An uncountable countable set
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From: Tony Orlow on 26 Oct 2006 11:25 Virgil wrote: > In article <453fb693(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> David Marcus wrote: >>> Virgil wrote: >>>> In article <453e824b(a)news2.lightlink.com>, >>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>> Virgil wrote: >>>>>> In article <453e4a85(a)news2.lightlink.com>, >>>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>>>> If the vase exists at noon, then it has an uncountable number of balls >>>>>>> labeled with infinite values. But, no infinite values are allowed i the >>>>>>> experiment, so this cannot happen, and noon is excluded. >>>>>> So did the North Koreans nuke the vase before noon? >>>>>> >>>>>> The only relevant issue is whether according to the rules set up in the >>>>>> problem, is each ball inserted before noon also removed before noon?" >>>>>> >>>>>> An affirmative confirms that the vase is empty at noon. >>>>>> A negative directly violates the conditions of the problem. >>>>>> >>>>>> How does TO answer? >>>>> You can repeat the same inane nonsense 25 more times, if you want. I >>>>> already answered the question. It's not my problem that you can't >>>>> understand it. >>>> It is a good deal less inane and less nonsensical than trying to >>>> maintain, as TO and his ilk do, that a vase from which every ball has >>>> been removed before noon contains any balls at noon that have not been >>>> removed. >>> Ah, you are forgetting the balls labeled with "infinite values". Those >>> balls haven't been removed before noon. Although, I must say I'm not too >>> clear on when they were added. >>> >> At noon > > Where in the original problem does it say anything like that? It doesn't. It specifically excludes noon as a time in the experiment by specifying that all balls are finitely numbered and all events are finitely before noon. Duh.
From: Tony Orlow on 26 Oct 2006 11:29 MoeBlee wrote: > Tony Orlow wrote: >>> But none of Robinson's non-standard numbers are cardinalities. >> No kidding. They actually make sense. > > You said you have not properly studied chapter II in the book - the one > that includes mathematical logic, model theory, and set theory (does it > not? I'll stand corrected if it doesn't). What are you going to say > when you find out that what you say makes sense rests on a foundation > of set theory that you say doesn't make sense? Or, if I'm incorrect > that Robinson's work in non-standard analysis doesn't presuppose basic > mathematical logic, model theory, and set theory, then I'll benefit by > being corrected in my admittedly cursory understanding of the matter. > > MoeBlee > Uh, if Robinson's thesis is built upon transfinite set theory, then that is evidence right there that it's inconsistent, since you have a smallest infinity, omega, but Robinson has no smallest infinity. Robinson doesn't use ordinals or cardinals that I've seen. He basically defines what a well-formed formula is in his system, which is a little more restrictive that some others, it seems, and uses the language to extend what can be said about finite n in N to include infinite n in *N. TOEKnee
From: Tony Orlow on 26 Oct 2006 11:41 David Marcus wrote: > Tony Orlow 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, > > At noon. When nothing happens? What balls are removed at noon? Can the vase become empty by a means other than ball removal? > >> and how? > > By removing all balls before noon, but leaving some balls in the vase at > every time before noon and after one minute before noon. > Uhhhhh......does that actually make sense to you? Can it? There are balls at every moment before noon, nothing happens at noon, but they are all gone suddenly? What causes the vase to be empty at noon, if not removals at noon, which cannot occur?
From: Tony Orlow on 26 Oct 2006 11:47 David Marcus wrote: > Tony Orlow 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? > > The sum is over all positive integers. There is no B_oo. I'm sticking to > standard Calculus notation. Does that change your answers below? Not really, but it's hard to tell with that notation whether you are including noon or not. > >>> 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. > > By "noon exists" do you mean there is a ball B_oo? There isn't. No kidding. That's why noon cannot be part of the experiment. > >>> 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 is the limit of V(t) as t approaches zero from the left. So, L = oo. > > I don't know what you mean by "lim(x->oo: oo-x)". We don't normally > define things like oo-x, for x an integer, in Calculus. Of course, the > most natural definition would be for oo-x to equal oo, in which case > your limit would also be oo. But, you say it is zero. If oo=oo then oo-oo=0. > >> 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. > > I'm afraid I don't know what I'm supposed to keep track of. In truth, I > thought that by calculating V, I was keeping track. But, neither of the > two quantities I can get from V, i.e., L and S, seem to consistently > match your value. > You are supposed to keep in mind the coupling of 10 additions with each subtraction, and note that this sum of balls cannot converge to 0 no matter how long you keep it up.
From: Tony Orlow on 26 Oct 2006 11:58
David Marcus wrote: > Tony Orlow wrote: >> David Marcus wrote: >>> Tony Orlow wrote: >>>> 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. > > Sorry, I don't follow. Were you answering my question? I gave you a > choice: > > 1. Disagree with the interpretation/translation > 2. Agree with the interpretation/translation, but disagree that V(0) = 0 > > Are you picking #1 or #2? I'll choose #2 on the grounds that 0 does not exist in the experiment and that V(0) is therefore without meaning. > >>> 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? > > No balls are added or removed at noon, but the vase becomes empty at > noon. Through some other mechanism than ball removal? > > If you consider the vase becoming empty to be "something" rather than > "nothing", then it is not true that nothing occurs at noon. If by > "nothing occurs at noon", you mean no balls are added or removed, then > it is true that nohting occurs at noon. And, if no balls are moved at noon, what causes the vase to become empty at noon? Evaporation? A black hole? > > The cause of the vase becoming empty at noon is that all balls are > removed before noon, but at all times between one minute before noon and > noon, there are balls in the vase. The fact that there are balls at all times before noon and that no balls are removed at noon imply that there are balls in the vase at noon, if it exists in the experiment at all to begin with. > > Let me ask you the same question regarding the following problem. > > Problem: For n = 1,2,..., let > > 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 B_n(t). What is V(0)? Answer: V(0) = 0. > > Considering that for all n we have A_n <> 0 and B_n <> 0 and that V(t) > is approaching infinity as t approaches zero from the left, what causes > V(0) to be zero? > The fact that you have no upper bound to the naturals. This is the same technique, essentially, which equates the naturals with, say, the evens, or squares of naturals, even though those are proper subsets of the naturals. You can draw a 1-1 correspondence between the balls in and out, sure. There's a bijection there. Infinite bijections do not given any notion of measure unless they are parameterized. Here, you can look at number of balls in the vase as a function of n or of t. In either case, the sum diverges. It is only in trying to consider the unbounded set as completed that you come to this silly conclusion. |