I wrote
I should have added that I mean for all the elements in S to be terms in the series. That is, every term in the series is an element of S, and every element in S is a term in the series.
Sorry, Tyrrell McAllister, I misunderstood what you were saying about ZenBeam’s post. I didn’t read your post closely enough, so I thought you were objecting to him grouping his terms by putting parentheses around them. You were really talking about rearrangements, and in that case you’re correct that you can’t rearrange an infinite series and say that you still have the same series, because it can converge to something else unless the series is absolutley convergent. I’m not ready to say what implications (if any) this has on ZenBeam’s argument.
I’m not so sure about this. For one, there seems to be a perfect map between the coins and the pieces of the cylinder. On one move, you break a piece of cylinder in two, and if you want, you could call one piece the “old piece” and the other the “new piece.” With the coins, you can stack the coins, and on one move you can place a new coin next to one of the old coins, and there is a pattern to where you place the new coin that can mimic the cylinder problem. If you wanted to, you could even keep shrinking the coins, so that the two problems would look identical.
I’m not sure why you’re mapping the cylinder pieces to base-2 “decimals” like .01, .1101, & .001001 insteading of mapping them to the corresponding base-2 integers, like 10, 1011, & 100100. Whichever map you choose (and my inclination is that the map to the integers is more fitting), it looks like you should be able to use the same map for either case.
All of this is irrelevant to the OP, of course.
I like your dragon example, Cabbage. I’ll have to think about it some more.
Let me try to explain my thinking in more detail. Consider a binary tree:
/\
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/\ /\
/ \ / \
/ \ / \
/ \ / \
/\ /\ /\ /\
/ \ / \ / \ / \
/\ /\ /\ /\ /\ /\ /\ /\
:: :: :: :: :: :: :: ::
__________________________________
In this tree, the branches at each level are half the length those on the preceding level, so the tree ends at the horizontal line at the bottom.
The tree has countably many branch points (nodes) because you can enumerate them by labeling the node on the first level 1; the nodes on the second level 2 and 3; the nodes on the third level 4, 5, 6, and 7; and so on.
But consider the set of all paths down the tree. This set is uncountable, because I can define a bijection f between these paths and the binary numbers between zero and one as follows:
[ul]f§ = 0.[symbol]e[sub]1[/sub]e[sub]2[/sub]e[sub]3[/sub]e[sub]4[/sub][/symbol]…,[/ul] where P is a path and, for each i = 1, 2, 3, . . . , I define [symbol]e[/symbol][sub]i[/sub] = 0 (resp. 1) if path P takes the left (resp. right) branch at the ith node on the path. (It is because each path passes through infinitely many nodes that I need to use the digits on the right of the decimal point, rather than on the left, to represent them).
The horizontal line at the bottom is supposed to be all the “limit points” you can get to by traveling one of these paths down the tree (I called these points on the line “terminal nodes” in my other post). There is exactly one of these points for each path, so there are uncountably many of these points. The horizontal line is also like the cylinder of gold in you scenario. Each terminal node is like a coin in the cylinder at the 2 hour mark.
Now, the binary tree in your scenario is arranged a little differently if we want the vertical axis to be time. Here, there is only one branch point at each level before the horizontal line. So we have something more like:
/\ 0 hours
/ \
/ \ |
/ \ |
/ \ |
/ \ |
/\ \ | Time
/ \ \ |
/ \ \ |
/ \ /\ |
/\ \ / \ |
/ \ /\ / \ |
/ \ / \ /\ \ \ /
/ \ / \ / \ /\ v
: : : : : : ::
________________________________ 2 hours
But my function f still works, and so still establishes the uncountability of the coins at 2 hours.
If Science Fiction has taught me anything[sup]1[/sup], it’s that in the next instant, the census taker turns to look at a large number of bodies charred beyond recognition, stretching to the horizon, saying “…so who are all those people.” [Cue the spooky music, fade to black.]
[sub]1. Here’s a free straight line.[/sub]
Um, at the risk of being obtuse and dim, it seems to me that the exact number of coins at the two hour mark is akin to some sort of critical point that cannot easily be analysed, like 1/x at x=0. Does it much matter how many coins there are at exactly the two hour point? Isn’t it of much more interest to decide on the state of things immediately before or immediately after the two hour point?
Not if we’re specifically interested in what happens at the two hour mark (or after, for that matter, since nothing has been said to happen after the first two hours, the dish will still be empty after two or more hours, but not before).
ZenBeam, proof by science fiction? Gimme a break. :rolleyes:
By definition of the problem, there are no other people. Are you claiming this process of people leaving and entering their houses is some kind of…(waving hands)…voodoo ritual causing other beings to manifest themselves in the village? Are you even considering my arguments, or are you just convinced that you are right and won’t listen to reason? Despite what you may think, I’ve already explained what’s wrong with your “refutation” involving series. Twice.
Here’s an idea, pretend these fictional people you bring up are, well, fictional. They don’t exist. All we have in the village are the numbered people, period. What do you say now?
The numbered people are fictional too. It sounds like you want me to assume the numbered people are real, and no science fiction allowed. This problem is ill-posed. You can’t have an infinite number of real people. The observable universe isn’t big enough. Once you go outside the observable universe, you’re into fiction. There are many other problems with trying to set up this problem in a physically real way.
Mathematically, your village is equivalent to the OP, and I’ve addressed this issue three times previously, and this is now the second time I’ve said that I adressed it three times previously. You want a fourth? If the number of chips in the dish, or people outside their houses, is undefined, there’s no contradiction in not being able to name one.
Here’s another setup exploring the name them issue, and I hesitate to bring it up, since this thread is full of them, and quite frankly this thread is wearing me out, but this one is pretty simple. Assume God has two dishes, A and B, and two chips, 0 and 1. At t=0, God puts chip 0 in dish A and chip 1 in dish B. At t=1 hour, He switches them, removing chip 0 from dish A, and placing it in dish B (to be precise, I’ll specify chip 0 is in dish A for t<1, and in dish B for t>=1), and simultaneously, chip 1 is likewise moved to dish A. At t=90 minutes, he switches them again, again at t=105 minutes, and so forth. At t=2, how many chips are in the union of the two dishes? This has to be two, right? How many chips are in each dish? Doesn’t the answer have to be 1? (I guess it could possibly be undefined.) If the answer is 1, at t=2, which chip is in which dish? This has got to be undefined.
While we’re here, what if I try to determine how many chips are in dish A? For every time chip 0 is put in dish A, I can tell you when it was removed, just like in the OP. Same for each time chip 1 is placed in dish A. Does that mean dish A is empty? The same argument applies to dish B. But the union of dishes A and B has two chips, so where are they?
You can describe this problem with two villagers and two huts, but mathematically it’s the same problem, with the same answers (or lack thereof). Physically, you don’t have the problem of an infinite number of people, but you do run into infinite velocities as t–>2, and quantum mechanically there’d be issues also.
I meant “fictional” in the context of the problem, of course. In the context of Lord of the Rings, Gandalf is real, but Darth Vader is fiction. The people/chips and so on are merely ways to help think of this purely mathematical problem, so who cares if I can have an infinite number of people, either. I certainly have an infinite number of integers, and I can phrase the problem in terms of those. Or are you saying that all abstract thought is meaningless, since it’s not real?
But it’s not undefined, every set has a cardinality. The cardinality of the collection of people in the village is omega (or aleph-naught, if you prefer.). The cardinality of the chips involved in the problem is, again, omega. The cardinality of any countably infinite set is omega. And since, in each case, our “universal” set of chips/people is countable, we certainly know that the number of chips/people left has either finite cardinality or omega cardinality.
I would agree that your example is not well defined. Since the chips switch at every step, it’s not defined which is where at the end.
This is a significant difference from the OP, where, once a chip has left, it never returns to the dish.
This really is nothing more than a simple application of mathematical induction. I can demonstrate that chip one is not in the dish, I can demonstrate that if chip n is not in the dish, then neither is chip n+1 (since it’s the very next to be taken out). Therefore, none of the chips numbered 1,2,3,… are in the dish. Finally, since there are no other chips we can conclude the dish is empty.
I’m getting a little tired of dealing with this thread as well, and if I can’t convince you of the above induction argument, there’s probably not much point in either of us prolonging it.
This is not a simple case of mathematical induction, because induction doesn’t always work so well when you have an infinite set. I argued by induction against the no chips left conclusion, on examples IIIC/IIID, where on each move you put in a chip, take out a chip, renumber the chip that you took out so that it has the same number as the other chip you’re supposed to put in, and then choose whether to put in the chip that was just in or the other chip. The conclusion from the “no chips left” crowd was that, if you always put the same chip back, you end up with infinitely many chips, but, if you always put the other chip in, you end up with 0. So the final result depends on your choices. But if you put in the new chip on the 1st move, that doesn’t change anything - there’s still 2 chips. If you put in the new chip on the nth move, that doesn’t change anything - there’s still n+1 chips. For any n, regardless of what you do on the first n moves, it’s still possible to end up with infinitely many chips. Your nth move does not matter. So now you can just do induction, or let n go to infinity, or say that if it holds for any n, it holds for all n. And so there are infinitely many chips no matter what you choose on each move. So there are infinitely many chips in the situation in the OP.
I don’t know if there’s a way to reconcile this argument (and the other ones for infinitely many chips) with your arguments for 0 chips. I’m suspecting this situation is more like the lightbulb problem or the 2-chip problem that ZenBeam just gave than I thought it was, and that the answer is undefined.
I’m going to be traveling for a little while, so I don’t know if I’ll be able to keep up with this thread very well. Unless anyone has some new insight, it’s fine with me if this thread dies.
I still like the idea of numbering by dots.
The original problem, expressed this way, goes something like this:
God puts a chip in the bowl with one dot.
God then, all at the same time, puts two chips in, with two and three dots respectively, and takes out the one with one dot.
He then puts in chips with four and five dots, and takes out the one with two.
He puts in chips with six and seven dots, and takes out the one with three.
Et cetera.
Now, are we agreed that this is equivalent to the original problem?
Now, how about this:
On the first move, God puts in a chip with one dot.
On the second and following moves, he takes out the lowest chip, doubles the amount of dots on it, and puts it back in, along with another chip with one more dot on it.
In this case, what He ends up with is a countably infinite number of chips in the bowl, each of which has a countably infinite number of dots on its surface. There is no integer in the bowl, but there are still chips with symbols on them. It’s just that none of the symbols is an integer any more.
In what way, if any, does this differ from the original problem?
Huh? I don’t follow you here. The whole point of mathematical induction is to deal with infinite sets. Anybody can deal with finite sets, just take each case individually…you never need induction for that. Induction is specifically for dealing with infinite sets.
And I thought I already mentioned the problem in example IIIC/D; we don’t have a direct correspondence between the numbers and the chips. In the original example, you could point at each chip and say, “That’s chip 1, that’s chip 2,…,that’s chip 425463,…”, and therefore, you could talk about the problem in terms of the chips, or in terms of the numbers. That’s not true of your example; there, you can talk about the chips, or you can talk about the numbers, but there’s no longer any guarantee of them behaving the same way, which is why in one case you have infinitely many chips in the dish, another case you have zero chips in the dish. And note that in both cases IIIC and IIID, you have zero integers in the dish, which has been my whole point all along.
Anyway, changing the subject, although I am kind of getting worn out by this never ending debate, I have enjoyed reading and thinking about everyone’s contributions. I tend to avoid Great Debates, and often debates in general, usually, because I always worry that someone may take it personally and be personally offended by something I say. I hope that hasn’t happened here. I’m glad to have had the discussion, and I hope ya’ll are to, but, yeah, I’m about ready to let it drop if everyone else is up for that. And I wish ya’ll a Merry Christmas and Happy New Year! 
Okay, I’ll give it a chance with another more formal presentation of the problem :
Given B[sub]0[/sub] = { 0 } (the null set, meaning the bowl starts empty), define
B[sub]i[/sub] = B[sub]i[/sub]’ - min(B[sub]i[/sub]’)
where min (A) = smallest element of A, and
B[sub]i[/sub]’ = { B[sub]i-1[/sub], 2i-1, 2i }
If you do not agree that that is a correct statement of the problem, explain why you feel it is inaccurate.
Consider this :
Proposition-
P[sub]j[/sub] : For any integer s < j, s is not a member of B[sub]j[/sub].
Proof by induction-
B[sub]1[/sub] = { 2 }. Clearly P[sub]1[/sub] is true.
P[sub]k[/sub] => P[sub]k+1[/sub]?
P[sub]k+1[/sub] : For any integer s < k+1, s is not a member of B[sub]k+1[/sub]
From the set-up, B[sub]k+1[/sub]’ = { B[sub]k[/sub], 2k+1, 2k+2 }
If P[sub]k[/sub] is true, min(B[sub]k[/sub]) is >= k.
Since 2k+1 and 2k+2 > k, min(B[sub]k+1[/sub]’) >= k. In either case of (>=), there are no integers < k+1 in B[sub]k+1[/sub].
Thus P[sub]k[/sub] => P[sub]k+1[/sub], and the proposition is true by induction.
Now, it is possible to show that the number of chips in the bowl increases without bound as i->inf and we approach t=2. However, it’s also been shown (several times now) that once we have completed the infinite task, there can be no integers in the bowl.
The fact that removing a countably infinite number of members from a countably infinite set will or won’t change the set’s cardinality depending on which members you remove may seem strange (and clearly it’s not well-understood). It would have been nice to be able to explain it, but like the rest, I’m not sure if it’s going to work.
I will agree and say that this is one of the best threads in a long while and I’ve enjoyed it immensely.
Allow me to clarify that last post. First, I had changed my indices, so P[sup]k[/sup] actually works for s <= k. Replace the appropriate statements with :
**If P[sub]k[/sub] is true, min(B[sub]k[/sub]) is >= k+1.
Since 2k+1 and 2k+2 > k+1, min(B[sub]k+1[/sub]’) >= k+1. In either case of (>=), there are no integers < k+2 in B[sub]k+1[/sub]. **
Additionally, It is not the intention of that proof to show that the bowl is empty at the end (i.e. for t = [2, oo ) ), because it doesn’t show that. Those proofs have already been given by Cabbage, Tyrell McAllister, et al. I just thought that by also observing similar behavior on the other side (leading up to 2), it might cause some light to dawn.
My main intention was to give the formulation I did. Hopefully this will also explain why the problem can’t be modeled as a series. A sequence of sets, maybe. BUT we are only concerned with a single set - the terminal set after the operations are completed.
I agree this has been a fun thread. I haven’t always enjoyed it, but I have on the whole.
I’m ready to let it lie too, but I think Chronos’s last post needs to be addressed: numbering with knock knock’s dots exposes a crucial element. For visuallization purposes, put all the dots in a line on the chip. 1 is represented by a dot at 1, 2 by dots at 1 and 1/2, 3 by dots at 1, 1/2, 1/3, and so forth. The symbol for infinity naturally falls out of this: dots at 1/n for all positive n. As Chronos points out, this symbol isn’t an integer. The only difference between Chronos’s first example and the OP seems to be the existance of a symbol for infinity which falls naturally from the notation for the integers. This doesn’t seem to be a sufficient difference for these two cases to have different answers.
Many of the proofs here show there are no chips with integers on them left in the dish at t=2. This doesn’t rule out chips with the symbol for infinity, which therefore must be ruled out some other way. Certainly, there are no such chips for t<2. Has anyone shown there must be no such chips for t = 2? It would seem this can’t be said to be obvious, certainly not when using the dot notation, since the numbers of dots on the chips in the dish are approaching infinity. The simplest argument, “no chips ever have an infinite number of dots for t<2, therefore there aren’t any at t=2” would seem to be equally as valid or invalid as saying “the number of chips is > 0 for t<2, therefore the number is > 0 for t=2”. I don’t recall seeing anything more rigorous than that. Also, that argument would apply equally to Chronos’s second example above, but I think everyone’s going to agree there are chips with infinitely many dots at the end in that case.