An uncountable countable set
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From: David Marcus on 24 Oct 2006 21:27 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. 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. 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. Problem 2. For n = 1,2,..., define A_n = -1/n, R_n = -1/(n+1). Then L = 1, S = 0, T = 1. Problem 3. For n = 1,2,..., define A_n = -1, R_n = -1/n. Then L = infinity, S = 0, T = 0. Tony, can you give us a general procedure to let us determine T given the A_n's and B_n's? -- David Marcus
From: David Marcus on 24 Oct 2006 21:30 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"? > 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. -- David Marcus
From: David Marcus on 24 Oct 2006 21:45 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? > 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). -- David Marcus
From: David Marcus on 24 Oct 2006 21:47 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? -- David Marcus
From: David Marcus on 24 Oct 2006 21:51
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? > 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. -- David Marcus |