An uncountable countable set
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From: Virgil on 25 Oct 2006 15:40 In article <453fb285(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Mike Kelly wrote: > > Randy Poe wrote: > >> > >> How is it, in your world, that when I specify times for all natural > >> numbered > >> balls, I am required to put in balls that don't have natural numbers? > > > > The problem is that Tony thinks time is a function of the number of > > insertions you've gone through. In order to "get to" any particular > > time you have to perform the insertions "up to" that point. He then > > thinks that if you want to "get to" noon, you have to have performed > > some "infinite" (whatever that means) iterations, where balls without > > natural numbers are inserted. That this is obviously not what the > > problem statement says doesn't seem to bother him. Nor that it's > > absolutely nothing like an intuitive picture of what time is. > > Time is ultimately irrelevant in this gedanken, but if it is to be > considered, the constraints regarding time cannot be ignored. Events > occurring in time must occupy at least one moment. How is time irrelevant when every action is specified by the time at which it is to occur? The only relevant question is "According to the rules set up in the problem, is each ball inserted at a time before noon also removed at a time before noon?" An affirmative answer confirms that the vase is empty at noon. A negative answer directly violates the conditions of the problem. How does TO answer this question? As usual, he avoids such relevant questions in his dogged pursuit of the irrelevant. > > > > > Obviously, time is an independent variable in this experiment and the > > insertion or removal or location of balls is a function of time. That's > > what the problem statement says: we have this thing called "time" which > > is a real number and it "goes from" before noon to after noon and, at > > certain specified times, things happen. There are only > > naturally-numbered balls inserted and removed, always before noon. > > Every ball is removed before noon. Therefore, the vase is empty. > > No, you have the concept of the independent variable bent. The number of > balls is related to the time by a formula which works in both directions. As time is a continuum and the numbers of balls in the vase is not, there is no way of inverting the realtionship in the way that TO claims. > > So, when does the vase become empty? Nothing can occur at noon, as far > as ball removals. AT every time before noon, balls are in the vase. So, > when does the vase become empty, and how? The vase is empty when every ball has been removed, and that occurs at noon. > > > > > If you follow the sequence of insertions and removals you never "get > > to" noon but this doesn't imply that noon is never reached, or that > > iterations involving non-naturally numbered balls occur. It just > > implies that all insertion and removal is performed before noon. > > > > Tony won't let himself understand this. He is delusional. His problem. > > > > I won't let myself accept self-contradictory conclusions. At least not unless they are TO's own personal self-contradictory conclusions. Like the existence of balls in a vase from which all balls have been removed. There is no > moment at which this can occur. The problem is perfectly modeled by a > divergent infinite series. No last ball can be removed without there > having been an negative number of balls previously. You solution fails. TO's assumption that there must be a last ball removed in order for all balls to have been removed is part and parcel of his persistent delusion that there must be a last (finite) natural number in order to have a set of all (finite) natural numbers.
From: Virgil on 25 Oct 2006 15:43 In article <453fb2e8(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> David Marcus wrote: > >>> Tony Orlow wrote: > >>>> David Marcus wrote: > >>>>> Tony Orlow wrote: > >>>>>> stephen(a)nomail.com wrote: > >>>>>>> Also, supposing for the sake of argument that there are "infinitely > >>>>>>> number balls", if a ball is added at time -1/(2^floor(n/10)), and > >>>>>>> removed > >>>>>>> at time -1/(2^n)), then the balls added at time t=0, are those > >>>>>>> where -1/(2^floor(n/10)) = 0. But if -1/(2^floor(n/10)) = 0 > >>>>>>> then -1/(2^n) = 0 (making some reasonable assumptions about how > >>>>>>> arithmetic > >>>>>>> on these infinite numbers works), so those balls are also removed at > >>>>>>> noon and > >>>>>>> never spend any time in the vase. > >>>>>> Yes, the insertion/removal schedule instantly becomes infinitely fast > >>>>>> in > >>>>>> a truly uncountable way. The only way to get a handle on it is to > >>>>>> explicitly state the level of infinity the iterations are allowed to > >>>>>> achieve at noon. When the iterations are restricted to finite values, > >>>>>> noon is never reached, but approached as a limit. > >>>>> Suppose we only do an insertion or removal at t = 1/n for n a natural > >>>>> number. What do you mean by "noon is never reached"? > >>>> 1/n>0 > >>> Sorry, I meant t = -1/n. So, I assume your answer is that -1/n < 0. > >>> > >>> But, I don't follow. Translating "-1/n < 0" back into words, I get "all > >>> insertions and removals are before noon". However, I asked you what > >>> "noon is never reached" means. Are you saying that "noon is never > >>> reached" means that "all insertions and removals are before noon"? > >> Yes, David. What else happens in this experiment besides insertions and > >> removals of naturals at finite times before noon? If the infinite > >> sequence of events is actually allowed to continue until t=0, then you > >> are talking about events not indexed with natural numbers, so you're not > >> talking about the same experiment. If noon is not allowed, and all times > >> in the experiment are finitely before noon, well, at none of those times > >> does the vase empty, as we all agree. This is why I am asking when this > >> occurs. It can't, given the constraints of the problem. > > > > Does your "Yes" at the beginning of your reply mean that you agree that > > "noon is never reached" means that "all insertions and removals are > > before noon". By "mean", I mean that that is what the words mean, not > > that the two statements are equivalent or deducible from each other. > > > > Yes, every event, every insertion or removal, happens at a specific time > before noon. At each of those times, the vase is non-empty. Nothing else > occurs, as far as insertions of removals. Is that clear enough? So, when > does the vase become empty, and how? The vase becomes empty in the usual way, by having everything in it removed. And the time at which that finally has occurred is noon. The only relevant question is "According to the rules set up by the problem, is each ball inserted before noon also removed before noon?" An affirmative answer confirms that the vase is empty at noon. A negative answer directly violates the conditions of the problem. How does TO answer?
From: Virgil on 25 Oct 2006 15:47 In article <453fb473(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David Marcus wrote: > > 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? When B_n(0) = 0 for all n, the sum is also 0. > > > > 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. That was given by the time hacks of the original problem statement. The only relevant question is "According to the rules set up in the problem, is each ball inserted before noon also removed before noon?" An affirmative answer confirms that the vase is empty at noon. A negative answer directly violates the conditions of the problem. How does TO answer?
From: Virgil on 25 Oct 2006 15:51 In article <453fb506$1(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > 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. As that is TO's thesis, he is the only one required to defend it. For the most of us who can see how silly it is to to claim that any balls remain after all of them have been removed, we are not required to argue in support of that silliness.
From: Virgil on 25 Oct 2006 16:07
In article <453fb590(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > t=0 is precluded by n e N and t(n) = -1/n. Not at all. TO would allow the vagaries of his intuition to stop clocks. But it doesn't happen, except in TO's dream worlds. > >> 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? The fact that each ball has already been removed before noon. At least if one follows the problem as given. |