Infinite Balls and Jugs [solution]

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Re: Infinite Balls and Jugs [solution]

Postby Poker » Wed Dec 21, 2016 4:11 pm UTC

kryptonaut wrote:
gmalivuk wrote:Your problem is your continued insistence that somehow balls like ω and ω+1 and 2ω just magically show up. All the balls we started with had natural numbers on them. So which natural number was originally on the now-lowest ball in the jug, which you claim is now numbered at least ω? When did it acquire its new label? How did this happen when all we've been doing is coloring some balls and removing balls one-by-one from the jug?

No, the problem is in treating ω as a cardinal number. N is an infinite set. If you set aside the numbers 1 to 10, what's left is still an infinite set, isomorphic to ω. If you set aside the number 1 million then what's left is still an infinite set isomorphic to ω. If you contrive to set aside the last number, by some procedure such as ucim's transfer jug, then what's left is an infinite set, isomorphic to ω. That last number, in this particular arrangement, can be given the label ω. If you contrive to set aside the last 10 numbers then they can be labelled ω,ω+1,ω+2,...ω+9. They have the property that there are an infinite number of numbers smaller than them. If you contrive to set aside a countably infinite set from the top end, as is done by the 'remove-lowest' variant of the original puzzle, then you still leave behind a set ω, and the infinite set that is set aside is also isomorphic to ω, it is {ω,ω+1,ω+2,...}. The actual numbers have the property of having an infinite number of numbers smaller than them.


First, as far as I can tell, it's not obvious that ucim's transfer jug contains a number at the end. Certainly, in the modify-number case it does, but there, as in so many examples before, the numbers are not fixed to the balls.

Second, you're saying there is a last number, yet there are numbers after it? If I'm understanding you correctly, anyway. If so, how could that number possibly be considered to be the last one?

kryptonaut wrote:
Xias wrote:Let's examine what is in the set at the beginning of step ω, and call a ball in that set bx. If x is finite, then bx would have been removed at tx, which is prior to midnight. It's midnight right now, so bx is not in the jug. So x is not finite. If x is infinite, then it must have been added at or after midnight, but it's midnight right now and we haven't added any balls yet. So x is not infinite. If x is neither finite nor infinite, then it does not exist.

At midnight, because of the construction of the supertask, it's not possible to declare which (non-finite) steps have or haven't happened.


What about at the point before any non-finite step?

kryptonaut wrote:
Xias wrote:To illustrate this, let's perform the supertask twice. The first time we do it, let's paint every ball removed blue, and every ball remaining in the jug red. So we have a blue ball for every natural number, and a red ball for every ball in the set of balls never removed (infinite numbers or unknown numbers or whatever). Dump out the red balls and start the supertask again.

You will find that there is never a time when there is both a red and a blue ball in the jug.
You will find that all blue balls are added before midnight, and no red ball is added before midnight.
You will find that all blue balls are removed before midnight, and no red ball is removed before midnight.

The only way, then, to end up with red balls in the jug is to have an empty jug at midnight, and then to add red balls at or after midnight. And since the puzzle does not prescribe adding anything at all at or after midnight, there is no reason for us to do anything to the empty jug that we have at midnight. Then it was also empty the first time we did it, and there were no balls to paint red in the first place.


I'm not 100% sure I follow your argument, or understand the procedure(s) you are describing. But dividing the infinite input set up into two infinite sets, and then repeating the process on one of those infinite sets can produce two more infinite sets.


I'm going to try to set up an equivalent (though not completely identical) example, and maybe that'll help. Xias, could you double-check and make sure my example really is equivalent, and that there's not something weird going on that I didn't expect?

We're going to run the supertask twice, with the same set of balls each time. The first time we run it, every time we put a ball in the jug we paint it red, and every time we take a ball out of the jug we paint it blue. So after the supertask has finished, every ball that got removed is blue, while if there are any balls left in the jug, they would be red. Now, we're going to empty the jug, so that any red balls are in the same place as the blue balls. Finally we take that whole set of balls - whether each one wound up blue or red - and run the whole process again, exactly the same as before, with exactly the same ordering as before, only now we're not going to paint any balls - we're just going to run the normal supertask, only the balls are now colored based on their final position.

Tell me, under what circumstances would there be a blue ball and a red ball in the jug at the same time?

kryptonaut wrote:Take two Hilbert hotels, A and B, where A is occupied and B is empty. Tell everyone in A to write their current room number on a piece of paper and put it in their pocket. Now fill each room Bn by taking the occupant of A1 and moving everyone in A down one room. A will still be full at the end of the supertask, as will B - but what number is written on the piece of paper held by the current resident of A1?
We have divided the original infinite set into two, one of which is numerically higher than the other. We could repeat the process, filling a new hotel C with the occupants of B, still leaving B full. And so on.


Can you prove A will still be full? Sure, we only remove one person at a time, but by the end we have removed an infinite number of people, so to suddenly claim that it's still infinite is not immediately obvious. Infinity minus infinity does not have to be infinity.

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Re: Infinite Balls and Jugs [solution]

Postby gmalivuk » Wed Dec 21, 2016 6:05 pm UTC

kryptonaut wrote:Take two Hilbert hotels, A and B, where A is occupied and B is empty. Tell everyone in A to write their current room number on a piece of paper and put it in their pocket. Now fill each room Bn by taking the occupant of A1 and moving everyone in A down one room. A will still be full at the end of the supertask
You can't just declare this without justification and then ask questions about it expecting to trip us up.

Why do you think A will still be full at the end of the supertask?

but what number is written on the piece of paper held by the current resident of A1?
Why should there be any such number? For all finite numbers, the person in A1 can't have that finite number written on their piece of paper because the person with that number eventually moved over to B. For all infinite numbers, the person in A1 can't have that infinite number written on their piece of paper because no one wrote any infinite numbers down in the first place.

What we conclude from this is that there is no one in A with a number in their pocket.

We have divided the original infinite set into two, one of which is numerically higher than the other.
No, we haven't. If you want to do that, you need to specify a different supertask.

For example, you could instruct everyone in an odd-numbered room to change their paper from 2k-1 to ω+k, and then instruct everyone in an even-numbered room to change their paper from 2k to k, and then have them move to hotels A and B such that k is in Bk and ω+k is in Ak. But this is definitely not the same supertask as the one you described.

To be honest I'm coming to the view that the problem is poorly specified
No, it's not. It's just specified in a way that's different from the alternative versions you're coming up with, but you can't seem to recognize that those are all different versions and so you keep expecting that they should have the same result.

However I find it strange that the argument that 'the jug sometimes ends up empty depending on how the balls are removed' does not square up with the observation that if the balls are unnumbered, or if no heed is paid to the numbering, that there is no way for the result to vary from one run of the experiment to the next. There will always be an infinite number of balls left in the jug.
I disagree with this. If "no heed is paid to the numbering" then you're declaring it to be a random choice, but this is underspecified because there is no uniform random distribution on a countably infinite set, and thus you haven't specified what the random choice is.

Therefore, you can't say what "there will always be" after the task, just like you can't say what the probability of picking 7 is when you don't first specify the distribution from which 7 is potentially to be picked.
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Re: Infinite Balls and Jugs [solution]

Postby Xias » Wed Dec 21, 2016 9:00 pm UTC

kryptonaut wrote:
gmalivuk wrote:Your problem is your continued insistence that somehow balls like ω and ω+1 and 2ω just magically show up. All the balls we started with had natural numbers on them. So which natural number was originally on the now-lowest ball in the jug, which you claim is now numbered at least ω? When did it acquire its new label? How did this happen when all we've been doing is coloring some balls and removing balls one-by-one from the jug?

No, the problem is in treating ω as a cardinal number. N is an infinite set. If you set aside the numbers 1 to 10, what's left is still an infinite set, isomorphic to ω. If you set aside the number 1 million then what's left is still an infinite set isomorphic to ω. If you contrive to set aside the last number, by some procedure such as ucim's transfer jug, then what's left is an infinite set, isomorphic to ω. That last number, in this particular arrangement, can be given the label ω. If you contrive to set aside the last 10 numbers then they can be labelled ω,ω+1,ω+2,...ω+9. They have the property that there are an infinite number of numbers smaller than them. If you contrive to set aside a countably infinite set from the top end, as is done by the 'remove-lowest' variant of the original puzzle, then you still leave behind a set ω, and the infinite set that is set aside is also isomorphic to ω, it is {ω,ω+1,ω+2,...}. The actual numbers have the property of having an infinite number of numbers smaller than them.


This is just not true.

We've shown you multiple times that you can't partition N that way. There is no element x in N such that there is an infinite subset of N where each element is less than x.

kryptonaut wrote:
Xias wrote:Let's examine what is in the set at the beginning of step ω, and call a ball in that set bx. If x is finite, then bx would have been removed at tx, which is prior to midnight. It's midnight right now, so bx is not in the jug. So x is not finite. If x is infinite, then it must have been added at or after midnight, but it's midnight right now and we haven't added any balls yet. So x is not infinite. If x is neither finite nor infinite, then it does not exist.

At midnight, because of the construction of the supertask, it's not possible to declare which (non-finite) steps have or haven't happened.


I can talk about what happens before midnight (which is all of the finite steps and none of the non-finite steps). For a non-finite step to occur, it must occur at or after midnight. So at midnight, before any non-finite steps have occured, the jug is empty. You are the only one claiming that non-infinite steps are going to occur, and such steps are not defined by the supertask.

kryptonaut wrote:I'm not 100% sure I follow your argument, or understand the procedure(s) you are describing. But dividing the infinite input set up into two infinite sets, and then repeating the process on one of those infinite sets can produce two more infinite sets.


Read Poker's restatement of my suggestion (thanks Poker). Once we have done the supertask once, all blue painted balls have finite numbers on them, and all non-finite labels are only adhered to red balls. Are you suggesting that by doing the supertask again, that blue balls with non-finite numbers on them suddenly appear?

I put forth the procedure in part to help illustrate to you that the argument is not based on us only having finite numbered balls. We can start with non-finite balls (in fact, exactly the non-finite balls that you think end up in the jug) waiting on the sidelines to be added. To go back to notation we were using earlier, I supposed that we had the set N U K, with N painted blue and K painted red. And the supertask never adds anything from K.

If you still don't follow the argument:

Spoiler:
In what domain of time are non-finite balls added to the jug?
In what domain of time are finite balls added to the jug?
In what domain of time are non-finite balls removed from the jug?
In what domain of time are finite balls removed from the jug?

When the first non-finite ball is added to the jug, what is in the jug?

Where and how does the supertask define what you do in the domain of time that is the answer to the first question in this spoiler block?


kryptonaut wrote:Take two Hilbert hotels, A and B, where A is occupied and B is empty. Tell everyone in A to write their current room number on a piece of paper and put it in their pocket. Now fill each room Bn by taking the occupant of A1 and moving everyone in A down one room. A will still be full at the end of the supertask, as will B - but what number is written on the piece of paper held by the current resident of A1?
We have divided the original infinite set into two, one of which is numerically higher than the other. We could repeat the process, filling a new hotel C with the occupants of B, still leaving B full. And so on.


As has been said, the A hotel is empty. There is no occupant in A1 whose paper you can check. If you then move everyone from B into C, then B is empty and C is full. You can't partition sets in the way you are describing.

kryptonaut wrote:To be honest I'm coming to the view that the problem is poorly specified.


It's specified precisely to end up with an empty jug. Here are some of the mathematically consistent justifications for there being an empty jug:

Spoiler:
For any ball put in the jug, there is a finite time before midnight in which it is removed.

The set of balls added is equal to the set of balls removed.

As we get closer and closer to midnight, the amount of time any ball spends in the jug approaches zero (thanks ucim)

The limsup and liminf of the sequence of sets of "balls in the jug" at each step are both the null set, therefore the limit of that sequence of sets is the null set.

For any ball still in the jug at midnight, the properties of such ball invariably lead to contradictions.


Here is a list of the apparent contradictions that an empty jug leads to:

Spoiler:
The number of balls in the jug at each step is increasing without bound.

The lowest number in the jug is increasing without bound.


Neither of those things are any basis for what is in the set at midnight, so they are not contradictions at all. The first is not a sufficient condition for the set at the limit to have an infinite cardinality (see my post about Kryptonaut's Conjecture). The second is not a sufficient condition for the lowest ball remaining in the jug to be infinite, since the "lowest numbered ball" at each step is a discrete object, not the output of a function.

So the only remaining contradiction of an empty jug is this:

Spoiler:
It doesn't agree with Kryptonaut's intuition.


As far as I can tell, this doesn't carry any mathematical weight.

kryptonaut wrote:However I find it strange that the argument that 'the jug sometimes ends up empty depending on how the balls are removed' does not square up with the observation that if the balls are unnumbered, or if no heed is paid to the numbering, that there is no way for the result to vary from one run of the experiment to the next. There will always be an infinite number of balls left in the jug.


You mean if we do different things we end up with different results? That surprises you?

kryptonaut wrote:The set-theory claim that the jug is empty after the 'remove-lowest' supertask is really a mathematical restatement of the observation that every finite ball that's added also gets removed, but whether you accept it or not, it does nothing to address the paradox - it simply restates it in mathematical terms.


And in purely mathematical terms there is no contradiction. So on what basis is there a contradiction, if not mathematical?

kryptonaut wrote:If the balls are unnumbered then you'll always get an infinite set at the end. If they are numbered and you don't pay attention to the numbers, the same thing happens, with absolutely no way to end up with an empty jug. But if they are numbered and you happen to move them in a particular way, suddenly something different is predicted.


Right (with regard to the last bit; the rest is underdefined). As I have said before: If I watch you add and remove balls, but I can't see the labels, then if you ask me what's in the jug the answer is "I don't know." If you then show me the contents of the jug, the contents that I see will depend on how you added and removed balls. Nothing you've described is at all contradictory. Just a little weird and uninituitive, as infinity tends to be.

kryptonaut wrote:That's the paradox which is not explained by the set-theory argument that the jug is empty. To resolve the paradox you need to explain how performing the task without caring about numbers could sometimes produce one result, and sometimes another. Either that or present a set-theory argument that always produces the same consistent result as the un-numbered result (which is my preferred approach.)


This is absurd. Why should doing different things have the same consistent result?

It doesn't even make sense to have this standard for a solution in the most trivial cases. If I remove the lowest even numbered ball, then what I am left with is the set of odd-numbered balls. If I remove the lowest odd-numbered ball, then what I am left with is the set of even-numbered balls. I added and removed the same number of balls, but the exact balls that I removed were different. So I end up with two different sets.

We have no reason to treat these two completely disjoint sets as "the same consistent result" just because they have the same cardinality. The are actually completely different results.

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Re: Infinite Balls and Jugs [solution]

Postby gmalivuk » Wed Dec 21, 2016 9:52 pm UTC

Xias wrote:
kryptonaut wrote:To be honest I'm coming to the view that the problem is poorly specified.
It's specified precisely to end up with an empty jug.
It is certainly so specified if we also define "jug state at midnight" to mean exactly "set-theoretic limit (if it exists) of the sequence of sets of balls in the jug at each step".

This is a perfectly reasonable definition, in my opinion, but I'm personally fine with people who don't like it, on the basis that it assumes a kind of continuity (the function from time to sets-of-balls-in-the-jug is continuous at time = midnight) which there may be no good reason to assume.

What I'm not fine with is kryptonaut's insistence that rejecting this definition means the problem is underspecified (it's not underspecified, it's just specified in a way kryptonaut doesn't like), or that it means infinite-numbered balls somehow show up at some point.
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Re: Infinite Balls and Jugs [solution]

Postby Ermes Marana » Thu Dec 22, 2016 2:56 am UTC

It's more intuitive if we put upper bounds on the speed and space involved. Instead, have the higher numbered balls get smaller so that the total volume of the infinitely many balls is finite, and none of the balls ever travel above some maximum speed.

Now the volume in the jug is a continuous function of time. Even though the number of balls in the jug is increasing, the volume in the jug is approaching 0, and reaches 0 at midnight. This is very similar to Zeno, where even though there are always infinitely many intervals left before the finish line we can look at distance as a continuous function of time all the way through and beyond the race.

What makes it unintuitive to me is when the balls are the same size, requiring infinite space and unbounded speed, and volume has an infinite discontinuity at midnight (increasing to infinity before midnight and then suddenly 0). But the infinite sum of continuous functions doesn't have to be continuous, so if I'm willing to accept infinite space and unbounded speed, it's not that big a leap to accept discontinuous volume-in-the-jug.


gmalivuk wrote:It is certainly so specified if we also define "jug state at midnight" to mean exactly "set-theoretic limit (if it exists) of the sequence of sets of balls in the jug at each step".

This is a perfectly reasonable definition, in my opinion, but I'm personally fine with people who don't like it, on the basis that it assumes a kind of continuity (the function from time to sets-of-balls-in-the-jug is continuous at time = midnight) which there may be no good reason to assume.



I don't think that is a reasonable definition. Why can't a ball arrive in the jug at exactly midnight? Technically, the puzzle could be considered underspecified because it doesn't say that doesn't happen, but it clearly intends that the balls are not going back in the jug. If necessary we can specify the paths of the balls to explicitly say they are not in the jug at midnight.

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Re: Infinite Balls and Jugs [solution]

Postby gmalivuk » Thu Dec 22, 2016 3:42 am UTC

Ermes Marana wrote:Why can't a ball arrive in the jug at exactly midnight? Technically, the puzzle could be considered underspecified because it doesn't say that doesn't happen
No, because if we define the state of the jug at midnight to be the limit of the states of the jug before midnight, then that does specify that nothing else happens at midnight. When we assume that the function (from time to jug-states) is left-continuous at midnight, then this implies that the state at midnight is completely determined by all the states strictly before midnight.
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Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Thu Dec 22, 2016 10:17 am UTC

Xias wrote:Right (with regard to the last bit; the rest is underdefined). As I have said before: If I watch you add and remove balls, but I can't see the labels, then if you ask me what's in the jug the answer is "I don't know." If you then show me the contents of the jug, the contents that I see will depend on how you added and removed balls. Nothing you've described is at all contradictory. Just a little weird and uninituitive, as infinity tends to be.

kryptonaut wrote:That's the paradox which is not explained by the set-theory argument that the jug is empty. To resolve the paradox you need to explain how performing the task without caring about numbers could sometimes produce one result, and sometimes another. Either that or present a set-theory argument that always produces the same consistent result as the un-numbered result (which is my preferred approach.)


This is absurd. Why should doing different things have the same consistent result?

It doesn't even make sense to have this standard for a solution in the most trivial cases. If I remove the lowest even numbered ball, then what I am left with is the set of odd-numbered balls. If I remove the lowest odd-numbered ball, then what I am left with is the set of even-numbered balls. I added and removed the same number of balls, but the exact balls that I removed were different. So I end up with two different sets.

We have no reason to treat these two completely disjoint sets as "the same consistent result" just because they have the same cardinality. The are actually completely different results.

You can describe a general case where the balls are un-numbered and everyone (as far as I know) agrees you end up with an infinite number of them. No possibility whatsoever of any other outcome. Yet by assigning each one a number and moving them in a special way, suddenly you (claim that you) actually don't end up with an infinite number of them after all. How do you see that as non-paradoxical?

To resolve that paradox you'd either need to show how sometimes the un-numbered balls might mysteriously vanish too, or show that there is in fact no special way of moving them that makes them mysteriously vanish.

If you do the supertask with un-numbered balls you end up with an infinite set of balls. If you label them with numbers you will end up with an infinite set of balls labelled with numbers - the particular numbers depend on how you select the ones to discard. If you don't have the right, or sufficient, labels to give to the balls, then that's a problem with the labels you've chosen - it doesn't mean there are no balls to receive the labels.

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Re: Infinite Balls and Jugs [solution]

Postby mward » Thu Dec 22, 2016 10:55 am UTC

kryptonaut wrote:
mward wrote:(C) For task C, imagine that the jug is divided into two halves, L and R, and initially all the balls are in the L half. The finite subtasks are TCn:
On step int((n-1)/10)+1 move ball n from L to R.
On step n take ball n (which will be in side R) out of the jug
All other steps: do nothing


Firstly, thanks for posting this description.

I have a problem with case C though, because of the number of iterations of each type of subtask. This is really the crux of my issue with the whole explanation - the two infinite supertasks are running in parallel but at different rates, so I don't think they are necessarily separable.

If you have a problem with case C, does that mean you have no problem with cases A and B?

In my previous post I ask a number of questions at each stage: these were primarily directed at you, kryptonaut. Please can you go through the post carefully and answer each of the questions. In particular: task B shows that just knowing the number of balls at each step tells you nothing about the number of balls at the end of the process. In task B the jug is empty at the end, but in the "do nothing" task the jug has ℵ0 balls at the end. In both cases, the sequence of "number of balls in the jug" is ℵ0, ℵ0, ℵ0, ...

The only way to know how many balls there are in the jug at the end is to calculate which balls are in the jug (i.e. calcluate the set of balls in the jug at the end) and then count them.

In task C there are not "two infinite supertasks running in parallel" but an infinte set of finite subtasks (to be precise, an ω-indexed ordered set of finite subtasks) TC1, TC2, ... all running in parallel. These subtasks are separable because each one acts on exactly one ball, and no two subtasks act on the same ball. For example, task TC123 moves ball 123 from L to R on step 13, and then on step 123 it moves ball 123 out of the jug. Each finite subtask moves a ball from L to R on one step and takes it out of R on another step. Each ball has one subtask assigned to it. So all the balls are accounted for. After running all of them in parallel, there is nothing in the jug at the end.

If you are happy that task B ends up with the jug empty, then why doesn't task C end up with the jug empty? (Task C is just task B with a subdivided jug). If the whole jug is empty, then each of the two halves of the jug must be empty. On the other hand, if there are balls in the jug at the end of task B, which ones are they? Why didn't the subtasks assigned to each of these balls take them out?

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Re: Infinite Balls and Jugs [solution]

Postby Demki » Thu Dec 22, 2016 1:38 pm UTC

kryptonaut wrote:You can describe a general case where the balls are un-numbered and everyone (as far as I know) agrees you end up with an infinite number of them. No possibility whatsoever of any other outcome. Yet by assigning each one a number and moving them in a special way, suddenly you (claim that you) actually don't end up with an infinite number of them after all. How do you see that as non-paradoxical?

I did not see such a case described, could you describe it for me?

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Re: Infinite Balls and Jugs [solution]

Postby gmalivuk » Thu Dec 22, 2016 1:48 pm UTC

The unnumbered (or unseen) version has been mentioned a couple of times, but kryptonaut is the only one who I remember claiming it always results in an infinite number of balls left in the jug.

The rest of us contend that the unnumbered version is definitely underspecified, and thus no conclusions can be drawn about its end state.
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Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Thu Dec 22, 2016 1:57 pm UTC

mward wrote:In task C there are not "two infinite supertasks running in parallel" but an infinte set of finite subtasks (to be precise, an ω-indexed ordered set of finite subtasks) TC1, TC2, ... all running in parallel. These subtasks are separable because each one acts on exactly one ball, and no two subtasks act on the same ball. For example, task TC123 moves ball 123 from L to R on step 13, and then on step 123 it moves ball 123 out of the jug. Each finite subtask moves a ball from L to R on one step and takes it out of R on another step. Each ball has one subtask assigned to it. So all the balls are accounted for. After running all of them in parallel, there is nothing in the jug at the end.

If you are happy that task B ends up with the jug empty, then why doesn't task C end up with the jug empty? (Task C is just task B with a subdivided jug). If the whole jug is empty, then each of the two halves of the jug must be empty. On the other hand, if there are balls in the jug at the end of task B, which ones are they? Why didn't the subtasks assigned to each of these balls take them out?

In TaskA you just dump all the balls out on step 1. By definition all the balls leave the jug.

In TaskB you perform ω steps removing one ball numbered n at each step, so the set of balls corresponding to N is removed.

In TaskC, divided into a set of parallel two-phased subtasks TCn (as described above), one per ball - the number of steps between each phase of the subtask TCn (approx 9n/10 steps) increases with n towards infinity. So you need a supertask to get all the subtasks started, but to actually complete all of the subtasks you'd need to run a subsequent supertask to wait for another ω steps to pass. And if you did run another supertask, you'd run into the same issue again at the end of it since you'd have been starting even more subtasks in the meantime.

Demki wrote:I did not see such a case described, could you describe it for me?

Is it that hard to imagine? For each step of the supertask add 10 balls and remove 1. What do you have at midnight?

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Re: Infinite Balls and Jugs [solution]

Postby gmalivuk » Thu Dec 22, 2016 2:48 pm UTC

kryptonaut wrote:
Demki wrote:I did not see such a case described, could you describe it for me?

Is it that hard to imagine? For each step of the supertask add 10 balls and remove 1. What do you have at midnight?
As we have explained to you repeatedly, it is impossible to determine. You have not specified which balls are removed, therefore you have not specified the information necessary to say what the limit is.

Even if you deny that supertasks are completable or that the set-theoretic limit is a good description of the state at midnight, the mathematically defined set-theoretic limit itself depends on which balls are added and removed, and if you don't tell us which balls, then we don't know what sets are in the sequence. And if we don't know what sets are in the sequence, how can we possibly know what the limit of the sequence is?

Forget supertasks entirely. Imagine you have a set with a countably infinite number of balls. Then remove infinitely many balls. How many balls remain?

It could be anything from zero to aleph-null, because I didn't tell you which balls you removed. Infinity minus infinity is indeterminate. In limits of continuous real-valued functions, you need to do more work (and know more about the functions) to figure out what a limit is when it initially has the form of infinity minus infinity. The same is true with the limits of sets.
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Re: Infinite Balls and Jugs [solution]

Postby Demki » Thu Dec 22, 2016 2:53 pm UTC

kryptonaut wrote:
Demki wrote:I did not see such a case described, could you describe it for me?

Is it that hard to imagine? For each step of the supertask add 10 balls and remove 1. What do you have at midnight?

In my point of view, it is underdefined.
I am unable to know for any given ball if it is in the jug or not at the end of the supertask, so I am unable to know what is left in the jug at midnight.

If you could answer these 2 questions:
1. What do you think is left at midnight?
2. Could you provide a formal proof for your answer to (1.)? Could you include the set of axioms you use in your proof and following standard rules of inference?

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Re: Infinite Balls and Jugs [solution]

Postby mward » Thu Dec 22, 2016 3:12 pm UTC

Kryptonaut: please answer the questions in my previous post.

kryptonaut wrote:In TaskB you perform ω steps removing one ball numbered n at each step, so the set of balls corresponding to N is removed.

So, are you agreed that after Task B the jug is empty? Even though, on every finite step the jug contains exactly ℵ0 balls?

Task C carries out exactly the same operations as B on the jug as a whole (balls are also moved from the L side to the R side, but stay inside the jug). So after task C the jug is also empty: correct? So the L and R parts of the jug are also empty after task C: correct?

kryptonaut wrote:In TaskC, divided into a set of parallel two-phased subtasks TCn (as described above), one per ball - the number of steps between each phase of the subtask TCn (approx 9n/10 steps) increases with n towards infinity. So you need a supertask to get all the subtasks started.

Every subtask TCn is a finite task, all subtasks start at the same time and run in parallel and do not interfere with each other (as for TAn and TBn). Each finite subtask TCn takes completes in the same time as the corresponding finite subtask TBn. If there is no problem with finite tasks TBn and task B: why should there be any problem with finite tasks TCn and task C?

Regarding the unnumbered balls problem: you agree that task B removes all the balls. But if I drop subtasks TB1 to TB10 inclusive, then the new task B' (consisting of finite subtasks TB11, TB12, ...) will leave exactly 10 balls in the jug (numbered 1 to 10). If I change the subtask TBn to remove ball 2n instead of ball n, then the new task B'' will leave all the odd numbered balls in the jug. In every case (B, B' and B'') the number of balls in the jug after n steps is precisely ℵ0. So, the number of balls remaining after the whole supertask depends on which balls are removed. As I said earlier, you cannot determine how many balls are left from the number of balls in the jug at each finite step. You can only determine how many balls are left by determiningwhich balls (if any) are left, and then counting them. This in turn mens that you need to know which balls are added and/or removed in each step. So "add 10 balls and remove 1 ball on each step" does not give enough information to determine the state of the jug after an infinite number of steps.

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Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Thu Dec 22, 2016 4:25 pm UTC

demki wrote:kryptonaut wrote:
Demki wrote:
I did not see such a case described, could you describe it for me?

Is it that hard to imagine? For each step of the supertask add 10 balls and remove 1. What do you have at midnight?

In my point of view, it is underdefined.
I am unable to know for any given ball if it is in the jug or not at the end of the supertask, so I am unable to know what is left in the jug at midnight.

I'm not asking about any given ball - they're all the same so there's no way to ask/answer that question. I'm asking how many there are.

demki wrote:If you could answer these 2 questions:
1. What do you think is left at midnight?
2. Could you provide a formal proof for your answer to (1.)? Could you include the set of axioms you use in your proof and following standard rules of inference?

1. an infinite number of balls
2. at each step 10 are added and 1 is removed, all in one action. 10-1=9. Resulting in 9 being added at each step. The number is linearly ever-increasing. The limit is infinite. Sum the series 9,9,9,9,...

mward wrote:Every subtask TCn is a finite task, all subtasks start at the same time and run in parallel and do not interfere with each other (as for TAn and TBn). Each finite subtask TCn takes completes in the same time as the corresponding finite subtask TBn. If there is no problem with finite tasks TBn and task B: why should there be any problem with finite tasks TCn and task C?

So how many steps does subtask TCn have to wait before it does it's first action, adding ball n? And how many steps does it wait after that before it does its second action, removing ball n? And what are the limits of those values?

The thing is, if you restrict everything to neverending steps and never allow the supertask to complete then I agree that every ball that enters the jug also exits it - the numbers keep on churning, although the jug is never empty. But the supertask does end somewhere. Wherever it ends, finite or infinite, there are subtasks TCwhatever that were started but not finished, so the balls that were introduced by those subtasks are not removed.

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Re: Infinite Balls and Jugs [solution]

Postby gmalivuk » Thu Dec 22, 2016 5:49 pm UTC

kryptonaut wrote:
demki wrote:kryptonaut wrote:
Demki wrote:
I did not see such a case described, could you describe it for me?

Is it that hard to imagine? For each step of the supertask add 10 balls and remove 1. What do you have at midnight?

In my point of view, it is underdefined.
I am unable to know for any given ball if it is in the jug or not at the end of the supertask, so I am unable to know what is left in the jug at midnight.

I'm not asking about any given ball - they're all the same so there's no way to ask/answer that question. I'm asking how many there are.
Yeah, but we cannot say how many there are without knowing which ones they are. And no, they are not entirely the same, because then they'd all be the same ball and that means just one ball.

demki wrote:If you could answer these 2 questions:
1. What do you think is left at midnight?
2. Could you provide a formal proof for your answer to (1.)? Could you include the set of axioms you use in your proof and following standard rules of inference?

1. an infinite number of balls
2. at each step 10 are added and 1 is removed, all in one action. 10-1=9. Resulting in 9 being added at each step. The number is linearly ever-increasing. The limit is infinite. Sum the series 9,9,9,9,...
What if all the balls are already in the jug, and you're just painting 10 blue each step before you remove the lowest one. Then you've sequentially removed all the balls from the jug, so how can there be infinitely many?

mward wrote:Every subtask TCn is a finite task, all subtasks start at the same time and run in parallel and do not interfere with each other (as for TAn and TBn). Each finite subtask TCn takes completes in the same time as the corresponding finite subtask TBn. If there is no problem with finite tasks TBn and task B: why should there be any problem with finite tasks TCn and task C?

So how many steps does subtask TCn have to wait before it does it's first action, adding ball n? And how many steps does it wait after that before it does its second action, removing ball n? And what are the limits of those values?
The limit of the number of steps between adding and removing a given ball is infinite, but the limit of the amount of time those steps take to complete is zero. Ball 10 stays in the jug for 10 steps which take 9 minutes, 58.8 seconds to complete. Ball 100 stays in the jug for 90 steps which take less than 0.2 seconds to complete.

The thing is, if you restrict everything to neverending steps and never allow the supertask to complete then I agree that every ball that enters the jug also exits it - the numbers keep on churning, although the jug is never empty. But the supertask does end somewhere. Wherever it ends, finite or infinite, there are subtasks TCwhatever that were started but not finished, so the balls that were introduced by those subtasks are not removed.
The "or infinite" is where you keep going wrong.

If it ends after an infinite number of steps, then there are no remaining TCwhatever tasks, because all the TCn tasks were completed and there are no TCx tasks for x outside the natural numbers.

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Edit: Suppose you have a sequence of continuous real-valued functions:
fn(x) = {0 for x≤0, 5nx for 0<x≤2-n, 5n(2-n+1-x) for 2-n<x≤2-n+1, and 0 for 2-n+1<x}
Then each graph traces out the long sides of an isosceles triangle with its vertex at (2-n, 5n2-n), with area (5/4)n.
The pointwise limit, f(x), of this sequence of functions is everywhere 0 (because for every x, there exists N such that n>N implies fn(x)=0).
The area under f(x) is 0, even though the area under fn(x) is (5/4)n, which grows without bound.
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Re: Infinite Balls and Jugs [solution]

Postby mward » Thu Dec 22, 2016 5:50 pm UTC

kryptonaut wrote:
demki wrote:If you could answer these 2 questions:
1. What do you think is left at midnight?
2. Could you provide a formal proof for your answer to (1.)? Could you include the set of axioms you use in your proof and following standard rules of inference?

1. an infinite number of balls
2. at each step 10 are added and 1 is removed, all in one action. 10-1=9. Resulting in 9 being added at each step. The number is linearly ever-increasing. The limit is infinite. Sum the series 9,9,9,9,...

But in my task B you have acknowledged that the jug is empty at the end, while at each finite step the number of balls is ℵ0. So, you have acknowledged that the limit of ℵ0, ℵ0, ℵ0, ... could be 0 (it could also be or ℵ0, or any finite number in between!)

kryptonaut wrote:So how many steps does subtask TCn have to wait before it does it's first action, adding ball n? And how many steps does it wait after that before it does its second action, removing ball n? And what are the limits of those values?

Recall that the definition of TCn is:
  • On step int((n-1)/10)+1 move ball n from L to R. (This is before step n when n > 1)
  • On step n take ball n (which will be in side R) out of the jug
  • All other steps: do nothing
Every single task TCn completes in a finite time: it is a finite task and after n steps it will do nothing. Given n you can compute the exact time before midnight at which it finishes.

You accept this as valid for TBn: why not for TCn?

kryptonaut wrote:But the supertask does end somewhere. Wherever it ends, finite or infinite, there are subtasks TCwhatever that were started but not finished, so the balls that were introduced by those subtasks are not removed.

The supertask has ended by midnight, and so has every single one of the finite tasks TCn.

Which finite subtask do you think will not finish?

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Re: Infinite Balls and Jugs [solution]

Postby elasto » Thu Dec 22, 2016 8:02 pm UTC

It's amazing this has gone on for so many pages. I think it's worth reiterating how simple the solution really is.

For every ball labelled < infinity, some task removes it from the jug and no subsequent task adds it back.
There are no balls labelled >= infinity (or any other kind of labelling - real, complex, surreal etc.), because all balls have natural numbers.

Therefore, if you could complete the infinity of tasks, once you have done so there are no balls left in the jug.

Infinities can be counter-intuitive but the math on this one seems crystal clear.

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Re: Infinite Balls and Jugs [solution]

Postby Xias » Thu Dec 22, 2016 8:50 pm UTC

kryptonaut wrote:You can describe a general case where the balls are un-numbered and everyone (as far as I know) agrees you end up with an infinite number of them. No possibility whatsoever of any other outcome.


Except for the half dozen users in this thread, the ones who are actually having a conversation with you, that are in agreement that we don't actually know that.

kryptonaut wrote:Yet by assigning each one a number and moving them in a special way, suddenly you (claim that you) actually don't end up with an infinite number of them after all. How do you see that as non-paradoxical?


There is no paradox when we do something in a specific way and that leads to a specific solution.

kryptonaut wrote:To resolve that paradox you'd either need to show how sometimes the un-numbered balls might mysteriously vanish too, or show that there is in fact no special way of moving them that makes them mysteriously vanish.


As long as you specify that all of the added balls are unique (that is, if a ball is ever removed, it is never added again) then I can index those balls. Then as long as you specify that we are removing balls at random, I can say that ball bn has probability 1 of being removed. This is because it has nonzero probability of being removed over an infinite number of events. Then the jug is empty.

Specifying those two things creates a different task, though (one that is the same as the third game in the original post that started this conversation; see the first few pages of this thread.) But you haven't specified those things. The problem is underspecified.

Since the problem is underspecified, we don't know what is a possible outcome, nor do we know the probability of anything happening. Then it's impossible to say what the state of the jug is. We could end up doing the exact same thing we do in the above scenario. We also could end up doing the exact same thing as the remove-the-lowest game. We could also end up doing the exact same thing as the remove-the-highest game. The jug could end up empty, or with infinite balls, or with any finite number of balls, and we have no idea which of those situations is possible or probable or anything. So we don't know.

You might as well be saying "At each step, I do some stuff. What has been done at midnight?"

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Re: Infinite Balls and Jugs [solution]

Postby ucim » Fri Dec 23, 2016 1:48 am UTC

elasto wrote:It's amazing this has gone on for so many pages. I think it's worth reiterating how simple the solution really is.
Yes.... there is another thread this reminds me of. I'll spare you which one it is, but it's the first one I posted in. :)

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Re: Infinite Balls and Jugs [solution]

Postby Xias » Fri Dec 23, 2016 2:28 am UTC

ucim wrote:
elasto wrote:It's amazing this has gone on for so many pages. I think it's worth reiterating how simple the solution really is.
Yes.... there is another thread this reminds me of. I'll spare you which one it is, but it's the first one I posted in. :)

Jose


For me it reminds me of Blue Eyes. The bulk of that thread is spent trying to convince people that yes, the Guru gives them relevant information, and no, it is not common knowledge that there are ninety-something blues on the island. As sure as you are to bring up the inductive proof, someone is there to claim that the inductive proof breaks down somehow.

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Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Fri Dec 23, 2016 12:52 pm UTC

elasto wrote:It's amazing this has gone on for so many pages. I think it's worth reiterating how simple the solution really is.

For every ball labelled < infinity, some task removes it from the jug and no subsequent task adds it back.
There are no balls labelled >= infinity (or any other kind of labelling - real, complex, surreal etc.), because all balls have natural numbers.

Therefore, if you could complete the infinity of tasks, once you have done so there are no balls left in the jug.

Infinities can be counter-intuitive but the math on this one seems crystal clear.

The key part is 'if you could complete the infinity of tasks'. Each task TCn says 'wait for m steps, move ball n, wait for another n-m steps, remove ball n.' Where m=int((n-1)/10)+1.

So to complete the task, performing both actions, there are two waiting phases - each of which tends to an infinite number of steps. The infinite set of tasks will all have performed their first action by step ω but you have to wait another ω steps before they have all completed their second action.

You can complete the first phases (moving balls) of the infinity of tasks, but you can't complete phase 2 of them all (removing the balls) without a second (different) supertask.

Why is TCn different from TBn? Well, precisely because it's divided into two halves. In setting up the task you assert the presence of two step numbers- m and n - so if the first halves ever all finish then at that point you will have asserted the presence of step numbers later than all those used in the first halves - ultimately, step numbers of the form ω+n where n=0,1,2,...

Yes, infinities can be counter-intuitive.

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Re: Infinite Balls and Jugs [solution]

Postby Poker » Fri Dec 23, 2016 1:42 pm UTC

There are still several other posts that require your attention, kryptonaut. Just saying that before I begin. I mean, I don't want it to seem like you're avoiding them.

kryptonaut wrote:The key part is 'if you could complete the infinity of tasks'. Each task TCn says 'wait for m steps, move ball n, wait for another n-m steps, remove ball n.' Where m=int((n-1)/10)+1.

So to complete the task, performing both actions, there are two waiting phases - each of which tends to an infinite number of steps. The infinite set of tasks will all have performed their first action by step ω but you have to wait another ω steps before they have all completed their second action.


Confirm one thing for me before I begin: when you say ω, do you mean the first infinite number when they are listed in order? I'll proceed assuming the answer to this is yes.

Tell me, what is the situation immediately before step ω? What balls have been moved, and what balls have been removed? Every ball is either finite or infinite. Every finite ball is removed at some finite step, which would be before step ω. Every infinite ball can only be moved and removed at an infinite step, and since ω is the first infinite step, none of the infinite balls can be moved in the first place. Therefore, even going by your logic, immediately before step ω, there are no balls that have been moved but not removed.

Where am I going wrong? I'll even make it easy for you, and lay it out step by step for you, point out which numbered step is mistaken:

    1. Every ball is either finite or infinite.
    2. Every finite ball is removed (and moved) at a finite step.
    3. Therefore, immediately before the first infinite step, every finite ball has been removed.
    4. Every infinite ball is moved (and removed) at an infinite step.
    5. Therefore, immediately before the first infinite step, no infinite ball has been moved.
    6. Therefore, immediately before the first infinite step, there are no balls that have been moved but not removed.

kryptonaut wrote:You can complete the first phases (moving balls) of the infinity of tasks, but you can't complete phase 2 of them all (removing the balls) without a second (different) supertask.

Why is TCn different from TBn? Well, precisely because it's divided into two halves. In setting up the task you assert the presence of two step numbers- m and n - so if the first halves ever all finish then at that point you will have asserted the presence of step numbers later than all those used in the first halves - ultimately, step numbers of the form ω+n where n=0,1,2,...

Yes, infinities can be counter-intuitive.


I think I've just realized what your issue may be. You're (perhaps subconsciously) taking the end of the supertask to be after some vaguely-defined infinite number of steps - of first halves of TCns, if you will, both finite and infinite - when in actuality the end of the supertask is merely after every finite step. It's not vaguely-defined at all. Every finite-numbered step? Before the end of the supertask. Every infinite-numbered step (assuming that even makes sense)? They would be after the end of the supertask. The supertask is defined for step-numbers in the natural numbers and the natural numbers only. The supertask consist of an infinite number of finitely-numbered TCns, each of which takes a finite time to complete. The TCns never reach a point within the supertask when they begin to take infinite time (which would be necessary for there to be balls that have been moved but not removed), because they all have finite numbers. The TCns never finish their "first half" at any finite point, because there are an infinite number of them - there is no last natural number. The first halves and second halves never "finish" per se because that would imply there was such a last step, but they do "become finished," and in a counter-intuitive way, they "become finished" at the same time.

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Re: Infinite Balls and Jugs [solution]

Postby gmalivuk » Fri Dec 23, 2016 1:46 pm UTC

kryptonaut, you keep getting bogged down with step numbers (and then you keep adding step numbers that never existed).

So define the tasks TCn in terms of time. You don't add ball n on step int((n-1)/10)+1, you add it at 10*2^-int((n-1)/10) minutes before midnight. You don't remove ball n on step n, you remove it at 20*2^-n minutes before midnight.

In this way it is mathematically implied that every ball must be removed before midnight. It doesn't finish "after" the balls are added, because the time between adding and removing balls keeps getting shorter.
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Re: Infinite Balls and Jugs [solution]

Postby kryptonaut » Fri Dec 23, 2016 2:59 pm UTC

Poker wrote:I think I've just realized what your issue may be. You're (perhaps subconsciously) taking the end of the supertask to be after some vaguely-defined infinite number of steps - of first halves of TCns, if you will, both finite and infinite - when in actuality the end of the supertask is merely after every finite step. It's not vaguely-defined at all. Every finite-numbered step? Before the end of the supertask. Every infinite-numbered step (assuming that even makes sense)? They would be after the end of the supertask. The supertask is defined for step-numbers in the natural numbers and the natural numbers only. The supertask consist of an infinite number of finitely-numbered TCns, each of which takes a finite time to complete. The TCns never reach a point within the supertask when they begin to take infinite time (which would be necessary for there to be balls that have been moved but not removed), because they all have finite numbers. The TCns never finish their "first half" at any finite point, because there are an infinite number of them - there is no last natural number. The first halves and second halves never "finish" per se because that would imply there was such a last step, but they do "become finished," and in a counter-intuitive way, they "become finished" at the same time.

In which case, the whole paradox boils down to saying "I want you to actually finish counting an infinite number of things (the discarded balls) without getting to an infinite number (the lowest-numbered ball remaining)." That seems like an impossible task to me.

Either you never finish counting, in which case the numbers never settle to a fixed set but there is an ever-increasing number of them - or you somehow do finish counting, in which case you have an infinite number of discarded balls and the first retained one has a number higher than any of the discarded balls. You can't finish counting an infinite number of things and get a finite number.

The fact that you can't describe the numbers on the balls using a specific numbering system does not, to my mind, mean that the balls do not exist. YMMV. But if the problem as posed is unanswerable then I think it's just time to move on.

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Re: Infinite Balls and Jugs [solution]

Postby gmalivuk » Fri Dec 23, 2016 3:19 pm UTC

kryptonaut wrote:In which case, the whole paradox boils down to saying "I want you to actually finish counting an infinite number of things (the discarded balls) without getting to an infinite number (the lowest-numbered ball remaining)." That seems like an impossible task to me.
Why do you think it's impossible to not get to a ball that was never in the jug to begin with?

Either you never finish counting, in which case the numbers never settle to a fixed set but there is an ever-increasing number of them
The numbers do approach fixed set, though. By the definition of set-theoretic limits, the limit of the sequence of sets of balls in the jug is the empty set.

or you somehow do finish counting, in which case you have an infinite number of discarded balls and the first retained one has a number higher than any of the discarded balls. You can't finish counting an infinite number of things and get a finite number.
There are no retained balls, infinite or otherwise. You finish counting the infinite number of balls that were in the jug to begin with, and then there aren't any more.

Everyone agrees that you can't finish counting an infinite number of things and get a finite number. That's why we all agree that there are no finite-numbered balls left in the jug.

Your problem is insisting that infinite-numbered balls show up at some point and remain in the jug after all the finite ones are gone.

(One of your problems, anyway. It seems you also keep conflating cardinals and ordinals. The numbers on the balls are ordinals. The count of balls one place or another are cardinals.)

The fact that you can't describe the numbers on the balls using a specific numbering system does not, to my mind, mean that the balls do not exist. YMMV. But if the problem as posed is unanswerable then I think it's just time to move on.
The balls may exist, but they can't be in the jug because there was no point at which any of them were put there.

Remember, in the left/right version, or the red/blue version, no balls are ever added to the jug. Finite-numbered balls all start out there, and then are removed one at a time (after being manipulated in some way within the jug). Infinite-numbered balls, if they exist, are never added to the jug, so they can't suddenly show up there once the finite-numbered balls are removed.
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Re: Infinite Balls and Jugs [solution]

Postby Xias » Fri Dec 23, 2016 9:14 pm UTC

Kryptonaut, let's try this.

Instead of removing balls at each step, we'll install timers on each ball such that when the timer runs out, the ball immediately vanishes and ceases to exist.

We set the timer so that for each ball n, it will evaporate and disappear at 20/2n minutes before midnight. So ball 1 will disappear at 11:50, ball 2 will disappear at 11:55, ball 3 will disappear at 11:57:30, etc.

Now, what happens if we don't interact with the jug at all? We sit and watch our pile of infinitely many balls, poofing away at an increasing rate, always leaving us with infinite more to go. But at midnight, they've all disappeared, because each one's timer was set to a time before midnight.

What happens if just before 11:50, we throw all of the balls in the jug, and then wait until midnight? The same as before: At midnight, all of the balls have disappeared.

What about if, at each time 20/2n minutes before midnight, we add balls 10n-9 through 10n to the jug? By the time we've reached midnight, we have added all of the balls to the jug... but the inevitable decay of each ball still leads to each ball having vanished by that time. It doesn't matter that we are doing this parallel task at a "different rate" than the vanishing of the balls; at midnight, every ball has vanished. We cannot avoid it. And adding timered balls to the jug in no way creates timerless balls for us to keep at the end.

Do you disagree with any of that?

Now, since in the original game once a ball is removed it is never interacted with again, there really isn't a conceptual difference between "removing balls" and "vanishing balls." And the time in which we remove a ball exactly corresponds to the time in which that ball vanishes in this variant. So the balls remaining in the jug in both games should be exactly the same.

Since I've been critical of you bringing up a seemingly endless number of variants, I did my best to make this one actually functionally identical to the original game - so that it is not a "variant" at all. The only difference is that instead of removing a ball, it vanishes. But, I welcome you to explain to me any fundamental conceptual difference between balls that vanish and balls that get removed, if you think there is one.

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Re: Infinite Balls and Jugs [solution]

Postby Poker » Sat Dec 24, 2016 3:54 am UTC

kryptonaut wrote:In which case, the whole paradox boils down to saying "I want you to actually finish counting an infinite number of things (the discarded balls) without getting to an infinite number (the lowest-numbered ball remaining).


You know, I did say:

Poker wrote:The first halves and second halves never "finish" per se because that would imply there was such a last step, but they do "become finished," ...


There is a difference. You use the term "finish" which seems to imply in your head (I'm just guessing here, correct me if I'm wrong) that there is a "last" step on which we "finish." This is not the case, because there is no last step - hence, the point of this being a supertask in the first place. The supertask "becomes finished" in the same way that the supertask of listing all of the natural numbers would "become finished." You never get to an infinite number because there is no infinite natural number. When you become finished, you simply become finished - you can't tell me what the last number you counted was because there is no last natural number, just like there is no last digit of pi. Counterintuitive? Perhaps, but this is infinity we're talking about.

Also, I wonder what your response is to the rest of my post, as well as the first line:

Poker wrote:There are still several other posts that require your attention, kryptonaut. Just saying that before I begin. I mean, I don't want it to seem like you're avoiding them.

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Re: Infinite Balls and Jugs [solution]

Postby elasto » Sat Dec 24, 2016 12:57 pm UTC

In which case, the whole paradox boils down to saying "I want you to actually finish counting an infinite number of things (the discarded balls) without getting to an infinite number (the lowest-numbered ball remaining).

It's really not so different to Zeno's paradoxes, which I presume you have no problem with.

There's an infinite number of reals in between 0 and 1; When you go from 0 to 1, what is the last real you encounter before hitting 1? There isn't one. Yet you still manage to get to 1 none-the-less.

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Re: Infinite Balls and Jugs [solution]

Postby mward » Thu Dec 29, 2016 3:08 pm UTC

kryptonaut wrote:Why is TCn different from TBn? Well, precisely because it's divided into two halves. In setting up the task you assert the presence of two step numbers- m and n - so if the first halves ever all finish then at that point you will have asserted the presence of step numbers later than all those used in the first halves - ultimately, step numbers of the form ω+n where n=0,1,2,...

Every task TCn is a finite task and completes at a specific time before midnight. In fact, the task TCn takes exactly the same amount of time as TBn and completes at exactly the same time as TBn. Therefore, since you acknowledge that all tasks TBn complete by midnight, and so supertask TB completes by midnight (and leaves the jug empty): by the same reasoning, all tasks TCn also complete by midnight and supertask TC completes by midnight and also leaves the jug empty.

The tasks TCn are all completely independent of each other: the "second part" of one task TCn does not have to wait for all the "first parts" of all the other tasks to complete before it can process.

By the start of step k, all tasks TBn for n < k have completed, as have all tasks TCn for n < k.
By step ω, all tasks TBn for n < ω have completed, which is all the tasks. So supertask TB has completed.
Similarly, by step ω, all tasks TCn for n < ω have completed, which is all the tasks. So supertask TC has completed.

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Re: Infinite Balls and Jugs [solution]

Postby Wildcard » Fri Dec 30, 2016 3:16 pm UTC

kryptonaut wrote:The fact that you can't describe the numbers on the balls using a specific numbering system does not, to my mind, mean that the balls do not exist. YMMV. But if the problem as posed is unanswerable then I think it's just time to move on.

You know, I think I said this much back on page 8.

The question poses a situation that is impossible in the physical world we live in.

Therefore, the rules of the physical universe may not be referred to in attempts at a solution.

In actual fact the knowledge itself that we have of the "rules" of the physical universe comes from observation and experimentation—no other source.

So we can more easily say: The question is not susceptible to experimentation.

What is it, then? It is a hypothetical—an abstraction. Which is fine, except that as it poses an impossibility by self-contradiction, it must be more precisely defined along with the rules under which it is to be interpreted before it can even be said to be a question susceptible to being answered.

Understand, we're not merely talking about an "impossibility" according to incomplete knowledge gathered through observation and experimentation. "Communication faster than the speed of a racing horse" would be such an "impossibility" if hypothetically proposed to cavemen.

Instead, we're talking about words that do not string together in such a way as to communicate clearly an observation which we can imagine observing. This places the question on a par with such imponderables as, "What would a horse look like if it were not a horse?" The words each individually have meanings but the entire statement is just pure nonsense.

"Suppose I have infinitely many balls." Even after this one sentence, this hypothetical is already no longer susceptible to observation or experimentation. It thus becomes a question in pure imagination, and as such we can quite reasonably imagine any outcome whatsoever, including the outcome, "All the ducks come." (Which ducks? All of them. Where do they come? They just do.)

In order for the question to be answerable in any way, it needs to be predicated on agreements as to how such things as infinities actually work. These agreements need to be communicated, of course, and they must have some bearing on the actual factors present in the question.

Of course, then there wouldn't be any argument about the outcome, would there?

So once again I will state: There is no single correct answer to this riddle. Any result must be imagined, and thus any answer is entirely dependent upon the rules set up by the person doing the imagining.

There may be an "answer which is agreed upon by the greatest quantity of mathematicians," or an "answer which is believed by the mathematician with the most papers to his name," but again, there is no objectively correct answer because there is no objectivity to this question in the first place.
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Re: Infinite Balls and Jugs [solution]

Postby Xias » Fri Dec 30, 2016 6:05 pm UTC

Wildcard: Are we allowed to make objective statements about the square roots of negative numbers, or is that too imaginary for you?

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Re: Infinite Balls and Jugs [solution]

Postby gmalivuk » Sat Dec 31, 2016 7:39 pm UTC

Wildcard wrote:The question is not susceptible to experimentation.

What is it, then? It is a hypothetical—an abstraction. Which is fine, except that as it poses an impossibility by self-contradiction, it must be more precisely defined along with the rules under which it is to be interpreted before it can even be said to be a question susceptible to being answered.

Understand, we're not merely talking about an "impossibility" according to incomplete knowledge gathered through observation and experimentation. "Communication faster than the speed of a racing horse" would be such an "impossibility" if hypothetically proposed to cavemen.

Instead, we're talking about words that do not string together in such a way as to communicate clearly an observation which we can imagine observing. This places the question on a par with such imponderables as, "What would a horse look like if it were not a horse?" The words each individually have meanings but the entire statement is just pure nonsense.

"Suppose I have infinitely many balls." Even after this one sentence, this hypothetical is already no longer susceptible to observation or experimentation. It thus becomes a question in pure imagination, and as such we can quite reasonably imagine any outcome whatsoever, including the outcome, "All the ducks come." (Which ducks? All of them. Where do they come? They just do.)

In order for the question to be answerable in any way, it needs to be predicated on agreements as to how such things as infinities actually work. These agreements need to be communicated, of course, and they must have some bearing on the actual factors present in the question.

Of course, then there wouldn't be any argument about the outcome, would there?

So once again I will state: There is no single correct answer to this riddle. Any result must be imagined, and thus any answer is entirely dependent upon the rules set up by the person doing the imagining.

There may be an "answer which is agreed upon by the greatest quantity of mathematicians," or an "answer which is believed by the mathematician with the most papers to his name," but again, there is no objectively correct answer because there is no objectivity to this question in the first place.

This whole thing could be said about pretty much any part of axiomatic mathematics.
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Re: Infinite Balls and Jugs [solution]

Postby mward » Sun Jan 01, 2017 1:06 pm UTC

Wildcard wrote:Instead, we're talking about words that do not string together in such a way as to communicate clearly an observation which we can imagine observing. This places the question on a par with such imponderables as, "What would a horse look like if it were not a horse?" The words each individually have meanings but the entire statement is just pure nonsense.
The infinite set of balls cannot be physically realised in this world, but can be logically realised, and even physically realised in a different world: eg one in which matter is infinitely divisible.

Wildcard wrote:"Suppose I have infinitely many balls." Even after this one sentence, this hypothetical is already no longer susceptible to observation or experimentation. It thus becomes a question in pure imagination, and as such we can quite reasonably imagine any outcome whatsoever, including the outcome, "All the ducks come." (Which ducks? All of them. Where do they come? They just do.)

As gmalivuk pointed out, the same is true for all mathematics. Yet we do mathematics with infinite sets and prove results which turn out to apply to the real world! What is an induction proof if not an example of a supertask? If Wildcard's assertions were correct, then the results of mathematics would be completely independent of the real world. The fact that the results apply to the real world then becomes an astonishing coincidence that demands explanation.

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Re: Infinite Balls and Jugs [solution]

Postby Yakk » Mon Jan 02, 2017 5:48 am UTC

You can rephrase (all?) (most?) infinitary logic in mathematics with finite practical consequences as finitary arguments that reach the same conclusion.

This tends to be awkward and not very illuminating.

The use of infinities makes the mathematics with practical applications easier, so long as you carefully choose how your infinities behave. If you do not, you get nonsense. So using previously known and vetted formal notions of infinity and not hand waving is important, or you are just blathering and probably using an inconsistent system implicitly. Now you can invent your own way to reason around infinity, but if you are not formalizing it you are almost certainly using an inconsitent one (and there is no easy way to check either). You become not even wrong.
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Re: Infinite Balls and Jugs [solution]

Postby PeteP » Mon Jan 02, 2017 2:45 pm UTC

mward wrote:As gmalivuk pointed out, the same is true for all mathematics. Yet we do mathematics with infinite sets and prove results which turn out to apply to the real world! What is an induction proof if not an example of a supertask? If Wildcard's assertions were correct, then the results of mathematics would be completely independent of the real world. The fact that the results apply to the real world then becomes an astonishing coincidence that demands explanation.

That is a weird argument imo. I wouldn't call defining the natural numbers an example of an supertask at no point does it require actually going through it step by step and obviously no number requires going through a supertask to reach it. But more importantly you can describe the "ends with infinite balls" argument for the infinite ball thing like a proof by induction, for any step x the step x+1 has more balls and at the end of step 1/at step 2 you have a positive number of balls so at no step greater than 2 will it ever become less than 1. In other words proof by induction is another thing that breaks down when supertasks get introduced so I find it a bit weird call it an example of one.


------------------
Anyway What I am actually here for is to answer a question of gmalivuk (can't be bothered to search for the post to quote it and can't be bothered to actually catch up to the thread again because honestly it is a very boring discussion but it was a direct question so I thought I should answer it), it was about what I think of zenos when I doubt the whole reaching infinity thing. I simply don't count infinite divisibility, I consider it more of a mathematical definition than anything. Is that inconsistent? Possible or maybe probably which is why I am unsure of my stand point. But anyway I simply don't count any infinity that mostly exist by mathematically definition, if it involves no infinite speed, no infinite objects that aren't infinitely small, no infinite discrete actions that have clear beginnings and ends (instead of being arbitrary divisions of something continous) which is an imprecise definition and if I really cared about the nature of infinity I should think harder about it but I don't. To transfer it to something similiar to the ball thing: Take a stick, define it to have infinite different numbered areas (for instance by halving) and feed it to a shredder and I won't object, define only the first area at the beginning and say that for every area shredded 10 more are defined and I would call it weird nonsense but won't object. Say that someone actually marks these 10 new areas with an infinitely thin knife and I will complain about the infinite speed required, the infinite precision (and of course sticks are made of atoms and can't be marked with infinite distinct cuts).

Yes it is not particularly thought out. *shrugs* Anyway since I don't want to catch up this will probably be my last post unless asked a question, well until I mostly forget about this thread at least.

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Re: Infinite Balls and Jugs [solution]

Postby Wildcard » Wed Jan 04, 2017 10:38 am UTC

mward wrote:What is an induction proof if not an example of a supertask?

PeteP has answered this very nicely. Proof by induction directly invalidates the agreed-upon solution to the original puzzle, so that's evidence in favor of my point, rather than a counterexample.

mward wrote:If Wildcard's assertions were correct, then the results of mathematics would be completely independent of the real world. The fact that the results apply to the real world then becomes an astonishing coincidence that demands explanation.

Mathematics are neither true nor false; they are useful or not useful.

Also, you are generalizing: "The results [of mathematics] apply to the real world" can be interpreted to mean that some mathematical results apply to the real world, which is obviously true and is something I have never denied; or it can be interpreted to mean that all mathematical results apply to the real world, which is of course utter nonsense.

Mathematics can exist which contradict other mathematics. For example, as PeteP so ably pointed out, a proof by induction indicates that the jug will never be empty. This particular type of mathematics—proof by induction—has application to the real world. It is therefore useful in predicting actual results in the physical universe.

The mathematics of infinity is certainly interesting and you can have a lot of fun with it. But it does not apply to the real world. (It certainly doesn't apply to balls and jugs, as it would contradict the proof by induction.)

Perhaps someday some new phenomena in the real world will be discovered which are most accurately and usefully modeled by means of supertasks and infinite cardinalities. Until then, it's purely an abstraction, no more intrinsically valid than any other abstraction whether consistent or not.

That doesn't mean you can't have fun with it. It doesn't mean that you can't communicate to other mathematicians about it, prove abstractions with it, or anything else. It does mean that claiming that supertasks, infinite cardinalities and, yes, even complex numbers have objective validity (or objective truth) is simply not true.
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Re: Infinite Balls and Jugs [solution]

Postby gmalivuk » Wed Jan 04, 2017 12:48 pm UTC

The limit of the set of balls in the jug is the empty set.

You can argue against defining the result of the supertask as this limit, but you can't deny that limits have real-world applications.

The proof by finite induction only proves that the jug is not empty at any finite step, which no one denies. However, we can *also* prove inductively that all finite balls are removed from the jug, *and* that no infinite-numbered balls are ever added.
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Re: Infinite Balls and Jugs [solution]

Postby mward » Wed Jan 04, 2017 1:03 pm UTC

Wildcard wrote:Also, you are generalizing: "The results [of mathematics] apply to the real world" can be interpreted to mean that some mathematical results apply to the real world, which is obviously true and is something I have never denied; or it can be interpreted to mean that all mathematical results apply to the real world, which is of course utter nonsense.

What I mean is: any mathematics which can be applied to the real world gives correct results in the real world. This is astonishing if mathematics is as arbitrary as you suppose! Some mathematics has yet to find a real-world application. In other cases, the application is discovered decades after the mathematics has been discovered. For example, infinitary logic was explored by Carol Karp in the 1960's and turned out to have many applications in the derivation and analysis of computer programs, starting in or around the 1980's.

Wildcard wrote:Mathematics can exist which contradict other mathematics. For example, as PeteP so ably pointed out, a proof by induction indicates that the jug will never be empty.

Standard induction (from n to n+1) proves a formula for all values of n up to but not including ω. It is true that the jug is non-empty for all steps strictly less than ω (and therefore for all times strictly before midnight). To get to ω (midnight itself) and beyond we need to use transfinite induction: which is used extensively in the analysis of computer programs. As one might expect, the transfinite induction limit case fails for the jug problem, so we cannot use transfinite induction to "prove" that the jug is non-empty at midnight.

Wildcard wrote:The mathematics of infinity is certainly interesting and you can have a lot of fun with it. But it does not apply to the real world. (It certainly doesn't apply to balls and jugs, as it would contradict the proof by induction.)
As discussed above, there is no contradiction with induction.
Wildcard wrote:Perhaps someday some new phenomena in the real world will be discovered which are most accurately and usefully modeled by means of supertasks and infinite cardinalities.

Computation is an example of just such a phenomena. Just to be clear: computers do not calculate supertasks as such, but the derivation of a computer program from a specification can be carried out using infinitary logic and transfinite induction, and the resulting program turns out to be correct when it is tested in the real world. See, for example, "Provably Correct Derivation of Algorithms Using FermaT" by Martin Ward and Hussein Zedan, Formal Aspects of Computing, September 2014, Volume 26, Issue 5, Pages 993--1031, doi:dx.doi.org/10.1007/s00165-013-0287-2

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Re: Infinite Balls and Jugs [solution]

Postby Xias » Wed Jan 04, 2017 4:23 pm UTC

Again, Wildcard, do you hold the same views about the roots of negative numbers? I can literally replace two lines in your original post and everything else holds:

Spoiler:
Wildcard wrote:The question poses a situation that is impossible in the physical world we live in.

Therefore, the rules of the physical universe may not be referred to in attempts at a solution.

In actual fact the knowledge itself that we have of the "rules" of the physical universe comes from observation and experimentation—no other source.

So we can more easily say: The question is not susceptible to experimentation.

What is it, then? It is a hypothetical—an abstraction. Which is fine, except that as it poses an impossibility by self-contradiction, it must be more precisely defined along with the rules under which it is to be interpreted before it can even be said to be a question susceptible to being answered.

Understand, we're not merely talking about an "impossibility" according to incomplete knowledge gathered through observation and experimentation. "Communication faster than the speed of a racing horse" would be such an "impossibility" if hypothetically proposed to cavemen.

Instead, we're talking about words that do not string together in such a way as to communicate clearly an observation which we can imagine observing. This places the question on a par with such imponderables as, "What would a horse look like if it were not a horse?" The words each individually have meanings but the entire statement is just pure nonsense.

"Suppose I take the square root of negative one." Even after this one sentence, this hypothetical is already no longer susceptible to observation or experimentation. It thus becomes a question in pure imagination, and as such we can quite reasonably imagine any outcome whatsoever, including the outcome, "All the ducks come." (Which ducks? All of them. Where do they come? They just do.)

In order for the question to be answerable in any way, it needs to be predicated on agreements as to how such things as square roots actually work. These agreements need to be communicated, of course, and they must have some bearing on the actual factors present in the question.

Of course, then there wouldn't be any argument about the outcome, would there?

So once again I will state: There is no single correct answer to this riddle. Any result must be imagined, and thus any answer is entirely dependent upon the rules set up by the person doing the imagining.

There may be an "answer which is agreed upon by the greatest quantity of mathematicians," or an "answer which is believed by the mathematician with the most papers to his name," but again, there is no objectively correct answer because there is no objectivity to this question in the first place.


(EDIT: I just saw your comment about complex numbers not having so-called "objective validity." This isn't true. Once we decided to start thinking about imaginary numbers, the rules surrounding them were objectively true and we discovered them. And then, many years later, our understanding of complex numbers began having real world applications in fields such as mechanical engineering. Everything you've said, when applied to imaginary numbers, would have been wrong if said about imaginary numbers - whether said today or in 1700.)

Regardless, it's not like we are making up new mathematical axioms to reach the empty jug solution. The arguments that there exists any ball in the jug lead to logical contradictions. There is no logical contradiction with an empty jug. It may appear that there is (the number of balls in the jug is not decreasing at every step) but the premise "if the number of balls in the jug is nonzero and not decreasing at every step, then the number of balls after infinite steps is nonzero" is a false premise. I say that it is a false premise based on existing and applicable mathematics.

You may not realize this, but mathematicians have a pretty good grasp on number theory, set theory, limits, and infinity, all in ways that are applicable to real life. And all of that knowledge is consistent with an empty jug, and vice versa. A nonempty jug isn't even consistent with itself.

Or, perhaps you are right, and the only reason why we believe that there exists no such thing as "a horse that is not a horse" is because we haven't experienced one in real life yet.
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