An uncountable countable set
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From: Tony Orlow on 25 Oct 2006 15:01 David Marcus wrote: > Tony Orlow wrote: >> David Marcus wrote: >>> Tony Orlow wrote: >>>> David Marcus wrote: >>>>> Tony Orlow wrote: >>>>>> David Marcus wrote: >>>>>>> Tony Orlow wrote: >>>>>>>> David Marcus wrote: >>>>>>>>> Not sure what you mean by "separate events". Suppose we put all the >>>>>>>>> balls in at one minute before noon and take them out according to the >>>>>>>>> original schedule. How many balls are in the vase at noon? >>>>>>>> empty. >>>>>>> Why? >>>>>> Because of the infinite rate of removal without insertions at noon. >>>>> OK. Just to recall, this vase has all the balls put in at one minute >>>>> before noon, then taken out on the usual schedule. How many balls are in >>>>> this vase at times before noon? >>>> Some supposedly infinite number, as only a finite number have been removed. >>> But, for this vase, at all times before noon, there are an infinite >>> number of balls in the vase. So, how does this vase become empty at >>> noon? >> Using the time singularity of the Zeno machine, where there is a >> condensation point in the sequence that allows an infinite number of >> iterations to occur in a moment. Luckily for the vase, no one is >> inserting extra balls on the same schedule. > > Tony, > > For n = 1,2,..., suppose we have numbers A_n and R_N (the addition and > removal times of ball n where time is measured in minutes before > noon). For n = 1,2,..., define a function B_n by > > B_n(t) = 1 if A_n <= t < R_n, > 0 if t < A_n or t >= R_n. Fine for each ball n. > > Let V(t) = sum{n=1}^infty B_n(t). Let L = lim_{t -> 0-} V(t). Let S = > V(0). Let T be the number of balls that you say are in the vase at > noon. You are summing B_n(t) to oo? > > Problem 1. For n = 1,2,..., define > > A_n = -1/floor((n+9)/10), > R_n = -1/n. > > Then L = infinity, S = 0, and T = undefined. I say that if noon exists, there are an infinite number of balls in the vase. n=oo -> L=T. > > Problem 2. For n = 1,2,..., define > > A_n = -1/n, > R_n = -1/(n+1). > > Then L = 1, S = 0, T = 1. Yeah, L=T again. > > Problem 3. For n = 1,2,..., define > > A_n = -1, > R_n = -1/n. > > Then L = infinity, S = 0, T = 0. L=lim(x->oo: oo-x) = 0 <> oo L=T > > Tony, can you give us a general procedure to let us determine T given > the A_n's and B_n's? > You can keep track of the points between your time vortexes and what's going on during those periods, for starters.
From: Tony Orlow on 25 Oct 2006 15:03 David Marcus wrote: > Tony Orlow wrote: >> David Marcus wrote: >>> Tony Orlow wrote: >>>> As each ball n is removed, how many remain? >>> 9n. >>> >>>> Can any be removed and leave an empty vase? >>> Not sure what you are asking. >> If, for all n e N, n>0, the number of balls remaining after n's removal >> is 9n, does there exist any n e N which, after its removal, leaves 0? > > I don't know what you mean by "after its removal"? Does this refer to a > specific time or to any time that comes later (or does it mean something > else)? If the ball is removed at time t = -1/n, what time corresponds to > "after its removal"? > Ugh. AM i building the Queen Mary from toothpicks? From t=-1/n until t<-1/n+1, how many balls are in the vase? Uh, 9n. >> If >> not, then no matter how many n e N you remove from the vase, even if you >> remove all of them, every removal leaves balls in the vase. Paradoxical? >> Sure. But it's easily explainable and resolvable once a proper measure >> is applied to the situation. Omega doesn't lend itself to proper >> measure. Infinite series do. Bijection loses measure for infinite sets. >> N=S^L and IFR preserve measure. >> >> So, how do you empty the vase? Ball removal? Every removal leaves balls >> in the vase, as is obvious. > No comment? Every ball removed, leaves balls in the vase. Speak to that.
From: Tony Orlow on 25 Oct 2006 15:06 David Marcus wrote: > Tony Orlow wrote: >> David Marcus wrote: >>> I don't agree that he is assuming that. I think he isn't reasoning >>> logically at all. The number of balls approaches infinity as time >>> approaches noon. If you imagine a vase filling up with an infinite >>> number of balls, it is rather hard to imagine them suddenly all >>> disappearing. Of course, mathematics isn't constrained by our >>> imagination. It relies on precise definitions and logic. And, functions >>> do not have to be continuous. >> So, David, you think the fact that balls leave the vase only by being >> removed one at a time, and the fact that at all times before noon there >> are balls in the vase, and the fact that at noon there are no balls in >> the vase, is consistent with the fact that no balls are removed at noon? > >> How can you not see the logical inconsistency of an infinitude of balls >> disappearing, not just in a moment, but at no possible moment? Are you >> so steeped in set theory that you cannot see that an unending sequence >> of +10-1 amounts to an unending series of +9's which diverges? What is >> illogical about that? > >> In your set-theoretic interpretation of the experiment there is a >> problem which makes your conclusion incompatible with conclusions drawn >> from infinite series, and other basic logical approaches. > > I gave a Freshman Calculus interpretation/translation of the problem (no > set theory required). Here is a suitable version: > > For n = 1,2,..., define > > A_n = -1/floor((n+9)/10), > R_n = -1/n. > > For n = 1,2,..., define a function B_n by > > B_n(t) = 1 if A_n <= t < R_n, > 0 if t < A_n or t >= R_n. > > Let V(t) = sum{n=1}^infty B_n(t). What is V(0)? > > I suppose you either disagree with this interpretation/translation or > you disagree that for this interpretatin V(0) = 0. Which is it? > t=0 is precluded by n e N and t(n) = -1/n. >> It is not that >> I don't understand how your logic works. It's that I see clearly that it >> doesn't, and I'm trying to precisely pin down exactly where the error >> is. It's not an easy task, since this transfinite theory is rather well >> crafted and tweaked over the years. However, there are clear reasons, >> once the matter is fully investigated, why the logic fails. The >> conclusion produces clear contradictions in terms of a time of emptying >> and the requirement at some point of a negative number of balls in the >> vase in order for it to empty at all, and it all derives from using the >> Zeno schedule to complete a sequence which has no end, hiding this fact >> in a time singularity at t=0. >> >> Very basic logic would hold that, if the vase is not empty at any time t >> such that -1<=t<0, and the vase is empty at t=0, then balls were removed >> at t=0, since that's the only way the vase can become empty. However, >> t=0 corresponds, according to the stated schedule, to infinite index n >> in the sequence, and an infinite label on a ball, which is not allowed, >> as per the experiment. Therefore, no ball can be removed at t=0, and the >> vase cannot become empty at that point, or at any point before. >> >> I asked you when you thought the vase became empty. You avoided the >> question, saying it was interesting, and then going on with your same >> tired formulation of the problem, as if I haven't followed the logic and >> pointed out the flaw in the approach. >> >> So, answer the question. When does this miracle of emptiness occur? > > Given my interpretation/translation of the problem into Mathematics (see > above) and given that the "moment the vase becomes empty" means the > first time t >= -1 that V(t) is zero, then it follows that the "vase > becomes empty" at t = 0 (i.e., noon). > Yes, now, when nothing occurs at noon, and no balls are removed, what else causes the vase to become empty?
From: Tony Orlow on 25 Oct 2006 15:06 David Marcus wrote: > Tony Orlow wrote: >> David Marcus wrote: >>> cbrown(a)cbrownsystems.com wrote: >>>> stephen(a)nomail.com wrote: >>>>> With the added surreal twist that the limit of the number >>>>> of unnumbered balls in the vase as we approach noon is 0, >>>>> but the number of unnumbered balls in the vase at noon is >>>>> infinite. :) >>>> I think his response, when I pointed this out to him, was either "Oh, >>>> shut up!" or "Whatever." >>> That is consistent with my suggestion that Tony is reasoning by >>> imagining a vase filling up. If you visualize the vase filling up in >>> your mind, you don't see the unnumbered balls in the picture. >> If you have an infinite ocean wit 10 liter/sec flowing in, and 1 >> liter/sec flowing out (and no evaporation), will it ever empty? No. Same >> difference. > > So, you confirm that the vase situation and the ocean situation are the > same, i.e., should have the same answer? > Indeed.
From: Tony Orlow on 25 Oct 2006 15:08
David Marcus wrote: > Tony Orlow wrote: >> Virgil wrote: >> < endless reiterations of the following > >>> The only question is "According to the rules set up in the problem, is >>> each ball which is inserted into the vase before noon also removed from >>> the vase before noon?" >>> >>> An affirmative answer confirms that the vase is empty at noon. >>> A negative answer violates the conditions of the problem. >>> >>> Which answer does TO choose? >> God, are you a broken record, or what? Let's take this very slowly. Ready? >> >> Each ball inserted before noon is removed before noon, but at each time >> before noon when a ball is removed, 10 balls have been added, and 9/10 >> of the balls inserted remain. Therefore, at no time before noon is the >> vase empty. Agreed? >> >> Events including insertions and removals only occur at times t of the >> form t=-1/n, where n e N. Where noon means t=0, there is no t such that >> -1/n=0. Therefore, no insertions or removals can occur at noon. Agreed? >> >> Balls can only leave the vase by removal, each of which must occur at >> some t=-1/n. The vase can only become empty if balls leave. Therefore >> the vase cannot become empty at noon. Agreed? > > Not so fast. What do "become empty" or "become empty at" mean? > "Not so fast"???? We've been laboring this point endlessly. The vase goes from a state of balledness to a state of balllessness starting at time 0. Balls have to have been removed for this transition to occur. >> It is not empty, and it does not become empty, then it is still not >> empty. Agreed? >> >> When you bring t=0 into the experiment, if anything DOES occur at that >> moment, then the index n of any ball removed at that point must satisfy >> t=-1/n=0, which means that n must be infinite. So, if noon comes, you >> will have balls, but not finitely numbered balls. In this experiment, >> however, t=0 is excluded by the fact that n e N, so noon is implicitly >> impossible to begin with. >> >> Have a nice lunch. > |