Like Chronos I would have thought the answer undefined. But is there something about this problem that makes the solution definite after all?
If you model the coconuts that the ape has at the end of each step as a set of numbers, then the standard notion of what it means for a sequence of sets to converge forces the limit to be the empty set. That’s the very natural model for a mathematician, and I’m not familiar with any objections to it that don’t basically boil down to not liking the conclusion. That’s a reasonable position to take, mind you, but if you’re going to come up with a different model, you need to do some arguing that it’s a reasonable way of looking at simpler situations.
Is there a known result for determining whether the limit of the size of sets is the size of the limit of the sets? That is, if S[sub]n[/sub] are a sequence of sets which converge to S, and the function #(X) is the number of elements in a set X, is there a condition ensuring that:
lim[sub]n -> infintiy[/sub] #(S[sub]n[/sub]) = #(S)?
I wonder, because clearly Hari Seldon’s problem doesn’t satisfy this equality. Maybe we can think of it as analogous to integration, and note that this problem would have no dominating function.
Assuming choice, this holds if and only if all of the S[sub]n[/sub] are subsets of some finite set.
But there are other models which are just as natural but which give different conclusions. For instance, you could consider the quantity of coconuts that the ape has as a function of time, and then take the limit as t -> 1 hour, under which model the ape has an infinite quantity of coconuts at the one hour mark.
And consider related problems: Suppose, for example, in problem B, that the monkey on each of his burglaries steals the second-lowest-labeled coconut from the ape. Well, in that case, it’s clear that the ape will always keep coconut #1, regardless of what else he has.
But now suppose, in problem C, that on every trip, the monkey grabs the two lowest coconuts in the ape’s cave, erases the markings on both, and then writes the markings back on, swapped, and now takes the (new) lowest coconut? Well, now the monkey still ends up with exactly the same set of coconuts as he had in B, and the ape still has that same coconut he had in B, but that coconut has had its number erased and re-written so many times that we can’t tell what the label is on it now.
There are similar re-labeling tricks the monkey could pull that would leave the ape with any set of coconuts, of any size. But all of these problems have, at any given time, each simian having the same quantity of coconuts as in the original problem. If they’re all the same on the time interval that’s actually specified, what is the justification for saying that they’re different outside of that time interval?
I’m not sure what you’re defining the individual sets S[sub]n[/sub] in Hari Seldon’s problem. The gorilla collects in sets that make up the integers, and so does the monkey. Both of the sets they collected “converge” to infinity.
I’d consider the problem as follows:
For any coconut x, the gorilla has placed it in his cave.
For any coconut y that was placed in the gorilla’s cave, the monkey has stolen it.
Therefore, for any coconut x, it is in the monkey’s possession.
Here’s a variation:
The gorilla isn’t quite as dumb as he looks. After the monkey has stolen ten coconuts, the gorilla notices. So the next time he goes to gather from the pile (and each time thereafter), he steals one coconut back from the monkey, starting with coconut 1. How many coconuts will the monkey have at the end of the hour?
That makes sense. It’s what dominated convergence would provide.
I’m imagining S[sub]n[/sub] as the set of coconuts the ape has after gathering for the n[sup]th[/sup] time. So S[sub]1[/sub] = {1,2,…,1000}, S[sub]2[/sub] = {2,3,4,…,2000}, S[sub]3[/sub] = {3,4,5,…,3000}, etc.
I can kinda see the analogy, but you’d have to be careful about how exactly you try to make it precise if you want to say it’s more than analogy.
I think you should ask my genius brother-in-law. He’ll just look at you, and wisely intone:
“One to one. Intuitively obvious. Don’t think so much.”
Why “only if”? Are you assuming the S[sub]n[/sub] and S are finite? Because, otherwise, one has, for example, the trivial counterexample where all the S[sub]n[/sub] and S are the same infinite set.
There’s so much going on in this thread, it’s hard to know where to begin. My area of research is number theory, and while my work is largely algebraic in nature, one cannot help but study analytic number theory eventually. Questions like “what is the ratio of even to odd numbers” or “what fraction of primes are congruent to 3 modulo 8” arise all the time in analytic number theory, perhaps most frequently in sieve theory.
Others have pointed out that every set that can be put in bijection with the natural numbers has the same cardinality. This is important, but useless for answering questions like the ones above. One way that a number theorist answers questions like these is by defining the natural density of a subset of the natural numbers. (I want to stress this is not the only way one can approach questions like these, but it is a natural way, and undoubtedly the way most common to a number theorist.) I thought about trying to define natural density here, but decided not to try mathematical notation here. I hope you’ll read the link.
Using this concept of natural density, I can show that the density of the even and odd natural numbers exist and are the same, namely the density of both is one-half. By an abuse of terminology, therefore, I can make statements like “half of the natural numbers are even, and half are odd,” where I am implicitly referring to their natural density, and if this is ambiguous or I need to be more rigorous, I will explicitly state that.
Yes, there is, assuming that by “converge to S” you mean “converge in measure.” Measure theory is the natural way to approach a question like this. Given a measure on a set S, that measure is continuous from below, so assuming you choose your subsets appropriately, then yes, you obtain the desired result. I must add that not every set is measurable, so conceivably this question would be meaningless from the point of view of measure theory. The natural numbers are measurable, though–the usual measure is the counting measure.
I somehow mentally attached a “for every sequence” to the original statement. Mea culpa.
Yes, there is. It is that each individual coconut ends up in the monkey’s cave. That is almost too obvious for words. If you disagree, name one that doesn’t. Notice that it crucially depends on the number and the order in which things happen, just like the ratio in the OP. Infinity is different.
You are presupposing a certain relation between the coconut’s state at the end of the process and the coconut’s state throughout the period prior to the end of the process, Hari. That connection is what people are taking issue with as underdetermined by the problem as specified.
Suppose John and Bill toss a coconut back and forth. John starts out with the coconut. After an hour, he gives it to Bill. After another half an hour, Bill gives it back to John. After another quarter hour, John gives it back to Bill. Etc. Two hours after the start, who has the coconut?
For every coconut, label if with the same number in red. At each step, switch the red label (n-1)*1000+1 with the red label of the coconut to be removed. So the first step, no relabeling is done, and red 1 is removed. The next step, after the ape collects coconuts 1001-2000, red 1001 is switched with red 2, and red 1001 is removed by the monkey. On the third step, red 2001 is removed, and so forth.
This can be done for every step. Labels 2,3,…,1000 are always on a coconut, and are never removed. Labels 1002,1003,…,2000 are also always on a coconut, and are never removed, etc.
If a coconut labeled red 2 is not in the Ape’s cave, on which step was it removed from the cave? Or failing that, if there is no coconut with a red 2, on which step did that occur? Neither of those occurs with any of the actions mentioned.
The assumption implicit here is that a property that holds after finitely many steps must necessarily hold after infinitely many steps, which is not something that you can ever claim without a good reason. It’s very easy to see that it’s false in general: consider the sequence of sets S_n = {1, 2, …, n}. Each set in the sequence is finite, but it doesn’t follow that the limit of the sequence is finite.
This is a different process, and we shouldn’t expect it to have the same outcome.
I guess since I’m in it now, I’ll ask about a related setup:
The Monkey has a bamboo pipe in his cave, infinitely long in one direction. It has an inside diameter of 1 inch, and it is marked off 1, 2, 3, … in 1 inch sections. There are mangoes, conveniently cylindrical shaped, 1 inch diameter and 1 inch long. The mangoes are also numbered 1, 2, 3, …
While the ape is collecting his first 1000 coconuts, the monkey picks up mango 1 and slides it into section 1 of the pipe. While the ape is collecting the next 1000 coconuts, the monkey slides mango 2 into section 1 of the pipe, pushing mango 1 into section 2. This continues, with the monkey always adding one mango to the pipe, sliding the ones already there to one section higher.
At the end of the hour, which mangoes are in the pipe?
It’s the same process. I’ve just added red labels to the coconuts.
That doesn’t disprove anything. The limit of S_n is still infinite.
The problem with all of this is that you are saying that 1000*infinity = infinity. And I still can’t understand why this is acceptable when doing higher math, but yet is considered an error when you use it to prove that 1 = 2. If a known application produces false results, then why do we assume it produces accurate results in unknown applications?
But the argument that the ape’s cave ends up empty must also make an assumption like that. The property “coconut 1 is in Monkey’s cave” is true after finitely many steps, certainly, but what makes you say that it’s true after infinitely many steps? That’s why I started off the discussion by saying flat-out that the situation at the end of the hour just plain isn’t specified.