Infinite Balls and Jugs [solution]

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Re: Infinite Balls and Jugs

Postby Wildcard » Sat Oct 17, 2009 9:30 am UTC

Furanku wrote:I really resent these problems.

If we look at the ratio of balls left in to balls taken out, every interaction, we get (10 balls in - the ball out)/(the ball out).
For n interactions, this is 9n/n.
Obviously, resolving this to n=infinity, we get 9(infinity/infinity), which cancels to 9(1/1), or 9.
So even after an infinite number of interactions, we still have a ratio of nine balls in to 1 taken out.

It just doesnt seem right to me to resolve this type of question with a "lul but i is subtracting infinity from infinity wich is 0 so there be nun" equivalent.
Lets say that we take out the lowest number ball at each interaction. Interaction 1, removed ball = 1, highest ball = 10. Interaction 10, removed ball = 10, highest ball = 100. For any arbitrarily large number of interactions, the number of balls remaining = 10n-n = 9n, lowest ball remaining: n+1, highest ball remaining = 10n. Suddenly claiming that there are 'no' balls remaining after an 'infinite' number of interactions may be mathematically possible, but since mathematics tends to have problems with the concept of infinity (or atleast in the methods you're using), not necessarily logically so.

As for what numbers are on the balls removed, it seems like an illogical question for a jar theoretically containing 'infinity' balls.


Of course, I'm obviously wrong with all of this and am going to be crucified because I didn't say 'probability', or show any understanding of statistics.
Sorry.

Huzzah!

No crucification. I thoroughly agree, especially with the part where you use bad grammar to convey your utter contempt for the preposterous claims of infinity-infinity = zero. And, you saved me a whole lot of typing, since you already said most of what I was going to say.

(What he said!)

I'll also add, though, that it's preposterous to keep on using "math" with a symbol in it that's undefined. Or to put it another way, to use operations on a number that they cannot be applied to. Infinity is undefined...just like ZERO. (Zero is defined, rather, but it's almost never specified. It's a useful fiction, but particularly when you're dealing in physics, you better be damn sure you know zero of WHAT, zero WHERE and zero WHEN. "A zero of apples on that table right now" does not equal "zero universes" and never will. So, like I said, unspecified. :P )

Edit: Sorry for the bump, didn't notice the date on the thread.
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Re: Infinite Balls and Jugs [solution]

Postby skeptical scientist » Sat Oct 17, 2009 7:05 pm UTC

Nice strawman, Wildcard. Not one person ever said that the jug is empty, "because ∞-∞=0." What we said was that the jug is empty because ever ball which is put into the jug is later removed. In other words, if the jug is not empty, there is some ball in the jug at the end, and since the balls are all numbered, that ball must have a number. But by looking at the number on the ball, you can figure out when it was removed, and conclude it can't wind up in the jug.

P.S. the symbol '∞' is quite well defined, thanks, and it's not at all preposterous to do mathematics using it.
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Re: Infinite Balls and Jugs [solution]

Postby Wildcard » Sat Oct 17, 2009 8:28 pm UTC

Skep: Okay, fair enough on infinity-infinity.

Here's the counter-argument to your explanation: after each single iteration, there are exactly 9 balls more in the jug than before that iteration. If the jug is to be empty at the end, you must point to at least one specific iteration wherein more balls are removed from the jug than added to the jug.

Another thing: relative sizes of infinity. infinity != infinity in every case. So for any given number N from 1 to infinity which was indeed removed from the jug, I contend that it is a meaningful and valid statement to say that there are 9 balls which can be found in the jug, with numbers (infinity + N), (2infinity + N), etc up to (9infinity + N). (And as an additional note, those 9 balls were not added at the same time as ball N was removed, but can all be found in the jug.)

Or, another argument: The jug had balls in it at some points. Only one ball is removed at any one step. If it is to be empty at the end, there must have been a ball that was removed which left the jug empty in that step. I ask you to tell me the number on the ball which was removed last. (And I propose that the answer is infinity, but that the balls (infinity + 1) to (10infinity) are left in the jug.)

The point being, you can't answer this question, which is exactly as valid as your question about naming a numbered ball that is still in the jug. This is why I say that math with infinity is ridiculous, because you can set up contradictory proofs with equal apparent validity. The only way to really come to an agreement which is consistent (that I see) is to accept that phrases like "2infinity + 4" actually have meaning.

Edit: And how do you get the infinity symbol anyways?
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Re: Infinite Balls and Jugs [solution]

Postby jestingrabbit » Sat Oct 17, 2009 9:14 pm UTC

Wildcard wrote:Or, another argument: The jug had balls in it at some points. Only one ball is removed at any one step. If it is to be empty at the end, there must have been a ball that was removed which left the jug empty in that step. I ask you to tell me the number on the ball which was removed last. (And I propose that the answer is infinity, but that the balls (infinity + 1) to (10infinity) are left in the jug.)

The point being, you can't answer this question, which is exactly as valid as your question about naming a numbered ball that is still in the jug. This is why I say that math with infinity is ridiculous, because you can set up contradictory proofs with equal apparent validity. The only way to really come to an agreement which is consistent (that I see) is to accept that phrases like "2infinity + 4" actually have meaning.


Let me ask you this. I have a one unit length ruler, on which 1/2, 1/4, 1/8 etc are marked. You move a pencil along the edge from 1 to 0. What is the last mark that the pencil touches?

You make a series of assumptions about how these sort of "infinite work" procedures must operate. List them all and we can talk. As it stands, the only rule that the "empty jug" proponents need is that you can trace what happens to any one object being moved around, and by examining what happens to the individual objects we can infer what happens to the system as a whole.
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Re: Infinite Balls and Jugs [solution]

Postby Wildcard » Sat Oct 17, 2009 9:55 pm UTC

The last mark is 0, of course. Alternatively, if you meant the last mark *before* zero, the answer is 2^(-infinity).

Unfortunately I will now be absent from xkcd forums for some time, so can't immediately continue the conversation.
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Re: Infinite Balls and Jugs [solution]

Postby jestingrabbit » Sat Oct 17, 2009 10:08 pm UTC

Wildcard wrote:2^(-infinity)


That isn't a mark I put on the ruler, so how does the pencil touch it?
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Re: Infinite Balls and Jugs [solution]

Postby skeptical scientist » Sat Oct 17, 2009 10:14 pm UTC

Wildcard wrote:The point being, you can't answer this question, which is exactly as valid as your question about naming a numbered ball that is still in the jug. This is why I say that math with infinity is ridiculous, because you can set up contradictory proofs with equal apparent validity. The only way to really come to an agreement which is consistent (that I see) is to accept that phrases like "2infinity + 4" actually have meaning.

This is nonsense. Talking about infinity is fine in any formal mathematical context. If you ever have contradictory proofs in such a formal context, it's because one of them is incorrect, and usually it's not hard to figure out which. If you think that the concept itself causes problems, it's probably because you don't understand it (as is further evidenced by your "2-∞ is the last mark before 0" comment). The disagreements involving this question do not involve the nature of infinity, but how to formalize, mathematically, a nonphysical situation.

If you formalize the question, say by saying that the set of balls at midnight is the limit* of the sets of balls at times approaching midnight, then there is no debate over what happens in that model. It is quite clear that the limit of the sets of balls is the empty set, and therefore the jug winds up empty. The only possible room for debate is over which model is correct. Usually in such situations we could decide by an experiment: the correct model is the one that produces the correct prediction. Here, of course, no experiment is possible, so we are left with no way to decide which model is correct.

———
*As mentioned earlier in the thread, if [imath](A_n)[/imath] is a sequence of sets, then [imath]\limsup_n A_n=\bigcap_N \bigcup_{n>N} A_n[/imath] is the set of elements infinitely often in [imath]A_n[/imath], [imath]\liminf_n A_n=\bigcup_N \bigcap_{n>N} A_n[/imath] is the set of elements eventually always in [imath]A_n[/imath], and if [imath]\liminf_n A_n = \limsup_n A_n[/imath], then we call it the limit of the sequence of sets.

Wildcard wrote:Edit: And how do you get the infinity symbol anyways?

I use alt+5 on my mac keyboard, but you can also use copy/paste from my post or anywhere else you can find the symbol if that's not an option. (Wikipedia's good in this regard - that's where I usually go if I want a greek letter to copy/paste into a post.)
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Re: Infinite Balls and Jugs [solution]

Postby Macbi » Sun Oct 18, 2009 7:41 am UTC

Alt 236 also makes a ∞. i.e.Hold down alt while typing 236 on the num pad, and then let it go.
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Re: Infinite Balls and Jugs [solution]

Postby quintopia » Tue Oct 20, 2009 5:40 am UTC

Or put \infty in a math tag

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Re: Infinite Balls and Jugs

Postby WarDaft » Mon Nov 09, 2009 12:53 am UTC

I contend that the jug has infinitely many balls in it with indescribable numbers on each. At least, indescribable as far as the natural numbers go.
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Re: Infinite Balls and Jugs [solution]

Postby skeptical scientist » Mon Nov 09, 2009 4:25 am UTC

Well, bully for you. Way to resurrect a three-week old thread to assert a conclusion that directly contradicts several other arguments already put forward in the thread, without addressing any of those arguments or providing any of your own. Do you have any reasoning to back up your assertion, and, if so, how do you address the arguments of those who disagree with you?
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Re: Infinite Balls and Jugs [solution]

Postby WarDaft » Mon Nov 09, 2009 3:40 pm UTC

Sure. (I admit my assertion was rather sensationalist, I was tired, and there are only a finite number of balls in the jug I believe)

In a simpler case, if we start with a jug that has a ball labeled 0, and on step one we remove it, destroy it, and add a ball labeled 1. On step two we remove that, destroy it, and add a ball labeled 2, and so on, with exactly one ball labeled with each finite number, then the same argument applies that every finite number is eventually removed, but for any step δ there is a ball labelled δ in the jug and no balls labeled less than δ. For all finite labeled balls to be removed, δ must be greater than every finite number. Since δ must be greater than every finite number, a greater than finite number of steps must be taken, and a greater than finite number of additions must be made. Since there is exactly one ball labeled with each finite number each of which we destroyed, then after a greater than finite number of additions, the number on the ball must be greater than any finite number. Since we selected only such a δ to remove all finite numbered balls, we have not removed any balls added with a greater than finite number on them. Since the addition is at the end of a continuously increasing sequence of finite numbers, one such appropriate label for the ball is the ordinal ω. If the addition had be made at the beginning, it would not.


On an even simpler level, consider a ball with, instead of a label, a counter on it. If on step δ it is provable that there are balls with labels greater than or equal to δ, then at midnight, the counter would read with a number greater than or equal to some equivalent of ω, otherwise, there is no appropriate number for it to display.


Of course, if you add ball a ball on steps evenly divisible by 2 and remove a ball on steps evenly divisible by 3, then you still have a vanishing set, because there are no balls labeled greater than or equal to ω and every ball less than ω has been removed at a step less than ω. It's not the vanishing I'm taking issue with, it's that we always have balls with labels greater than the step number in the jug.
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Re: Infinite Balls and Jugs [solution]

Postby skeptical scientist » Mon Nov 09, 2009 4:05 pm UTC

So, where did this ω-labeled ball come from? There was never a label with 'ω' on it at any finite stage, and yet somehow it materializes at the end of the process?
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Re: Infinite Balls and Jugs [solution]

Postby WarDaft » Mon Nov 09, 2009 4:16 pm UTC

The union of all finite steps is step ω is it not? If we do not take the union of all finite steps, how do we remove all finite numbers? Does this step not also add balls?

I fail to see why it's absence in any finite stage is a reason for it's total absence when it's also not empty in any finite stage.
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Re: Infinite Balls and Jugs [solution]

Postby Token » Mon Nov 09, 2009 4:30 pm UTC

WarDaft wrote:For all finite labeled balls to be removed, δ must be greater than every finite number.

What is δ here? In the previous sentence, you had it being an arbitrary finite number, but now it's not clear what you mean.
Since there is exactly one ball labeled with each finite number each of which we destroyed, then after a greater than finite number of additions, the number on the ball must be greater than any finite number.

Again, the number on which ball? If you mean "the ball that's left in the jar", you've got to show that such a thing exists first.
On an even simpler level, consider a ball with, instead of a label, a counter on it. If on step δ it is provable that there are balls with labels greater than or equal to δ, then at midnight, the counter would read with a number greater than or equal to some equivalent of ω, otherwise, there is no appropriate number for it to display.

Why do you dismiss the "no appropriate number for it to display" option out of hand?
WarDaft wrote:The union of all finite steps is step ω is it not? If we do not take the union of all finite steps, how do we remove all finite numbers? Does this step not also add balls?

I fail to see why it's absence in any finite stage is a reason for it's total absence when it's also not empty in any finite stage.

If it's in the jar, then it must have been added at some point. If it was never added, it can't be in the jar.
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Re: Infinite Balls and Jugs [solution]

Postby WarDaft » Mon Nov 09, 2009 4:45 pm UTC

What is δ here? In the previous sentence, you had it being an arbitrary finite number, but now it's not clear what you mean.
δ is the step number.
Again, the number on which ball? If you mean "the ball that's left in the jar", you've got to show that such a thing exists first.
That's a circular argument. The argument that it isn't there is that it can't have a number, yet if it can have a number I must prove it is there to have it?
Why do you dismiss the "no appropriate number for it to display" option out of hand?
What number is greater than all finite numbers but smaller than all numbers greater than all finite numbers? That's an empty set is it not?
If it's in the jar, then it must have been added at some point. If it was never added, it can't be in the jar.
It's added on step ω, unless we re-define the problem to have step ω to only remove balls and not add them. If the step is less than ω, there are still finite numbers.
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Re: Infinite Balls and Jugs [solution]

Postby Token » Mon Nov 09, 2009 5:25 pm UTC

WarDaft wrote:
What is δ here? In the previous sentence, you had it being an arbitrary finite number, but now it's not clear what you mean.
δ is the step number.

The step number of some hypothetical step on which all balls have been removed? You've said any such step must have a number greater than finite number, but there is no such step, so on no step have all the balls been removed. This is perfectly fine, as we're talking about what happens after all the steps.
That's a circular argument. The argument that it isn't there is that it can't have a number, yet if it can have a number I must prove it is there to have it?

No, the argument that it isn't there is that if there were a ball in the jar, it would have to have been added at some point but not removed, but there is no such ball.
What number is greater than all finite numbers but smaller than all numbers greater than all finite numbers? That's an empty set is it not?

Yes - continue this line of reasoning.
If it's in the jar, then it must have been added at some point. If it was never added, it can't be in the jar.
It's added on step ω, unless we re-define the problem to have step ω to only remove balls and not add them. If the step is less than ω, there are still finite numbers.

Wait... you defined "step ω" to be the "union of all finite steps", right? Therefore, if it's added in step ω, it must have been added in some finite step, surely?
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Re: Infinite Balls and Jugs [solution]

Postby WarDaft » Mon Nov 09, 2009 5:44 pm UTC

The step number of some hypothetical step on which all balls have been removed? You've said any such step must have a number greater than finite number, but there is no such step, so on no step have all the balls been removed. This is perfectly fine, as we're talking about what happens after all the steps.
No, it's any step number, like saying step N or step X. Are you saying it's better to map the countdown to midnight to the class of all possible numbers, finite and infinite? If it is, I'll concede the point.
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Re: Infinite Balls and Jugs [solution]

Postby Token » Mon Nov 09, 2009 5:50 pm UTC

WarDaft wrote:No, it's any step number, like saying step N or step X.

I get this, but then you clearly reduced the scope of δ in the sentence I quoted, and then somehow didn't conclude that such a step number doesn't exist.
Are you saying it's better to map the countdown to midnight to the class of all possible numbers, finite and infinite? If it is, I'll concede the point.

I'm saying no such thing. You're the one apparently asserting the existence of steps with an infinite number.
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Re: Infinite Balls and Jugs [solution]

Postby WarDaft » Mon Nov 09, 2009 6:33 pm UTC

I'll assume for the moment that I am wrong and I just don't realize how yet. Earlier you asked why I dismissed it out of hand for the existence of a number a counter could display after midnight for rules where on step K the counter would read less than K.

If we have a counter that reads K after K increments, are you saying there is no appropriate number for it to display after our midnight singularity?
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Re: Infinite Balls and Jugs [solution]

Postby Dutchflyboy123 » Mon Nov 09, 2009 11:16 pm UTC

Maybe I'm simplifying the problem too much, but this is how I would solve it:
FIrst about the time, it doesn't really matter how late it is, as it forms a series with an infinite number of elements, the time doesn't bring any other information. Now to calculate the number of left balls:

At iteration n:
Lowest ball in urn (at step 1, lowest is 2: put in 1-10, took out 1): [math]n+1[/math]
Highest ball in urn: [math]10*n[/math]
Number of balls in the urn: [math]10*n-(n+1)[/math]
If we take the limit:
[math]lim_{x->+\infty} {[10*n-(n+1)]} = lim_{x->+\infty} {(9*n-1)} = +\infty[/math]
This is when you take the lowest ball out every time. However, I can also state that the number of balls remaining in the urn is the final number of balls put in minus the final number of balls I took out, which gives:
[math]lim_{x->+\infty} {(10*n)}-lim_{x->+\infty} {(n+1)} = +\infty - \infty = undefined[/math]
Contradiction!
I think the problem is that you can't state the following as a certainty:
[math]lim_{x->+\infty} {(a+b)} = lim_{x->+\infty} {(a)} + lim_{x->+\infty} {(b)}[/math]
For example:
[math]lim_{x->+\infty} {(x^2-x)} \neq lim_{x->+\infty} {(x^2)} - lim_{x->+\infty} {(x)}[/math]
The left side clearly tends to infinity, while the right side is undefined. The question is more whether the action of putting the balls is tied to taking one out (then the number of balls in the urn tends to infinity) or if the counting is done at the end. Because of this the solution changes if you understand it differently: which means that the problem isn't stated completely enough. Both solutions are correct, they are just answers to different problems (at least, in my opinion).

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Re: Infinite Balls and Jugs [solution]

Postby skeptical scientist » Mon Nov 09, 2009 11:45 pm UTC

Why would either of those limits matter? I contend that to see what balls end up in the jug, we need only follow each ball. Every single ball is added to the jug exactly once, and then removed some number of steps later, and therefore every ball ends up outside the jug, so the jug is empty. No numerical computation of limits is required, helpful, or even meaningful.
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Re: Infinite Balls and Jugs [solution]

Postby Dutchflyboy123 » Tue Nov 10, 2009 12:11 am UTC

Well, that's exactlymy point, you can look at this problem in two ways:
1. You put the balls in and take out right away. This means that at each iteration, nine extra balls go in. This means the urn will be filled with an infinite number of balls.
2. You look at what happens to each ball. But with this method you have a problem. What you're doing is not permitted with infinite series, as you're removing part of the information (that 10 balls are put in for each ball that is taken out). It's like the following series(btw, I think it was already mentioned before): 1+(-1)+1+(-1)+... = ? What you're saying is the following:
[math](1-1)+(1-1)+... = 0[/math]
As every ball goes in and every ball goes out, there are no balls left. But this isn't the right solution. Grouping numbers is not allowed when looking at infinite sequences, because you are in fact modifying the sequence, hence the need of exact mathematics and thus, limits. Just to show why it's not allowed, if I take the same sequence again, I could also say that:
[math]1+(-1+1)+(-1+1)+... = 1[/math]
As you see, just by moving the part I observe I changed the result. You can't say that just because every ball is removed eventually, the urn will be empty, that's one of the dangers with infinity. (The right solution for the sequence is:
[math]1+(-1)+1+(-1)+... = 1/2[/math]
but don't ask me why).

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Re: Infinite Balls and Jugs [solution]

Postby jestingrabbit » Tue Nov 10, 2009 12:50 am UTC

The standard question for someone who believes that there is a ball in the urn at midnight is "You reach into the urn and pull out a ball. What number is on the ball?"

I don't think that Grandi's series is the least bit significant to the discussion. When you make that comparison you are "removing part of the information": that the balls are distinct and identifiable with a unique number. Seeing what happens to every ball is the only way to make sense of this situation imo.
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Re: Infinite Balls and Jugs [solution]

Postby WarDaft » Tue Nov 10, 2009 5:08 am UTC

Well, depending on the naming rules... ω, ω+1, or some other number. Some naming rules do not permit any names, greater or less than ω even if we assume ω steps is a given. (My add one at multiples of two, remove one at multiples of three case)

The only question I can see with the phrasing we have, is are there 9 balls, or ω balls in the jar at noon (I vote 9)... OR... is my understanding of set theory so wrong that saying ω is not the same as saying after all finite numbers in sequence. If so, I would love to know what the difference is.


Also I see people stating it's empty like it's agreed upon fact, I was under the impression that it's not actually resolved yet.
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Re: Infinite Balls and Jugs [solution]

Postby skeptical scientist » Tue Nov 10, 2009 5:35 am UTC

It's resolved to my satisfaction that it is empty. Others disagree, so it's unresolved in the sense that there is not universal agreement on an answer.

ω is the set of all natural numbers, and also the first infinite ordinal. If there were a first stage after stages 1, 2, 3, etc., then it would make sense to call it the ωth stage. However, there are only stages for each natural number in this scenario, and the question is what is left after all of those infinitely many stages are past; there is no ωth stage, or at least no such stage when anything happens.
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Re: Infinite Balls and Jugs [solution]

Postby phlip » Tue Nov 10, 2009 6:00 am UTC

WarDaft wrote:Also I see people stating it's empty like it's agreed upon fact, I was under the impression that it's not actually resolved yet.

It's resolved to the same extent that, say, 0.999...=1 or the Monty Hall problem are... in formal theory (set theory for this puzzle, real numbers for the 0.999...=1 thing, Bayesian probability for the Monty Hall problem), the answers are clear-cut, resolved, and agreed upon. In naive intuition, they're unresolved, and can get different answers depending on the intuition of the person involved.

I see no problem with stating the formal result as though it were aggreed-upon fact, even if the intuitive result is different.

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Re: Infinite Balls and Jugs [solution]

Postby WarDaft » Tue Nov 10, 2009 6:44 am UTC

No, if I Google this, I get at least one link to university professors arguing about it at least as recently as 2000. I'm asking if it was resolved since then.
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Re: Infinite Balls and Jugs [solution]

Postby skeptical scientist » Tue Nov 10, 2009 7:02 am UTC

WarDaft wrote:No, if I Google this, I get at least one link to university professors arguing about it at least as recently as 2000.

Where?
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Re: Infinite Balls and Jugs [solution]

Postby WarDaft » Tue Nov 10, 2009 7:11 am UTC

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Re: Infinite Balls and Jugs [solution]

Postby Dutchflyboy123 » Tue Nov 10, 2009 7:22 am UTC

Well, my point is made by someone who is probably much better at math than I am (the professor in the document WarDaft posted):
It is well-known, however, that a rearrangement of an infinite series leaves the sum unchanged if the series is absolutely convergent, but not necessarily in other cases


But just to convince you, do you agree that the number of balls in the urn at any moment can be calculated with:
[math]10*n - (n+1)[/math]
(the number of balls put in minus the number of balls that have ben taken out)?

Because if you take the graph of the function, you see it's a line. It doesn't tend to zero at all:
Image

But I might have made some error somewhere, that's the problem with infinity.

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Re: Infinite Balls and Jugs [solution]

Postby jestingrabbit » Tue Nov 10, 2009 10:16 am UTC

I'll take Littlewood's resolution over someone who published in the Missouri journal of mathematical sciences any day.

Frankly, I can't believe that crap got published. Second sentence in the last para on page 2 begins "the instant before noon the urn contains an infinity of distinct balls". WTF happened to the Archimedean principle? On page three we have the mysterious appearance of "the ball labeled [imath]\omega[/imath]". There is no omega in the natural numbers. The gibberish that proceeds on page 4 trying to justify this ludicrous inclusion is... more than simply perplexing.

That paper is not even wrong.
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Re: Infinite Balls and Jugs [solution]

Postby Token » Tue Nov 10, 2009 11:25 am UTC

jestingrabbit wrote:That paper is not even wrong.

"Not even wrong" implies sound reasoning, but on incorrect or unverifiable premises. It's just wrong.

This is also a lesson on evaluating mathematical papers on their merit, not on the perceived credentials of their author(s). If you're interested in that kind of thing, I recommend this guy's other selected works.
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Re: Infinite Balls and Jugs [solution]

Postby WarDaft » Tue Nov 10, 2009 3:54 pm UTC

Oh well. I wasn't putting it forth as an argument that it was full, just that there still appeared to be high level debate. If not, that's fine.


In the mean time, I think I isolated the root of why I believe it is not empty.

Say we have the simplest case, a counter that switches on and displays 0 with the first tick, and increases to the next smallest number at every subsequent tick that has gone past in the singularity event.
To interpret which tick we are past, we construct a set from every number that has appeared on the counter after it has ticked.

When we have {0} we know 1 interval has gone by.
When we have {0,1} we know 2 intervals have gone by.
In general, when we have [0,X) we know X intervals have gone by.

So is it, or is it not, correct to say [0,ω) after midnight? All of the finite numbers have gone by, so we cannot say [0,n) for any natural n. Is everything I know based on a lie and there is something between these two concepts?

If I am not a complete idiot, then why can there is a ball labeled n at each interval [0,n), and not a ball labeled ω at [0,ω)? Is there something in the problems wording I am missing that precludes its existence?
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Re: Infinite Balls and Jugs [solution]

Postby Macbi » Tue Nov 10, 2009 4:58 pm UTC

You can't have a ball with ω printed on it, because the balls can't possibly know what Greek symbols mathematicians have adopted to stand for infinite ordinals! No, seriously, what do you expect the ω-ball to look like? I don't necessarily think you're wrong about there being some kind of balls left at the end, but saying that they know Greek is a stretch.
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Re: Infinite Balls and Jugs [solution]

Postby WarDaft » Tue Nov 10, 2009 5:03 pm UTC

Why should they know the Arabic numeral system either then?

The logical interpretation is that someone's writing on them, they can know Greek, or some other system for writing the equivalent in a way intelligible to others.
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Re: Infinite Balls and Jugs [solution]

Postby Macbi » Tue Nov 10, 2009 5:07 pm UTC

Because we made the ball with the Naturals on before starting the problem (or at least during it), whereas the ω-ball has magically appeared in the jug.
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Re: Infinite Balls and Jugs [solution]

Postby WarDaft » Tue Nov 10, 2009 5:27 pm UTC

Can we at least assume they are uniquely labeled (except in some proposed alterations which involve moving or changing labels) in a way useful and intelligible to us, such that even if we cannot possibly notate the method the balls use we can still determine a functional equivalent from some other form of notation we traditionally employ? I believe that still leaves my questions in-tact.
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Re: Infinite Balls and Jugs [solution]

Postby mike-l » Tue Nov 10, 2009 6:38 pm UTC

WarDaft wrote:Can we at least assume they are uniquely labeled (except in some proposed alterations which involve moving or changing labels) in a way useful and intelligible to us, such that even if we cannot possibly notate the method the balls use we can still determine a functional equivalent from some other form of notation we traditionally employ? I believe that still leaves my questions in-tact.


The point is that the problem explicitely states where ball n comes from, for all natural n. So there is a ball labeled 1,000,000,000, there is a ball labeled with a googolplex, there is a ball labeled with Graham's number. But there is no ball labeled omega.

As for your sequence counter, let me correct your notation.

After step X, your sequence counter is [imath][0,X] \cap \mathbb{N}[/imath]. Notice that X is included in the interval. The union of all steps leaves you with [imath][0,\infty)\cap \mathbb{N}=\mathbb{N}[/imath]. Infinity, or omega, or whatever you want to make up to call it, isn't in the set.
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Re: Infinite Balls and Jugs [solution]

Postby WarDaft » Tue Nov 10, 2009 6:51 pm UTC

That's a different counter. Either that, or I am still a complete idiot and intersecting {0,1,2,3} with {0,1,2,3,...} gives us {0,1,2}, representing the ordinal 3, for the time between step 3 and step 4.

Furthermore, the question as posed does not specify that we are only dealing with natural numbers, it simply gives us examples which are both natural numbers and ordinal numbers, and we have infinitely many (we will presume uniquely) labeled balls.

Now, if we were to explicitly declare that only balls with natural numbers are included under any circumstances, then it is of course empty, as we cannot add balls with ordinals even if it were appropriate. If we had, without question, ω+3 steps, but were only allowed to add each natural number once, then we are defining it to be empty regardless of it's circumstances at anything but finite steps. We can run out of balls, but still want to add some given the other rules of the process. If we only had balls labeled up to 50, it would also be empty at steps 51 plus, despite the rules telling us we should have put enough in such that it would not be empty at step 51.

If we permit the addition of balls with ordinal numbers, then we do not have a case where the other rules are telling us to add a ball that we do not have available to us, and it is only empty if the other rules never ask us to add a ball with a number greater than or equal to ω.

When I read the question, I am not given the impression that the availability of balls is our limiting factor. If it is explicitly incorporated into the question, it becomes trivially obvious.
Last edited by WarDaft on Tue Nov 10, 2009 7:24 pm UTC, edited 2 times in total.
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