## Infinite Balls and Jugs [solution]

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

Kryptonaut, let's explore your variant a bit. I like it quite a bit, because we can make justified claims about irrational numbers in the set (0,1) that we can't necessarily make about 100000.... (or 0.000....1, for that matter).

If I'm not mistaken, we end up going through every rational number between 0 and 1. So let's start with a collection of balls, with all of the real numbers in (0,1).

Now let's divide them up into two piles, one with all of the balls with rational numbers Q = Q∩(0,1) and the other with irrational numbers R = Qc∩(0,1). Do you agree that the intersection of Q and R is the null set? Do you agree that their union is (0,1)?

Now let's paint all of the balls in the Q pile blue, and all of the balls in the R pile red. Do you agree that it follows that every ball is painted with exactly one color?

Now let's do the process you described. You asserted that in the end, you have a jug filled with irrational-numbered balls. So then all of the balls must be red. Yet, throughout the process, you inserted and removed only blue balls. In fact, at any given step, the net number of blue balls in the jug went up by 9.

So you are claiming that something about the process does two things:

1. At midnight, all blue balls have been removed, despite the number of blue balls in the jug monotonically increasing.
2. At midnight, all red balls end up in the jug, despite all being outside of the jug at any time before midnight.

I agree with (1), for exactly the same reason that the jug is empty in the original puzzle. Lets talk about (2). Can you think about the following questions? I don't necessarily want an answer/explanation for each one, but your thoughts on any of them that you find particularly enlightening would be useful.

a. Is there any time in which there is both a red ball and a blue ball in the jug?
b. Is there any step in which there is both a red ball and a blue ball in the jug?
c. If the red balls are all destroyed before the game begins, is there any time in which the game would halt (due to being unable to follow the instructions) while there are still blue balls in the jug?
d. If the red balls are all destroyed before the game begins, is there any step in which the game would halt while there are still blue balls left in the jug?
e. If the red balls are all destroyed before the game begins, and the game halts due to error but only after there are no blue balls left in the jug, at what time/step does this occur?
f. If the red balls are all destroyed before the game begins, and the game does not halt due to error, are any balls left in the jug at midnight?
g. If the answer to (f) is yes, what color are they?
h. If the answer to (f) is no, is this not identical to the original puzzle?

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

ucim wrote:
kryptonaut wrote:These infinitely long representations can be generated by the infinite process described in the puzzle, since we never stop subdividing the intervals between existing numbers. However none of them exists during the process even though they are there at the end of it.

But if they don't exist during the process then they are not available to be removed from the jug at any point...

Not at any finite point. But (if we do this every 1/2^n seconds), then at the infinitessimal interval before the end of the supertask, they are all created and removed. And if we do this without "cleverly crunching time", then they are all created and removed "at infinity" (another way of saying the task never ends, so we never have to face this, which is the whole point of crunching time).

No, at any time before the end of the supertask, there exist only finite-length numbers. At the end of the supertask there exist only infinite-length numbers which are not removed because they are not natural numbers

ucim wrote:One thing that may help you get your mind around this... intuitively you are adding ten balls and taking away one, and ten minus one equals nine. It seems like you keep adding balls. But when you do it an infinite number of times, you have something like (ten times infinity) minus (one times infinity) and you do not get (nine times infinity), because infinity is like unto the sea - it is big and does what it wants.

Hmmm, I don't think you can prove or disprove much with that argument - or rather you can prove or disprove anything you like To my mind, the problem is that 'at infinity' we are dealing with balls with infinite numbers which are not permitted in the natural numbers, so at that point we run out of balls. In normal time we wouldn't have to worry about it, but in crunched time we do. It's the fact that the size of N is not a member of N that makes the problem paradoxical.

Xias wrote:If I'm not mistaken, we end up going through every rational number between 0 and 1. So let's start with a collection of balls, with all of the real numbers in (0,1).

Now let's divide them up into two piles, one with all of the balls with rational numbers Q = Q∩(0,1) and the other with irrational numbers R = Qc∩(0,1). Do you agree that the intersection of Q and R is the null set? Do you agree that their union is (0,1)?

Now let's paint all of the balls in the Q pile blue, and all of the balls in the R pile red. Do you agree that it follows that every ball is painted with exactly one color?

Yes, the union of the sets is (0,1). I'm not convinced you can paint them different colours though, because I don't think you can necessarily identify which set a ball belongs in by inspection, only by seeing whether the decimal expansion terminates which will potentially take infinite time. But let's pretend we can do it.

Spoiler:
Xias wrote:So you are claiming that something about the process does two things:

1. At midnight, all blue balls have been removed, despite the number of blue balls in the jug monotonically increasing.
2. At midnight, all red balls end up in the jug, despite all being outside of the jug at any time before midnight.

I agree with (1), for exactly the same reason that the jug is empty in the original puzzle. Lets talk about (2). Can you think about the following questions? I don't necessarily want an answer/explanation for each one, but your thoughts on any of them that you find particularly enlightening would be useful.

a. Is there any time in which there is both a red ball and a blue ball in the jug?
b. Is there any step in which there is both a red ball and a blue ball in the jug?
c. If the red balls are all destroyed before the game begins, is there any time in which the game would halt (due to being unable to follow the instructions) while there are still blue balls in the jug?
d. If the red balls are all destroyed before the game begins, is there any step in which the game would halt while there are still blue balls left in the jug?
e. If the red balls are all destroyed before the game begins, and the game halts due to error but only after there are no blue balls left in the jug, at what time/step does this occur?
f. If the red balls are all destroyed before the game begins, and the game does not halt due to error, are any balls left in the jug at midnight?
g. If the answer to (f) is yes, what color are they?
h. If the answer to (f) is no, is this not identical to the original puzzle?
(spoilered for brevity)

1) Yes, I agree there are no blue balls left after an infinite number of steps, because
2) They have been replaced by red ones.

It is not possible to say when this happens other than 'at infinity (midnight)', for exactly the same reason you could not say when the labels with appended zeros became infinitely long.

I think if we are dealing with R then the time interval we are dealing with includes midnight, whereas with N we can only talk about what happens strictly before midnight.

So my answers/comments are:
a. At midnight, when infinite steps occur in zero time, all blue balls are removed and replaced with red. It is not possible to separate the events in time.
b. It is not possible to say - you can loosely think of infinite 'steps-numbered-infinity' happening at midnight when everything changes. Just like there is no step when the 'append zeros' labels become infinitely long.
c. If there are no red balls then the game ends (or becomes ill-defined) at midnight when it becomes impossible to generate the irrational numbers. The number of blue balls is the same as the number of 'append zero' labels there are immediately before they become infinite in length. I don't think this number is defined.
d. Without red balls the game would not be able to proceed through infinite steps. That's all that can be said.
e. through h. Without red balls, the game is ill-defined because it cannot proceed to infinity. It can approach arbitrarily close but it can't get there because N does not contain numbers with infinite digits. If it can't get to infinity then we cannot talk about what happens at or after midnight.

Trying to resolve the puzzle with N is ill-defined because we need to include infinite numbers which are not part of the problem space.

Let's define K as the set of infinite quantities represented by infinite sequences of digits.
The set of numbers needed for the puzzle is the union of N and K. Using the mapping I described this can be shown to be the same cardinality as R. The balls remaining at midnight correspond to the values in K. These are not natural numbers.

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

Kryptonaut, can you justify any of your conclusions mathematically? In particular, these:

Spoiler:
- Operating on the interval (0,1) somehow closes the upper bound of {SUMi=1n(1/2)^n : n ∈ N (from "if we are dealing with R then the time interval we are dealing with includes midnight";

- That infinite steps happen at midnight;

- That those steps are infinitely numbered, and that something happens during those steps in which balls are added;

- That the set of balls during the infinite, infinitely numbered steps, is exactly Qc ∩ (0,1), as opposed to any other arbitrary set (since you say that (2) happens "at infinity" when infinite, infinitely numbered steps happen);

- That such steps are less "ill-defined" than those in the original puzzle, which are well-defined in the set-based proof that I offered;

- That you can't perform an infinite number of steps on the set Q ∩ (0,1), despite it having a cardinality of infinity;

- That NK maps to R; that is to say, the number defined as 10000... with infinitely many digits maps to a real number in (0,1), and that some infinitely large number maps to pi/4.

I think that is a nearly comprehensive list of things that you've said, that I cannot think of any mathematical justification for. I'm not trying to be rude; I just think that you've tied yourself up in knots of absurdity trying to account for your intuition telling you that the jug cannot empty.

For comparison, these are my conclusions:

Spoiler:
(1) is true. (2) is false. The game is exactly equivalent to the original, because the set Q ∩ (0,1) is countable.

My justification is the set-based proof I provided, which proves the only controversial thing about the empty jug:

Spoiler:
The limit of the cardinality of a set is not necessarily the same as the cardinality of the limit of a set.

My set-based proof can be generalized by replacing the set N with any countable set G = {gm: mN} and replacing all instances of a natural number m with gm.

I hate to keep tooting my own horn about the set-based proof. I'm just unsure why you either don't believe the proof is sound,or don't believe it corresponds exactly to the puzzle. EDIT: Actually, in order for all of your conclusions to hold, my proof must not be sound regardless of whether it corresponds to the puzzle exactly or not.

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

I'll try to clarify my argument. My degree was in physics not maths so my terminology may be inaccurate but I'm fairly convinced that my reasoning is sound.

The root of the paradox is that it is being discussed in terms of natural numbers which can be arbitrarily large but explicitly not infinite, and yet we are asked to consider what happens at/after infinity. We can't use a natural number to evaluate the infinite sum 1+1+1+1..., and likewise we can't use natural numbers to explore what happens after infinite applications of the rule in the puzzle.

The example of the ever-increasing label illustrates this, where a zero is appended to the label on a ball at each step. You are happy to accept that this produces a label with infinite length. But adding a zero is multiplying by ten, so the infinite-length label represents n*10^∞ which is not in the natural numbers. Applying the rule a finite number of times gives a number in N but applying it an infinite number of times does not.

However we can define a set K where each element is an infinite sequence of digits. Treated as numbers these are infinitely large, and although they are distinct they cannot be ordered. However they are larger than all members of N. (I showed in my previous post that this set has the same cardinality as the set of irrational reals, as can be seen by imagining a '0.' written in front of each.) Taking the union of K with N we can now express quantities corresponding to any sequence of digits, finite or infinite, with the caveat that ordering is undefined for the infinite sequences. Now applying the rule any finite number of times will give us members of N but applying it an infinite number of times will leave us with members of K which are all effectively infinite but distinct.

I suspect that the problem with your set-theory proof is that it deals with things like limk→∞Ak whereas we need to actually evaluate A which we can't do using natural numbers. Using the limit works for all finite steps, all times before midnight, but fails at midnight when the steps have become infinite.

I'll agree that at midnight no natural-numbered balls remain, but I contend that because we have reached infinity, balls with infinite numbers from the set K are left behind. Without considering these balls we are not justified in discussing what happens at midnight, we can only discuss what happens strictly before midnight. So the puzzle has to be framed in these terms or else it's like asking "What's the biggest number and you can't just say infinity because that's not a number". We are asking what happens at infinity, se we have to deal with the infinite.

I'll address some of your queries:
Spoiler:
Xias wrote:- That such steps are less "ill-defined" than those in the original puzzle, which are well-defined in the set-based proof that I offered;

- That you can't perform an infinite number of steps on the set Q ∩ (0,1), despite it having a cardinality of infinity;

- That N∪K maps to R; that is to say, the number defined as 10000... with infinitely many digits maps to a real number in (0,1), and that some infinitely large number maps to pi/4.

- The ill-defined steps are the ones that result in infinitely big numbers - which we can't represent as naturals but which the rule generates at infinity.

- If you take a rational number and append an infinite sequence of digits (other than all zeros) you end up with an irrational number. This is basically the proof that the set of reals is larger than the set of rationals.

- Every sequence of digits after the decimal point in a real number (0,1) either terminates with an infinite sequence of zeros or continues to infinity. The terminating sequences correspond to unique members of N and the non-terminating sequences correspond to unique members of K. One of the infinite members of K which begin 78539816... maps to pi/4. (In my example mapping I reversed the digits of the sequence - this avoids dealing with leading vs trailing zeros and also fills the number-line in a clearly discernible pattern, showing that after an infinite number of iterations every gap would be filled, which is the proof that the set has the same cardinality as R. One of these reversed infinite sequences maps to pi/4 but it's impossible to say which, just as it's impossible to give the last digits of pi/4.)

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

Where did the infiniteth step comes from? The problem does not mention any step that does not correspond to a natural number.
Consider this 2 scenarios:
1: the jug starts(step 0) with an object that is labeled 0, at each step n, where n is a natural number, remove the object labeled with a string of n 0s and insert an object labeled with a string of n+1 0s.
After countably infinite steps, ordered by the natural ordering of the natural numbers(that is has the same order as the ordinal omega), we have that every object that was at some step in the jug, eventually was removed in another finite step, thus the jug ends up empty(there are no infiniteth steps)

2: the jug starts(step 0) with an object that is labeled 0, at each step n, where n is a natural number, apped the label of the object in the jug with a 0.
After countably infinite steps, ordered by the natural ordering of the natural numbers(that is has the same order as the ordinal omega), we have an object in the jug(we never removed it), and it is labeled with a countably infinite string of 0s. Each 0 in the label was added in some finite step(the first one in step "0"). There is no infiniteth step, there is not infiniteth 0, each 0 corresponds to a natural number.

Do you have a problem with either of the arguments?
Last edited by Demki on Thu Nov 17, 2016 3:38 pm UTC, edited 1 time in total.

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

I have a problem with case 1 (assuming a typo and you meant n+1).

If you are happy with the phrase "after countably infinite steps" then why are you not happy with a string of "countably infinite zeros" for the label? No, you can't find the length of the label in the natural numbers any more than you can find the cardinality of the natural numbers in the set of natural numbers - and that's the whole point. If you confine yourself to natural numbers then the phrase "after countably infinite steps" is meaningless. You can't talk about what happens at infinity. If you want to do that then you have to work with the set of natural numbers with infinities that I described earlier. With natural numbers you can only talk about the limit as you approach infinity.

There is no step when everything becomes infinite, but at midnight infinite steps have been performed. Weird.

Let's take a scenario 3: The jug starts with a ball labeled {"0",0} and at every step a zero is appended to the string and the number is incremented. What do we have at midnight? Is the string infinitely long? What's the number?

Just for fun let's have a scenario 4: the jug starts at position 0 and every step n we move it by 2^-n metres. How far has the jug moved at midnight?

Demki wrote:Where did the infiniteth step comes from? The problem does not mention any step that does not correspond to a natural number.

The problem asks what happens after an infinite number of steps have been taken, which implies a step that does not correspond to a natural number. The fact that the problem is expressed using natural numbers but the answer cannot be, is what makes it appear paradoxical.

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

Yes there was that typo.
Your 'problem' with my scenario 1 is that you failed to read "at every step n, where n is a natural number", that is, we only take steps for natural numbered steps, and yes, there are countably infinite such steps.
Where does you supposed "countably infinite label" come from in screnario 1, which explicitly states only natural numbered steps?
Having done infinite steps does not mean there needs to be an infiniteth step, just like the set of natural numbers is infinite without containing any "infinite" elements.
I also explicitly stated the steps are ordered as the ordinal omega, if you want to also include the "check what's in the jug" as a step, I'd have ordered the steps as the ordinal omega+1, with the checking step being the last, so no modifying of the jug's contents is done for any step that isn't a natural number.

I'd say scenario 3 is divergent, just like the infinite series 1+1+1+1+... is divergent, or like sin(1/x) is divergent at x=0. or the sequence (1,2,1,2,1,.. ) is divergent.

In scenario 4, the jug has moved 1 meter(if we start moving at step 1), this does not require any infiniteth step, the sequence of the distance the jug is traveling is convrergent, and converges to 1, by the mathematical sense that for any real number e>0, we can find a natural number N such that for all n>N we have the jug's distance from having traveled 1 meter be less than e.

I think that if you have a problem with performing infinite steps without having an infiniteth step, then you should have a problem with the ordinal omega being infinite.

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

Ok sorry, perhaps I misread your scenario 1 somewhat.

If you want to talk about countably infinite steps then you can't talk about 'after' and expect the result to be a natural number. You can keep going for ever, removing one thing and putting something different back in, but you can't talk about the end of that process in terms of natural numbers because the length of that process is not a natural number. Does that make things clearer?

I am not pointing to a step and saying "that's step number infinity", but I am saying that after an infinite number of steps have been performed, well, an infinite number of steps have been performed, so the number is infinitely big. Now this is not representable in the set of naturals, so it's fair to say there is no ball whose label contains a natural number of zeros after performing infinite steps.

But you can't have your cake and eat it - either you deny the possibility of a ball with an infinitely long label/infinitely big number and accept that you can't talk about what happens at (or after) infinity/midnight, or you permit balls to have infinitely long labels (meaning they are numbered using a superset of N) and then you can examine what happens at infinity/midnight. I think the latter is a more interesting resolution of the paradox.

Saying a situation is divergent seems like a cop-out, it really just means the outcome is not representable in N - which is exactly the case in this puzzle.

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

There's the thing, in scenario 1 you can't have a ball with a label, because we only ever add balls with labels that have natural numbers on them, there is no magic smoke that produces that 'infinitely big' label in scenario 1

In scenario 2, however, we basically map the natural numbers to a string on the label on the ball, in a 1 to 1 correspondence. So since the natural numbers are infinite, so must be the label on the ball.

Divergence is not a cop out, it simply states that there is no limit. Yes, you could label 1+2+3+4+... as -1/12 because it fits some useful function, but that doesn't make 1+2+3+4+... equal -1/12.

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

Kryptonaut, why are the balls from the set K left behind? It seems to me that if the set of all balls that have been added to the jug at midnight is equal to NK, then it still seems that the set of all balls that have been removed from the jug at midnight is also equal to NK. Granted, I'm still not convinced that the set is NK rather than merely N. But it seems like if we accept that, then to remain consistent the jug must still be empty.

Furthermore, I think it may be fruitless to continue coming up with additional examples of alternative games to discuss, because it seems like you disagree on the outcome of the most trivial game:

Spoiler:
At each time step n as defined, place ball n in the jug. At midnight, the jug has exactly every ball in N and none outside of N.

The conclusion here is consistent with all of the rules of abstract algebra, yet cannot be true if what you are saying is true.

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

Demki wrote:Divergence is not a cop out, it simply states that there is no limit. Yes, you could label 1+2+3+4+... as -1/12 because it fits some useful function, but that doesn't make 1+2+3+4+... equal -1/12.

In this case, I think the more logical answer to SUM1->infinity(n) is that the answer is infinite, which is outside the scope of natural numbers. At every step along the way the running total is a natural number but when all the steps are done you've left the realm of natural numbers. So if you want to stay in that realm you have to take limits as n tends to infinity and so on, and then you can say that the sum tends towards infinity too. But if you want to talk about what actually happens at infinity then you have to allow the sum to become something other than a natural number.

Xias wrote:Furthermore, I think it may be fruitless to continue coming up with additional examples of alternative games to discuss, because it seems like you disagree on the outcome of the most trivial game:

Spoiler:At each time step n as defined, place ball n in the jug. At midnight, the jug has exactly every ball in N and none outside of N.

The conclusion here is consistent with all of the rules of abstract algebra, yet cannot be true if what you are saying is true.

As midnight approaches, the set of balls approaches N - but actually at midnight the set also includes infinity. Either that or midnight never arrives, take your pick.

Let's try an analogy. We know that the set of reals is fundamentally bigger than the set of rationals - that there are numbers like e which are members of the reals but not the rationals. But if we sum the rational numbers 1/(n!) for all n from 0 up to infinity we get exactly e. There is no single step where you can say we've reached e, at each successive step you get a rational number which is a closer and closer approximation, but after an infinite number of steps you find you're there in the land of real numbers. Same thing with natural numbers - after you've added an infinite number of them together you can find yourself in the land of infinite numbers even though at no stage of the journey did you add an infinite number to anything.

Can you see that moving from the rationals to the reals is at least in some way analogous to moving from the naturals to the N∪K?

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

kryptonaut wrote:Can you see that moving from the rationals to the reals is at least in some way analogous to moving from the naturals to the N∪K?

I agree completely that the sum of the numbers on all of the balls will be a nonnatural number. But we aren't finding the sum of the balls. We are finding the set of a union of sets of discrete elements. {1} + {2} is not {3}, but {1, 2}. What you are doing is comparing the behavior of the cardinality of the set with the contents of the set itself, which is not something you can do in mathematics. The behavior will be different because they are different things.

More analogous is that you have an infinite number of balls at midnight (the cardinality of the set of balls), despite having a finite number of balls at each step before midnight. I'm absolutely with you on that one.

The leap you are taking from that to then say that the union of every one of infinite partitions of a set does not equal the set itself is not a conclusion that is justified mathematically. If you have any number of partitions of a set, infinite or not, and combine them together, you get the set itself. This follows from the definition of partitions, and the definition of union.

EDIT: And I forgot to say that you didn't acknowledge the first point that I made. If the set of all balls added is N∪K, then why is the set of ball removed not also N∪K?

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

Xias wrote:The leap you are taking from that to then say that the union of every one of infinite partitions of a set does not equal the set itself is not a conclusion that is justified mathematically. If you have any number of partitions of a set, infinite or not, and combine them together, you get the set itself. This follows from the definition of partitions, and the definition of union.

I'm not sure I have much to say on that. What I'm saying boils down to the assertion that you can't fully enumerate N since it is infinite yet does not contain infinity.

Consider a set of objects, one for every element of N. Absolutely all of them. Each object is labelled with its corresponding number. You can let the n'th object have size 2^-n centimeters so you can hold the whole infinite set in your hand. Now, does this set of objects contain an object X with an infinitely large label? If it does, then X cannot correspond to an element in N since no such number exists in N. But if X does not exist, then there is an object Y with the largest non-infinite label (since you are holding the complete set) - but that would imply a largest natural number, which also does not exist. So the premise that you can enumerate every single natural number leads to a contradiction.

Now you can avoid this contradiction by approaching but never quite getting the complete set, but this would make supertasks impossible to complete. Alternatively you can allow for infinitely big numbers by numbering the objects using a superset of the natural numbers that also contains infinity. This is a set with greater cardinality than that of N

Xias wrote:EDIT: And I forgot to say that you didn't acknowledge the first point that I made. If the set of all balls added is N∪K, then why is the set of ball removed not also N∪K?

K is the set that is required to terminate the infinite series of steps, to complete the supertask. The elements of K are infinite in magnitude, greater than any element of N. They are added at midnight, no more steps are performed after they are added. It is not possible to describe the step before they were added any more than you can point to the step before e is created from the infinite sum of 1/n! At the end of the process the real number e exists, during the process rational approximations to e exist.

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

kryptonaut wrote:Consider a set of objects, one for every element of N. Absolutely all of them. Each object is labelled with its corresponding number. You can let the n'th object have size 2^-n centimeters so you can hold the whole infinite set in your hand. Now, does this set of objects contain an object X with an infinitely large label? If it does, then X cannot correspond to an element in N since no such number exists in N. But if X does not exist, then there is an object Y with the largest non-infinite label (since you are holding the [i]complete set)[/i] - but that would imply a largest natural number, which also does not exist. So the premise that you can enumerate every single natural number leads to a contradiction.
Emphasis mine. The emphasized statement is incorrect.

You are holding the entire set. The set contains only elements with finite length labels. So, you cannot be holding an element with an infinitely long label.

The set contains an infinite number of elements, labeled by the integers. Therefore, there is no largest* element, as there is no largest integer.

It doesn't matter that you can "hold it in your hand" or not. This is one of the differences between (the concept of) infinity and finite numbers.

There can be an infinitely large set of unique finite integers (i.e. N) without requiring that there exist an "infinite" integer. Be sure your head is fully around this before going further; attempts to plaster this idea on common objects only obscures things, and if you don't have this property (of infinite sets) down, this will allow the problem to lead you down the rabbit hole and away from understanding.

You can certainly enumerate a countably infinite set, the answer however is not finite. The cardinality of an infinite set is not the same as its largest element. Ordinal numbers are not the same as cardinal numbers; although this is not apparent with finite sets, it is important with infinite sets.

*(element with a label corresponding to the largest integer)

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

Let me put it another way then. The set of natural numbers covers the interval [1, infinity) but when dealing with a completed supertask we need to explore [1, infinity] because infinite steps have been completed.

By excluding the possibility of infinite numbers from the result, you confine yourself to steps which approach but never reach the end of the supertask.

ucim wrote:You can certainly enumerate a countably infinite set, the answer however is not finite. The cardinality of an infinite set is not the same as its largest element. Ordinal numbers are not the same as cardinal numbers; although this is not apparent with finite sets, it is important with infinite sets.

Is that not exactly what I was saying? At the end of the supertask you have actually enumerated the countably infinite set, and the answer is not finite. At every stage before the end of the supertask, the answer was finite, but at the end it is not. The final answer cannot be expressed using natural numbers but that doesn't mean the answer doesn't exist. It just needs infinite numbers to express it.

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

kryptonaut wrote:Let me put it another way then. The set of natural numbers covers the interval [1, infinity) but when dealing with a completed supertask we need to explore [1, infinity] because infinite steps have been completed.

Which infinity? There are more than one. There are an infinite "number" of infinities, each different and bigger than the other one. And then, there are the big ones.

There is no such interval as [1, infinity] because on the other side of the ] there is an even bigger infinity. Always. And not just "infinity plus one" but gargantuan monstrosities that make the biggest image of infinity you can imagine cower in comparison.

A supertask never finishes (ordinality), but it does become finished (cardinality).

In a similar vein, probability zero events can happen. (Select a real number between 3 and 4. The chance of picking pi is zero, but pi is in the set and can be picked.)

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

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

Sure, there are an infinite number of infinities - perhaps I should have specified Aleph_null or omega or something - but the point is you seem to be excluding all infinities which is making it impossible to express the outcome of the supertask.

ucim wrote:A supertask never finishes (ordinality), but it does become finished (cardinality).

I've tried to consistently point out that there is no last finite step even though there is an endpoint, precisely because it is a supertask.

The point is that when the supertask has finished you have enumerated the countably infinite set N and so the result is not in N.

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

How can Scenario A and Scenario R have different ends to their supertasks? They're the same after each step, so they must have the same limit, if such a limit exists. Specifically, claims keep being made about how in Scenario A, the jug must be empty, because no non-naturally-numbered ball is ever added to the jug, and every natural number is taken out. Why doesn't the same argument hold for Scenario R? No non-naturally-numbered ball is ever created, since at each time step each ball is labeled with a finite string of digits, and yet every natural number is *10'd away eventually.

It seems to me that for all the talk of intuitions being led astray by infinity, that people are letting their intuitions do the talking on the results of Scenarios A and R. The warring intuitions that there must be an infinite number of balls in the jug and that every ball must be removed from the jug are apparently ranking at different priorities in the two scenarios? If these two intuitions are warring in your mind, and you think Scenarios A and R can't possibly have the same limit, then I offer you a third choice: maybe the supertask has no limit, that it's poorly defined what happens "at infinity"?
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.

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

If you ask me, scenarios A and R do end up the same - with an infinite number of balls with infinite numbers on them.

Those who believe that the infinite numbers cannot be generated assert that the balls disappear instead.

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

kryptonaut wrote:
Xias wrote:But if X does not exist, then there is an object Y with the largest non-infinite label (since you are holding the complete set) - but that would imply a largest natural number, which also does not exist. So the premise that you can enumerate every single natural number leads to a contradiction.

This argument is just not true. I'm not sure how to convince you of it, but there can exist both no largest finite label, and no infinite label, without contradiction. That's true of set N, so why you are claiming that it can't be true of our set of objects that is mapped to N is unclear to me.

kryptonaut wrote:
Xias wrote:EDIT: And I forgot to say that you didn't acknowledge the first point that I made. If the set of all balls added is N∪K, then why is the set of ball removed not also N∪K?

K is the set that is required to terminate the infinite series of steps, to complete the supertask. The elements of K are infinite in magnitude, greater than any element of N. They are added at midnight, no more steps are performed after they are added. It is not possible to describe the step before they were added any more than you can point to the step before e is created from the infinite sum of 1/n! At the end of the process the real number e exists, during the process rational approximations to e exist.

That doesn't address my question at all, so perhaps I wasn't clear in asking it.

I'm assuming that we both agree that the two games where you play with balls in N and balls in (0,1) are identical. I contend that they are identical on the side of emptying the jug, while you contend that they are identical on the side of the jug ending up with balls not in N and not in Q∩(0,1), respectively.

If you are correct, then there is something stopping us from adding all of N, ten by ten, to the jug unless we also add K. Yet in order for K to remain in the jug, then you were somehow able to remove all of N, one by one, from the jug without removing K. How do we do this?

And if adding all of Q∩(0,1), ten by ten, to the jug forces us to also add all of Qc∩(0,1) at midnight, then how are we able to remove all of Q∩(0,1), one by one, from the jug without also removing Qc∩(0,1)? Even if you are correct about adding K and Qc∩(0,1) at midnight being the only way to resolve the game, it follows directly from that that K and Qc∩(0,1) are at once removed at midnight.

So either of the following must be the case:

- It is possible to describe a supertask that removes only balls N and Q∩(0,1), respectively, and it is therefore possible to describe a supertask that adds only them as well, and the combination of these two supertasks leaves the jug empty; or
- It is not possible to describe a supertask that adds only N or Q∩(0,1) balls to the jug without adding K or Qc∩(0,1) respectively as well. Then any supertask that removes N also removes K; and any supertask that removes Q∩(0,1) also removes Qc∩(0,1), and when you combine the supertasks you end up with an empty jug.

In either case, the jug is empty.

kryptonaut wrote:The point is that when the supertask has finished you have enumerated the countably infinite set N and so the result is not in N.

What do you think the "result" of an enumeration is? The enumeration of N results in the set N. Not a number. Not infinity. A set.

Cauchy wrote:How can Scenario A and Scenario R have different ends to their supertasks? They're the same after each step, so they must have the same limit, if such a limit exists. Specifically, claims keep being made about how in Scenario A, the jug must be empty, because no non-naturally-numbered ball is ever added to the jug, and every natural number is taken out. Why doesn't the same argument hold for Scenario R? No non-naturally-numbered ball is ever created, since at each time step each ball is labeled with a finite string of digits, and yet every natural number is *10'd away eventually.

It seems to me that for all the talk of intuitions being led astray by infinity, that people are letting their intuitions do the talking on the results of Scenarios A and R. The warring intuitions that there must be an infinite number of balls in the jug and that every ball must be removed from the jug are apparently ranking at different priorities in the two scenarios? If these two intuitions are warring in your mind, and you think Scenarios A and R can't possibly have the same limit, then I offer you a third choice: maybe the supertask has no limit, that it's poorly defined what happens "at infinity"?

By scenario A and R, you mean the original puzzle and the append-0 puzzle, respectively, right?

Your error occurs at the bolded statement. There is no reason to believe that a property that exists at every finite step also exists after an infinite number of steps.

The balls in scenario A are all removed, but in scenario R all remain and their numbers are replaced by non-natural numbers with infinite digits. It is not my intuition that tells me this; it is mathematics:

Scenario A:

Spoiler:
The math is in normal colored text. The mathematical statements are true irrespective of the balls in a jug game. The blue text is the equivalent condition or quality of the ball game that corresponds to the mathematical statement.

Let α1 = {1}, α2 = {2}, and so on; αn = {n}. This corresponds to the ball removed from the jug at step n.

Let β1 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, β2 = {11, 12, 13, 14, 15, 16, 17, 18, 19, 20}, and so on: βn = ∪i=0→9{10n-i}.This corresponds to the balls added to the jug at step n.

Then let
Ak = ∪i=1→kαi. This is the set of all balls that have been removed from the jug after step k.
Bk = ∪i=1→kβi. This is the set of all balls that have been added to the jug after step k.

Then
limk→∞Ak = limk→∞Bk = N. At midnight, all of the natural numbered balls have been added to the jug. Likewise, all of the natural numbers have been removed from the jug.

Now, it's clear that Ak⊆Bk for all k∈N. The set of balls that have been removed at step k is always only balls that have at some point been added to the jug. And it is a proper subset, because there are always balls that have been added that have yet to be removed.

It's also clear that

|Bk\Ak| < |Bk+1\Ak+1| k∈N. The number of balls in the jug is always increasing.

And finally,
|Bk\Ak| = 9k for all k in N, The number of balls in the jug at step k is 9k.
then limk→∞|Bk\Ak| = limk→∞9k = ∞ The limit of the number of balls in the jug at step k is infinity.

However, once you take k to infinity, if C is the set of all balls remaining at midnight:

C = ∪i=1→∞βi \ ∪i=1→∞αi = N \ N = 0. The difference between the set of balls added and the set of balls removed at midnight is the null set.

Then
|limk→∞Bk \ limk→∞Ak| = |N \ N| = 0. The number of balls in the jug at infinity is zero.

This shows that when evaluating what the set B \ A is at midnight, you get a result for which the cardinality is zero, despite the cardinality of Bk\Ak being divergent. Which limit takes precedence? Since cardinality is a function of the set itself, you have to evaluate what the set is to determine what the cardinality is, not the other way around. The limit of the cardinality does not matter, the limit of the set does.

Scenario B:

Spoiler:
While the sets have the same elements at every step, all balls with numbers divisible by 10 are not defined as the corresponding natural number, they are defined as a natural number multiplied by some power of ten. This makes no difference at finite steps, but after infinite steps the balls you have in the jug do not correspond to N anymore, but to a set of all natural numbers not divisible by ten appended with infintie zeroes.

I could spend a great deal of time formally defining the set of balls in the jug at the general step n. However, that would be number/symbol salad that would be difficult and distracting to read. It would also open me up to typos and other errors that have to do with trying to type it out, and not with the definition of the set itself. So instead, I'll just show a few cases:

Step 1: {2, 3, 4, 5, 6, 7, 8, 9, 1*10^1}
Step 2: {3, 4, ... , 9, 1*10^1, 11, ... , 19, 2*10^1}
Step 3: {4, ... , 9, 1*10^1, 11, ... , 19, 2*10^1, 21, ... , 29, 3*10^1}
Step 10: {all of the elements that existed at step 9 except for 1*10^1, 91, 92, ... , 99, 1*10^2}

When you evaluate the limit of this set, you end up with a set of {1*10^∞, 2*10^∞, 3*10^∞, ...} defined for all n not divisible by 10.

The intuitive contradiction of "But the set never has any infinite numbers in it before midnight" does not stop it from having this result. In the same way, the infinite sum of 1/2n evaluates to 2, despite always evaluating to a number less than 2 for any finite n.

If you don't believe me that the set can be different simply because the same numbers 10, 20, 30, etc. are evaluated differently, consider the series:

1, 2, 3, ...

What is the next number in the series? It depends on what each term is defined as. If it is defined as n, then the next term is 4. If it is defined as an+2=an+1+an, then the next term is 5. Just because in both cases the third term evaluates to 3 does not mean that they are the same series. The same goes for two series that are the same for every finite term, if the terms are defined as to have different limits.

Another example: y = 1 and y = x/x define the same set of points except for when x = 0. The definition of the relationship of y to x is what matters, not their equivalence at every other point.

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

Xias wrote:Your error occurs at the bolded statement. There is no reason to believe that a property that exists at every finite step also exists after an infinite number of steps.

And yet you believe that so vehemently in Scenario A, that no ball with an "infinite label" is added because there's no finite step in which it's added. This strikes me as somewhat hypocritical.

However, once you take k to infinity, if C is the set of all balls remaining at midnight:

C = ∪i=1→∞βi \ ∪i=1→∞αi = N \ N = 0. The difference between the set of balls added and the set of balls removed at midnight is the null set.

Why does C describe the set of balls in the jug at midnight? Certainly the set of balls (if such a set exists) is D = limn→∞(∪i=1→nβi \ ∪i=1→nαi), but it's a leap to say that you can pass this limit through the set difference operator, one I'd want to see you justify.

Scenario B:

While the sets have the same elements at every step, all balls with numbers divisible by 10 are not defined as the corresponding natural number, they are defined as a natural number multiplied by some power of ten. This makes no difference at finite steps, but after infinite steps the balls you have in the jug do not correspond to N anymore, but to a set of all natural numbers not divisible by ten appended with infintie zeroes.

I could spend a great deal of time formally defining the set of balls in the jug at the general step n. However, that would be number/symbol salad that would be difficult and distracting to read. It would also open me up to typos and other errors that have to do with trying to type it out, and not with the definition of the set itself. So instead, I'll just show a few cases:

Step 1: {2, 3, 4, 5, 6, 7, 8, 9, 1*10^1}
Step 2: {3, 4, ... , 9, 1*10^1, 11, ... , 19, 2*10^1}
Step 3: {4, ... , 9, 1*10^1, 11, ... , 19, 2*10^1, 21, ... , 29, 3*10^1}
Step 10: {all of the elements that existed at step 9 except for 1*10^1, 91, 92, ... , 99, 1*10^2}

When you evaluate the limit of this set, you end up with a set of {1*10^∞, 2*10^∞, 3*10^∞, ...} defined for all n not divisible by 10.

BUT I CAN WRITE THE NUMBERS IN SCENARIO A THE SAME WAY! Your "defined differently" argument has no bearing if the definitions give the same result for every finite step, since the limit only cares about results during finite steps.

The intuitive contradiction of "But the set never has any infinite numbers in it before midnight" does not stop it from having this result. In the same way, the infinite sum of 1/2n evaluates to 2, despite always evaluating to a number less than 2 for any finite n.

And yet apparently this intuition is spot-on during Scenario A?

If you don't believe me that the set can be different simply because the same numbers 10, 20, 30, etc. are evaluated differently, consider the series:

1, 2, 3, ...

What is the next number in the series? It depends on what each term is defined as. If it is defined as n, then the next term is 4. If it is defined as an+2=an+1+an, then the next term is 5. Just because in both cases the third term evaluates to 3 does not mean that they are the same series. The same goes for two series that are the same for every finite term, if the terms are defined as to have different limits.

I'm honestly not sure if you realize that two series that are the same at every finite term *must* have the same limit, if such a limit exists.

Another example: y = 1 and y = x/x define the same set of points except for when x = 0. The definition of the relationship of y to x is what matters, not their equivalence at every other point.

These have the same limit as x approaches 0.

So what you're telling me is that if the lights went out during each step, so that you couldn't see the exact process by which the jug progressed from step k to step k+1 (let's say someone else is working Thomson's lamp at double speed or something), then you'd answer "it's impossible to know" to the question of what happens at the end of the supertask.

Here's a math fact, since we're apparently listing them: if f(n) = g(n) for all integers n, and if lim_{x->infty} f(x) and lim_{x->infty} g(x) both exist as real number limits, then lim_{x->infty} f(x) = lim_{x->infty} g(x) = lim_{n->infty} f(n), where the last limit is a limit in integers. For this example, that means that if you claim Scenarios A and R can't have the same outcome, then you're really arguing that one or more of their limits can't exist.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.

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

Just a quick question as it's late here. (Incidentally Scenario A was originally the 'Append zero' and Scenario R the 'Replace a ball' scenario, you have them the other way round.)

In your description of the 'Add a zero' scenario, you are stating that
Xias wrote:When you evaluate the limit of this set, you end up with a set of {1*10^∞, 2*10^∞, 3*10^∞, ...} defined for all n not divisible by 10.

You seem happy that you produce infinitely long labels.

Can you explain why you don't get exactly the same limit in the 'Replace a ball' scenario? We are taking the limit, to infinity, of an increasing sequence of numbers, the number on the balls added. The limit at infinity of this sequence is a bunch of infinities for exactly the same reason. No, these are no longer natural numbers, because we have reached the limit of infinity just as you did with the labels. The reason these balls are not removed is because they are only there after the supertask has ended, there are no further steps at which they could be removed.

Another question - why does everyone assume that the balls are numbered with natural numbers? Sure, we are using them to count/number the balls, but these numbers could just as easily be taken from a superset of the naturals. That could be the reals or it could be N∪K or perhaps something else entirely. This is in fact what my explanation boils down to. We are counting with numbers that correspond with naturals all the way until the end of the supertask. Once the supertask has ended we have balls numbered with infinite numbers. Those numbers are not in N, but can you give a reason why they should be?

The whole essence of the supertask is to explore what happens not as we approach infinity, but when we allow ourselves to actually be there. In order to do that we have to use numbers outside the realm of N

This is analogous to adding an infinite sequence of rationals - when we have finished the task we may discover we no longer have a rational number. To make it clearer, lets say that at each step n we put a ball in the jug labelled with SUM0..n1/n! and remove the ball that's currently in there. The number on the ball is always rational until the supertask has ended, when the number has become e, an irrational with an infinitely long representation and not present in the set of rationals. And once the supertask is ended there are no further rationals that get added, and no further removals of the ball, so it stays in the jug. But on the way every rational in the sequence SUM0..n1/n! has been added, and removed. We don't have a rational number at the end, but that doesn't mean we don't have a number at all, or that we don't have a ball to put it on.

Ok, so that was more than a quick question Xias
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### Re: Infinite Balls and Jugs [solution]

Cauchy wrote:
Xias wrote:Your error occurs at the bolded statement. There is no reason to believe that a property that exists at every finite step also exists after an infinite number of steps.

And yet you believe that so vehemently in Scenario A, that no ball with an "infinite label" is added because there's no finite step in which it's added. This strikes me as somewhat hypocritical.

Is the union of all partitions of a set not the set itself? The only way for an infinite label to be added is if the procedure calls for it. My argument is not that "It doesn't happen at a finite step, so it doesn't happen at infinity." My argument is that "it doesn't happen by definition."

Cauchy wrote:
However, once you take k to infinity, if C is the set of all balls remaining at midnight:

C = ∪i=1→∞βi \ ∪i=1→∞αi = N \ N = 0. The difference between the set of balls added and the set of balls removed at midnight is the null set.

Why does C describe the set of balls in the jug at midnight? Certainly the set of balls (if such a set exists) is D = limn→∞(∪i=1→nβi \ ∪i=1→nαi), but it's a leap to say that you can pass this limit through the set difference operator, one I'd want to see you justify.

Is the set of balls in the jug at midnight not the set of all balls added, minus the set of all balls removed?

I can think of no way to define the limit of such a sequence of sets except for this: Take any element x in D = limn→∞ (Bn/An). Then there exists M in N such that x is in Bn/An for all n>M.

So applying that to our definitions of Bn and An, choose x in D. Then there exists M, etc. Then for all n>M, x is in Bn but not in An. Choose n*>M. Then x is in Bn* but not in An*. Then x is in βi for some i<=n*. x is also in αx, and 10(i-1) < x <= 10i. Then x is in An for all n>x. Then x is in An for all n>max({x, n*}). Then there exists some m>n* for which x is in both Bm and Am, and therefore not in Bm\Am. This is a contradiction. Therefore, x is not in D. Therefore, D is the null set.

Cauchy wrote:
Scenario B:

While the sets have the same elements at every step, all balls with numbers divisible by 10 are not defined as the corresponding natural number, they are defined as a natural number multiplied by some power of ten. This makes no difference at finite steps, but after infinite steps the balls you have in the jug do not correspond to N anymore, but to a set of all natural numbers not divisible by ten appended with infintie zeroes.

I could spend a great deal of time formally defining the set of balls in the jug at the general step n. However, that would be number/symbol salad that would be difficult and distracting to read. It would also open me up to typos and other errors that have to do with trying to type it out, and not with the definition of the set itself. So instead, I'll just show a few cases:

Step 1: {2, 3, 4, 5, 6, 7, 8, 9, 1*10^1}
Step 2: {3, 4, ... , 9, 1*10^1, 11, ... , 19, 2*10^1}
Step 3: {4, ... , 9, 1*10^1, 11, ... , 19, 2*10^1, 21, ... , 29, 3*10^1}
Step 10: {all of the elements that existed at step 9 except for 1*10^1, 91, 92, ... , 99, 1*10^2}

When you evaluate the limit of this set, you end up with a set of {1*10^∞, 2*10^∞, 3*10^∞, ...} defined for all n not divisible by 10.

BUT I CAN WRITE THE NUMBERS IN SCENARIO A THE SAME WAY! Your "defined differently" argument has no bearing if the definitions give the same result for every finite step, since the limit only cares about results during finite steps.

I acknowledged that you can write the numbers the same way. Why do you think that you can caps lock your way into changing my mind?

You stating matter-of-factly that the "limit only cares about results during finite steps" contradicts the actual math that shows that the two limits converge to different sets. They follow the same path for natural n, but for different reasons, and are therefore different sets at infinity.

Cauchy wrote:
The intuitive contradiction of "But the set never has any infinite numbers in it before midnight" does not stop it from having this result. In the same way, the infinite sum of 1/2n evaluates to 2, despite always evaluating to a number less than 2 for any finite n.

And yet apparently this intuition is spot-on during Scenario A?

What are you talking about? The intuitive answer to the original game is "The balls are always increasing, so it can never empty." I have shown the opposite to be true mathematically. In what way am I invoking intuition?

Cauchy wrote:
If you don't believe me that the set can be different simply because the same numbers 10, 20, 30, etc. are evaluated differently, consider the series:

1, 2, 3, ...

What is the next number in the series? It depends on what each term is defined as. If it is defined as n, then the next term is 4. If it is defined as an+2=an+1+an, then the next term is 5. Just because in both cases the third term evaluates to 3 does not mean that they are the same series. The same goes for two series that are the same for every finite term, if the terms are defined as to have different limits.

I'm honestly not sure if you realize that two series that are the same at every finite term *must* have the same limit, if such a limit exists.

I was a bit unclear. The bolded part should be

"The same goes for two series of sets that are the same set for every finite term of the series, if the terms within the sets are defined as to have different sets as their limits."

One series of sets is found by taking the difference between the sets that are terms of two different series. This is all well defined. Since the terms in both series converge to the same set, their difference is the null set at infinity.

In the other series of sets, each set is a set of functions. Each of those functions is defined in a way that the function diverges. The functions that serve as its terms evaluate to the same values as the terms in the first set, but the limit of the series is a set of infinite limits of functions.

Perhaps I'm wrong, though. Formalizing it is too difficult, and perhaps formalizing it would show the error. If that's the case, it still wouldn't change the fact that the jug is empty in the original game; at best, the append-0 game may not resolve to a consistent answer. But that doesn't matter; the append-0 game was a fun afterthought - a way to introduce more intuitive contradictions to the mathematical result of the original game.

Cauchy wrote:So what you're telling me is that if the lights went out during each step, so that you couldn't see the exact process by which the jug progressed from step k to step k+1 (let's say someone else is working Thomson's lamp at double speed or something), then you'd answer "it's impossible to know" to the question of what happens at the end of the supertask.

Yes? I'm not sure why that's somehow noteworthy. If the process is well-defined in a way that empties the jug, then the jug is empty. If it's well-defined in a way that leaves infinite balls in the jug (like putting ball n in the jug at step n, and never removing a ball) then there are infinite balls in the jug. If the process is ill-defined because I'm blindfolded or distracted or someone is giving me a seizure with the lights, then what's the problem with saying I don't know?

Cauchy wrote:Here's a math fact, since we're apparently listing them: if f(n) = g(n) for all integers n, and if lim_{x->infty} f(x) and lim_{x->infty} g(x) both exist as real number limits, then lim_{x->infty} f(x) = lim_{x->infty} g(x) = lim_{n->infty} f(n), where the last limit is a limit in integers. For this example, that means that if you claim Scenarios A and R can't have the same outcome, then you're really arguing that one or more of their limits can't exist.

We're dealing with sets. Typically, a function that increases without bound diverges. A set that increases without bound (by adding n at each step n) converges to N. If you would like to justify why a rule of limits of functions should apply to limits of functions of sets, I'm all ears.

kryptonaut wrote:Just a quick question as it's late here. (Incidentally Scenario A was originally the 'Append zero' and Scenario R the 'Replace a ball' scenario, you have them the other way round.)

Thanks for the correction. I switched to just saying "original" and "append-0" to refer to them clearly.

kryptonaut wrote:In your description of the 'Add a zero' scenario, you are stating that
Xias wrote:When you evaluate the limit of this set, you end up with a set of {1*10^∞, 2*10^∞, 3*10^∞, ...} defined for all n not divisible by 10.

You seem happy that you produce infinitely long labels.

Can you explain why you don't get exactly the same limit in the 'Replace a ball' scenario? We are taking the limit, to infinity, of an increasing sequence of numbers, the number on the balls added.

We're not taking a limit of a sequence of numbers. We are taking a limit of a sequence of sets. The sets are well-defined as containing only natural numbers. When we remove no balls, the limit of that sequence of sets is the set of natural numbers, since those sets are well-defined as being partitions of the set of natural numbers. When we remove the lowest balls, the sequence of sets is a difference of two sets which both converge to the set of natural numbers.

When we append-0, we are taking the limit of a sequence of sets that are defined as containing functions on the natural numbers. f1(n) is evaluated to 10 from n=1 to 9, 100 from n=10 to 99, 1000 from n = 100 to 999, etc. At step n, the set is {f1, f2, ... , f9n}. So as n approaches infinity, you end up with infinite sets of functions. Each of those functions are functions of n, so we take their limit as well and they all diverge. So we end up with a set of infinite limits of functions that diverge to infinity.

kryptonaut wrote:Another question - why does everyone assume that the balls are numbered with natural numbers?

From the original problem thread:

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

This is typically taken in mathematics to mean they are numbered with the natural numbers starting with 1. If someone wishes to include non-natural numbers, they say so.

kryptonaut wrote:Once the supertask has ended we have balls numbered with infinite numbers. Those numbers are not in N, but can you give a reason why they should be?

I don't accept the premise. The union of all partitions of the set of natural numbers is the set of natural numbers. We don't have any infinite numbers because infinite numbers are not in N.

kryptonaut wrote:The whole essence of the supertask is to explore what happens not as we approach infinity, but when we allow ourselves to actually be there. In order to do that we have to use numbers outside the realm of N

This is analogous to adding an infinite sequence of rationals - when we have finished the task we may discover we no longer have a rational number. To make it clearer, lets say that at each step n we put a ball in the jug labelled with SUM0..n1/n! and remove the ball that's currently in there. The number on the ball is always rational until the supertask has ended, when the number has become e, an irrational with an infinitely long representation and not present in the set of rationals. And once the supertask is ended there are no further rationals that get added, and no further removals of the ball, so it stays in the jug. But on the way every rational in the sequence SUM0..n1/n! has been added, and removed. We don't have a rational number at the end, but that doesn't mean we don't have a number at all, or that we don't have a ball to put it on.

This will be the third time that I'm going to ask you to reconcile the apparent contradiction in this logic, and I likely won't respond to you again until you at least provide your thoughts on it, because it is crucial to me understanding why you think the way you do. I'm okay with spending as much time on this conversation as you are; but only so long as I feel like you are willing to address my questions when I ask you to.

In the supertask that you describe, you claim that eventually (at midnight, if you will) you add a ball with the number that is the limit of the numbers on the previous balls. I'm willing to grant that in order to proceed. You claim that this is a necessary consequence of performing the supertask. I am also willing to grant this. However, at the same time you claim that the sub-supertask of removing balls has an end result of removing all balls in the set of balls added, except for e.. How is the supertask capable of resulting in an infinite sequence of balls labeled SUM0..n1/n! being removed from the jug, but incapable of adding the same infinite sequence of balls without also adding e?

The same goes for adding and removing balls in N, but being left with K. How is N removed without also removing K?

Do you not understand the question that I'm asking? You're saying at once that "The supertask cannot add balls from N without adding balls from K. Then the supertask removes balls from N, leaving K in the jug." Do you see the blatant contradiction?

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

Xias wrote:The only way for an infinite label to be added is if the procedure calls for it. My argument is not that "It doesn't happen at a finite step, so it doesn't happen at infinity." My argument is that "it doesn't happen by definition."

But the procedure does call for it. It calls for balls with number n to be created, for all n up to and including infinity. The 'including' part is essential, it is what makes this a supertask. And after infinite steps have been taken the supertask has ended so nothing more gets removed.

Xias wrote:In the supertask that you describe, you claim that eventually (at midnight, if you will) you add a ball with the number that is the limit of the numbers on the previous balls. I'm willing to grant that in order to proceed. You claim that this is a necessary consequence of performing the supertask. I am also willing to grant this. However, at the same time you claim that the sub-supertask of removing balls has an end result of removing all balls in the set of balls added, except for e.. How is the supertask capable of resulting in an infinite sequence of balls labeled SUM0..n1/n! being removed from the jug, but incapable of adding the same infinite sequence of balls without also adding e?

The same goes for adding and removing balls in N, but being left with K. How is N removed without also removing K?

Do you not understand the question that I'm asking? You're saying at once that "The supertask cannot add balls from N without adding balls from K. Then the supertask removes balls from N, leaving K in the jug." Do you see the blatant contradiction?

I understand the question perfectly, but I don't understand why you can't follow my reasoning. This is a supertask. It follows all steps up to and including infinity, then stops (at midnight). Now, when it has performed an infinite number of steps, it has created a ball numbered exactly e. The supertask ends because by definition it has to perform all steps from 1 up to and including infinity. We are not at liberty to investigate the 'penultimate' step, but at every step a ball is added and a different one removed. After adding and removing rationally-numbered balls an infinite number of times we are left with an irrationally numbered ball numbered e, then the supertask ends. Logically the supertask has to end anyway - we can't make the number any closer to e because it is exactly e already.

The same argument goes for naturally numbered balls - after adding and removing increasing naturally-numbered balls an infinite number of times we are left with infinitely numbered balls, then the supertask ends.

There is no contradiction - the supertask ends after infinite steps, and after infinite steps the numbers on the balls have reached the infinite limit of N. The only tricky part to get one's head around is that we cannot investigate the transition between the finite steps and the state where we've performed infinite steps. This would be like trying to evaluate infinity minus one, or asking what is the number that's almost e but differs only in the last decimal place.

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

Why should a limit of a sequence care about the specific formula you happen to use to define the terms of the sequence? It's certainly not true for sequences of real numbers, so why should it suddenly become true for sequences of sets? This is what I can't understand about your wanting Scenarios A and R (I'm sorry that I got them backwards) to behave different ways.

I bothered to actually look up the set-theoretic limit, and it says that both Scenarios A and R converge to the empty set, since the set-theoretic limit is effectively the pointwise limit if such a limit exists (and it exists in both scenarios). If you want to argue this, you have to argue that supertasks don't represent the limit of their finite steps.
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Xias
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### Re: Infinite Balls and Jugs [solution]

Cauchy wrote:Why should a limit of a sequence care about the specific formula you happen to use to define the terms of the sequence? It's certainly not true for sequences of real numbers, so why should it suddenly become true for sequences of sets? This is what I can't understand about your wanting Scenarios A and R (I'm sorry that I got them backwards) to behave different ways.

I bothered to actually look up the set-theoretic limit, and it says that both Scenarios A and R converge to the empty set, since the set-theoretic limit is effectively the pointwise limit if such a limit exists (and it exists in both scenarios). If you want to argue this, you have to argue that supertasks don't represent the limit of their finite steps.

Suppose we have a sequence of sets, where Sn[/b] is defined as {f[sub]1(n), f2(n)}, where f1(n) = (-1)^n, f2(n) = (-1)^(n+1). Is the limit of this sequence {1, -1}? Or is the limit of the sequence {lim f1, lim f2}?

kryptonaut wrote:
Xias wrote:The only way for an infinite label to be added is if the procedure calls for it. My argument is not that "It doesn't happen at a finite step, so it doesn't happen at infinity." My argument is that "it doesn't happen by definition."

But the procedure does call for it. It calls for balls with number n to be created, for all n up to and including infinity. The 'including' part is essential, it is what makes this a supertask. And after infinite steps have been taken the supertask has ended so nothing more gets removed.

Infinity is not a number that n can be, supertask or not.

kryptonaut wrote:I understand the question perfectly, but I don't understand why you can't follow my reasoning.

You clearly don't understand the question, because all you have done is restate the contradiction. Perhaps it will be easier if you respond point by point?

1. You say that in order to add the balls from N to the jug, you must also add the balls from K. So at midnight, you have added the balls from NK. Is this right?

2. You then say that at midnight all of the natural numbers have been removed, and all of the balls from K remain. Is this right?

3. So then, one at a time, you were able to remove exactly all of the balls from N from the jug. Do you agree?

4. How did you remove exactly all of the balls from N from the jug? By your reasoning - not mine! - you can't do that.

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

Xias wrote:Infinity is not a number that n can be, supertask or not.

I disagree, n can take an infinite value. Just like an infinite sum of rationals can be irrational, an infinite sum of finite numbers (or a number that has been increased an infinite number of times) can be infinite. You seem to accept this in the case of the ever-growing label, but not in the case of the ever-incrementing number - it puzzles me how you differentiate between the two.

Out of interest, what value do you think n has at the end of the supertask? If it is finite then the supertask by definition has not completed.

Xias wrote:You clearly don't understand the question, because all you have done is restate the contradiction. Perhaps it will be easier if you respond point by point?

1. You say that in order to add the balls from N to the jug, you must also add the balls from K. So at midnight, you have added the balls from N∪K. Is this right?

2. You then say that at midnight all of the natural numbers have been removed, and all of the balls from K remain. Is this right?

3. So then, one at a time, you were able to remove exactly all of the balls from N from the jug. Do you agree?

4. How did you remove exactly all of the balls from N from the jug? By your reasoning - not mine! - you can't do that.

I'll assume we're talking about the variant of the puzzle that removes the lowest numbered ball. After n finite steps we all agree that there are 9n balls in the jug, numbered from n+1 to 10n

1. The numbers are just numbers, they don't know what set they are in. Before midnight the numbers we are adding all have the property of being finite positive integers that we choose to call natural numbers. At midnight, after an infinite number of steps, the numbers no longer have the 'finiteness' property that we demand from numbers in the set we call natural numbers - they are now infinitely big. I have chosen to put numbers with these properties into a set called K, but the numbers don't care. They just do what they do, they are what they are. After increasing a number an infinite number of times the number is infinitely big, and we choose not to classify it as a natural number.

2. At midnight after an infinite number of steps have been taken, the numbers on the balls in the jug have increased an infinite number of times so are all infinite. The number of balls in the jug has also increased an infinite number of times, so there are an infinite number of balls in the jug, all bearing an infinite number. The balls are not tagged in some way with an N or a K to say what set they are in - they just bear a number which may be finite or infinite. If it's finite we can classify it as a natural number (although we could just as validly classify it as a real if we chose to), otherwise if it's infinite we can say it is a member of K

3. The numbers ended up being increased an infinite number of times so we no longer classify them as belonging to N

4. At every finite step there are only finite numbered balls in the jug. After infinite steps these numbers have been increased to infinite values so are no longer classified as belonging to N

The simplified version of the puzzle surely makes it easier to visualise, where we start with ball 0 in the jug and then at step n we add ball n and remove ball n-1. So after each step we have just one ball in the jug, numbered n. I assert that at midnight after an infinite number of steps we are left with a ball carrying an infinite number. We don't call that a natural number, but the number doesn't worry about what we call it, it is just a number which happens to be infinite. What do you think we are left with at midnight, and why?

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

kryptonaut wrote:
Xias wrote:Infinity is not a number that n can be, supertask or not.

I disagree, n can take an infinite value. Just like an infinite sum of rationals can be irrational, an infinite sum of finite numbers (or a number that has been increased an infinite number of times) can be infinite. You seem to accept this in the case of the ever-growing label, but not in the case of the ever-incrementing number - it puzzles me how you differentiate between the two.

Out of interest, what value do you think n has at the end of the supertask? If it is finite then the supertask by definition has not completed.

n does not have a value at the end of the supertask. n is shorthand for a variable term in a set of natural numbers. At the end of the supertask, n has been every natural number. That doesn't mean n is now a non natural number. n isn't being called upon anymore because the supertask is over.

With the sum to e, if you ask "for which value of n is the sum equal to e?" The answer is "none." There is no value of n for which the sum is e. n is not infinity. The limit of the sum is e as n approaches infinity, but that is a different statement. We may have jumped from a series of rational numbers to an irrational number, but that has nothing to do with n. It's not a "thing" to keep existing at midnight; it's a representation of the enumeration done on a set in order to make it into a sequence. (See the rest of my post).

kryptonaut wrote:At midnight, after an infinite number of steps, the numbers no longer have the 'finiteness' property that we demand from numbers in the set we call natural numbers - they are now infinitely big.

This just isn't true. If we removed no balls, and merely added balls at every step, then we would have balls that correspond exactly with n. The only justification for why there would be infinitely large numbers that you have provided is because you said so.

If we have a set of balls, labeled with each natural number, and no other numbers, we can put them in a pile. Then we can separate the pile a number of ways: into even numbers and odd numbers, into numbers less than 101 and more than 99, into primes and composites. We can separate it infinitely as well, into individual piles, or piles of ten consecutive numbers, or a pile for each set of powers of the same prime number (and a pile with everything else). This is called partitioning.

If you then scoop them all back up at once, you have the set of all natural numbers again. Or, if you're some sort of time-god, you can order them and pick them up one at a time over an eternity, and in the end you have the set of all natural numbers again. If you're capable of a supertask, then you can pick them up one at a time over a finite time defined by that supertask, and end up with exactly the set of natural numbers in your hands.

You keep making reference to a different concept entirely, which is that if you start at 1 and keep counting upward forever, you diverge to infinity. But it's not the same thing. Don't get me wrong, it applies to the cardinality, and it applies to the max function, but both of those are functions of the elements within the set itself.

There are multiple different things here that we can observe and ask questions about. Lets look at the sum to e.

1. We can examine the sequence of sums and find that the limit of this sequence is e. This is what you keep referencing.

2. We can examine the sequence of terms that sum to e. This is a sequence that gets smaller and smaller, eventually converging to zero.

3. We can examine the infinite series which sums the sequence in (2). This is an expression of infinite terms that evaluates to e. In fact, it is representative of the limit of the sequence in (1), but is separate from the sequence itself.

4. We can examine the set of terms in (1). This set has a supremum of e.

All of these things are related, but not identical.

You are conflating many of the properties of each. When we define the sequence in (1), we are indexing elements in (4) so as to create a sequence that converges to the supremum of the set. e is not in the set. The act of indexing (enumerating) the set is to assign to each term a value of n for all n. e is not in the set, so it does not get indexed..

If we indexed the set in the same way as in (1), e would still not be in the set. If we then partitioned that set into infinite subsets, each with one term from the original set, e would still not be in the set. If we then created a supertask that took the union of every partitioned subset one at a time, we would end up with the original set again, and e would still not be in the set. We created a sequence of sets, which is different from the sequence in (1), and which converges to the set in (4). e is not in this new sequence, nor in any of the sets that make up its terms, nor in the set it converges to.

Likewise, the set of natural numbers forms a sequence that diverges to a non-natural number. But if you take the set apart, and then put it back together using a supertask, you don't end up with a set that includes any nonnatural numbers. You end up with a sequence of sets that converge to the set of natural numbers, and no nonnatural number occurs in that sequence, nor an any of the sets that make up its terms, nor in the set that it converges to.

All of this is acceptable, consistent, and correct in mathematics. You are conflating series and sets in a way that is not justified, and again, I believe you are making up these relationships of rules to justify your intuition telling you that the jug can't be empty. It's certainly not consistent with mathematics. If the jug is not empty, it's not for any reason you have given.

Maybe I'm doing the same thing with regard to the append-0 game. Cauchy seems to think so, and I'm doing my best to investigate why I think what I'm saying is justified. And I might turn out to be wrong. So please don't take this as an insult.

As for your last question: we have a sequence of sets which converges to an empty set. So the jug is empty at midnight.

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

Xias wrote:Suppose we have a sequence of sets, where Sn is defined as {f1(n), f2(n)}, where f1(n) = (-1)^n, f2(n) = (-1)^(n+1). Is the limit of this sequence {1, -1}? Or is the limit of the sequence {lim f1, lim f2}?

It's {-1, 1}. Maybe a more enlightening example would be that if R_n = {1/n}, then the set-theoretic limit of R_n is empty and is specifically not {0}.

Again, you can argue that the set-theoretic limit is not what a supertask finishes with, but that's really the only line of attack here.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.

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

Xias wrote:
kryptonaut wrote:Out of interest, what value do you think n has at the end of the supertask? If it is finite then the supertask by definition has not completed.

n does not have a value at the end of the supertask. n is shorthand for a variable term in a set of natural numbers. At the end of the supertask, n has been every natural number. That doesn't mean n is now a non natural number. n isn't being called upon anymore because the supertask is over.

Ok, that was the programmer in me talking. Let me ask it this way - was the greatest value of n that was used to perform the supertask finite or infinite? I contend that the supertask is defined to run ad infinitum - to infinity, the ineffable upper limit of the natural numbers - so n ultimately takes the value of that limit which is infinite by definition. If the greatest n was finite then we could have performed more steps, so the supertask wasn't over yet.

Xias wrote:
kryptonaut wrote:At midnight, after an infinite number of steps, the numbers no longer have the 'finiteness' property that we demand from numbers in the set we call natural numbers - they are now infinitely big.

This just isn't true. If we removed no balls, and merely added balls at every step, then we would have balls that correspond exactly with n. The only justification for why there would be infinitely large numbers that you have provided is because you said so.

Not 'because I said so,' but because the creator of the puzzle said so when setting it up as a supertask.

To be clear: we are not evaluating something as n approaches infinity. The whole purpose of the supertask is to allow us to consider the situation when n actually has become infinite.

If we remove no balls then yes, we have balls that correspond exactly with n. In this case the supertask is specifically set up to allow n to become infinite, so we have balls that correspond with infinite n.

Xias wrote:If we have a set of balls, labeled with each natural number, and no other numbers, we can put them in a pile. Then we can separate the pile a number of ways: into even numbers and odd numbers, into numbers less than 101 and more than 99, into primes and composites. We can separate it infinitely as well, into individual piles, or piles of ten consecutive numbers, or a pile for each set of powers of the same prime number (and a pile with everything else). This is called partitioning.

If you then scoop them all back up at once, you have the set of all natural numbers again. Or, if you're some sort of time-god, you can order them and pick them up one at a time over an eternity, and in the end you have the set of all natural numbers again. If you're capable of a supertask, then you can pick them up one at a time over a finite time defined by that supertask, and end up with exactly the set of natural numbers in your hands.

This argument is also true if we have a set of balls labelled with infinite numbers, or some of each, or anything you like.

Xias wrote:e is not in the set.

e is not in the set if it is enumerated for any finite n. The supertask is a mechanism that allows us to investigate the set at infinite n, at which point e is in the set.

You seem determined to label every step with a finite natural number even though the supertask by definition takes an infinite number of steps. This is leading to contradictions.

If you accept that the supertask has ended then n must have reached the upper limit of all finite numbers, which by definition is an infinite value. If you insist that n be finite then there is always at least one more step to be done and the supertask is impossible to complete.

I think you have a philosophical choice:
a) to deny that infinite numbers can arise at the upper limit of the natural numbers, and therefore to reject the possibility of the supertask ending, or
b) to accept and explore the possibility of performing an infinite number of steps, and then try to work out the properties of the infinite numbers that arise.
I am taking the latter approach - do you think there are other options that don't lead to self-contradictions?

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

kryptonaut wrote:
Xias wrote:
kryptonaut wrote:Out of interest, what value do you think n has at the end of the supertask? If it is finite then the supertask by definition has not completed.

n does not have a value at the end of the supertask. n is shorthand for a variable term in a set of natural numbers. At the end of the supertask, n has been every natural number. That doesn't mean n is now a non natural number. n isn't being called upon anymore because the supertask is over.

Ok, that was the programmer in me talking. Let me ask it this way - was the greatest value of n that was used to perform the supertask finite or infinite?

All values of n are finite by definition. There is no "greatest value of n." Since such a number does not exist, it has neither the property of being finite nor infinite. You might as well be asking "Is the greatest number in the set N odd or even?"

kryptonaut wrote:I contend that the supertask is defined to run ad infinitum - to infinity, the ineffable upper limit of the natural numbers - so n ultimately takes the value of that limit which is infinite by definition. If the greatest n was finite then we could have performed more steps, so the supertask wasn't over yet.

There are infinitely many values for n to take without ever taking a value of infinity. The supertask does infinite tasks, one for each value of n, of which there are infinitely many possible values, all of which are within the set N, which is an infinite set.

Saying that the greatest n must be infinite because it being finite leads to a contradiction is like saying that the greatest number in N must be odd because being even is a contradiction.

kryptonaut wrote:
Xias wrote:
kryptonaut wrote:At midnight, after an infinite number of steps, the numbers no longer have the 'finiteness' property that we demand from numbers in the set we call natural numbers - they are now infinitely big.

This just isn't true. If we removed no balls, and merely added balls at every step, then we would have balls that correspond exactly with n. The only justification for why there would be infinitely large numbers that you have provided is because you said so.

Not 'because I said so,' but because the creator of the puzzle said so when setting it up as a supertask.

To be clear: we are not evaluating something as n approaches infinity. The whole purpose of the supertask is to allow us to consider the situation when n actually has become infinite.

With supertasks they are the same! You are attributing something to supertasks that nobody else ever does. You've made it up. In order to do something for every n in N, you need an infinite number of steps that are also indexed by N. The supertask only serves to define a time for every n in N such that the sequence of time converges, so that the infinitely many steps (of n taking every value in N) can occur in finite time. "At midnight" is the limit of the series of time as n approaches infinity, so to judge what exists "at midnight" we also look at the limit of the series of whatever the supertask is doing as n approaches infinity.

From wikipedia: "In philosophy, a supertask is a countably infinite sequence of operations that occur sequentially within a finite interval of time."

Countably infinite means that it can be indexed by N. Infinity can't be indexed by N, nor can any infinitely large number. So your requirement that a supertask involves steps outside of N is completely a creation of your own mind. The infinite cardinality of N allows us to do infinite things without actually calling on an infinitely large value.

kryptonaut wrote:
Xias wrote:e is not in the set.

e is not in the set if it is enumerated for any finite n. The supertask is a mechanism that allows us to investigate the set at infinite n, at which point e is in the set.

You seem determined to label every step with a finite natural number even though the supertask by definition takes an infinite number of steps. This is leading to contradictions.

The set is enumerated for infinite n in the sense that there are infinitely many finite values for n to take. And e is still not in the set. For all values of the infinitely many values of n, every single one, e is not in the set.

I can label every single step with a natural number and still have an infinite number of steps, since there are infinitely many natural numbers. So this contradiction is not with how supertasks are defined; it's with a rule that you have made up all on your own.

kryptonaut wrote:If you accept that the supertask has ended then n must have reached the upper limit of all finite numbers, which by definition is an infinite value. If you insist that n be finite then there is always at least one more step to be done and the supertask is impossible to complete.

n only ever takes a finite value, but takes infinitely many of them, since it takes every value in n. n doesn't "reach" the limit as n approaches infinity. What n does do is take the value of every element in N. Which it can do, because for every such value there is a time defined in the supertask.

I'm not insisting that n "is" anything, let alone a finite value. n takes on infinite values, all of which are finite. There isn't "at least one more step to be done" because n takes all values in N. Asking what value n has at midnight is a nonsense question, because n has finished taking the values of all of the natural numbers and we are not calling it anymore. It is neither infinite nor finite at midnight because it has no value at all.

kryptonaut wrote:I think you have a philosophical choice:
a) to deny that infinite numbers can arise at the upper limit of the natural numbers, and therefore to reject the possibility of the supertask ending, or
b) to accept and explore the possibility of performing an infinite number of steps, and then try to work out the properties of the infinite numbers that arise.
I am taking the latter approach - do you think there are other options that don't lead to self-contradictions?

I do, since everything I've said in this post is consistent with itself. There are no self-contradictions, only contradictions with something in your mind, which I can only assume is based on some misunderstanding of the properties of N, or what "n" means, or what a supertask is, or some combination of the three.

" to deny that infinite numbers can arise at the upper limit of the natural numbers " - I have denied no such thing. There is no reason to consider limn->infinity n because the supertask is defined only for all n in N, and has nothing to do with the sequential limit of n itself.

"to accept and explore the possibility of performing an infinite number of steps" - This is the most ironic thing, since it is you who refuses to "accept and explore the possibility" of performing a step for every natural number, which is an infinite number of steps, which does not include infinity or any infinitely large numbers.

How about this third philosophical choice:

c) Understand how supertasks are actually defined.

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

kryptonaut wrote:If you accept that the supertask has ended then n must have reached the upper limit of all finite numbers, which by definition is an infinite value. If you insist that n be finite then there is always at least one more step to be done and the supertask is impossible to complete.
I know of no such definition in generally accepted mathematics. There is no upper limit of all finite numbers.

Consider the supertask (ok, a hypertask, but that doesn't matter here) of piling all the non-negative real numbers that are less than one on a stack, in order, using the rule "at t=k, the real number k is placed on the stack if it is nonnegative and less than one". When the supertask is finished, will 1 be on the stack? You certainly have reached 1, the same way the jug supertask "reaches" infinity. Yet, in neither case is this supremum actually added to the stack.

kryptonaut wrote:b) to accept and explore the possibility of performing an infinite number of steps, and then try to work out the properties of the infinite numbers that arise.
I am taking the latter approach - do you think there are other options that don't lead to self-contradictions?
Yes, and they will become apparent when you take the simplest cases of supertasks. More complex supertasks are really those simple tasks in disguise. Strip off the disguise and see it for what it hides. The hypertask I proposed is "even harder" (it's an uncountable set) but I'll bet it's easier for you to get your mind around, because there's no disguise. Don't be misled by the disguise a supertask wears - you never reach infinity even if you do an infinite number of steps.

You can't get there from here.

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

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

Cauchy wrote:
Xias wrote:Suppose we have a sequence of sets, where Sn is defined as {f1(n), f2(n)}, where f1(n) = (-1)^n, f2(n) = (-1)^(n+1). Is the limit of this sequence {1, -1}? Or is the limit of the sequence {lim f1, lim f2}?

It's {-1, 1}. Maybe a more enlightening example would be that if R_n = {1/n}, then the set-theoretic limit of R_n is empty and is specifically not {0}.

Again, you can argue that the set-theoretic limit is not what a supertask finishes with, but that's really the only line of attack here.

Okay, after a some reflection I'll abandon the "it's a set with a different limit even though it's the same set at every n" argument as I can't really support it.

I do think there's something else going on with the append-0 case, and I'll walk you through my process:

The original paradox of adding ten, removing the lowest, is that there is a conflict between the ideas of "the number of balls in the jug is always increasing" and "Every ball in the jug is eventually removed." They seem like they can't both be true at the same time. The pencil-eraser analogy posted much earlier in the thread inspired me to come up with an analogous situation involving purely mathematical objects that is well-defined and comes to the same result, but has an obvious answer. (If you are unfamiliar, the pencil-eraser analogy is one where at every step you make ten marks on a ruler, at points defined in the supertask as to approach the end of the ruler; meanwhile, you are dragging the eraser behind it in a way that erases one mark at each step. At the end, both the pencil and eraser have reached the end of the ruler).

The result was my set theory proof, which doesn't so much prove that the jug is empty; but more so proves that "the cardinality of the limit of a sequence of sets is not necessarily equal to the limit of the sequence of cardinalities of each set in the sequence." I could have used any number of different constructions of sets to prove that, but I chose to construct sets that were analogous to the balls in the game, and what they were doing. Once we have that statement, there is no longer a contradiction. If we are asking about what is in the jug at midnight, we are asking about the set that the sequence converges to; and that cardinality is 0 despite the sequence of cardinalities diverging.

With the append-0 case, there's a new paradox introduced, which is that there is a conflict between "The sequence of sets converge to the null set" and "no ball is ever removed from the jug."

I can see a few ways to resolve this:

1. The sequence of sets in the first game do not converge to the null set; I don't believe that this is true.

2. The balls in the jug stop existing because there are no natural numbers that can be put on them, and so we end up with an empty jug in both cases; this is immensely unsatisfying to me.

3. There is something different about the sets in the append-0 game that causes them to converge to a set that is not the null set.

I thought (3) is the case, and I still do. Perhaps I'm grasping at something that isn't there, but if you'll entertain me for a moment I think I have a better explanation than I had before.

In the original case, each ball and label are attached to each other throughout the whole game. I don't mean physically attached with adhesive; I mean they are linked by definition. There is no need to even have labels, as long as the balls have the identity of the natural number they are associated with. This is what makes the set of balls placed into the jug correlate perfectly with N.

In the append-0 case, the balls and labels are not the same. The ball labeled 100 at the end of step 10 still has the identity of the first ball added to the jug. If we take all of the balls that will be added (all of the balls labeled with natural numbers not divisible by 10) and index them, then ball a1 is a ball that takes on the label 1, 10, 100, etc. But it's still ball an. We can then construct a set of all of the balls that will be added and call it A, indexed by N. Then the set of balls added in the supertask is A, and the set of balls removed is the null set, so the set of balls remaining at midnight is A.

Why does this set have any importance compared to the set of what the labels on the balls are?

Because I can ask two different questions about the balls in the jug at any step n: a) what is the set of labels contained on the balls? and b) what is the set of initial labels on each of the balls?

The answers to both of these at, for example, step 2, is: a) the set of natural numbers from 3 to 20; and b) the union of the set of natural numbers from 1 to 9 with the set of natural numbers from 11 to 20 19 (thanks Demki).

In the original puzzle, that question always has the same answer due to the identity of the balls being attached to the label; in the append-0 case, they are different and they are different in a way that is well-defined.

So, I think that we can resolve the paradox in this way: We have a sequence of sets of balls that is different from the sequence of sets of balls in the first game; and a sequence of sets of labels which is the same as the sequence of sets of labels in the original game. At midnight, we have an infinite set of balls, which each corresponds with a value in A, that were never removed. We also have a sequence of sets of labels which converges to the null set, so we could say that there are no labels. Or, we could examine instead the sequence of values that each label took, and find that each of those sequences diverge, and upon close inspection to the rules of the game, diverge because they each take a value equal to n with an infinite number of zeroes. I'm partial to the latter, which i think is okay to do since the labels are not constrained to natural numbers in the same way that balls with unchanging labels are. However, I'm not opposed to accepting that there are no labels either, if there is such a constraint. What matters is, the sets of balls aren't actually the same at each step n, and the paradox is resolved.
Last edited by Xias on Mon Nov 21, 2016 7:53 pm UTC, edited 1 time in total.

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

Xias wrote:The answers to both of these at, for example, step 2, is: a) the set of natural numbers from 3 to 20; and b) the union of the set of natural numbers from 1 to 9 with the set of natural numbers from 11 to 20.

I believe you mean from 11 to 19

I'd like to add that the 'game' I proposed earlier better illustrates this, where you have just 1 ball in the jug with label of n 0s at step n. Which further illustrates that in the append-0 game the labels are not 'attached' to the balls, since the set of balls in the jugs stays at 1 ball throughout, while the label is different at every step.

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

@kryptonaut - How about this game:

There is a source jug S, a transfer jug T, and a destination jug D, all of which start out empty.

For phase one, we place a ball into S and label it with the next natural number. The first ball added will have the number 1 on it. We iterate this in supertask fashion adding a new ball every half-of-the-previous-interval-of-time.

1: Would you agree that at the end of the first supertask, S is full, contains a countably infinite number of balls, and contains exactly one ball labeled with each natural number? In short... that S is isomorphic to the set of natural numbers N?

1a: If not, why not? Also if not, would you be willing to posit the existence of such a filled jug for the second supertask, no matter how it was created?

For phase two, we remove the balls corresponding to the first ten natural numbers from jug S and place them into jug T.

2: Would you agree that S still has a countably infinite number of balls, corresponding to the natural numbers except for {1,2,3,4,5,6,7,8,9,10} ?
2a: Would you agree that T has ten balls in it?

... if not, why not?

For phase three, we remove the lowest numbered ball from S and place it in to T, and then remove the lowest number ball from T, and place it in D. We iterate this in supertask fashion, halving the interval every time.

3: During the supertask, how many balls are in S? ... in T? ... in D?
3a: Would you agree that during the supertask, the number of balls in S remains infinite, and the number of balls in T is constant, and the number of balls in D is finite? If not, why not?
3b At (or after) the end of the supertask, how many balls are in S? ... in T? ... in D?
3c: Would you agree that there are zero balls in S, zero balls in T, and a countably infinite number in D, isomorphic to N?
If not, why not?
If so, "how'd that happen?"

With your head around this, now think about the original problem, which is almost the same except that if formulated this way, T will be monotonically increasing rather than constant. If that's a problem for you, but a constant T is not, then what is the difference that makes your head asplode?

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

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

Ok, I took some time to think after the last few posts - I had indeed used some invalid arguments in an attempt to support an intuitive picture. Apologies. However I still can't accept the conclusion that the jug can end up empty But on thinking about the puzzle I was bothered by the realisation that if the 'remove the lowest ball' part of each step is moved out to a subsequent supertask (so first we add all the balls ten at a time, then remove them all one at a time) it seems clear that the jug should be empty. So what's the difference if the removal and addition are performed simultaneously instead of in two separate supertasks?

I then asked myself - what if the jug initially started with 1000 balls numbered 1...1000, and then the supertask step n was simply to add one ball numbered 1000+n and remove ball numbered n. Following the 'set-theory proof' would lead to the jug ending up empty, and it just seems completely illogical to be able to make things disappear by the power of maths. And with the same starting conditions, but renumbering n to 1000+n at step n, it seems to make it obvious that we'll end up with 1000 balls in the jug.

So, on reflection, my argument is this: I think that the reasoning that 'infinite balls are added and infinite balls are taken away, therefore nothing is left' is flawed. By adding infinite balls you destroy information about what they were added to, unless you account for it at each step of the supertask. (This is analogous to the 'proofs' that 1=2 that involve division by zero at some point.)

(If there are 1000 guests currently in Hilbert's Hotel, and an infinite number of new guests check in, and then an infinite number check out, you can end up with any number you like in the hotel depending on how you perform the checking out. The trip to infinity and back destroys information about the finite number of original occupants.)

1000+∞ = ∞
∴0=1000
????
profit!

I think a valid approach that takes into account the initial conditions as well as the process and end result, is to look for things that are invariant from one step to the next, and thus for all steps right through the supertask.

In the 1000-ball example, one invariant would be the number of balls in the jug. Another would be the difference between the highest and lowest number. Another is that the lowest number always increases. No step can change these things, so at the end of the supertask none of the steps would have changed them, and there would still be 1000 sequentially-numbered balls, although we could not say what the numbers actually are other than that they are 'infinitely big'.

For the 'add ten balls and remove the highest one' game, after step n we have 9n balls with all numbers from 1 to 10n except multiples of 10. An invariant is that the lowest number is always 1, another is that the set contains no multiples of 10, and another is that the size of the set is always larger after a step than before it. I'm not sure how theoretically sound it would be to perform arithmetic using n as it approaches infinity, but I think with just these three invariants we can say that we'll end up with an infinitely large set containing all natural numbers except the multiples of 10.

For the contentious 'add ten balls and remove the lowest one' game, after step n we are left with 9n balls numbered n+1 through 10n. An invariant is that the ratio of (highest number) to (lowest number minus one) is always 9. Another invariant is that the size of the set of balls is always larger after a step than before it. Another is that the lowest number is always increased after each step. The logical conclusion is that we end up with an infinite number of sequentially numbered balls, with infinitely high numbers. I'm not sure at the moment what more we can say about the numbers.

Does this sound like a more fruitful approach to resolving the paradox? It certainly seems to produce more intuitive outcomes.

@ucim - thanks for not giving up on trying to convince me! I've responded to your post in the spoiler, in light of the above reasoning:
Spoiler:
1: yes, I agree.
2: yes, I agree.
2a: yes to this too.
3: S: countably infinite. T: 10. D: n
3a: I agree with these statements.
3b: S: empty, T: 10, D: countably infinite
3c: For the reasons given above, the number of balls in T is invariant throughout the supertask so remains 10 at the end. The lowest number on these balls has been constantly increasing so at the end they are 'infinitely high'. D and S are as you suggest.

We nearly agree ConMan
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### Re: Infinite Balls and Jugs [solution]

I feel like anyone who thinks that the two processes (add and remove balls versus add them and relabel them) should have the same result will also have trouble accepting the fact that given a suitable convergent series (like, say, 1 - 1/2 + 1/3 - 1/4 + 1/5 - ...) you can make it sum to anything you want just - including positive and negative infinity - by adding it up in a different order.
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I want to learn this smile, perfect it, and then go around smiling at lesbians and freaking them out.

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

Re: the simple game, (There is a source jug S, a transfer jug T, and a destination jug D, all of which start out empty...) above:

kryptonaut wrote:3b: S: empty, T: 10, D: countably infinite
3c: For the reasons given above, the number of balls in T is invariant throughout the supertask so remains 10 at the end. The lowest number on these balls has been constantly increasing so at the end they are 'infinitely high'. D and S are as you suggest.

Ok. Forget about the more complex games... let's just concentrate on this one.

During the supertask, yes, there are ten balls in T. They are a different ten balls at every step (being completely replaced after ten steps) but they are always ten during each of the finite! steps. However, when you do an infinite number of steps, finite invariance does not hold.

1: Only finite numbers are ever added to T.
1a: There are no infinite numbers in T, ever. N does not contain infinite numbers. It also does not contain emoji or ham sandwiches, even though a ham sandwich is better than complete happiness (proof upon request).

2: Every finite number gets removed from T (and placed into D) at some point. Every. One. Of. Them.
2a: There is no greatest integer. Conceptually, this can be hard to accept, especially if we arrange circumstances to make it so, which is what a supertask does. So, the task of (one by one) moving integers from one set to the other never ends because there is no end point. No final step. No biggest number. Halving the time interval only makes your head explode, it doesn't generate a final step. As an analogy, consider the function y=1/x. For every (we'll stick to integers) y, as y goes from 1 to infinity starts at one and increases without bound, there is a corresponding value of x. When does x reach zero?

It never does, because y never reaches infinity. This is true even though x=0 clearly exists, and the entire graph exists all at once.

Consider also the graph of tan x. There is a discontinuity at zero. Even though x passes smoothly through zero, tan x does not. Again, the entire graph exists all at once, and yet I don't think you have a problem with this discontinuity.

With the simple supertask described above, there is a discontinuity at t=1, which is where the number of steps equals infinity has increased without bound, something we've made concrete by imagining the halving of the time interval. But just like nothing is "moving along the graph" of tan x, there is also nothing "moving along the supertask". There is a discontinuity at t=1 and voom! Jug T is empty.

Cogitate on that for a while until you are as comfortable with it as you are with the discontinuity of y=tan x or of y=1/x.

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

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

kryptonaut wrote:So, on reflection, my argument is this: I think that the reasoning that 'infinite balls are added and infinite balls are taken away, therefore nothing is left' is flawed. By adding infinite balls you destroy information about what they were added to, unless you account for it at each step of the supertask. (This is analogous to the 'proofs' that 1=2 that involve division by zero at some point.)

(If there are 1000 guests currently in Hilbert's Hotel, and an infinite number of new guests check in, and then an infinite number check out, you can end up with any number you like in the hotel depending on how you perform the checking out. The trip to infinity and back destroys information about the finite number of original occupants.)

1000+∞ = ∞
∴0=1000
????
profit!

That is true when dealing with real numbers; infinity is not a real number, and leads to contradictions when you try to perform arithmetic on it like it is one.

What we have here, though is a set; and a countably infinite set is just as "real" as a finite set, in terms of what can be done with set arithmetic.

If you take the set of natural numbers from 1 to 1000 (call it A), and then add the set of natural numbers greater than 1000 (call it B), you end up with the entire set of natural numbers. Then if you take away the entire set of natural numbers, you end up with the null set. Within the context of set arithmetic, this is fine. If we forget for a moment that we're working with sets, we might get distracted by the cardinalities:

|A| + |B| = 1000 + Aleph_0; = Aleph_0;
∴0=1000

But that is only a contradiction if we forget that we are working with sets, of which the cardinality is a function of, and not with cardinality itself.

Now, this only works because the sets are well-defined; so we know that when we add sets A and B we get N and not some other set. If A and B were defined differently, we could get N, some other set, or no well-defined set at all. But A and B defined as they are sum (well, union) to N. So when we pass into the Hilbert's Hotel game, assuming that the hotel has a countably infinite number of rooms, we can create an bijection from the rooms to N; this bijection can be considered "numbering the rooms." So when we have a finite set of current guests in the hotel, and a countably infinite number of people come in and fill the hotel, the cardinality of both the set of incoming guests and the set of all of the rooms in the hotel are both "infinite" - that is, aleph_0.

Then if a set of guests that also has cardinality aleph_0 leaves, we can't say for sure how many guests are left. As you said, it depends on how we check them out. If we check out the guest corresponding to every room, then we are left with the null set. If we check out everyone who came in the night before, we have 1000 left. If we check out every even-numbered room, then we have an infinite number of guests left. But what we are left with is defined by who checks out; and if that is well-defined, we can be certain of who is left. If the guests who leave are not well-defined, we aren't certain of who is left.

kryptonaut wrote:I think a valid approach that takes into account the initial conditions as well as the process and end result, is to look for things that are invariant from one step to the next, and thus for all steps right through the supertask.

In the 1000-ball example, one invariant would be the number of balls in the jug. Another would be the difference between the highest and lowest number. Another is that the lowest number always increases. No step can change these things, so at the end of the supertask none of the steps would have changed them, and there would still be 1000 sequentially-numbered balls, although we could not say what the numbers actually are other than that they are 'infinitely big'.

For the 'add ten balls and remove the highest one' game, after step n we have 9n balls with all numbers from 1 to 10n except multiples of 10. An invariant is that the lowest number is always 1, another is that the set contains no multiples of 10, and another is that the size of the set is always larger after a step than before it. I'm not sure how theoretically sound it would be to perform arithmetic using n as it approaches infinity, but I think with just these three invariants we can say that we'll end up with an infinitely large set containing all natural numbers except the multiples of 10.

The problem you run into here is that there is yet another invariant: that the balls in the jug are always naturally numbered. We can't just cherry-pick which invariant qualities at all steps n are going to pass from the supertask to the end. Remember that the sequence of sums of factorials that converges to e is invariantly rational numbers. So to decide which ones hold and which ones don't, we actually have to examine what the set is doing.

You mention the "remove the highest" game, and come to the conclusion that it is a set containing all natural numbers except the multiples of 10. In that case, the "all balls are naturally numbered" invariant holds. But the only reason we know that it holds is because the set we converge to is a set that we understand to be {1, ... , 9, 11, ... , 19, ... etc.} which has all natural numbers. There's no conflict with our intuition, so we just accept that it is all natural numbers.

The construction of that set is similar to the construction of the sets in the "remove the lowest" game. Instead of the alpha sets being defined as {1}, {2}, {3}, etc. or {n} for all n, they are defined as {10}, {20}, {30}, etc. or {10*n} for all n. And the union of all of those sets converges to the set of all natural numbers that are multiples of 10. Let's call that set T. The set of all balls added is still N, and the set of all balls removed is T, so the set remaining at midnight is N \ T. And N \ T is well defined. That's how we get the solution that we both agree on.

The same goes for the original game, only instead of the balls removed converging to T, they converge to N. N is well defined, and so is N \ N.

kryptonaut wrote:For the contentious 'add ten balls and remove the lowest one' game, after step n we are left with 9n balls numbered n+1 through 10n. An invariant is that the ratio of (highest number) to (lowest number minus one) is always 9. Another invariant is that the size of the set of balls is always larger after a step than before it. Another is that the lowest number is always increased after each step. The logical conclusion is that we end up with an infinite number of sequentially numbered balls, with infinitely high numbers. I'm not sure at the moment what more we can say about the numbers.

I believe that the entire paradox is boiled down to the contradiction between that and converging to an empty set. Yet, when we stop thinking about balls and jugs and think purely about mathematical concepts, we find that we can construct two sets for which all of the same properties hold. The difference of the sets is increasing in cardinality for all n; has the property that the ratio of the max to the min-1 is 9 for all n; and always has 9n terms defined as {n+1, n+2, ... , 10n}. And yet, the limit is the null set, which has none of those properties. If these can all be true for a bunch of mathematical constructions, then the balls having the same properties is not a contradiction.

All of that being said, I think that we are justified in saying that in the 1000-ball case you presented, we still end up with the null set; since we can construct a sequence of sets that have the same properties and show that they converge to the null set without contradiction, there's no reason to believe there is a contradiction when working with the balls. There is certainly a contradiction with our intuition: the number of balls remains 1000 for all finite steps, and there is no step in which this wavers. Yet, the union of the sets {1, 2, ... , 1000} with {1001}, {1002}, ... , {1000 + n} for all n, results in N, and the removal of {1}, {2}, ... , {n} for all n also results in N, and their difference is the null set despite the cardinality being 1000 for all n.

We can't take the statement "If the cardinality remains constant c for all n, then it is c at the limit" for granted. We have to prove it to be true. Instead, we actually prove it to be false by contradiction, by showing a sequence of sets that has constant cardinality for all n but has a different cardinality at the limit.

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

Ok, let's try this.

In the case of the 1000 ball problem, another invariant is that the lowest number on the remaining set of balls is always strictly greater than n. Since n takes on all countable finite values, this means that we can describe the total set of balls added by midnight as {1,2,3,... ω,ω+1,ω+2,...ω+999} having ordinal ω+1000

Similarly in the 'remove the highest' game, the total set of balls added by midnight is {1,2,3,... ω,ω+1,ω+2,... } having ordinal ω+ω.

We can match the balls in the infinite first parts of each of these sets to the infinite set N of balls which are removed {1,2,3,...}

So:
In the '1000' puzzle this leaves us with a finite set {ω,ω+1,...ω+999} having ordinal 1000
In the 'remove highest' puzzle we are left with an infinite set {ω,ω+1,ω+2,...} having ordinal ω Return to “Logic Puzzles”

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