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Topic: Impish Pixie revisited (Read 7731 times) |
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rmsgrey
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Re: Impish Pixie revisited
« Reply #25 on: Jul 23rd, 2005, 9:38am » |
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But there are two different functions being evaluated. One is the number of balls in the jar (as a function of time) and the other is the set of balls in the jar (as a function of time).
The first tends to infinity as time tends to infinity; the second to the empty set. The fact that a certain relationship between them holds at all finite times which fails to hold in the limit tells us that taking the size of a set and then taking a limit can give a different answer if done in the reverse order.
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Eigenray
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Re: Impish Pixie revisited
« Reply #26 on: Jul 23rd, 2005, 1:01pm » |
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My two cents:
Mathematically, it's a question of convergence, and therefore the answer depends on what you choose for a state space -- not just as a set, but as a topological space. Loosely speaking, lim xn = x exists iff xn will eventually be as close to x as you want given enough time. Of course this depends on what "close" means; having a topology allows us to say "for every neighborhood U of x, there is an N such that xn is in U for all n>N." (Strictly speaking, not all notions of convergence come from a topology. For example, a.e. convergence is not topological.)
If we only care about the number of balls in the urn, we can, in the obvious way, take our state space to be N. Thus if xn is the state after the nth step, then x1=1, x2=2, and in general xn=n. With the usual (discrete) topology on N, lim xn does not exist (a sequence in a discrete space converges iff it is eventually constant). The question is then meaningless: if I write the number 1 on a piece of paper, erase it, write 2, erase it, write 3, erase it, etc., what's on the paper when I'm done with it?
A better space is Y=N U {infinity} = omega+1, the one-point compactification of N. (This topology agrees in this case with the order topology on Y.) Note that Y is not discrete: a sequence in Y converges to a finite number iff it is eventually constant, but converges to infinity iff it leaves every compact (i.e., finite, i.e., bounded) subset of N. Thus lim xn = infinity, and there are an infinite number of balls in the limit.
But what if we want more information? We could instead use as a state space the power set of the natural numbers, P(N), which we can think of as the set of functions N->{0,1}, i.e., the cartesian product of infinitely many copies of {0,1}. If, for each k, the urn contains xk balls labelled k, then we say the urn is in state x=(x1,x2,...), and that is the only information we keep track of. Now let xn be the state after the n-th step, i.e.,
x1=(0,1,0,0,0,....)
x2=(0,0,1,1,0,0,0,...)
x3=(0,0,0,1,1,1,0,0,...),
or whatever you like. The standard way to topologize P(N) is to let 2={0,1} be discrete, and give P(N)=2omega the product topology. It is left as an exercise to the reader to show that if xn [in] 2[omega] is the characteristic function of the set Xn [subset] N, then lim xn exists iff limsup Xn = liminf Xn (cf. Icarus's discussion of limits of sets). With this topology, lim xn = (0,0,0,...), i.e., the urn is empty in the limit. To see this, any open set U around (0,0,...) is of the form U1 x U2 x ..., where only finitely Un are {0}, and the rest are 2. Thus if N=max { n : Un={0} }, then xn [in] U for n>N.
Now the "contradiction" is easily explained: if f:2omega->Y is the function given by f(x1,x2,...)=x1+x2+... (note that the infinite sum always converges in Y), then we've just seen that lim f(xn) = infinity, while f(lim xn)=0. Thus f is not continuous, and that's all there is to it.
Of course, f is continuous if we give P(N) a different topology: take as open the sets f-1(U) where U is open in Y. We've added enough open sets to make f continuous; moreover, lim xn=x iff lim f(xn)=f(x). So we haven't added enough: P(N) is no longer Hausdorff, and "the" limit is no longer unique; any x with f(x)=infinity will do.
Exercise: is there a topology on P(N) such that f is continuous and lim xn exists and is unique?
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| « Last Edit: Jul 23rd, 2005, 1:08pm by Eigenray » |
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jeffypop
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Re: Impish Pixie revisited
« Reply #27 on: Jul 25th, 2005, 4:06pm » |
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Okay, now that I've gone and re-read the entire Original Thread, which I should have done to begin with, I see that my previous posts were garbage and already addressed (sorry about that). 
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Icarus
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Re: Impish Pixie revisited
« Reply #28 on: Jul 28th, 2005, 7:23pm » |
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If you now understand something you didn't before, then your questions were worthwhile.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Slappy
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Re: Impish Pixie revisited
« Reply #29 on: Aug 27th, 2005, 5:29pm » |
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I think the answer is
| hidden: | | a googol is the highest number before infinity so can be thought of as infinity - 1. |
| hidden: | | adding two and removing 1 each minute will always result in an infinite # of balls |
| hidden: | | removing 1 from infinity will always be a googol |
I'm brand new here so pardon me if this has been answered in another thread
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BNC
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Re: Impish Pixie revisited
« Reply #30 on: Aug 27th, 2005, 10:49pm » |
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on Aug 27th, 2005, 5:29pm, Slappy wrote:
a googol is the highest number before infinity so can be thought of as infinity - 1.
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Sorry, but no. A googol is only a lowly 10^100, which is in no way even close to infinity.
As for previous posts, do try to take the time to read the original thread.
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| « Last Edit: Aug 27th, 2005, 11:21pm by BNC » |
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Icarus
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Re: Impish Pixie revisited
« Reply #31 on: Aug 28th, 2005, 3:35pm » |
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Indeed. A "googol" is 10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,0 00,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, considerably smaller than infinity. a googol + 1 is 10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,0 00,000,000,000,000,000,000,000,000,000,000,000,000,000,000,001.
Which still has only 101 digits. Infinity, if you could write it out, would have an infinity of digits. googol is not even the largest natural number that gets used. For a common method of data encryption, the standard practice is to create numbers of about 200 digits consisting of the product of 2 primes (to decode the data, you have to know how to factor the 200 digit number).
There are many larger natural numbers that have been named and studied. One such set includes the "googol-plex" - 1 followed by a googol of zeros, the "googol-plex-plex" - 1 followed by a googol-plex of zeros, etc. The "Ackerman numbers" are larger yet.
But even these are not the highest natural number - there is no highest. For any natural number that you give me, I can give you a higher number by simply adding one. But no matter how high these go, infinity is beyond all of them. The infinite cannot be reached by adding finite numbers together.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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meatbalz
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Re: Impish Pixie revisited
« Reply #32 on: Feb 10th, 2006, 5:58am » |
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sorry to resurrect this topic but after reading through both threads relative to the problem, i'm not sure i understand the reasoning that leads to the answer there are no balls left "after the task is completed"
at no point in the entire set of steps is there a step which involves removing more balls than you put in in that same step.
if you look at the problem going backwards in time (using the 1/2, 1/4, 1/8... time intervals), then the step immediately prior to the task being completed involves two balls being added and one being taken away.
hence after the last step before the task is completed, we are left with at least one ball still in the bucket.
also, one of the posts stated that "when the process is over, all natural numbers have been removed from the bucket".
how can this be reconciled with the fact that, in the step immediately prior to the process being completed, two balls were added and only one removed?
please do not post replies such as "read the thread again", respond to the points i made in my post.
thanks 
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rmsgrey
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Re: Impish Pixie revisited
« Reply #33 on: Feb 10th, 2006, 7:21am » |
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on Feb 10th, 2006, 5:58am, meatbalz wrote:sorry to resurrect this topic but after reading through both threads relative to the problem, i'm not sure i understand the reasoning that leads to the answer there are no balls left "after the task is completed"
at no point in the entire set of steps is there a step which involves removing more balls than you put in in that same step.
if you look at the problem going backwards in time (using the 1/2, 1/4, 1/8... time intervals), then the step immediately prior to the task being completed involves two balls being added and one being taken away.
hence after the last step before the task is completed, we are left with at least one ball still in the bucket.
also, one of the posts stated that "when the process is over, all natural numbers have been removed from the bucket".
how can this be reconciled with the fact that, in the step immediately prior to the process being completed, two balls were added and only one removed?
please do not post replies such as "read the thread again", respond to the points i made in my post.
thanks |
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You're assuming that there is a last step...
For every step, two specific balls are added, and one specific ball removed, so which two balls are added on the last step? and which ball is removed? If you're going to reverse the process, you have to be able to do the first step of the reversed process...
The problem is that there's a discontinuity between finite and infinite - there is no finite state that has any infinite state as a direct successor, so you can't say "well, just before..." or "the last step to go infinite does..."
Ultimately, the answer to this problem rests on what assumptions you choose to make about infinite processes. The standard assumptions made give the answer that the first infinite state has an empty bucket. If you want to use different assumptions, you can get different answers, but the standard assumptions have been chosen because they generally work more nicely.
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