then... well, every element of S is in exactly one of A and B, and it's already in A, so it can't be in B. None of the elements of S can be in B, so B must be empty. This is true regardless of whether S is finite or infinite, it doesn't matter. Because we're not doing any fancy mapping stuff here, we're just talking about set equality.

Just to back this up, I have here the definition of a partition from A Book of Abstract Algebra, 2nd Edition, Charles C. Pinter:

Spoiler:

By a partition of a set A we mean a family {A_{i} : i ∈ I} of nonempty subsets of A such that (i) If any two classes, say A_{i} and A_{j}, have a common element x (that is, are not disjoint), then A_{i} = A_{j}, and (ii) Every element x of A lies in one of the classes.

So phlip isn't just making it up, I'm not just making it up, and ucim hasn't been just making it up when we say that you can't partition N in the way you are describing.

So when you say things like this:

Spoiler:

kryptonaut wrote:But if you contrive to partition N somewhere 'partway along' or 'so many steps from the end' then omega is the ordinal of the first element of the second partition - this is because the first partition is countably infinite even if it's a 'fraction' of the original whole. It's the act of partitioning it that creates the omega.

kryptonaut wrote:By introducing a partition into N with an unbounded number of elements before it, we create two sets, at least one of which is infinite. The second set by definition then has to start at an infinite number.

kryptonaut wrote:This is what I mean when I say that partitioning the set N can create an element with an infinite number. It is a number you can't count up to, because its specification is a function of the size of N.

you're just not talking about anything we can actually do with mathematics.

Regarding set partitioning: Take an infinite set S={1,1,1,1,....} Append another 1. Or a hundred of them, or an infinite number. You have the same set. Do that backwards, you have partitioned S into an infinite set and another set. Take an infinite set N={1,2,3,...} Append an infinite value. Or a hundred of them, or an infinite number. You have the same set. Do that backwards, you have partitioned N into an infinite set and another set. Give every guest in the Hilbert hotel a red key. Insert another infinite number of guests with blue keys, one at a time by shuffling the red-key guests up one room each step. You have partitioned N into an infinite blue-key set starting at 1 and an infinite red-key set starting at an infinite value.

To those who question where infinite-numbered balls come from: We are dealing with an infinite task that runs to completion. Whilst at every finite step the numbers on the balls are finite, at the end they must be infinite otherwise the task has not ended. If the numbers are finite then there are more of them to come - it's not midnight yet. Personally I have no problem accepting that infinite numbers appear after counting for an infinite number of steps. If you don't like thinking that N contains these infinite values when it's fully evaluated, then either: (A) the supertask can never end and the whole problem is impossible, or more interestingly (B) this concept of N is not the correct set to model the balls with, try using a set that includes infinite values because they are required if the task is to end. There was no stipulation in the puzzle statement that the balls be given finite numbers. If that requirement existed then the puzzle would be unsolvable - as I said earlier it'd be like counting your fingers without using the concept of 10.

A thought on convergence: If the puzzle is to have a meaningful solution, its state after infinite steps should be convergent in the sense that any further steps leave it unchanged. Having an empty set in the jug is not a convergent state because at the next step nine more balls are added, leaving it in a different state. However having an infinite set of infinite-numbered balls in the jug is convergent, because appending 10 infinite-numbered balls and removing 1 leaves the same infinite set, and appending the removed ball to the now-infinite 'discard' set leaves that set unchanged too.

If you try to model the infinite problem without using infinite numbers then the model will fail, leading to the paradoxical situation that because you can't describe the remaining balls, the only remaining assumption is that they don't exist.

If you allow infinite numbers in the model, then there really is no paradox.

kryptonaut wrote:↶ Take an infinite set S={1,1,1,1,....} Append another 1. Or a hundred of them, or an infinite number. You have the same set. Do that backwards, you have partitioned S into an infinite set and another set.

That is not an infinite set. That is a finite set. With one element. That element is 1.

This... is honestly like one of the first things that comes up when you learn what a set is? Like, possibly the second thing, after "order doesn't matter"? A set can only contain the same element once. You can add the same element multiple times, but the redundancy does nothing.

kryptonaut wrote:↶ Take an infinite set N={1,2,3,...} Append an infinite value. Or a hundred of them, or an infinite number. You have the same set.

You... absolutely do not have the same set.

N does not contain any elements which are infinite. "N with an infinite value appended" does. They are not the same. There exists an element in one that is not in the other. They are different things.

They can be mapped to each other, they're the same size, but they're not the same set.

If I snuck into your house and secretly replaced your bed with 0.6m^{3} of goat feces, would you be satisfied? It's the same size, so it must therefore be the same object!

kryptonaut wrote:↶ Give every guest in the Hilbert hotel a red key. Insert another infinite number of guests with blue keys, one at a time by shuffling the red-key guests up one room each step. You have partitioned N into an infinite blue-key set starting at 1 and an infinite red-key set starting at an infinite value.

These aren't even almost the same sets. On one hand, you have sets of either keys or people, depending on exactly which part of the description you're looking at, and on the other hand you have a set of numbers. These don't even have any elements in common, let alone being identical sets. It's also really bad customer service to refer to your hotel guests merely as numbers.

kryptonaut wrote:↶ We are dealing with an infinite task that runs to completion. Whilst at every finite step the numbers on the balls are finite, at the end they must be infinite otherwise the task has not ended.

This continues to be abject nonsense. I don't think I can find yet another way of trying to rephrase "N contains infinitely many numbers, but no number contained within N is itself infinite, this is not a contraction" in the hopes of it somehow breaking through, so I'll just say it again, but with a passive-aggressive framing device around it that gets kinda meta at the end.

kryptonaut wrote:↶ If the puzzle is to have a meaningful solution, its state after infinite steps should be convergent in the sense that any further steps leave it unchanged.

For certain types of objects in certain types of limits in certain types of situations, this is decent intuition! The sequence of numbers formed by iterating a continuous R->R function, for instance, if it converges, must converge to a fixed point of the function. In other situations, it doesn't hold. For example, when the function your iterating is discontinuous, (say, because you're operating on discrete objects like sets, and not a continuum like numbers). Oh hey, how about that, it's that thing that we're doing. Surprise!

kryptonaut wrote:Regarding set partitioning: Take an infinite set S={1,1,1,1,....} Append another 1. Or a hundred of them, or an infinite number. You have the same set. Do that backwards, you have partitioned S into an infinite set and another set. Take an infinite set N={1,2,3,...} Append an infinite value. Or a hundred of them, or an infinite number. You have the same set. Do that backwards, you have partitioned N into an infinite set and another set.

If you append an infinite value to N you do not have N anymore because N does not have infinite values.

I don't know what you mean by "do that backwards." There is no way to remove the last term of N because there is no last term.

You can partition N into an infinite set and another set. You cannot partition N in the way you are describing. There is no element x in N for which there are an infinite number of elements less than x. The only subset of N that has infinite sequential integers starting with 1 is N itself, and N - N is the null set.

kryptonaut wrote:Give every guest in the Hilbert hotel a red key. Insert another infinite number of guests with blue keys, one at a time by shuffling the red-key guests up one room each step. You have partitioned N into an infinite blue-key set starting at 1 and an infinite red-key set starting at an infinite value.

No. Since you cannot partition N in the way you are describing, your conclusion is not correct. Even if the hotel had infinite rooms, and even if we painted all finite rooms blue and all infinite rooms red, there is no way for a red key to ever be assigned a red room. The number of rooms available to red-keys is zero at midnight, and the set of rooms assigned to red keys is the null set.

kryptonaut wrote:To those who question where infinite-numbered balls come from: We are dealing with an infinite task that runs to completion. Whilst at every finite step the numbers on the balls are finite, at the end they must be infinite otherwise the task has not ended. If the numbers are finite then there are more of them to come - it's not midnight yet.

The jug being empty resolves this in a way that is consistent not only with itself and with every variation of the puzzle, but is also consistent with the mathematics - a quality that your conclusion lacks.

kryptonaut wrote:Personally I have no problem accepting that infinite numbers appear after counting for an infinite number of steps. If you don't like thinking that N contains these infinite values when it's fully evaluated, then either: (A) the supertask can never end and the whole problem is impossible, or more interestingly (B) this concept of N is not the correct set to model the balls with, try using a set that includes infinite values because they are required if the task is to end. There was no stipulation in the puzzle statement that the balls be given finite numbers. If that requirement existed then the puzzle would be unsolvable - as I said earlier it'd be like counting your fingers without using the concept of 10.

I pointed out earlier that in mathematics we can use something like {1, 2, 3, ...} as a shorthand for natural numbers. But even so, the balls don't have to be explicitly stated to have natural numbers since the sequence of times are only defined in a way that can be indexed to the natural numbers. So even if we had balls labeled with fractions or infinite numbers, we would never actually interact with those balls because there is no step defined for us to operate on them. Ucim has gone over this many times and in a better way than I have.

The only basis you have for claiming that the process must use infinite balls to end is that you have decided that the jug must have infinite balls in it and are looking for validation. Every bit of mathematics we can use leads us to the conclusion that the supertask ends and we have an empty jug.

kryptonaut wrote:A thought on convergence: If the puzzle is to have a meaningful solution, its state after infinite steps should be convergent in the sense that any further steps leave it unchanged.

Not only is this assertion untrue, but it's revealing of your approach to this problem that you are willing to make up a rule about convergence that we all must follow, while you ignore all of actual mathematical definitions of limits that say that the jug is empty.

kryptonaut wrote:If you allow infinite numbers in the model, then there really is no paradox.

Except for every contradiction with mathematics and set theory. There is also the problem that I've pointed out to you many times, which is that it requires the act of simultaneously removing balls to change the balls added. This is something you still have yet to resolve. I'll ask it again:

You are saying that removing balls somehow changes the balls that are added. I have yet to see any mathematical or even philosophical justification for that.

Since you may have forgotten what I'm asking you here:

A) When we only add balls ten at a time sequentially in order, we end up with exactly N in the jug at midnight. B) If we start with N in the jug, and remove balls one at a time sequentially in order, then we end up with an empty jug at midnight. C) YOUR CLAIM: If we do exactly what we did in (A), only we simultaneously remove balls one at a time sequentially in order in the same manner we followed in (B), then we end up with infinite balls in the jug with infinite labels on them at midnight.

If (C) is true, then the act of performing (B) simultaneously with performing (A) somehow adds more and different balls (N U J with J non-empty) than does performing (A) alone (only and exactly N). How does removing balls as in (B) change the balls added? This is some "Observer Effect"-level phenomenon that you have yet to justify.

EDIT:

phlip wrote:

kryptonaut wrote:↶ Take an infinite set S={1,1,1,1,....} Append another 1. Or a hundred of them, or an infinite number. You have the same set. Do that backwards, you have partitioned S into an infinite set and another set.

That is not an infinite set. That is a finite set. With one element. That element is 1.

I think this might be illustrative of why Cauchy's assertion that the balls in "remove the lowest" and "append-0" are the same set is misguided. If I put balls 1 and 2 in the jug, and then change the label on ball 2 to "1", I now have two balls with the same label on them. The only way to express this in set theory is to acknowledge that the balls are still {b_{1}, b_{2}} and the set of labels is {1}, but those sets are not equivalent.

Xias wrote:A) When we only add balls ten at a time sequentially in order, we end up with exactly N in the jug at midnight. B) If we start with N in the jug, and remove balls one at a time sequentially in order, then we end up with an empty jug at midnight. C) YOUR CLAIM: If we do exactly what we did in (A), only we simultaneously remove balls one at a time sequentially in order in the same manner we followed in (B), then we end up with infinite balls in the jug with infinite labels on them at midnight.

If (C) is true, then the act of performing (B) simultaneously with performing (A) somehow adds more and different balls (N U J with J non-empty) than does performing (A) alone (only and exactly N). How does removing balls as in (B) change the balls added? This is some "Observer Effect"-level phenomenon that you have yet to justify.

We have differing opinions on whether a fully instantiated infinite set of numbers includes infinite numbers or not. So whatever I answer here will only lead to us going round in circles.

If the puzzle is modeled as simply a count of balls in the jug (9n) and a lowest number (n+1) then the model indicates that you'll end up with an infinite number of infinitely numbered balls.

If the puzzle is modeled as balls laid out on a number line stretching away from us, at each step getting longer but further away, then the model indicates you'll end up with an infinitely long line infinitely far away, literally a dot on the horizon.

If the puzzle is modeled as an unnumbered set of balls, you end up with an infinite number of them.

If the puzzle is modeled with labeled balls where the labels are rewritten, you end up with an infinite number of infinitely long labels.

If the puzzle is modeled using a set of numbers which you define as finite, is it any surprise it leads to a paradox? Really, that is the problem here. The assumption that the numbers are all finite leads to a contradiction (that the set {n+1,n+2,...10n} is empty), therefore the assumption is incorrect.

kryptonaut wrote:We have differing opinions on whether a fully instantiated infinite set of numbers includes infinite numbers or not. So whatever I answer here will only lead to us going round in circles.

Then let's try a different tactic. We are not merely dealing with a generic "fully instantiated infinite set of numbers" - we are dealing with a particular set: the natural numbers. So let's talk about those. How do you define the natural numbers?

I'll respond to the rest of your post as well, but the point above is what I want to explore the most, because I feel like it might be a productive path to follow.

kryptonaut wrote:If the puzzle is modeled as simply a count of balls in the jug (9n) and a lowest number (n+1) then the model indicates that you'll end up with an infinite number of infinitely numbered balls.

The model is incomplete, because it ignores the fact that balls are removed.

kryptonaut wrote:If the puzzle is modeled as balls laid out on a number line stretching away from us, at each step getting longer but further away, then the model indicates you'll end up with an infinitely long line infinitely far away, literally a dot on the horizon.

The model is a poor description of the original puzzle, because in this model balls are never removed - they are endlessly renumbered.

kryptonaut wrote:If the puzzle is modeled as an unnumbered set of balls, you end up with an infinite number of them.

The model is incomplete, because the removal of the balls is ill-defined. We need to trace the path of each individual ball to determine how many are never removed.

kryptonaut wrote:If the puzzle is modeled with labeled balls where the labels are rewritten, you end up with an infinite number of infinitely long labels.

The model is a poor description of the original puzzle, because in this model balls are never removed - they are endlessly renumbered.

kryptonaut wrote:If the puzzle is modeled using a set of numbers which you define as finite, is it any surprise it leads to a paradox? Really, that is the problem here. The assumption that the numbers are all finite leads to a contradiction (that the set {n+1,n+2,...10n} is empty), therefore the assumption is incorrect.

If the puzzle is modeled using balls that are not just removed but systematically removed, is it any surprise the previous models do not necessarily hold?

(Edit: Also, what are you talking about? He never defined that set of numbers as finite.)

kryptonaut wrote:We have differing opinions on whether a fully instantiated infinite set of numbers includes infinite numbers or not. So whatever I answer here will only lead to us going round in circles.

And my opinion is consistent with set theory and abstract algebra, and yours is not. So the circle we've been in has been me and others trying to explain set theory to you, and you denying it.

kryptonaut wrote:If the puzzle is modeled using a set of numbers which you define as finite, is it any surprise it leads to a paradox? Really, that is the problem here. The assumption that the numbers are all finite leads to a contradiction (that the set {n+1,n+2,...10n} is empty), therefore the assumption is incorrect.

I have varying degrees of agreement and disagreement with the other conclusions you've asserted, but this one I'll address. It does not lead to a paradox in the "this statement is a lie" sense. It leads to a paradox in the "infinity doesn't match our intuition" sense. Hilbert's Hotel, in its original and unadulterated form, is the same sort of paradox: Rearranging the guests in a full hotel typically cannot open any additional rooms, let alone an infinite number of them. But I can mathematically show you that the set can be manipulated that way, and I can mathematically show you that the jug is empty.

It is not contingent on the labels. The labels are nothing but mechanisms that allow us to track individual balls and ensure that we remove them in a particular order. You could forgo labels entirely as long as you are able to track the balls. What matters is the steps, which are defined in such a way as to index an infinite number of balls and define a sequence with them. When you've done that, the index set for the balls is N.

If you had access to balls labeled with all of the real numbers, and then set up the task to index and sequence the balls labeled {1, 1/2, 1/3, 1/4, ...}, the outcome would be the same. If you instead set up the task to index and sequence the balls labeled {1, 1/2, 1/4, 1/8, ...} the outcome would be the same. The task could index and sequence the balls labeled {n^{1/2}: n ∈ N} the outcome would be the same. The task could index and sequence a random arrangement of balls, and so long as b_{n} was clearly defined, then the outcome would be the same. The particular labels don't matter.

What matters is that we have a sequence of times. There are only times defined for finite steps. There are infinitely many such steps. The index set is N. So our sequence is defined as {t_{i} : i ∈ N}. Then, when you add balls, you also index the balls. Merely adding balls without any removal gets us the sequence of balls {b_{n}: n ∈ N}. From there, we can define how we remove the balls. This is a subsequence {b_{n_k}}, where n_k is defined by our removal process. If we remove the lowest even ball, then n_k = 2k. If we remove the lowest odd ball, then n_k = 2k-1. If we remove the highest ball, then it happens to be n_k = 10k since we are adding balls ten at a time. If we remove the lowest ball, then n_k = k, and {b_{n_k}} = {b_{k}} = {b_{n}} - and it follows that the subsequence of removing balls is the same as the original sequence of adding balls. It doesn't matter what label b_{n} has on it.

Now we can specify what we are actually doing at every step. In the original variant, we add ten balls at a time. So at time t_{i}, we add the set of balls {b_{n} : n ∈ {10i-9, 10i-8, ... 10i}}. What is the union of all such sets for all i? {b_{n}: n ∈ N}. This is because the union of {10i-9, 10i-8, ... 10i} for all i in N is N. No matter how we label the balls, the set of all balls added at midnight is indexed by N!

Then we specify what we are removing. If we remove the highest ball, we end up with the ball removed at time t_{i} being ball b_{10i}. If we remove the lowest even ball, we get b_{2i}; lowest odd ball we get b_{2i-1}; lowest ball greater than 5 we get b_{5+i}. If we remove the lowest ball, we get b_{i}. Note that when I say "highest," "lowest," "lowest even," etc. I'm referring to the index, not the label. If we had labeled the balls with {1, 1/2, 1/4, 1/8}, then the ball labeled 1/4 would be an "odd" ball.

Now, at midnight, the set of all balls removed is as follows:

Spoiler:

Remove the highest: we take the union of all sets {10i} for all i, which gives us {10, 20, 30, ...}. Let's call this set H. Remove the lowest even: we take the union of all sets {2i} for all i, which gives us {2, 4, 6, ...}. Let's call this set E. Remove the lowest odd: we take the union of all sets {2i-1} for all i, which gives us {1, 3, 5, ...}. Let's call this set O. Remove the lowest number greater than 5: we take the union of all sets {5+i} for all i, which gives us {6, 7, 8, ...}. Let's call this set G. Remove the lowest: we take the union of all sets {i} for all i, which gives us {1, 2, 3, ...}. Let's call this set N, since that's what it is.

So what we have now is a set of balls added which is equal to {b_{n}: n ∈ N}, and a set of balls removed which is equal to {b_{n}: n ∈ D}, where D is whichever subset of N we got in the previous spoiler. Then the set of balls remaining in the jug at midnight is equal to {b_{n}: n ∈ D^{c}}. You'll find that D^{c} is, in the following cases:

Spoiler:

Remove the highest: D^{c} = H^{c} = {1, 2, ..., 9, 11, 12, ... 19, 21, 22, ...}. Remove the lowest even: D^{c} = E^{c} = {1, 3, 5, ...}. Remove the lowest odd: D^{c} = O^{c} = {2, 4, 6, ...}. Remove the lowest number greater than 5: D^{c} = G^{c} = {1, 2, 3, 4, 5}. Remove the lowest: D^{c} = N^{c} = {}.

Again, the numbers in each of these sets are the indices of the balls in the jug, not the labels. {b_{n}: n ∈ E^{c}} when b_{n} = 1/n, is equal to {1, 1/3, 1/5, 1/7, ...}. When b_{n} = n^{2}, we get {1, 4, 9, 16, ...} edit: the complement of this set, which is {2, 3, 5, 6, 7, 8, 10, ...}. And so on. We just happen to, in the original game, treat b_{n} as equal to n.

It shouldn't take much convincing to tell you that if we remove the highest, that we end up with balls 1-9, 11-19, etc., or all balls not divisible by 10 (since we took out every ball divisible by 10). It also shouldn't take much convincing to tell you that if we remove even balls we get odd balls, and vice versa. This is consistent with the partitioning of N into those sets, as you have pointed out. When we look at D^{c} in the greater-than-5 game or the remove-the-lowest game, however, your reasoning leads us to believe that the set of balls added is not equal to D^{c}. So either something magical happens in those games, which you have yet to justify (and I'm certain that you can't), or the set of balls remaining is not equal to D^{c} in any game, which you have yet to justify (and I'm certain that you can't).

The following is a note on the append-0 game. This is not for you, kryptonaut, since you refuse to accept any of the previous arguments. But it is for completeness, and to show that the apparent contradiction is not a contradiction at all.

Spoiler:

Despite appending a 0 to the label at each step, the balls added are the exact same as the balls remaining in the jug in the remove-the-highest game. At every step, we add {b_{n}: n ∈ H}. At every step, we remove nothing. At every step, we take a ball and append a 0 to the label.

If you believe, as Cauchy does, that changing the label changes the index, then I disagree. We are merely changing the label to match the ball of a different index. Treating the label as the index in this way introduces inconsistencies with changing labels in other ways (such as adding dots), and seems to be dependent on word play and trickery based on the arbitrary association of the digits 1 and 0 with the number ten.

What is consistent is to treat the ball b_{1} as b_{1} always throughout the supertask, and to consider the action performed at step i as follows: For i not divisible by 10, append a 0 to the label on ball b_{i}. For i divisible by 10, i = 10j, append a 0 to the label on ball b_{j}. This not only ends up with a jug at all finite steps that is consistent with what we would expect, but it makes it clear that we end up with balls {b_{n}: n ∈ H^{c}} remaining in the jug.

The label on any given ball can be defined by a function l(b_{n},i) from {b_{n}: n ∈ N}×N to R. Then l(b_{1},i) = {10^{k} when 10^{k}<=i<=10^{k+1}}. l can be defined for each b_{n} in a similar way. So what label is on a given b_{n} at midnight? The answer is lim_{i→infinity}l(b_{n},i). Since l is divergent, this is undefined, infinity, etc. We can interpret that limit to mean the labels are "a finite sequence of digits followed by an infinite sequence of 0s" just because that's the best interpretation of the limit in this case, but really the limit is undefined.

If instead we replace "append a 0" with "add a dot" then the label is no longer defined by the same function l (since the range is no longer R, but some description of a number of dots). We can look at the "number of dots" function d(b_{n},i) from {b_{n}: n ∈ N}×N to R, though, and find that it also has a limit of infinity. It is also equivalent to the number of 0s appended, which further supports the interpretation of the limit: we get every natural number not divisible by 10, with an infinite number of dots.

This does not contradict the solution to the original, remove-the-lowest game. Since the labels do not change in that game, lim_{i→infinity}l(b_{n},i) = n, and lim_{i→infinity}d(b_{n},i) = 0 (since we do not append any dots or 0s). There is no reason to take the limit of some arbitrary variable like "the index of the lowest ball in the jug" because the set at midnight is not dependent on that limit.

Poker wrote:(Edit: Also, what are you talking about? He never defined that set of numbers as finite.)

Kryptonaut means a set of numbers, for which all numbers in the set are defined as finite. Not a finite set of numbers.

EDIT: A further note on this:

kryptonaut wrote:The assumption that the numbers are all finite leads to a contradiction (that the set {n+1,n+2,...10n} is empty), therefore the assumption is incorrect.

Since the limit of the sequence of sets {n+1,n+2,...10n} as n approaches infinity is actually empty, the contradiction you are trying to point out doesn't actually exist. So the assumption is not disproven by contradiction. On the other hand, assuming that the set is nonempty leads to a contradiction (since the limit of the sequence of sets is empty) and therefore that assumption must be incorrect, right?

Or are you merely asserting that the math is wrong if it contradicts your intuition? That's not how proof by contradiction works.

Again, if you don't understand limits supremum and infimum, that's one thing and it's totally fine. Denying them with confidence is not going to get you anywhere.

You said you're a physicist, right? Try this: "Special Relativity assumes that the speed of light is constant c relative to all frames of reference, and that assumption leads to a contradiction (that if I were moving at c/2, the speed of light would still be c relative to me rather than c/2). Therefore, the assumption is incorrect." What do you say to that other than repeat over and over that the speed of light in a vacuum is a universal constant in all reference frames?

kryptonaut wrote:Take an infinite set S={1,1,1,1,....}

That's not an infinite set. That's an infinite amount of drivel to describe a finite set - a set with one element, to wit: the first counting number.

kryptonaut wrote:Take an infinite set N={1,2,3,...} Append an infinite value. [...] You have the same set.

You now do not have set N. You have set K. The cardinality of the two sets is the same (aleph null) but the two sets are not identical, to wit: one has an infinite value as an element, the other does not. You could do the same thing with emoji or ham sandwiches.

Sets and partitions just do not work this way.

"To the question of where infinite-numbered balls come from:"

kryptonaut wrote:Whilst at every finite step the numbers on the balls are finite, at the end they must be infinite otherwise the task has not ended.

No. At the end there are an infinite number of balls, all of them finite. There are, after all, an infinite number of finite numbers, and there is no highest finite number.

There is no highest number yet you can still have all of them in a set.

Similarly, the set {1/1, 1/2, 1/3, 1/4...} all contain only finite numbers less than or equal to 1, but strictly greater than zero. If you put all of them in a set (set F, to give it a name), zero is not in that set. And if you devise a supertask to count them all and line them up one by one and paint them red and make them juggle monkeys, zero will still not be in that set.

kryptonaut wrote:If you don't like thinking that N [the set of natural numbers] contains these infinite values...

... then you are not doing math, because N is defined to be the set of positive (or in some cases non-negative) finite integers. If you start with that definition, and then "don't like it", well, then you don't like it. Not liking it doesn't make it false.

Again, stop with the jugs and labels and dots and all the other stuff that obfuscates the problem you are having. Go back to partitioning {1,2,3...} into L and R, and into L, C, and R. Stick with the simplest case that shows the issue if you want to either understand it or convince us.

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 wrote:We have differing opinions on whether a fully instantiated infinite set of numbers includes infinite numbers or not. So whatever I answer here will only lead to us going round in circles.

kryptonaut wrote:If you don't like thinking that N contains these infinite values when it's fully evaluated [...]

OK then, let's make this simple. Name one.

Name one infinite number which is an element of N, the set of all positive integers.

Poker wrote:Then let's try a different tactic. We are not merely dealing with a generic "fully instantiated infinite set of numbers" - we are dealing with a particular set: the natural numbers. So let's talk about those. How do you define the natural numbers?

I ask "why are we dealing with an infinite set of finite numbers if the goal is to disprove the intuitive view that we end up with infinitely big numbers?" If you want to convincingly disprove that view then you can't start out by assuming it's false.

Poker wrote:

kryptonaut wrote:If the puzzle is modeled as simply a count of balls in the jug (9n) and a lowest number (n+1) then the model indicates that you'll end up with an infinite number of infinitely numbered balls.

The model is incomplete, because it ignores the fact that balls are removed.

Really? In what way does saying that the lowest number is n+1 at step n ignore the fact that balls 1 to n have been removed at that step? And how does saying the balls in the jug correspond to a set {n+1,n+2,...10n} express that in any different manner? The difference is that in the set-theory analysis you have started out assuming that all the numbers are finite - and then, surprise! - there are no infinite numbers in the answer. Well you haven't really proved anything there.

Xias wrote:And my opinion is consistent with set theory and abstract algebra, and yours is not. So the circle we've been in has been me and others trying to explain set theory to you, and you denying it.

Your reasoning may well be correct within a framework of strictly finite numbers. But I suggest that's an inappropriate framework for this problem.

Xias wrote:

kryptonaut wrote:If the puzzle is modeled using a set of numbers which you define as finite, is it any surprise it leads to a paradox? Really, that is the problem here. The assumption that the numbers are all finite leads to a contradiction (that the set {n+1,n+2,...10n} is empty), therefore the assumption is incorrect.

I have varying degrees of agreement and disagreement with the other conclusions you've asserted, but this one I'll address. It does not lead to a paradox in the "this statement is a lie" sense. It leads to a paradox in the "infinity doesn't match our intuition" sense. Hilbert's Hotel, in its original and unadulterated form, is the same sort of paradox: Rearranging the guests in a full hotel typically cannot open any additional rooms, let alone an infinite number of them. But I can mathematically show you that the set can be manipulated that way, and I can mathematically show you that the jug is empty.

Well, it leads to a paradox in the '9n=0 for some finite n>0' sense. You can mathematically show that the jug is empty of finite-numbered balls.

Your discussion on the unimportance of the numbers themselves - that it's the ordering and indices that count - is all valid. The ordinals are what matters. But don't you see that that very argument also applies to the expanding-label scenario? The label "100" or "1 with two dots" or whatever you like to write it as, is treated as an ordinal that places it one place higher than the label "99". It has ordinal 100 in this ordering of the infinite set of possible labels. When it acquires another zero or dot or whatever, it has ordinal 1000. What happens when it gets infinitely long? Well, I'd say it has ordinal ω plus something, giving it an infinite (but orderable) value. I don't think it ceases to exist.

Exactly the same sequence of ordinals that applies to the labels also applies to the balls with regular numbering as specified in the puzzle.

The 'n' that indexes the steps is an ordinal - first step, second step, third step, etc. At midnight, I contend that n actually takes on the value ω - we have finally finished performing an infinite number of steps. That's what ω is, it's the ordinal that comes after all the finite ordinals. Now you can simply say that the balls in the jug have ordinals ω+1 to ω+9ω which equates to an infinite set (mappable to N if you want to) of infinite numbered balls. The first one is a number so high that you'll never count up to it.

ucim wrote:That's not an infinite set. That's an infinite amount of drivel to describe a finite set - a set with one element, to wit: the first counting number.

Yeah, I wish I hadn't written that particular example now

ucim wrote:Again, stop with the jugs and labels and dots and all the other stuff that obfuscates the problem you are having. Go back to partitioning {1,2,3...} into L and R, and into L, C, and R. Stick with the simplest case that shows the issue if you want to either understand it or convince us.

I have tried to illustrate my thoughts with many alternative scenarios and have been told not to deviate from the original puzzle. Discussing a model where you start out with an assumption that all the numbers taking part are finite will not help to prove or disprove the assertion that infinite numbered balls end up in the jug.

----------------------------------------

Model the last discarded ball as n, the number of balls in the jug as 9n, the lowest number in the jug as n+1. Let n->∞. You get infinite balls in the jug, with infinitely big numbers.

If you try to model that with a set of finite numbers and discover that no finite number is left in the jug, all you have proved is that your model contains only finite numbers.

Poker wrote:Then let's try a different tactic. We are not merely dealing with a generic "fully instantiated infinite set of numbers" - we are dealing with a particular set: the natural numbers. So let's talk about those. How do you define the natural numbers?

I ask "why are we dealing with an infinite set of finite numbers if the goal is to disprove the intuitive view that we end up with infinitely big numbers?" If you want to convincingly disprove that view then you can't start out by assuming it's false.

Why are we dealing with an infinite set of finite numbers? Because the set of balls we start out with is an infinite set of finite numbers, by the very definition of the original problem. If balls with infinitely-big numbers exist, they have to come from somewhere. In the append-0 problem (or other variations) they come from infinitely relabeling numbers. What mechanism exists for creating infinite numbers in the original problem?

kryptonaut wrote:

Poker wrote:

kryptonaut wrote:If the puzzle is modeled as simply a count of balls in the jug (9n) and a lowest number (n+1) then the model indicates that you'll end up with an infinite number of infinitely numbered balls.

The model is incomplete, because it ignores the fact that balls are removed.

Really? In what way does saying that the lowest number is n+1 at step n ignore the fact that balls 1 to n have been removed at that step? And how does saying the balls in the jug correspond to a set {n+1,n+2,...10n} express that in any different manner? The difference is that in the set-theory analysis you have started out assuming that all the numbers are finite - and then, surprise! - there are no infinite numbers in the answer. Well you haven't really proved anything there.

The difference is, the count of balls in the jug (9n at every finite step) is not a most basic fact. The presence or absence of any given ball is.

kryptonaut wrote:

Xias wrote:And my opinion is consistent with set theory and abstract algebra, and yours is not. So the circle we've been in has been me and others trying to explain set theory to you, and you denying it.

Your reasoning may well be correct within a framework of strictly finite numbers. But I suggest that's an inappropriate framework for this problem.

Again, because by the very definition of the original problem all the balls start out with only finite numbers, it is not only an appropriate framework, it is the most appropriate framework, at least for how the ball start. If you want the balls to drift beyond that framework, you need to provide a mechanism for producing them, one that can exist within the original problem as stated, where balls are not created or altered in any way.

kryptonaut wrote:

Xias wrote:

kryptonaut wrote:If the puzzle is modeled using a set of numbers which you define as finite, is it any surprise it leads to a paradox? Really, that is the problem here. The assumption that the numbers are all finite leads to a contradiction (that the set {n+1,n+2,...10n} is empty), therefore the assumption is incorrect.

I have varying degrees of agreement and disagreement with the other conclusions you've asserted, but this one I'll address. It does not lead to a paradox in the "this statement is a lie" sense. It leads to a paradox in the "infinity doesn't match our intuition" sense. Hilbert's Hotel, in its original and unadulterated form, is the same sort of paradox: Rearranging the guests in a full hotel typically cannot open any additional rooms, let alone an infinite number of them. But I can mathematically show you that the set can be manipulated that way, and I can mathematically show you that the jug is empty.

Well, it leads to a paradox in the '9n=0 for some finite n>0' sense.

Which is equivalent to the "infinity doesn't match our intuition" sense when you think about it. Going back to Hilbert's Hotel, if you multiplied each guest's room by 10, you could fit 9 additional guests for every 1 original guest, so the number of new guests would be 10n, but it's the same as n, so 10n=n, so subtracting n from both sides, 9n=0. (The whole subtracting n from both sides might not be a valid step, but if it's not a valid step for me, then it's not a valid step for you either, thus resolving the paradox anyway.)

kryptonaut wrote:You can mathematically show that the jug is empty of finite-numbered balls.

Your discussion on the unimportance of the numbers themselves - that it's the ordering and indices that count - is all valid. The ordinals are what matters. But don't you see that that very argument also applies to the expanding-label scenario? The label "100" or "1 with two dots" or whatever you like to write it as, is treated as an ordinal that places it one place higher than the label "99". It has ordinal 100 in this ordering of the infinite set of possible labels. When it acquires another zero or dot or whatever, it has ordinal 1000. What happens when it gets infinitely long? Well, I'd say it has ordinal ω plus something, giving it an infinite (but orderable) value. I don't think it ceases to exist.

Exactly the same sequence of ordinals that applies to the labels also applies to the balls with regular numbering as specified in the puzzle.

The 'n' that indexes the steps is an ordinal - first step, second step, third step, etc. At midnight, I contend that n actually takes on the value ω - we have finally finished performing an infinite number of steps. That's what ω is, it's the ordinal that comes after all the finite ordinals. Now you can simply say that the balls in the jug have ordinals ω+1 to ω+9ω which equates to an infinite set (mappable to N if you want to) of infinite numbered balls. The first one is a number so high that you'll never count up to it.

The difference is, in the expanding-label scenario, there is a mechanism for producing these infinite labels. If such a method exists in the original scenario, you have not made that method clear to us.

kryptonaut wrote:

ucim wrote:Again, stop with the jugs and labels and dots and all the other stuff that obfuscates the problem you are having. Go back to partitioning {1,2,3...} into L and R, and into L, C, and R. Stick with the simplest case that shows the issue if you want to either understand it or convince us.

I have tried to illustrate my thoughts with many alternative scenarios and have been told not to deviate from the original puzzle.

The reason is that your alternative scenarios come with ready-made ways to modify the labels, thus allowing them to get infinitely-big in the first place. Everyone accepts this. What none of us has accepted so far is that such a method exists for the original problem, where the labels do not change by definition.

kryptonaut wrote:Discussing a model where you start out with an assumption that all the numbers taking part are finite will not help to prove or disprove the assertion that infinite numbered balls end up in the jug.

I'm repeating myself in this post, but that is exactly how the original problem is stated - as us starting with a set of finite numbers. Now, it is not necessarily impossible for balls with infinite numbers to suddenly appear, but if they do, they must have come from somewhere, since they did not start out existing. Any expanding-label variation has an obvious method for creating them, by applying the label-expansion. In the original problem and the original problem only, what mechanism is used to create balls with infinite labels?

Poker wrote:In the original problem and the original problem only, what mechanism is used to create balls with infinite labels?

By starting an infinite task and actually finishing it. We set out to count to infinity and we got there. We didn't just tend towards it, we really did it. The whole thing. Done.

Impossible you say? Well then your watch stopped before midnight.

Poker wrote:In the original problem and the original problem only, what mechanism is used to create balls with infinite labels?

By starting an infinite task and actually finishing it. We set out to count to infinity and we got there. We didn't just tend towards it, we really did it. The whole thing. Done.

Impossible you say? Well then your watch stopped before midnight.

And how exactly did that cause new balls to come into existence?

kryptonaut wrote:I have tried to illustrate my thoughts with many alternative scenarios and have been told not to deviate from the original puzzle.

Mainly because your deviations are complications and obfuscations. If you want to make progress (either in understanding or in convincing) you have to reduce down to the absolute simplest problem that still displays the interesting behavior. This prevents distraction, and right now you (and all of us) are very distracted.

The simplest scenario that displays the interesting behavior (our differing understanding of the result) is the pure LR and LCR partition of the set of natural numbers N. In this scenario, you agree that R ends up empty, but U (the union of C and R) contains an element that was never in the original set that was partitioned. If you can rigorously explain where this ham sandwich came from, all the rest of the scenarios will be solved. If you cannot, all the rest of the scenarios will still contain this same misunderstanding under the hood. This is what you need to concentrate on. Nothing else.

kryptonaut wrote:Discussing a model where you start out with an assumption that all the numbers taking part are finite will not help to prove or disprove the assertion that infinite numbered balls end up in the jug.

But that is the crux of the question. It's not an "assumption". It's a stated fact of the original conditions: "the set of natural numbers", which is a set with infinite cardinality, all of whose members are finite. Further, they all have finite ordinality to boot, since the supertask moves the balls (or partitions the set) in the natural order of the natural numbers. That is the original problem, as stated. Given this original problem, I (and others here) contend that neither infinite numbers, nor emoji, nor ham sandwiches, are created during the supertask.

Now, either they are (and I'm wrong), or they aren't (and you're wrong).

If they aren't, then the jug (or set C) must be empty at the end, even though there's no step where that happens, since there are an infinite number of steps.

If they are, then how is it that turkey sandwiches and purple rain aren't also created?

kryptonaut wrote:By starting an infinite task and actually finishing it. We set out to count to infinity and we got there. We didn't just tend towards it, we really did it. The whole thing. Done.

Yanno, when I say N, the set of all natural numbers, I also "got there". Every natural number is in the set, instantly. The whole thing. Done. And there isn't a non-finite number in the set. And when I go through them, one by one (increasingly fast), when I'm done, the set hasn't changed. This means there still isn't a non-finite number in the set.

kryptonaut wrote:If you try to model that with a set of finite numbers and discover that no finite number is left in the jug, all you have proved is that your model contains only finite numbers.

Well, that doesn't even need to be proven if it's defined as the starting condition (which it is). Since this is true, the jug (or set C) must be empty, despite people's mistaken intuition about infinity that is derived from experience with finite entities. There is no actual paradox. What there is is a failure of intuition. That is enlightenment. That is the first step in getting one's head around "never stop counting".

(And we haven't even begun to discuss the big infinities.)

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.

The puzzle starts "Suppose I have infinitely many balls, numbered 1,2,.. and so on."

It says nothing about finite numbers, or natural numbers. Just "infinitely many balls". If you actually count them all, it's no surprise you get an infinite number. If you try to model it using finite numbers then it's no surprise that it falls over.

kryptonaut wrote:The puzzle starts "Suppose I have infinitely many balls, numbered 1,2,.. and so on."

It says nothing about finite numbers, or natural numbers. Just "infinitely many balls". If you actually count them all, it's no surprise you get an infinite number. If you try to model it using finite numbers then it's no surprise that it falls over.

I'll put aside for the moment the fact that "1,2,.. and so on" is basically always used to represent the natural numbers. I'll also put aside for the moment the fact that the only way it "falls over" is with regard to intuition based on sets of finite cardinality.

Let me see if I have your argument straight.

A. There are infinitely many balls. B. If you count all of the balls, you get an infinite number of them.

For your argument to be relevant, it would need to continue something like the following:

C. Every ball is numbered with the number you are at when you count it. D. Therefore, the number of balls, which is infinite, must be included within the infinitely many balls.

If I have your argument wrong somewhere let me know, but I'll proceed assuming this is essentially your argument. Now, watch what happens if I replace all of the balls with natural numbers:

A'. There are infinitely many natural numbers. B'. If you count all of the natural numbers, you get an infinite number of them. C'. Every natural number is numbered with the number you are at when you count it. D'. Therefore, the number of natural numbers, which is infinite, must be included within the infinitely many natural numbers.

D' is false, while A'-C' are true. Your argument does not necessarily hold. The only way for you to assume your argument is in any way valid is to assume that from the beginning, we have more than the natural numbers.

And now I return back to what I said earlier in the post when I said:

above I wrote:I'll put aside for the moment the fact that "1,2,.. and so on" is basically always used to represent the natural numbers.

You claim this is not one of those times. Are you making that claim just because the consequences of it being the natural numbers go against your finite-set-based intuition? Or is there some more concrete reason?

kryptonaut wrote:Well, it leads to a paradox in the '9n=0 for some finite n>0' sense.

My set-based proof from many pages ago resolves this paradox.

The intuition is that since the cardinality of the set of balls in the jug is increasing without bound - and is in fact 9n for all finite n - that the cardinality of the set at the limit must be the same as the limit of the cardinality. Put formally, I will call this Kryptonaut's Conjecture:

Kryptonaut's Conjecture: For any sequence of sets {A_{n}}_{n=1→infinity,} |lim_{n→infinity} A_{n}| = lim_{n→infinity} |A_{n}|.

If Kryptonaut's Conjecture is true, then we have: |A_{n}| = f(n) ⇒ |lim_{n→infinity} A_{n}| = lim_{n→infinity}f(n). |J_{n}| = 9n |lim_{n→infinity} J_{n}| = lim_{n→infinity} 9n = infinity.

I will show that Kryptonaut's Conjecture is not true, by contradiction.

Assume that Kryptonaut's Conjecture is true. Consider the sequence of sets {B_{n}}_{n=1→infinity,}, where B_{n} = {n+1, n+2, n+3, ... , 10n}. lim_{n→infinity} B_{n} = {}. Therefore, |lim_{n→infinity} B_{n}| = 0. |B_{n}| = 9n. Then, by Kryptonaut's Conjecture, |lim_{n→infinity} B_{n}| = lim_{n→infinity} 9n = infinity.

This is a contradiction. Therefore, Kryptonaut's Conjecture is false.

I just showed that, without any consideration for balls and jugs, a sequence of sets with cardinality that increases without bound does not necessarily have a cardinality of infinity at the limit. So, when we consider balls and jugs, what rule exactly are you saying that an empty jug contradicts? It can't be Kryptonaut's Conjecture, since that isn't a rule at all.

In fact, there is no contradiction, and the paradox has been resolved.

Perhaps you don't believe me that lim_{n→infinity} B_{n} = {}, despite that being the mathematical result of taking the limit. Then I'll try again:

Assume that Kryptonaut's Conjecture is true. Consider the sequence of sets {B_{n}}_{n=1→infinity,}, where B_{n} is an increasing set of numbers greater than (1/2)^{n} and less than (9/10)^{n} such that |B_{n}| = n. We can choose any such numbers, but let's pick in the following manner:

What this does is choose n equally spaced points between (1/2)^{n} and (9/10)^{n}, exclusive.

Since at the limit, (1/2)^{n} and (9/10)^{n} are both equal to 0 (thanks Demki), and there are no points that are both greater than 0 and less than 0, then it follows that lim_{n→infinity} B_{n} = {}. This is also true by liminf and limsup. Therefore, |lim_{n→infinity} B_{n}| = 0.

|B_{n}| = n. Then, by Kryptonaut's Conjecture, |lim_{n→infinity} B_{n}| = lim_{n→infinity} n = infinity.

This is a contradiction. Therefore, Kryptonaut's Conjecture is false.

So once again, Kryptonaut's Conjecture is not a justification for a contradiction, there is no contradiction, and the paradox has been resolved.

kryptonaut wrote:The puzzle starts "Suppose I have infinitely many balls, numbered 1,2,.. and so on."

It says nothing about finite numbers, or natural numbers. Just "infinitely many balls". If you actually count them all, it's no surprise you get an infinite number. If you try to model it using finite numbers then it's no surprise that it falls over.

The way that we count them is prescribed by the puzzle. We place them ten at a time in order at specified times. Those times are specified to be midnight - 20*(1/2)^{n}. This is only defined (mathematically) for finite n, of which there are infinitely many. We don't have a step at midnight (infinite n), nor do we have balls with infinite numbers on them (infinite n).

The limit may be midnight, but we know that because we follow the rules of limits - and the same rules of limits are the rules that tell us that the jug is empty.

Last edited by Xias on Sat Dec 10, 2016 4:43 am UTC, edited 1 time in total.

Xias wrote:Since at the limit, (1/2)^{n} and (9/10)^{n} are both equal to 1...

You mean 0

@kryptonaut: "1,2... and so on" is used to refer to the set of natural numbers greater than 0. If people want to specify some set that contains infinitely big numbers(such as the superreals), they say so explicitly. If people want to specify some proper class that contains infinitely big numbers(such as the class of ordinals and the class of cardinals), they say so explicitly. If they don't say that explicitly, they are using non-standard notation without informing about it, and are communicating badly.

Xias wrote:Since at the limit, (1/2)^{n} and (9/10)^{n} are both equal to 1...

You mean 0

Yes, thanks for that.

Demki wrote:@kryptonaut: "1,2... and so on" is used to refer to the set of natural numbers greater than 0. If people want to specify some set that contains infinitely big numbers(such as the superreals), they say so explicitly. If people want to specify some proper class that contains infinitely big numbers(such as the class of ordinals and the class of cardinals), they say so explicitly. If they don't say that explicitly, they are using non-standard notation without informing about it, and are communicating badly.

I want to reiterate that how we are interpreting the statement "1, 2, and so on" doesn't actually matter. Even if the problem specified that we had some extra balls lying around with infinitely large numbers or fractions or Ucim's missing ham sandwiches, the supertask would never actually call on those numbers because there is no infinitely large or fractional or edible step that we take.

I want that clarity because Kryptonaut has, a few times now, implied that our conclusion is based on us just not having balls with infinitely large numbers on them. As though we might all grant that the jug had infinite balls at midnight if only we had them in the first place, but gosh darn it the puzzle didn't give them to us! That is not the argument.

Xias wrote:I want to reiterate that how we are interpreting the statement "1, 2, and so on" doesn't actually matter. Even if the problem specified that we had some extra balls lying around with infinitely large numbers or fractions or Ucim's missing ham sandwiches, the supertask would never actually call on those numbers because there is no infinitely large or fractional or edible step that we take.

Thanks - that's good to keep in mind. However, the simplest version that displays the "interesting behaviour" is N. If it's not there to begin with, it can't be there in the end, and if that gives xir trouble, the more subtle case of unreachable numbers will not be any easier.

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.

ucim wrote:However, the simplest version that displays the "interesting behaviour" is N. If it's not there to begin with, it can't be there in the end, and if that gives xir trouble, the more subtle case of unreachable numbers will not be any easier.

Jose

Don't get me wrong; I'm all about reducing the puzzle to the simplest terms that still illustrate the behavior.

However, what if there were some convention we all were using to say that there were no balls labeled greater than 10? Then there would be merit to arguing that either the task creates balls greater than 10, or that it fails after step 1 because there are no balls to add, or that the jug being empty at step 10 is merely a trick based on our weird restriction. So then when we argue that there exist only finite-numbered balls to work with, how do we show that it's not the same thing?

Only having natural numbered balls is not a necessary condition of the jug being empty, so I think that passionately defending that point only clouds what is really going on, rather than illuminating it.

Below I present a method which I believe solves all the supertask puzzles. First, a resolution of the paradox:

Poker wrote:The difference is, the count of balls in the jug (9n at every finite step) is not a most basic fact. The presence or absence of any given ball is.

This is the key to resolving the paradox:

The limit of the count of the balls in the jug does not necessarily equal the count of the limit of the balls in the jug.

More generally, for any sequence of sets An, which has a set-theoretic limit lim(An), and any function f:

lim(f(An)) does not necessarily equal f(lim(An))

Take the simplest case, where balls are moved one at a time from jug A to jug B where at step n ball n is moved. After the supertask, every ball has been moved. Jug A is empty and jug B contains all the balls.

But at each step n, jug A still contains an infinite number of balls: {n+1, n+2, ...}. The limit of a sequence of infinite numbers is an infinite number: it cannot be zero. But this paradox is resolved by realising that the limit of the sizes of a sequence of sets is not necessarily the same as the size of the limit.

So there is no contradiction with jug A containing infinitely many balls on every step, but being empty at the end of the supertask. Similarly, there is no contradiction in the original puzzle with a jug containing an increasing finite number of balls on each step and ending up empty.

To calculate the output of any supertask it is sufficient to break down the supertask into a set of finite tasks running in parallel. A "finite task" is a task which completes within a finite number of steps and then does nothing for the remaining infinite number of steps.

(1) For the original puzzle: On step n, ball n has been moved into the jug and is then moved out, and is never moved again. So the ball ends up outside the jug. No balls other than finitely numbered balls are ever moved into the jug. The "finite task" for each ball is to move it into the jug, move it out again, and then do nothing for the remaining (infinite) number of steps. Every ball ends up outside the jug. So the jug ends up empty. No labels are modified, so the "finite task" for each label is "do nothing".

(2) For the "relabel the balls" version of the puzzle: Each ball is placed into the jug and never removed. So at the end of the task, the jug contains ω balls. The labels on the balls never settle down to a final, unchanged, value. So the operation applied to each label is itself a supertask. To break down this supertask into finite tasks we need to model each label as a set and examine the elements of this set: (2a) If we model the label as a number, then the (finite) label n is the set {0, 1, 2, ..., n-1}. On each step, the label on the lowest-numbered ball is multiplied by 10. This adds a load of numbers to the set forming the label. These numbers are added and then never removed. Consider the label on the ball originally labelled m. For each finite number n there is a step at which n is added (or was originally present) to the label in ball m and is never subsequently removed. Each of these tasks (adding the number n to the label on the ball originally labelled m) is a finite task. At the end of the supertask, every ball has a label consisting of the set of all finite numbers. This set is the ordinal number ω. So at the end of the task we have ω balls each labelled ω. (2b) If we model the label as a sequence of digits, then on each step we shift the digits to the left and set the lowest order digit to zero. For each digit n and each ball m at some finite point in the process, this digit on this ball is set to zero and never subsequently changed. So at the end of the process we have ω balls each labelled with an infinite sequence of zeros. (2c) To my mind, neither of (2a) and (2b) is an accurate model of the operation on the ball which is described as "appending a zero to the label". This description implies that the existing label is not overwritten, but something is "tacked on" to the end of it. To model "appending a zero" we need two sequences of digits. One sequence runs to the left and starts out storing a representation of the original number on the ball. The other sequence runs to the right and starts out empty. The "right sequence" represents the zeros that are appended. At some finite point in the supertask, for ball m the left sequence still contains a digit representation of the number m, while the right sequence has a zero added in the nth position. This zero is never removed. Adding this zero is a finite task. At the end of the process, therefore, we have ω balls each labelled with a different finite number in the left sequence and an infinite sequence of zeros in the right sequence. To summarise: the operation applied to the left sequence is a finite task: "do nothing". The operation applied to each digit of the right part of the label is also a finite task: at some finite point, the digit is set to zero and subsequently never changed. So the effect of the supertask is to set every one of the infinite sequence of digits on the right to zero. (Mathematically a sequence of digits is a function from an ordinal number to the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. The right sequence starts out as the empty set {}, then becomes { 0 |=> 0 }, then { 0 |=> 0, 1 |=> 0}, and so on. For every integer n, at some point the mapping n |=> 0 is added to the set and never subsequently removed).

(3) We can model Thomson's lamp by adding ball 1 to the jug on step 2n-1 and removing it on step 2n. For ball 1 there is no final resting place after which it is never moved. But this supertask cannot be divided into finite tasks, so the supertask cannot be computer, and so the result is indeterminate. Logically, the position of the ball on step ω is not determined by the set of previous steps.

(4) For Hilbert's hotel where a guest is moved infinitely often (say, from room n to some room greater than n on each step): the guest's location never settles down to a fixed value, so the supertask of detemining the guest's final resting place is undefined. However, in a modern hotel the clerk hands the guest a keycard and a piece of paper with their room number written on it. Suppose the clerk computes the supertask first by writing on the paper, instead of repeatedly moving the guest. In this case, we can compute the result which appears on the paper handed to the guest: At each stage the number on the paper is the set {0, 1, 2, ..., n-1}. The clerk adds n to this set, then n+1 and so on. The finite task for any given element m is that m is eventually added to the set (which is the room number written on the paper) and is never removed. At the end of the supertask, the set is {0, 1, 2, ...} which is precisely the number ω. The guest reads the paper and asks "where is room number ω?" The clerk reads the paper and realises he has made a mistake: there is no "room number ω" in the hotel. So the clerk has to find a different solution to his guest allocation problem.

I think that this method can be applied to any supertask: and will either calculate the result, or will prove that the result is indeterminate. I would love to be proved wrong, however, so if you have a counterexample, please post it!

Poker wrote:You claim this is not one of those times. Are you making that claim just because the consequences of it being the natural numbers go against your finite-set-based intuition? Or is there some more concrete reason?

I claim there are numbers that take an infinite number of steps to count to, for the very reason that we have to take an infinite number of steps to count to them.

Xias wrote:Kryptonaut's Conjecture: For any sequence of sets {An}n=1→infinity, |limn→infinity An| = limn→infinity |An|.

Spoiler:

First off, I have not claimed anything for "any sequence of sets", so for you to then go on and argue about a different sequence of sets is not proving anything about the case in point.

Xias wrote:What this does is choose n equally spaced points between (1/2)n and (9/10)n, exclusive.

Since at the limit, (1/2)n and (9/10)n are both equal to 0 (thanks Demki), and there are no points that are both greater than 0 and less than 0, then it follows that limn→infinity Bn = {}. This is also true by liminf and limsup. Therefore, |limn→infinity Bn| = 0.

Doesn't it strike you as contradictory that you specify the points as being in an exclusive range and then go on to say that 'at the limit' they take on the value of the endpoint of the range? Sure, for every finite n the values are between the endpoints. Are you now saying that it's ok to evaluate at the limit of n=infinity? If so, what stops you arguing exactly the same way for {n+1,...10n} at the limit? You can't have it both ways.

You have asserted several times that liminf and limsup of B_{n}={n+1,n+2,...10n} both equate to {}. I've said I was unsure about this conclusion - here's why:

For lim_sup you need to evaluate L_{sup} = ∩_{n≥1}∪_{j≥n}B_{j} But ∪_{j≥n}B_{j} = {n+1,n+2,...} So L_{sup} = ∩_{n≥1} {n+1,n+2,...} Which for any finite limit to n is clearly not {}, and in fact as n increases and values 'fall off' the bottom end of the intersection, the limit can only be seen as making us evaluate lim_{n->∞} B_{n} So we're right back where we started.

For lim_inf you have L_{inf} = ∪_{n≥1}∩_{j≥n}B_{j} Here ∩_{j≥n}B_{j} is empty for any finite limit to j sufficiently far above a particular n, but that 'sufficiently far' criterion is an ever-increasing margin tending to infinity as n tends to infinity, and the result is not necessarily {} at the limit of n, where we'd once again have to evaluate lim_{n->∞} B_{n}

So we're faced with the situation that L_{sup} ≠ L_{inf} for any finite limit to n (so no conclusion can be drawn about convergence), and for an infinite limit we're back where we started without having proved anything.

Xias wrote:The way that we count them is prescribed by the puzzle. We place them ten at a time in order at specified times. Those times are specified to be midnight - 20*(1/2)n. This is only defined (mathematically) for finite n, of which there are infinitely many. We don't have a step at midnight (infinite n), nor do we have balls with infinite numbers on them (infinite n).

The limit may be midnight, but we know that because we follow the rules of limits - and the same rules of limits are the rules that tell us that the jug is empty.

Then you're avoiding the whole essence of the puzzle which is to force us not to merely consider "What happens as midnight approaches" but "What happens after midnight has arrived." The theory of infinite ordinals is expressly designed to cover this kind of situation. If you avoid considering them you are not addressing the problem.

mward wrote:I think that this method can be applied to any supertask: and will either calculate the result, or will prove that the result is indeterminate. I would love to be proved wrong, however, so if you have a counterexample, please post it!

I like your reasoning, particularly as you seem prepared to at least admit to the presence of ordinals greater than or equal to ω . Clearly I disagree with your conclusion in case (1), the original 'remove-the-lowest' puzzle. My reasoning, in your framework, is that although at the end of the task ω balls have been added and removed, for each ball removed 9 were added and never removed, leaving 9ω (which equals ω) behind. When accounting for the balls added and removed, it is critical to note that the input set is 1..10n and the discards are 1..n - and although both of these can individually be mapped to N, crucially they cannot be mapped at the same time during the supertask - there is no index that can step through 1..n and 1..10n pairing elements off against each other 1:1. The final n+1..10n will always be left unmatched. These are the balls left in the jug.

That is also why the task can't be broken down into 'add all the balls then remove all the balls', which would destroy any information about the relative rates of the two tasks. For the same reason lim_{n->∞} 2n/n can't be broken down into (lim_{n->∞} 2n)/(lim_{n->∞} n)

kryptonaut wrote:I claim there are numbers that take an infinite number of steps to count to...

Even if true, it does not imply that taking an infinite number of steps will allow you to count to that number.

kryptonaut wrote:...for the very reason that we have to take an infinite number of steps to count to them.

But you haven't shown that we actually count to them after taking an infinite number of steps. You are assuming the consequent ("begging the quesiton").

Nobody here disputes the existence of ω any more than the existence of i or the square root of 2. We're just claiming that, even after ω steps, you don't actually reach ω. You never "reach" ω.

As a simple illustration, consider the (ordered) set {1,3,5,7...2,4,6,8...} and a supertask of moving the numbers into and out of a transfer jug, one at a time. When does 2 make it into the jug?

Spoiler:

My answer: Never. The original jug ends up with all the even numbers, and the destination jug ends up with all the odd numbers. The transfer jug is empty.

responding to mward, kryptonaut wrote:The final n+1..10n will always be left unmatched. These are the balls left in the jug.

How long do those balls remain in the jug? As n approaches infinity, what happens to the amount of time these balls remain in the jug?

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.

Poker wrote:You claim this is not one of those times. Are you making that claim just because the consequences of it being the natural numbers go against your finite-set-based intuition? Or is there some more concrete reason?

I claim there are numbers that take an infinite number of steps to count to, for the very reason that we have to take an infinite number of steps to count to them.

Your reasoning (A) seems fairly circular, the only "escape point" for that circular reasoning that I can see being the claim that there are more than 0 balls left at the end, and (B) is not at all obvious why it should be used in this instance but not others.

kryptonaut wrote:You have asserted several times that liminf and limsup of B_{n}={n+1,n+2,...10n} both equate to {}. I've said I was unsure about this conclusion - here's why:

For lim_sup you need to evaluate L_{sup} = ∩_{n≥1}∪_{j≥n}B_{j} But ∪_{j≥n}B_{j} = {n+1,n+2,...} So L_{sup} = ∩_{n≥1} {n+1,n+2,...} Which for any finite limit to n is clearly not {}, and in fact as n increases and values 'fall off' the bottom end of the intersection, the limit can only be seen as making us evaluate lim_{n->∞} B_{n} So we're right back where we started.

In order for x to exist, it must not "fall off" as you put it. Therefore, x must be a part of all ∪_{j≥n}B_{j}. By the very definition of a union of sets, this means that x must be a member of some B_{j} (and, in fact, for each n, some j>n).

Now, x is either finite or infinite. For completeness' sake, here's the argument for if x is finite: x must be a member of some B_{j}, but if x is finite, for n>x, x is not a member of ∪_{j≥n}B_{j} since x<n<j. So x cannot be in the lim_sup if x is finite.

What happens when x is infinite? First, x cannot be part of B_{j} for any finite j, since since each finite B_{j} only contains finite values. Therefore, if x is indeed in the lim_sup, since there must be some B_{j} that x is a member of, j must also be infinite. Then j also cannot be a natural number. Therefore, if you want to claim that x is in the lim_sup, you must also claim that the process used to define lim_sup must itself use a set of numbers that extends beyond the natural numbers. Is this something you are prepared to claim?

kryptonaut wrote:For lim_inf you have L_{inf} = ∪_{n≥1}∩_{j≥n}B_{j} Here ∩_{j≥n}B_{j} is empty for any finite limit to j sufficiently far above a particular n, but that 'sufficiently far' criterion is an ever-increasing margin tending to infinity as n tends to infinity, and the result is not necessarily {} at the limit of n, where we'd once again have to evaluate lim_{n->∞} B_{n}

Much like in lim_sup, in order for x to exist within lim_inf, by the very definition of a union of sets, x must be a member of some ∩_{j≥n}B_{j}. As you said, for any finite j, ∩_{j≥n}B_{j} is the empty set. Therefore, in this case as well, j would need to be infinite. Therefore, if you want to claim that x is in the lim_inf, you must also claim that the process used to define lim_inf must itself use a set of numbers that extends beyond the natural numbers. Is this something you are prepared to claim? (I suppose another possibility would be to debate the definition of a union of sets, though I doubt that's the direction you intend to head.)

kryptonaut wrote:When accounting for the balls added and removed, it is critical to note that the input set is 1..10n and the discards are 1..n - and although both of these can individually be mapped to N, crucially they cannot be mapped at the same time during the supertask - there is no index that can step through 1..n and 1..10n pairing elements off against each other 1:1.

And this inability to be mapped at the same time persists after the supertask because... why, exactly?

(Typing this post on my phone, so forgive me not using as much fancy formatting as I normally use...)

kryptonaut wrote:For lim_sup you need to evaluate L_{sup} = ∩_{n≥1}∪_{j≥n}B_{j} But ∪_{j≥n}B_{j} = {n+1,n+2,...} So L_{sup} = ∩_{n≥1} {n+1,n+2,...} Which for any finite limit to n is clearly not {}, and in fact as n increases and values 'fall off' the bottom end of the intersection, the limit can only be seen as making us evaluate lim_{n->∞} B_{n} So we're right back where we started.

The "finite limit" you mention is completely irrelevant, as the definition of L_sup is not a limit. Indeed, it wouldn't make sense to use a limit to define it, as we're doing this to define what limits of sequences of sets even are... If you defined L_sup as a limit, it would be a circular definition.

Instead, L_sup is just an intersection of a infinite family of sets. Not a limit of a sequence of finite intersections... but just, take the whole infinite family, do a single intersection, and done. Both intersection and union are already defined to work in exactly the way you'd expect over infinite (even uncountable, though we're not doing that here) families of sets.

So, for convenience sake, let's name the intermediate unions, U_n. So each U_i is the union of B_n after i, and then L_sup is the intersection of all U_n.

As you noted, U_i is just "the set of all natural numbers strictly greater than i". And there's one of these sets for every natural number i.

So... for something to be an element of L_sup, it must be a member of every U_i (that's what an intersection means). So it has to be a natural number that's strictly greater than every natural number. Which obviously doesn't exist, even if you still disagree about the definition of N.

Even more specifically, note that i is not an element of U_i, so in particular any element of L_sup would need to at least be a natural number that is not equal to any natural number.

Say L_sup is not empty. Say it contains some x. Well: if x isn't a natural number, then it isn't in any U_n set, let alone all of them, so that's right out. But if x is a natural, then it's not in U_x (and even if there's some sort of off-by-one error in the sets I haven't checked, it's also not in U_x+1), so it's not in every U_n either. So, by contradiction the L_sup must be empty.

Hopefully one of those arguments helps it click for you.

kryptonaut wrote:For lim_inf you have L_{inf} = ∪_{n≥1}∩_{j≥n}B_{j} Here ∩_{j≥n}B_{j} is empty for any finite limit to j sufficiently far above a particular n, but that 'sufficiently far' criterion is an ever-increasing margin tending to infinity as n tends to infinity, and the result is not necessarily {} at the limit of n, where we'd once again have to evaluate lim_{n->∞} B_{n}

In the same way, L_inf is not a limit, it's just a single union of an infinite family of sets.

If we define I_n as the intermediate intersections, you've already noted that I_j is empty for every natural j. Which is all the j.

In order for the union L_inf to be non-empty, it must contain some element x. And in order for that to happen, there must be some I_j that contains x. Not just vague "if you go to infinity it feels like there sorta is something there in the limit"... there needs to be a specific I_j which contains it.

It's like when I was talking about L_sup above... Intersections are the reverse - in order for a value to not be in the set, you need to be able to point to a specific set that it's not in. And that's what I did, by showing how i is not an element of U_i.

So now, we need to do the same here... in order for L_inf to be non-empty there needs to be some specific I_j which is non-empty. Can you name one?

Xias wrote:Kryptonaut's Conjecture: For any sequence of sets {An}n=1→infinity, |limn→infinity An| = limn→infinity |An|.

First off, I have not claimed anything for "any sequence of sets", so for you to then go on and argue about a different sequence of sets is not proving anything about the case in point.

You seem to think that there is a contradiction when we claim that a sequence of sets that are getting bigger and bigger suddenly become zero at the limit. What does that contradict?

kryptonaut wrote:

Xias wrote:What this does is choose n equally spaced points between (1/2)n and (9/10)n, exclusive.

Since at the limit, (1/2)n and (9/10)n are both equal to 0 (thanks Demki), and there are no points that are both greater than 0 and less than 0, then it follows that limn→infinity Bn = {}. This is also true by liminf and limsup. Therefore, |limn→infinity Bn| = 0.

Doesn't it strike you as contradictory that you specify the points as being in an exclusive range and then go on to say that 'at the limit' they take on the value of the endpoint of the range? Sure, for every finite n the values are between the endpoints. Are you now saying that it's ok to evaluate at the limit of n=infinity? If so, what stops you arguing exactly the same way for {n+1,...10n} at the limit? You can't have it both ways.

What are you even talking about?

I defined a sequence of sets of rational numbers that are within a certain range at each step. The way I defined the sequence meant that the elements of the set couldn't ever take the value of the endpoints. So when the range becomes zero because the endpoints are equal to eachother, there is no more range to be picked.

I could have defined it to include the endpoints. Then at midnight, there would be exactly one element in the set, and that element would be zero.

What stops me from arguing the same way for {n+1,...10n} at the limit? Nothing! I am arguing the same way for {n+1,...10n} at the limit!

kryptonaut wrote:You have asserted several times that liminf and limsup of B_{n}={n+1,n+2,...10n} both equate to {}. I've said I was unsure about this conclusion - here's why:

For lim_sup

...

For lim_inf

I feel that others have sufficiently explained why you are not correct here.

kryptonaut wrote:Then you're avoiding the whole essence of the puzzle which is to force us not to merely consider "What happens as midnight approaches" but "What happens after midnight has arrived."

Oh! Of course! If only I hadn't been arguing this whole time that what happens as midnight approaches doesn't matter! We can evaluate what happens after midnight has arrived, which is that the jug is empty.

It is especially ironic for you to argue this, since the entire point of the puzzle is to reference a particular solution to the Ross-Littlewood problem. If we add ten balls and remove one ball at every step, the jug will intuitively be full at midnight. However, if we remove balls in a particular and systematic way, we end up with an empty jug, leading to the Ross-Littlewood Paradox. This is the essence of the puzzle.

kryptonaut wrote:That is also why the task can't be broken down into 'add all the balls then remove all the balls', which would destroy any information about the relative rates of the two tasks. For the same reason lim_{n->∞} 2n/n can't be broken down into (lim_{n->∞} 2n)/(lim_{n->∞} n)

lim_{n->∞} f(n)/g(n) can be broken down into lim_{n->∞} f(n) / lim_{n->∞} g(n) if both lim_{n->∞} f(n) and lim_{n->∞} g(n) exist and lim_{n->∞} g(n) is nonzero.

In the same sense, lim_{n->∞} ( f(n) - g(n) ) can be broken down into lim_{n->∞} f(n) - lim_{n->∞} g(n) if both lim_{n->∞} f(n) and lim_{n->∞} g(n) exist.

Since in this game, the sets analogous to both lim_{n->∞} f(n) and lim_{n->∞} g(n) are both N, we can break it apart and end up with {}.

Or, on the other hand, we could remember that limits of functions and limits of sequences of values are not directly analogous to limits of sequences of sets. Then for a limit of a sequence of sets, we must abandon our intuition regarding limits of functions and limits of sequences of values, and find new rules to follow which are rigorously and consistently defined. Which leads us to taking the limsup and liminf and we get {}.

Choose whichever you like; the jug is empty either way.