Infinite Balls and Jugs [solution]

A forum for good logic/math puzzles.

Moderators: jestingrabbit, Moderators General, Prelates

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Thu Nov 10, 2016 12:03 pm UTC

phlip wrote:There are an infinite number of naturals, but each one individually is a finite number.


And at midnight after an infinite number of steps all the natural-numbered balls have been removed, leaving 'only' the infinitely many infinite-numbered balls behind.

How about looking at the problem as:
Evaluate 9N (number of balls left) in the limit as N->infinity
and:
Evaluate N+1 (number on the lowest-numbered ball) in the limit as N->infinity

Demki
Posts: 199
Joined: Fri Nov 30, 2012 9:29 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby Demki » Thu Nov 10, 2016 12:35 pm UTC

kryptonaut wrote:
phlip wrote:There are an infinite number of naturals, but each one individually is a finite number.


And at midnight after an infinite number of steps all the natural-numbered balls have been removed, leaving 'only' the infinitely many infinite-numbered balls behind.

How about looking at the problem as:
Evaluate 9N (number of balls left) in the limit as N->infinity
and:
Evaluate N+1 (number on the lowest-numbered ball) in the limit as N->infinity

Both sequences diverge, thus their limit is not any natural number.
If you are talking about limits to an infinite cardinal, you have to explain what you mean by that

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Thu Nov 10, 2016 1:51 pm UTC

Demki wrote:
kryptonaut wrote:
phlip wrote:There are an infinite number of naturals, but each one individually is a finite number.


And at midnight after an infinite number of steps all the natural-numbered balls have been removed, leaving 'only' the infinitely many infinite-numbered balls behind.

How about looking at the problem as:
Evaluate 9N (number of balls left) in the limit as N->infinity
and:
Evaluate N+1 (number on the lowest-numbered ball) in the limit as N->infinity

Both sequences diverge, thus their limit is not any natural number.
If you are talking about limits to an infinite cardinal, you have to explain what you mean by that


Well yes, hence the number of balls remaining at midnight is not any natural number- but neither is it zero. It is infinity, or it would be if the whole scenario were feasible.

Demki
Posts: 199
Joined: Fri Nov 30, 2012 9:29 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby Demki » Thu Nov 10, 2016 2:35 pm UTC

kryptonaut wrote:
Demki wrote:
kryptonaut wrote:
phlip wrote:There are an infinite number of naturals, but each one individually is a finite number.


And at midnight after an infinite number of steps all the natural-numbered balls have been removed, leaving 'only' the infinitely many infinite-numbered balls behind.

How about looking at the problem as:
Evaluate 9N (number of balls left) in the limit as N->infinity
and:
Evaluate N+1 (number on the lowest-numbered ball) in the limit as N->infinity

Both sequences diverge, thus their limit is not any natural number.
If you are talking about limits to an infinite cardinal, you have to explain what you mean by that


Well yes, hence the number of balls remaining at midnight is not any natural number- but neither is it zero. It is infinity, or it would be if the whole scenario were feasible.

But there's the thing, the cardinality of the limit of a sequence of sets is not necesarily the limit of the sequence of cardinalities, similar to the example I gave earlier in the thread with the 'pi=4', where the arclength of the limit of a sequence of sets of points is not the limit of the sequence of arclengths

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Thu Nov 10, 2016 5:08 pm UTC

Ok, after an infinite number of steps there are no balls left with natural numbers because it's always possible to specify the time when a particular numbered ball was removed.

But (since we've entered the Realm of Infinity) that doesn't preclude there being an infinite number of balls with infinitely large numbers on them, and this is a much more satisfying response than to assert that there are no balls left simply because they cannot be enumerated.

User avatar
ucim
Posts: 6860
Joined: Fri Sep 28, 2012 3:23 pm UTC
Location: The One True Thread

Re: Infinite Balls and Jugs [solution]

Postby ucim » Thu Nov 10, 2016 5:36 pm UTC

kryptonaut wrote:But (since we've entered the Realm of Infinity) that doesn't preclude there being an infinite number of balls with infinitely large numbers on them,
Yes it does. Those balls never existed. The jug is filled with an infinite number of balls, each with a unique natural number on it. Surreals are not natural numbers, neither are the transcendentals. At midnight, I'm not going to magically find a ball labeled pi either.

kryptonaut wrote:If you are prepared to consider an infinite number of steps having been performed by midnight, then you also need to accept an infinite number of infinitely-numbered balls being in the jug as a logical conclusion.
Not at all, unless I also need to accept an infinite number of balls labeled with emoji. Where did they come from?

kryptonaut wrote:My argument is that for any ball number N that is removed, there are always 9N balls with higher numbers left in the jug
No, not "always". Only during the time before midnight. There are an infinite number of times before midnight, but this doesn't represent all of time (or all of Time).

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
Posts: 363
Joined: Mon Jul 23, 2007 3:08 am UTC
Location: California
Contact:

Re: Infinite Balls and Jugs [solution]

Postby Xias » Thu Nov 10, 2016 10:22 pm UTC

Kryptonaut, my post here offers a set-based proof. Basically, if you consider the set of "balls put in the jug" and the set of "balls removed from the jug," the cardinality of both sets approaches infinity, which, as we are cautioned, can't be arithmetically manipulated in a consistent way. However, the elements of each set are defined in a way such that each set approaches the set of natural numbers, and the difference between them, then, becomes the null set. Since the cardinality is defined by the contents of the set, the cardinality is discontinuous at infinity, but is defined as zero for the null set that exists at infinity.

Also, you continue to assert that there are balls with infinitely large numbers on them, but I'm unsure what process you imagine created them. There isn't an "infinitely large step" to go with them being placed in the jug. "It's an exercise left to the reader" is not an explanation.

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Fri Nov 11, 2016 5:26 pm UTC

Let's try a slight variation on the question, and see what happens.

Imagine in the first step we add nine balls numbered 0.1, 0.2, 0.3, ... 0.9 and then remove the lowest numbered ball as before.
In the second step we add nine balls numbered 0.91, 0.92, 0.93, ... 0.99 and remove the lowest numbered ball (0.2)
In the third step we add nine balls numbered 0.991, 0.992, 0.993, ... 0.999 and remove ball 0.3
and so on.

So as before, for each ball we can tell what step it was added (by the length of its decimal representation after the decimal point) and what step it was removed (the sum of the digits in its decimal representation). So what's in the jug at midnight 'after' an infinite number of steps?

Does there exist a ball (or an infinite collection of balls) numbered 0.9999 recurring? If not, why not? Is it in the jug? If not, at which step was it removed? And if it was removed, what happened to the balls that were added 'after' it was added (since it can't have been the last, as there is no final step)?

I think this question is unanswerable.

Demki
Posts: 199
Joined: Fri Nov 30, 2012 9:29 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby Demki » Fri Nov 11, 2016 7:28 pm UTC

Since you only add balls labeled with a finite string of digits, there is no point at which a ball with an infinite string of digits is added, so 0.999... is never added to the jar, and thus is not in the jar.

mward
Posts: 123
Joined: Wed Jun 22, 2011 12:48 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby mward » Fri Nov 11, 2016 7:35 pm UTC

kryptonaut wrote:Does there exist a ball (or an infinite collection of balls) numbered 0.9999 recurring? If not, why not? Is it in the jug? If not, at which step was it removed? And if it was removed, what happened to the balls that were added 'after' it was added (since it can't have been the last, as there is no final step)?

As you said, you can tell when a ball was added by the length of its decimal representation. The ball 0.9999 recurring was never added, so it cannot be in the jug at the end of the process.

All the added balls had a finite number of non-zero decimals, and each one was eventually removed. So, as before, the jug is empty.

Note that one way to get a ball labelled 0.999 recurring in the jug without putting it there in a single step, is to apply an infinite number of operations to a ball. Eg start with a ball labelled 0.9 in the jug. On each step, take the ball out of the jug, add a 9 to the end of its (finite) decimal expansion and put it back.

At the end of every step, the ball has label 0.999...9 with a finite number of 9s

At midnight (after an infinite number of steps) the ball has label 0.9999 recurring.

Xias
Posts: 363
Joined: Mon Jul 23, 2007 3:08 am UTC
Location: California
Contact:

Re: Infinite Balls and Jugs [solution]

Postby Xias » Fri Nov 11, 2016 8:54 pm UTC

Kryptonaut, let's make an analogy of your example. It is loosely inspired by a previous analogy in the thread.

You have a ruler, a pencil, and an eraser.

At the first step, you divide up the entire length of the ruler into ten equal segments, and mark them one at a time, moving from one end of the ruler to the other. You also start dragging the eraser from the beginning to the end, without lifting the eraser, such that when the step is completed you have erased only the first line.

At each successive step, you divide up the farthest undivided segment from the eraser, and mark each of them. You drag the eraser forward, and erase the next unerased mark.

As you approach midnight, the pencil is always moving toward the end of the ruler, and the eraser is always moving toward the end of the ruler, and the steps are defined in such a way as they both converge on the end of the ruler at midnight. So, at midnight, both the pencil and the eraser have reached the end. The entire ruler has been erased. Yet the number of marks at each step was monotonically increasing.

You'll find that this translates perfectly to your analogy, as each number you defined corresponds to a length on the ruler, each ball put in the jug corresponds to one of those lengths being marked, and each ball removed from the jug corresponds to a mark being erased.

You'll also find that there was never a step in which you would have marked the point "0.999...", since that corresponds to the end of the ruler (since 0.999... = 1) and you only ever marked divisions of the ruler itself. You also never marked any point outside of the ruler for the eraser to miss, nor did you mark any point behind the eraser that had already been erased.

--

My set-based proof applies perfectly. Let the set {0.1, 0.2, ... , 0.9, 0.91, 0.92, ... , 0.99, 0.991, 0.992, ... , ...} be called D.

Now define a set αn as equal to each set of balls placed in the jug at step n: α1 = {0.1, 0.2, ... , 0.9}, α2 = {0.91, 0.92, ... , 0.99}, etc.

Now define a set βn as equal to the set of balls removed from the jug at step n: β1 = {0.1}, β2 = {0.2}, β9 = {0.9}, β10 = {0.91}, β18 = {0.99}, β19 = {0.991}, etc.

Let Ak be equal to the set of all balls that have been placed into the jug at step k, and Bk be equal to the set of all balls that have been removed from the jug at step k. Or, Ak is the union of α1, α2, ... , αk, and Bk is the union of β1, β2, ... , βk:

Ak = ∪i=1→kαi
Bk = ∪i=1→kβi

Then at any step k, the balls remaining in the jug is the set found by taking any element in set Bk from set Ak:

C = {All balls left in jug at step k} = Ak \ Bk = ∪i=1→kαi \ ∪i=1→kβi

Now, if you decide to take the cardinality of Ak - Bk, it will be 8k, and as k approaches infinity, 8k also approaches infinity.

However, notice that as k approaches infinity, Ak and Bk both approach the same set, which is D:

limk→∞Ak = limk→∞Bk = D

Then the set of balls in the jug at midnight is

C = ∪i=1→∞βi \ ∪i=1→∞αi = D - D = 0 (the null set)

And the cardinality of the null set is zero, so there are no balls left in the jug.

This is where the paradox is: If we take the limit of the cardinality, we get infinity; but if we take the cardinality of the limit, we get zero. How do we reconcile this?

Well, the first thing we do is recognize that cardinality is defined by the contents of the set, not the other way around. Furthermore, the cardinality can be discontinuous: If I said that at step 100, you dump out the jug, then the number of balls (the cardinality of the set of balls) in the jug at step 100 would be zero, despite the limit of the number of balls (the limit of the cardinality of the set of balls) in the jug as we approach step 100 from the left is 800. Yet you would hopefully recognize that the since the cardinality of the set of balls is based on what balls are in the jug, we have to look at the actual balls in the jug rather than the limit of the cardinality, to find the actual cardinality.

That last bit may be obvious to you because I explicitly defined a step where we dump out the balls. There is no such step in the given problem, which is why you've attached yourself to what the limit of the cardinality is doing. However, like how the set of balls left in the jug at step 100 in my example is defined by the step in which I dumped them out, the set of balls left in the jug in the original problem is defined by the process and that set is the null set. And the cardinality of the null set is zero, and the number of balls left in the jug at midnight is zero.

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Sat Nov 12, 2016 2:10 pm UTC

These arguments are quite convincing, at least insofar as they say there are no natural-numbered balls left.

But consider the following scenario:

Imagine a 1x1x1 cube, labeled '0'
At each step N (starting at N=1, timed in halving intervals up to midnight as before) we take away cube number N-1 and replace it with 8 new sub-cubes, each with one eighth the volume (so taking up the same total space as the original), and label the new sub-cubes 8N-7, 8N-6, 8N-5...8N
Continue to midnight.

Start: 0
Step 1: 1,2,3,4,5,6,7,8
Step 2: 2,3,4,5,6,7,8,9,10,11,12,13,14,15,16
Step N: N,N+1,N+2,...8N

So for any numbered cube it's possible to say when it was created and when it was removed from the structure . The total number of cubes at the end of step N is 7N+1.

But the total volume of the cubes is always 1x1x1

What is left at midnight?

I suggest there is a kind of 'cube of dust' made from an infinite number of infinitesimally small cubes, none of which bears a natural number, and whose total volume sums to 1.

I can't reconcile the cube just disappearing simply because nobody can assign a cube to any given number.

There is a disagreement between the argument above and the set-theory argument - so either one argument is wrong, or the original situation they are attempting to describe is somehow not properly specified. Which of these is the case?

elasto
Posts: 3757
Joined: Mon May 10, 2010 1:53 am UTC

Re: Infinite Balls and Jugs [solution]

Postby elasto » Sat Nov 12, 2016 2:30 pm UTC

kryptonaut: Your example strikes me as very similar to the proof that pi = 4

Put in simple terms, there can be a discontinuity at infinity: Something can be true for every step but not true in the limit (though the rigorous disproof of pi = 4 is slightly more nuanced than that)

Image

Xias
Posts: 363
Joined: Mon Jul 23, 2007 3:08 am UTC
Location: California
Contact:

Re: Infinite Balls and Jugs [solution]

Postby Xias » Sat Nov 12, 2016 4:35 pm UTC

kryptonaut wrote:There is a disagreement between the argument above and the set-theory argument - so either one argument is wrong, or the original situation they are attempting to describe is somehow not properly specified. Which of these is the case?


I'm not sure there is a disagreement. What determines the area of the cube at any given time?

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Sat Nov 12, 2016 8:31 pm UTC

Xias wrote:
kryptonaut wrote:There is a disagreement between the argument above and the set-theory argument - so either one argument is wrong, or the original situation they are attempting to describe is somehow not properly specified. Which of these is the case?


I'm not sure there is a disagreement. What determines the area of the cube at any given time?


The total volume of all the cubes is 1 after every step, so I fail to see how it can suddenly become zero at midnight, just because it's not possible to locate a cube with any given natural number.

Regarding the set-theory argument, is it not possible to construct a set containing an infinite number of infinite values? What is the union of this set with the infinite set of all natural numbers? And if you take this union and compare it with the set of natural numbers, how do they differ?

elasto wrote:Your example strikes me as very similar to the proof that pi = 4


I think the problem with this 'proof' is that the tangent to the angular shape never approaches that of the circle, so the final shape after infinite reversing of corners is not a circle even though it may look like one

Demki
Posts: 199
Joined: Fri Nov 30, 2012 9:29 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby Demki » Sat Nov 12, 2016 9:00 pm UTC

kryptonaut wrote:
elasto wrote:Your example strikes me as very similar to the proof that pi = 4


I think the problem with this 'proof' is that the tangent to the angular shape never approaches that of the circle, so the final shape after infinite reversing of corners is not a circle even though it may look like one

The limit of the sequence of sets of points is indeed a circle. If you don't agree with that, do you also believe that the limit of the sequence of numbers (0.9,0.99,0.999,0.9999,...) isn't 1? Because for all numbers in the sequence, they are non integer, yet the limit is an integer!

elasto
Posts: 3757
Joined: Mon May 10, 2010 1:53 am UTC

Re: Infinite Balls and Jugs [solution]

Postby elasto » Sat Nov 12, 2016 10:44 pm UTC

Demki wrote:If you don't agree with that, do you also believe that the limit of the sequence of numbers (0.9,0.99,0.999,0.9999,...) isn't 1? Because for all numbers in the sequence, they are non integer, yet the limit is an integer!

That is a great example of how the limit of a sequence can have a different property to every element in that sequence. Every element has the property of being non-integer, yet the limit has the property of being integer.

Likewise, with your sequence of cubes, every element has the property of having volume 1, yet the limit has volume 0.

Another example might be the sequence "1-1+1-1+1-..."
Odd elements have the sum 1 (eg 1-1+1)
Even elements have the sum 0 (eg 1-1+1-1)

However, in the limit, the sequence has the sum 1/2. This can be show by the following:
S = 1-1+1-1+1...
S = 1-(1-1+1-1...)
S = 1-(S)
2S = 1
S = 1/2

So every element of the sequence has an integer sum (zero or one), but the limit has a sum which is neither zero nor one but in fact a fraction!

User avatar
ucim
Posts: 6860
Joined: Fri Sep 28, 2012 3:23 pm UTC
Location: The One True Thread

Re: Infinite Balls and Jugs [solution]

Postby ucim » Sat Nov 12, 2016 10:46 pm UTC

kryptonaut wrote:Regarding the set-theory argument, is it not possible to construct a set containing an infinite number of infinite values? What is the union of this set with the infinite set of all natural numbers? And if you take this union and compare it with the set of natural numbers, how do they differ?


As I understand it, the surreal integers would fill your requirement of a set containing an infinite number of infinite values. We'll call this set S.
The natural numbers we'll call set N.

The intersection of N and S is the null set.

The union of N and S doesn't have a name, but we'll give it one. They will be the kryptunaut numbers, and be denoted as set K.

The union of N and S is K.

Now, the cardinality of N is aleph null. I think the cardinality of S is also aleph null, based on how I understand their construction.

If so, the cardinality of K is aleph null also. (Similarly, the odds and the evens have the same cardinality as each other, and as the natural numbers)

If we compare the two sets K and N, they differ by S.

If we compare the cardinality of K and N, they are identical. Aleph null.

Infinity minus infinity can be pretty much anything.

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
Posts: 363
Joined: Mon Jul 23, 2007 3:08 am UTC
Location: California
Contact:

Re: Infinite Balls and Jugs [solution]

Postby Xias » Sun Nov 13, 2016 12:46 am UTC

kryptonaut wrote:The total volume of all the cubes is 1 after every step, so I fail to see how it can suddenly become zero at midnight, just because it's not possible to locate a cube with any given natural number.


But how do you know it's 1 after every step? You're at step n, and the volume is 1, and then you perform step n+1. How do you find the area?

You find it by calculating the total area of all of the cubes, which are defined in the procedure. It happens that the procedure has been defined so that at the end of every step, the volume is 1. It also happens that the procedure has been defined so that at midnight, the volume is 0.

kryptonaut wrote:Regarding the set-theory argument, is it not possible to construct a set containing an infinite number of infinite values? What is the union of this set with the infinite set of all natural numbers? And if you take this union and compare it with the set of natural numbers, how do they differ?


It is possible to construct such a set, sure. But there is no reason to construct such a set because nothing in the procedure tells you to do so. If there were such a reason to construct such a set, you might end up with infinite balls: For example, there is a variant on the problem where you put the next 9 balls into the jug, then append the lowest ball with a 0. This ends up with the contents of the jug being the same at every step, but at midnight, there have been no balls removed. There are also no natural numbered balls remaining in the jug. On every ball there is natural number that does not end in 0, followed by an infinity of 0s. All of those balls have "infinite" numbers on them, because creating them is defined in the procedure.

In the original problem, and your block variant, there is no such set created. The set alpha-n is the set of balls put in at step n, and the union of every alpha-n from 1 to infinity is every single ball placed in the jug, and is also N. And the same can be said for beta-n. The sets described in the proof are exactly identical to the balls in the problem. So what do you think the balls are doing that the sets aren't?

ucim wrote:Infinity minus infinity can be pretty much anything.


Arithmetically, yes. But that doesn't push us to throw our hands up just because infinity isn't a number we can perform arithmetic on:

limx→∞x - limx→∞x = undefined

limx→∞(x-x) = 0.

User avatar
ucim
Posts: 6860
Joined: Fri Sep 28, 2012 3:23 pm UTC
Location: The One True Thread

Re: Infinite Balls and Jugs [solution]

Postby ucim » Sun Nov 13, 2016 5:11 am UTC

Xias wrote:limx→∞(x-x) = 0.
Well, that's not infinity minus infinity. If you subtract an actual infinity from another actual infinity, you can get duck soup if you want to.
Spoiler:
Ok, not actual duck soup, but you can get any finite number and some infinities too, depending on how the problem is defined.
I was just illustrating that trying to solve the problem using arithmetic on infinity does not work.

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.

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Sun Nov 13, 2016 10:27 am UTC

Xias wrote:It is possible to construct such a set, sure. But there is no reason to construct such a set because nothing in the procedure tells you to do so. If there were such a reason to construct such a set, you might end up with infinite balls: For example, there is a variant on the problem where you put the next 9 balls into the jug, then append the lowest ball with a 0. This ends up with the contents of the jug being the same at every step, but at midnight, there have been no balls removed. There are also no natural numbered balls remaining in the jug. On every ball there is natural number that does not end in 0, followed by an infinity of 0s. All of those balls have "infinite" numbers on them, because creating them is defined in the procedure.


This is where I am struggling to follow the argument, I think - and I guess that was the whole point of the original post. :)

It seems that there are two procedures (remove ball N and add 10N...10N+9, or renumber ball N to 10N and add 10N+1...10N+9) which produce the same set at each step, and yet the argument is that they converge to different values. How does the set somehow encode the method of its creation?

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

elasto
Posts: 3757
Joined: Mon May 10, 2010 1:53 am UTC

Re: Infinite Balls and Jugs [solution]

Postby elasto » Sun Nov 13, 2016 12:44 pm UTC

kryptonaut wrote:It seems that there are two procedures (remove ball N and add 10N...10N+9, or renumber ball N to 10N and add 10N+1...10N+9) which produce the same set at each step, and yet the argument is that they converge to different values. How does the set somehow encode the method of its creation?

The set doesn't encode the method of its creation, you have two different methods that happen to output the same value at every finite step, but happen to differ in the limit.

The reason why is that one method adds and removes balls, while the other only ever adds balls. And it can be shown that with the first method, every ball that gets added eventually gets removed, while that obviously doesn't happen in the second case.

(Yes, infinity is frustratingly counterintuitive, you only have to check into Hilbert's Hotel to see that)

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Sun Nov 13, 2016 6:39 pm UTC

elasto wrote:The reason why is that one method adds and removes balls, while the other only ever adds balls. And it can be shown that with the first method, every ball that gets added eventually gets removed, while that obviously doesn't happen in the second case.

(Yes, infinity is frustratingly counterintuitive, you only have to check into Hilbert's Hotel to see that)


But... why is writing a zero on the end of a number N not treated as removing number N from the set? Clearly that's what it does, and clearly after the operation, N has been removed from the set of numbers even though the physical ball has not been removed from the jug. So, looking at the physical balls we can say none has been removed, but looking at the numbers written on them it still seems that all the natural-numbered numbers have been removed by midnight.

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

[I've enjoyed staying at the Hilbert Hotel many times - it seems so crowded but the management always find me a nice room. :)]

Xias
Posts: 363
Joined: Mon Jul 23, 2007 3:08 am UTC
Location: California
Contact:

Re: Infinite Balls and Jugs [solution]

Postby Xias » Sun Nov 13, 2016 7:05 pm UTC

ucim wrote:
Xias wrote:limx→∞(x-x) = 0.
Well, that's not infinity minus infinity. If you subtract an actual infinity from another actual infinity, you can get duck soup if you want to.
Spoiler:
Ok, not actual duck soup, but you can get any finite number and some infinities too, depending on how the problem is defined.
I was just illustrating that trying to solve the problem using arithmetic on infinity does not work.

Jose


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

limx→∞(x-x) = 0 is not infinity minus infinity, just like |N \ N| is not infinity minus infinity, even though |N| - |N| is.

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

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


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

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

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

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

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

1: Zero balls outside, infinite balls inside.

2. Infinite balls outside, zero balls inside.

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Sun Nov 13, 2016 8:58 pm UTC

Thanks for keeping on explaining your thoughts to me, I'm enjoying this discussion :)

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

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

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

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

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

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

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


I'll comment on this first, although I disagree with your conclusions in the previous scenarios.
If you are saying that 1*10^1, is distinct from the natural number 10, then how are you justified in calling it the lowest numbered ball when its turn comes around? Surely it can only be compared with ball 11 if it is of the same type, ie a naturally-numbered ball?

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

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

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

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

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

1: Zero balls outside, infinite balls inside.

2. Infinite balls outside, zero balls inside.

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

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


Here I agree with you on scenario 1, but not 2 or 3. I would say that in case 2 there could be zero balls inside, but also there could be infinite balls inside. And in case 3 similarly there would be infinite balls in B but there could also be infinite balls in A.

My reasoning is based on the Hilbert's Hotel scenario, where the infinite hotel is already full and yet there is still space for an infinite number of new guests. In which case, an infinite number of guests could check out but still leave the hotel full (just reverse the security-camera's recording of the guests arriving). Similarly I'd argue that extracting infinite balls from a jug may leave it empty, or it may leave it full.

Is there a flaw in this reasoning?

User avatar
ucim
Posts: 6860
Joined: Fri Sep 28, 2012 3:23 pm UTC
Location: The One True Thread

Re: Infinite Balls and Jugs [solution]

Postby ucim » Sun Nov 13, 2016 9:56 pm UTC

kryptonaut wrote:Here I agree with you on scenario 1, but not 2 or 3. [...] based on the Hilbert's Hotel scenario [...] an infinite number of guests could check out but still leave the hotel full (just reverse the security-camera's recording of the guests arriving). Similarly I'd argue that extracting infinite balls from a jug may leave it empty, or it may leave it full.

Is there a flaw in this reasoning?
Yes.

Whether the hotel is left empty or full depends on the manner in which an infinite number of guests leave, and just which guests they are. In the ball scenario, it is specified, and we have an answer. Specify it differently, and we have a different answer.

Hilbert hotel: Starts full. Every 1/(2^n) as n goes from 1 to infinity seconds, a guest leaves. At the end of one second, after an infinite number of guests have left, how many remain?
Case 1: The guest in room n leaves.
Case 2: The guest in room 2n leaves.
Case 3: The guest in room n+k leaves (k a positive integer)
Imagine other well-defined cases if you like; you will get a definite answer for each case, but it will be potentially a different answer in each case.

Ditto the balls. You are correct that extracting infinite balls may leave it empty, full, or in between. But extracting infinite balls in the given manner will leave the jar empty.

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
Posts: 363
Joined: Mon Jul 23, 2007 3:08 am UTC
Location: California
Contact:

Re: Infinite Balls and Jugs [solution]

Postby Xias » Mon Nov 14, 2016 3:53 am UTC

kryptonaut wrote:I'll comment on this first, although I disagree with your conclusions in the previous scenarios.
If you are saying that 1*10^1, is distinct from the natural number 10, then how are you justified in calling it the lowest numbered ball when its turn comes around? Surely it can only be compared with ball 11 if it is of the same type, ie a naturally-numbered ball?


It is still the natural number 10, but why it's the natural number 10 is different. So it is indistinct in terms of what the set looks like at any given point, but is distinct in how it operates at infinity.

The lines

f(x) = x
g(x) = (x^2)/x

are the same at every point except for x = 0. The fact that they are the same at all other points doesn't say anything about what they are at the point x = 0. Their value at x = 0 is determined by the function. So you could say that while the values of f(2) = g(2), really what you are saying is that f(2) = 2 and g(2) = (2^2)/2, and both of those happen to evaluate to the same number for that x.

User avatar
Wildcard
Candlestick!
Posts: 253
Joined: Wed Jul 02, 2008 12:42 am UTC
Location: Outside of the box

Re: Infinite Balls and Jugs [solution]

Postby Wildcard » Mon Nov 14, 2016 10:24 am UTC

I'll just leave this here....

Why isn't finitism nonsense?

Also, this:

PeteP wrote:Making the statement of "what if the endless task ended" indirectly by declaring your are halving the time each iteration to create an endpoint and making the statement that the action that ended was removing every ball by using the label and the right order does makes it more convulted but that does not automatically make the result more interesting.
...
I suppose I should just accept the premise, but "if impossible thing is done then you can draw weird conclusion from impossible thing being true" still isn't an interesting result.
There's no such thing as a funny sig.

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Mon Nov 14, 2016 12:59 pm UTC

Xias wrote:It is still the natural number 10, but why it's the natural number 10 is different. So it is indistinct in terms of what the set looks like at any given point, but is distinct in how it operates at infinity.

The lines

f(x) = x
g(x) = (x^2)/x

are the same at every point except for x = 0. The fact that they are the same at all other points doesn't say anything about what they are at the point x = 0. Their value at x = 0 is determined by the function. So you could say that while the values of f(2) = g(2), really what you are saying is that f(2) = 2 and g(2) = (2^2)/2, and both of those happen to evaluate to the same number for that x.


Hmmm. I kind of see your point in the (x^2)/x example. But as far as I can see, the labels on the balls are just symbols which have a 1:1 correspondence with the natural numbers, and what those symbols are and how they got there shouldn't matter.

Let's call scenario A the case where 9 new balls are added at step N and then ball N has a zero Appended to its label.
Let's call scenario R the case where 10 new balls are added at step N and then the ball labelled N is Removed.

Your analysis says that in scenario A we end up with infinite balls with infinitely long labels, and in scenario R we end up with an empty jug because every natural number has been removed.

Ok, let's say that at step N we put red balls into the jug according to scenario R, and blue balls according to scenario A, and tape each red ball to its corresponding blue ball at the end of the step. You agree that after each step there is the same set of red and blue balls, so this joining of corresponding balls must be possible, surely? But now what happens at midnight? Do all the blue balls vanish leaving the infinite red balls with dangling bits of tape on them?

Actually I'm coming to the view that the paradox lies within the definition of the step, and whether or not it is treated as 'atomic'. If you see the step purely in terms of pre- and post-step state then it seems clear that scenario A and scenario R are the same. If you permit yourself to peek inside the step and look at its workings then it becomes unclear where the process converges to, like the +1-1+1-1... infinite sum or like Thomson's lamp - it becomes possible to argue that every ball is destroyed, just as it's possible to say that new balls are always added.

Reducing the problem to a less divergent version with one ball in the jug before and after each step:
At step N, add ball N and remove ball N-1
or:
At step N, remove ball N-1 and add ball N (if that's not the same as the above)
or:
At step N, change the number on the ball's label from N-1 to N

Do you end up with a ball with an infinite label, or no ball? Or is it a meaningless question to pose?

User avatar
ucim
Posts: 6860
Joined: Fri Sep 28, 2012 3:23 pm UTC
Location: The One True Thread

Re: Infinite Balls and Jugs [solution]

Postby ucim » Mon Nov 14, 2016 2:46 pm UTC

kryptonaut wrote:But as far as I can see, the labels on the balls are just symbols which have a 1:1 correspondence with the natural numbers, and what those symbols are and how they got there shouldn't matter.

This is not true. Illustration (but probably not rigourous proof) below:

There are three jars - one labeled F, labeled N and one labeled S.

In the jar labeled F there are balls that are already labeled with the natural numbers.
1: How many are in there?
2: Does it contain a ball that has an infinite number of zeros after it?

The jar labeled N is empty.
1: How many are in there?
2: Does it contain a ball that has an infinite number of zeros after it?

The jar labeled S is empty except for one ball, labeled with a 1 (the first natural number).
1: How many are in there?
2: Does it contain a ball that has an infinite number of zeros after it?

Every 1/2^nas n goes from 1 to infinity seconds you relabel the ball in S by appending a zero to it, and you remove the corresponding ball from F and place it in N.

At the end of a second,
1: how many balls are in N and how many balls are in S?
2: is there a ball with an infinite number of zeros in N? ... in S? ... in F?

My answer is that there is a ball with an infinite number of zeros in S, but there is no such ball in F or in N, because it didn't "pre-exist". The monster ball label (the one with an infinite number of zeros) is not a member of the set of natural numbers. So, this construction method, in an infinite number of steps, constructed a number that is not a natural number.

Therefore, there is not the 1:1 correspondence supposed above.

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.

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Mon Nov 14, 2016 5:17 pm UTC

Ok, perhaps when I said '1:1 correspondence with the natural numbers' I should have said '1:1 correspondence with the numbers generated by the alternative method'. So the label '3*10^2' (or however you want to represent it) corresponds to the number 300 just as the label '17' corresponds to the number 17. Every ball has a unique label which can be regarded as representing the number of that particular ball.

If you assert that there is some process creating a label with infinite zeroes in the append-zero-to-the-lowest-number scenario, then surely exactly the same process will be creating a ball with an infinite number (hence an infinitely long decimal representation) in the add-10-balls-and-remove-the-lowest scenario? Either way, an infinite number of steps have to be performed to generate this ball/these balls.

And if there are balls with infinite numbers then they won't necessarily ever be removed by the 'remove the lowest' step, which is why I suggest there will be an infinite number of infinitely-numbered balls in the jug at midnight.

User avatar
ucim
Posts: 6860
Joined: Fri Sep 28, 2012 3:23 pm UTC
Location: The One True Thread

Re: Infinite Balls and Jugs [solution]

Postby ucim » Mon Nov 14, 2016 6:17 pm UTC

kryptonaut wrote:If you assert that there is some process creating a label with infinite zeroes in the append-zero-to-the-lowest-number scenario, then surely exactly the same process will be creating a ball with an infinite number (hence an infinitely long decimal representation) in the add-10-balls-and-remove-the-lowest scenario?
No, because balls aren't being created in that one. They are being "taken" from the pre-existing set of balls that have (natural) numbers already written on them, and there are no balls with an infinite number of zeros in that set.

In the one scenario, you get all the natural numbers. All of them. All infinity of them (but they are all each finite).
In the other scenario, you create at least one unnatural number, because it's about creating, not selecting.

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
Posts: 363
Joined: Mon Jul 23, 2007 3:08 am UTC
Location: California
Contact:

Re: Infinite Balls and Jugs [solution]

Postby Xias » Mon Nov 14, 2016 6:35 pm UTC

Kryptonaut, let's take another look at Hilbert's Hotel, and turn it into a supertask with infinite steps: At step 1, you move the patron in room 1 to room 2, from room 2 to room 4, from room 4 to room 8, etc, then place incoming patron 1 to room 1. At step 2, you move the patron in room 3 to room 6, from room 6 to room 12, etc, then place incoming patron 2 into room 3. You repeat this, and at every step n you move the patron in room (2n-1)(2^(k-1)) to room (2n-1)(2^k), for all k in N, then put patron n into room 2n-1. At midnight, what you have done is moved every current patron from whatever room x they were in to room 2x, opening up all of the odd rooms for the new patrons. You now have an infinite number of natural-numbered rooms, each with a patron sleeping inside.

Now let's say you tell your weekend manager about this strategy, but he misunderstands. Over the weekend, another bus of numbered patrons rolls in, and your weekend manager tries to repeat what you did, but makes a mistake: Instead, at step 2, he moves the patron in room 2 to room 4, 4 to room 8, etc, ending up with room 2 available for patron 2. Then step 3 he moves the patron in room 3 to room 6, room 6 to room 12, etc., ending up with room 3 available for patron 3. Basically, at each step, instead of making the next odd-numbered room available by moving people into even numbered rooms, he makes the next natural-numbered room available.

At the end, where are the original patrons? They've each moved "up" to double their previous room an infinite number of times, leading them all to be assigned to a room number {n*(2^∞)}. Every natural numbered room has a new patron in it. You could say that if the hotel has only natural numbered rooms, they are now all standing outside in the rain with no room to go to. I'm okay with granting that the hotel also had available an infinite set of infinite-numbered rooms that all correspond to the final numbers for the original patrons. But then the question is, what went on in those rooms on the night you were in charge? They were never called on in the first place, because the process you used never created a patron with an infinite-number room assignment.

I've put my direct responses to some of your comments in spoilers to help keep the page readable:

Spoiler:
kryptonaut wrote:Hmmm. I kind of see your point in the (x^2)/x example. But as far as I can see, the labels on the balls are just symbols which have a 1:1 correspondence with the natural numbers, and what those symbols are and how they got there shouldn't matter.


But it does matter, because when you look at the individual ball that started with the label 1, what happens to it in both games? In scenario R, it gets removed at step N. In scenario A, it gets repeatedly appended with a 0 ad infinatum. Surely you don't believe that difference can be ignored?

If you define the function R(n) to generate the set of balls in the original game at step n, and function A(n) to generate the set of balls in the append-0 game, then at all finite values for n they look identical. But they are both getting there differently, just like f(x) and g(x) in my example. If you want to know what the values are at infinity, you have to look at what R(n) and A(n) actually are, not what their outputs are at any given finite value of n.


Spoiler:
kryptonaut wrote:Ok, let's say that at step N we put red balls into the jug according to scenario R, and blue balls according to scenario A, and tape each red ball to its corresponding blue ball at the end of the step. You agree that after each step there is the same set of red and blue balls, so this joining of corresponding balls must be possible, surely? But now what happens at midnight? Do all the blue balls vanish leaving the infinite red balls with dangling bits of tape on them?


Are you constantly untaping and retaping in this process? Because eventually you'll want to remove blue ball 10 (the blue ball with the 10th natural number on it) while red ball 10 (the red ball with 1*10^1 on it) stays behind and gets reattached to blue ball 100 (the blue ball with the 100th natural number on it). All of the blue balls do not vanish at midnight, since nothing happens at midnight; but at midnight they have vanished, and the red balls are left with no corresponding natural numbers to be adhered to.


Spoiler:
kryptonaut wrote:like the +1-1+1-1... infinite sum or like Thomson's lamp - it becomes possible to argue that every ball is destroyed, just as it's possible to say that new balls are always added.

Reducing the problem to a less divergent version with one ball in the jug before and after each step:
At step N, add ball N and remove ball N-1
or:
At step N, remove ball N-1 and add ball N (if that's not the same as the above)
or:
At step N, change the number on the ball's label from N-1 to N

Do you end up with a ball with an infinite label, or no ball? Or is it a meaningless question to pose?
[/quote][/quote]

In Thompson's Lamp, you have one lamp and one toggle switch. If you have a lamp and toggle switch for every natural number, the game is different, and if you've defined your steps in such a way that each toggle gets turned on and off exactly once, then the game terminates with every toggle in the off position.

For example, take the first game but instead of putting the balls in and out of the jug yourself, you let the Hilbert Hotel newcomers do it while they wait for their room to become available. You give to each patron a natural numbered ball n, which has the key which corresponds to their eventual room number 2n-1 attached to it. At each step, the appropriate 10 patrons place their ball/key in the jug, and then you call the next patron n to tell them that room 2n-1 is now available. They reach into the jug, pull out ball n and their key attached to it, and head to their room. At midnight, all of the hotel patrons have placed their key in the jug and then pulled it back out again.

The key (lol) factor that makes this puzzle solvable is that we are working with discrete objects. This is what allows the set proof to work.


Spoiler:
kryptonaut wrote:If you assert that there is some process creating a label with infinite zeroes in the append-zero-to-the-lowest-number scenario, then surely exactly the same process will be creating a ball with an infinite number (hence an infinitely long decimal representation) in the add-10-balls-and-remove-the-lowest scenario? Either way, an infinite number of steps have to be performed to generate this ball/these balls.

And if there are balls with infinite numbers then they won't necessarily ever be removed by the 'remove the lowest' step, which is why I suggest there will be an infinite number of infinitely-numbered balls in the jug at midnight.


The process that creates them is "appending a zero." That process does not take place in scenario R. In scenario R you start off with exactly the set of natural numbers N, each of which is labeling a ball. There are no infinite numbers in N. No additional balls are created in the scenario, and no numbers change for any of the balls, so at midnight you are still left with the set of natural numbers N, each of which is labeling a ball. And when you look at where they are, they are all outside of the jug.

If you take each number in N and append a 0 to it, you end up with numbers that are still in N, though not N itself. If you take each number in N and append infinite zeroes to the end, you end up with numbers that are not in N. Like in the alternative Hotel game, those numbers do not occur or appear anywhere in scenario R.

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Tue Nov 15, 2016 1:42 pm UTC

Thank you for taking the time to provide these informative responses, they are much appreciated. :)

I'm not sure if I'm being dense, or if I have just failed to appreciate some semantic subtlety in the argument, or if I am missing some crucial mathematical point - but I'm still puzzled.

Why is it that in scenario A (where you append zeroes to the labels) you are allowed to assert that you create infinitely long labels which don't correspond to natural numbers, whereas in scenario R (where you remove one ball and add some balls with higher numbers) you are not allowed to assert that you add balls with infinitely high numbers that don't correspond to natural numbers?

To create an infinitely long label in scenario A you have to have appended an infinite number of zeroes, as a result of performing an infinite number of steps. But performing an infinite number of steps in scenario R must also increase the number of the highest ball in the jug an infinite number of times. Why are these not comparable?

Would you say that there is a fundamentally different outcome if you:
1) Start with a single ball labelled '1' and then at each step append a '1' to the label (so counting in unary)
or:
2) Start with a single ball numbered 1 and at each step replace the ball with the next numbered ball

So (1) would end with an infinitely long label, and (2) would end with an infinitely high number which would require an infinitely long label to write it down on (in unary if you like). In neither case is the number a natural number, because by doing something an infinite number of times we've broken out of the realm of natural numbers in the same way that the sum of the infinite sequence (1,1,1,...) is infinite.

Does that make sense?

User avatar
ucim
Posts: 6860
Joined: Fri Sep 28, 2012 3:23 pm UTC
Location: The One True Thread

Re: Infinite Balls and Jugs [solution]

Postby ucim » Tue Nov 15, 2016 2:56 pm UTC

kryptonaut - consider an unrelated but perhaps illustrative question.

Every 1(2^n) second you place a ball from the jar filled with natural numbers into a jug that started out empty. At the end of a second (after an infinite number of steps) you have added an infinite number of balls. Which one was the last one you added?

You must have finished the task, because an entire second has elapsed and it's Miller time.
You must not have finished the task because there are an infinite number of steps, and you can't leave until you've accomplished them all.

When you get your head around this one and are comfortable with the apparent contradiction, then re-examine your other question:

Given that there's a digit (zero in this case) added to the label for every step in scenario A, how long is the label? Does such a natural number exist?
Given that scenario R relies on a supply of balls already labeled with the natural numbers (and nothing else), where would a ball with an infinite number of digits come from? Where would a ball labeled in emoji come from?

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.

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Tue Nov 15, 2016 8:26 pm UTC

Yes, those questions do help to crystallise things.

I think my confusion has been/still is caused by the set of naturals being infinite but not containing infinity.

ucim wrote:Every 1(2^n) second you place a ball from the jar filled with natural numbers into a jug that started out empty. At the end of a second (after an infinite number of steps) you have added an infinite number of balls. Which one was the last one you added?

You must have finished the task, because an entire second has elapsed and it's Miller time.
You must not have finished the task because there are an infinite number of steps, and you can't leave until you've accomplished them all.


I want to say 'infinity' but that isn't allowed, which makes it feel a bit like a trick question - I guess the answer has to either be
"there is no last number" or
"the last number is the size of the set of natural numbers" or
"infinity, and the initial set was something more than the natural numbers"

If asked "what's the biggest number (and you can't say infinity)?" there is no correct answer that I can think of.

ucim wrote:Given that there's a digit (zero in this case) added to the label for every step in scenario A, how long is the label? Does such a natural number exist?
Given that scenario R relies on a supply of balls already labeled with the natural numbers (and nothing else), where would a ball with an infinite number of digits come from? Where would a ball labeled in emoji come from?


The length of the label could be finite if each digit is half the width of the previous one :)

As for the number of zeroes - whatever the answer is to the first question, surely that's also the number of zeroes on the label? If the label is infinitely long then we can also answer "infinity" to the first question, which is not allowed. So the number of zeroes on the label must be countable but not infinite, so the number written on the label is a natural number even if we can't say what it is.

Or is it all just an impossible question?

Xias
Posts: 363
Joined: Mon Jul 23, 2007 3:08 am UTC
Location: California
Contact:

Re: Infinite Balls and Jugs [solution]

Postby Xias » Tue Nov 15, 2016 8:54 pm UTC

kryptonaut wrote:Thank you for taking the time to provide these informative responses, they are much appreciated. :)

I'm not sure if I'm being dense, or if I have just failed to appreciate some semantic subtlety in the argument, or if I am missing some crucial mathematical point - but I'm still puzzled.


I'm having a great time talking about it, so don't worry about it at all.

kryptonaut wrote:Why is it that in scenario A (where you append zeroes to the labels) you are allowed to assert that you create infinitely long labels which don't correspond to natural numbers, whereas in scenario R (where you remove one ball and add some balls with higher numbers) you are not allowed to assert that you add balls with infinitely high numbers that don't correspond to natural numbers?

To create an infinitely long label in scenario A you have to have appended an infinite number of zeroes, as a result of performing an infinite number of steps. But performing an infinite number of steps in scenario R must also increase the number of the highest ball in the jug an infinite number of times. Why are these not comparable?


The set of natural numbers is well defined, and pretty well understood. It is an infinitely large set, so it has an infinite number of elements; however, each individual element is finite. It contains 10^n for every n in itself, as well as 10^10^n, 10^10^10^10^10^n, and 10^...^10^n, with any natural number of 10s.

It does not, however, contain 10^infinity.

So if you take a 1 and append a 0 to it any (finite) number of times, you will still have a natural number, but once you do it an infinite number of times, you get a number that is no longer natural.

kryptonaut wrote:Would you say that there is a fundamentally different outcome if you:
1) Start with a single ball labelled '1' and then at each step append a '1' to the label (so counting in unary)
or:
2) Start with a single ball numbered 1 and at each step replace the ball with the next numbered ball

So (1) would end with an infinitely long label, and (2) would end with an infinitely high number which would require an infinitely long label to write it down on (in unary if you like). In neither case is the number a natural number, because by doing something an infinite number of times we've broken out of the realm of natural numbers in the same way that the sum of the infinite sequence (1,1,1,...) is infinite.

Does that make sense?


That is an excellent question that really illustrates where the disconnect here is.

It can't be a natural number, and I think we are both in agreement on that. The question becomes one of "does the act of adding and removing balls in the described manner (the process) allow for a non-natural numbered ball to exist?" I don't think it does.

The number "created" in (1) is somewhat understandable. It's not really a number we can ever truly see, but we know what rules it follows and what rules it doesn't. And I can define it mathematically in a way that is consistent with the process defined. If you ask me if it possesses one quality or another, I can give you justified answers.

In (2), however, I think the only reason for there to be a ball left in the jug is because "intuitively, it must be so." But it's a mistake to rely on intuition when working with infinity, because our intuition is often wrong. Instead, examine what number would be on such a ball. What properties does it hold? Can you say anything about it at all? I don't think you can get anywhere in that pursuit: It can't be a natural number, because then it would be the largest natural number, which is a contradiction. So you say it's an non-natural number. But we didn't start with any non-natural numbers, and nothing about the puzzle changes any properties of the number on any ball.

If there is a non-natural numbered ball in the jug despite this inconsistency, then surely there would be an infinite number of such balls in the jug if we never removed any ball at all. But if instead we, at step n, placed ball n in the jug and removed nothing, would you take any issue with me saying "at midnight, every ball in the jug corresponds to a natural number?" I think that there would be no controversy.

But then, what about removing balls changes the qualities of the balls yet to be removed? Does the ball materialize only to satisfy your intuition?

kryptonaut wrote:As for the number of zeroes - whatever the answer is to the first question, surely that's also the number of zeroes on the label? If the label is infinitely long then we can also answer "infinity" to the first question, which is not allowed. So the number of zeroes on the label must be countable but not infinite, so the number written on the label is a natural number even if we can't say what it is.

Or is it all just an impossible question?


We are justified in saying that there are an infinite number of digits on the label. We are not justified in offering the same answer for what ball is in the jug in the other scenario.

If you add 1/2^n infinitely, you get 1. How many additions did you make to get to 1? Infinitely many. At what n did you get to 1? There is no answer.

The answers correspond at all finite points, but not at infinity. But what happens at infinity is not contingent on what happens at finite numbers, so that's not really a problem.

EDIT: To clarify, at any given natural number n, the cardinality of the set {1, 2, ... , n} and the largest element in the set {1, 2, ... , n} are the same value, n, at any finite point. However, at infinity, the cardinality is Aleph_0, while the largest element doesn't exist. When you ask "How many digits are there?" you are asking what the cardinality of the set is. When you ask "what is the number on the ball" in the alternative game, you are asking what the largest number is. Both of those numbers correspond for every finite number n, but not at infinity.

User avatar
ucim
Posts: 6860
Joined: Fri Sep 28, 2012 3:23 pm UTC
Location: The One True Thread

Re: Infinite Balls and Jugs [solution]

Postby ucim » Wed Nov 16, 2016 2:18 am UTC

kryptonaut wrote:I think my confusion has been/still is caused by the set of naturals being infinite but not containing infinity.
Infinity is not a natural number, it's an idea. It's the idea of "never-stop-counting" that you may have used as a kid. There are many kinds and sizes of "never-stop-counting", and new sets have been created using some of them as elements. But they are not natural numbers. The set of natural numbers is infinite, and it does not contain infinity.

So, to my first question, the correct answer is: There is no last number. In parallel to that, what (real) number is closest to zero? There is none, because you can always get closer. Infinities are fun, aren't they!

As to my second question:
kryptonaut wrote:The length of the label could be finite if each digit is half the width of the previous one :)
Clever. But yeah, how many zeros?
kryptonaut wrote:As for the number of zeroes - whatever the answer is to the first question, surely that's also the number of zeroes on the label?
If you are adding (appending) a zero at every step, and there are an infinite number of steps, then you have a ball whose label has an infinite number of zeros. But since there is no natural number with an infinite number of (trailing) zeros, that ball, although created in scenareo A, does not exist in scenario R. It's a numeral which does not represent a number. Thus the difference.

Some infinities are actually bigger than other ones. Are you familiar with the proof that there are more (real) numbers between zero and one than there are integers?

Also give some thought to Xias's (ending) comment on cardinally vs largest element. It's well explained.

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.

User avatar
kryptonaut
Posts: 79
Joined: Mon Nov 07, 2016 4:06 pm UTC

Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Wed Nov 16, 2016 2:54 pm UTC

Thanks ucim and Xias, your last two posts prompted me to read the Wikipedia entry for supertasks and their apparent paradoxes, and to re-read about Cantor and the various cardinalities of infinite sets.

I've come to a conclusion regarding this particular paradox that I find intellectually satisfying.

I think that although the set of natural numbers may be sufficient to describe a process, it is not necessarily sufficient to state its outcome. For example the outcome of 1+1+1+1... is not in N.

For the case of the balls in the jug question, I think we need something with the cardinality of R for the outcome even though N is sufficient to state the problem.

Let's say that instead of adding/removing a ball numbered n at any point, we modify n to become the real number rn (where 0<rn<1) obtained by reversing the decimal digits of n and placing a decimal point in front of them. So the first few counting numbers would map to 0.1, 0.2, 0.3, ... 0.9, 0.01, 0.11, 0.21, ... 0.91,0.02,0.12, ... 0.89, 0.99, 0.001, 0.101, ...

The rules of the game now say that at step n we add ten new balls numbered r10n-9 ... r10n and remove ball rn. We can see that these balls map to the natural-numbered balls in the original game.

Now it's clear that any finite natural number n has a corresponding distinct real number rn, but there also exist irrational real numbers with infinitely long representations that don't have a corresponding natural number. 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, therefore they correspond to the numbered balls left at midnight. There are infinitely many of them, but mapping them back to natural numbers is not possible since they would each be a distinct infinite sequence of digits, which is not in N

[Edit: I guess an implication of this is that in the original puzzle if the balls are constrained to having natural numbers, then they will run out before the task ends]

Thoughts?

User avatar
ucim
Posts: 6860
Joined: Fri Sep 28, 2012 3:23 pm UTC
Location: The One True Thread

Re: Infinite Balls and Jugs [solution]

Postby ucim » Wed Nov 16, 2016 5:40 pm UTC

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).

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.

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.


Return to “Logic Puzzles”

Who is online

Users browsing this forum: No registered users and 12 guests