An uncountable countable set
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From: Tony Orlow on 13 Oct 2006 12:57 Han de Bruijn wrote: > Virgil wrote: > >> http://en.wikipedia.org/wiki/ZFC >> Axiom of infinity: There exists a set x such that the empty set is a >> member of x and whenever y is in x, so is S(y). So, we can interpret the empty set as 0, the origin, and then define successor any way we want. IF we define the successor of n as n+1, then we get the naturals. If we define the successor as 1-1/2(1-n), then we get our Zeno moments. The inductive set produced depends on what the null set represents and how successor is defined. > > Which is actually the construction of the ordinals. Right? > > Suppose we write 0 = { } > then 1 = { { } } = { 0 } > and 2 = { { } , { { } } } = { 0, 1 } > and 3 = { { }, { { } }, { { }, { { } } } } = { 0, 1, 2 } > > http://www.jboden.demon.co.uk/SetTheory/ordinals.html > > So the axiom of infinity says that you can get everything from nothing. > This is contradictory to all laws of physics, where it is said that you > pay a price for everything. E.g. mass and energy are conserved. > > Han de Bruijn > Han, you can't really be looking for conservation of energy or momentum or mass in abstract mathematics, can you? This axiom basically defines the infinite linear inductive set. Given this method of generation, there should be things we can say about the set, no? Tony
From: Tony Orlow on 13 Oct 2006 13:07 Randy Poe wrote: > Tony Orlow wrote: >> Randy Poe wrote: >>> Tony Orlow wrote: >>>> Randy Poe wrote: >>>>> Tony Orlow wrote: >>>>>> David Marcus wrote: >>>>>>> Virgil wrote: >>>>>>>> In article <452d11ca(a)news2.lightlink.com>, >>>>>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>>>>> >>>>>>>>>> I'm sorry, but I can't separate your statement of the problem from your >>>>>>>>>> conclusions. Please give just the statement. >>>>>>>>> The sequence of events consists of adding 10 and removing 1, an infinite >>>>>>>>> number of times. In other words, it's an infinite series of (+10-1). >>>>>>>> That deliberately and specifically omits the requirement of identifying >>>>>>>> and tracking each ball individually as required in the originally stated >>>>>>>> problem, in which each ball is uniquely identified and tracked. >>>>>>> It would seem best to include the ball ID numbers in the model. >>>>>>> >>>>>> Changing the label on a ball does not make it any less of a ball, and >>>>>> won't make it disappear. If I put 8 balls in an empty vase, and remove >>>>>> 4, you know there are 4 remaining, and it would be insane to claim that >>>>>> you could not solve that problem without knowing the names of the balls >>>>>> individually. >>>>> That's a red herring. It's not the name of the ball that's relevant, >>>>> but whether for any particular ball it is or isn't removed. >>>> The "name" is the identity. It doesn't matter which ball you remove, >>>> only how many at a time. >>>> >>>>>> Likewise, adding labels to the balls in this infinite case >>>>>> does not add any information as far as the quantity of balls. >>>>> No, but what the labels do is let us talk about a particular >>>>> ball, to answer the question "is this ball removed"? >>>> We care about the size of the collection. If replacing the elements with >>>> other elements changes the size of the set, then you are doing more than >>>> exchanging elements. >>>> >>>>> If there is a ball which is not removed, whatever label >>>>> is applied to it, then it is still in the vase. >>>> How convenient that you don't have labels for the balls that transpire >>>> arbitrarily close to noon. You don't have the labels necessary to >>>> complete this experiment. >>>> >>>>> If there is a ball which is removed, whatever label is >>>>> applied to it, then it is not in the vase. >>>> If a ball, any ball, is removed, then there is one fewer balls in the vase. >>>> >>>>>> That is >>>>>> entirely covered by the sequence of insertions and removals, quantitatively. >>>>> Specifically, that for each particular ball (whatever you >>>>> want to label it), there is a time when it comes out. >>>>> >>>> Specifically, that for every ball removed, 10 are inserted. >>> All of which are eventually removed. Every single one. >>> >> Every single one, > > Yes. > >> each after another ten are inserted, of course. > > And I can tell you the time that each of those is removed. > >> Come on! > > Come on yourself. You *know* there is a removal time > associated with every ball. > > - Randy > I know that at no time have all the balls previous inserted been removed, but only 1/9th of them, since 1 is removed for every 10 inserted. What is the flaw in that logic?
From: Tony Orlow on 13 Oct 2006 13:10 David Marcus wrote: > Virgil wrote: >> In article <452ef6e8(a)news2.lightlink.com>, >> Tony Orlow <tony(a)lightlink.com> wrote: >> >>> Virgil wrote: >>>> In article <452e8f61(a)news2.lightlink.com>, >>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>> >>>>> Virgil wrote: >>>>>> In article <452e5862$1(a)news2.lightlink.com>, >>>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>>> >>>>>>> David Marcus wrote: >>>>>>>> Virgil wrote: >>>>>>>>> In article <452d11ca(a)news2.lightlink.com>, >>>>>>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>>>>>> >>>>>>>>>>> I'm sorry, but I can't separate your statement of the problem from >>>>>>>>>>> your >>>>>>>>>>> conclusions. Please give just the statement. >>>>>>>>>> The sequence of events consists of adding 10 and removing 1, an >>>>>>>>>> infinite >>>>>>>>>> number of times. In other words, it's an infinite series of (+10-1). >>>>>>>>> That deliberately and specifically omits the requirement of >>>>>>>>> identifying >>>>>>>>> and tracking each ball individually as required in the originally >>>>>>>>> stated >>>>>>>>> problem, in which each ball is uniquely identified and tracked. >>>>>>>> It would seem best to include the ball ID numbers in the model. >>>>>>>> >>>>>>> Changing the label on a ball does not make it any less of a ball, and >>>>>>> won't make it disappear. If I put 8 balls in an empty vase, and remove >>>>>>> 4, you know there are 4 remaining, and it would be insane to claim that >>>>>>> you could not solve that problem without knowing the names of the balls >>>>>>> individually. Likewise, adding labels to the balls in this infinite case >>>>>>> does not add any information as far as the quantity of balls. That is >>>>>>> entirely covered by the sequence of insertions and removals, >>>>>>> quantitatively. >>>>>> >>>>>> If, as in the original problem, each ball is distinguishable from any >>>>>> other ball, TO should be able to tell us which ball or balls, if any, >>>>>> remain in the vase at noon. >>>>>> >>>>>> Suppose, for example, the balls are all of different sizes, with each in >>>>>> sequence being only 9/10 as large as its predecessors. Then each >>>>>> iteration consists of putting that largest 10 balls that have not yet >>>>>> been in the vase into it and then taking the largest ball in the vase >>>>>> out. >>>>>> >>>>>> Since the balls are well ordered by decreasing size, any non-empty set >>>>>> of them must have a largest ball in it. So what is the largest ball in >>>>>> the vase at noon, TO? >>>>> It's obviously going to be infinitesimal. >>>> But since none of the actual balls are infinitesimal (for every n in N, >>>> (9/10)^(n-1) is finitesimal, not infinitesimal) that means there are no >>>> balls in the vase at noon. >>> What is the smallest ball inserted? >> That is a "Have you stopped beating your wife?" question, which assumes >> a condition contrary to fact. >> >> As there is no last ball inserted, there is no smallest ball inserted, >> nor any last ball removed, nevertheless, all balls are inserted, and all >> balls are removed, and all before noon. > > Virgil, > > Do you think it helps for you to respond to every post that Tony makes > and in each response to respond to multiple separated paragraphs of > Tony's post? Wouldn't it be better to just have a small number of > threads with Tony and in each thread just make one comment in each of > your posts? > I'm the cat to Virgil's dog. Yappedy yap yap. Meow, Tony
From: David Marcus on 13 Oct 2006 13:43 Tony Orlow wrote: > David Marcus wrote: > > How about this problem: Start with an empty vase. Add a ball to a vase > > at time 5. Remove it at time 6. How many balls are in the vase at time > > 10? > > > > Is this a nonsensical question? > > Not if that's all that happens. However, that doesn't relate to the ruse > in the vase problem under discussion. So, what's your point? Is this a reasonable translation into Mathematics of the above problem? "Let 1 signify that the ball is in the vase. Let 0 signify that it is not. Let A(t) signify the location of the ball at time t. The number of 'balls in the vase' at time t is A(t). Let A(t) = 1 if 5 < t < 6; 0 otherwise. What is A(10)?" -- David Marcus
From: imaginatorium on 13 Oct 2006 13:44
Tony Orlow wrote: > imaginatorium(a)despammed.com wrote: > > Tony Orlow wrote: > >> David Marcus wrote: > > > > <snip-snop: the valiant shall see> > > > >> Time is actually irrelevant. > > > > If you are trying to determine the limit of the sequence of operations, > > time does appear to be irrelevant, yes. > > Individual operations are indistinguishable at noon. You must take the > limit as the number of iterations approach oo. Then what do you get? Why > do you have a conflict between looking at it in terms of iterations vs. > time? Because of the clever little Zeno machine. Nice obfuscation. I'm not sure what you mean by "obfuscation" - just because you are confused doesn't mean anyone is trying to confuse you. Consider my blue sliver - if you think of sliding along it, almost vertically, closer and closer yet never quite touching the y-axis, this is a journey that never ends. But sliding along the x-axis under the sliver will surely reach (0, 0), as long as you don't put the brakes on. There is no obfuscation here - though like most of maths there is certainly something you have to think carefully about. Anyway,... > >> ... The sequence is measured in iterations as > >> n->oo, and the number of balls in the vase at iteration n is represented > >> by sum(x=1->n: 9). The limit of this sum as x diverges also diverges in > >> linear fashion. > > > > Certainly does. I mean that sum from x=1 as x increases 2, 3, 4, ... > > without limit of (10-1) diverges. > > Right, and that characterizes the salient features of the gedanken. > > > > > Let me ask you another question, Tony, as I don't think you answered > > the last one. > > I don't see any previous question at this point, but I'm relatively sure > I answered what was asked. > > Here is an argument, ending with a conclusion I don't > > personally swallow. Can you tell me at what point it goes wrong? > > (Or do you think it is valid?) You don't think it's valid - good. > > Consider the function step0: R -> R mapping x to 0 if x<0 and mapping x > > to 1 if x>=0. > > A discontinuous function at x=0. Right. Well, slightly more precise to say "a function with a discontinuity at x=0" I think. > > > > FWIW, we can write this function in a C-like way (taking 'TRUE' and > > 'FALSE' to have the numeric values 1 and 0 respectively), so it is just > > a simple expression: > > > > step0(x) = (x>=0) > > > > OK, for n a positive integer, now consider the sequence of values of > > step0(p) for p=-1, -1/2, -1/3, ... -1/n, ... without end > > > > For any n, -1/n < 0, therefore step0(-1/n) = 0. > > > > So the sequence of values is simply the constant sequence 0, 0, 0, 0, > > .... without end > > > > The limit of a constant sequence of values is the single value itself. > > > > Therefore lim(n->oo) step0(-1/n) = 0 > > > > By the Orlovian limit-swapping axiom, therefore: > > > > step0(lim(n->oo) -1/n) = 0 > > > > But lim (n->oo) -1/n = 0. > > > > Thus step0(0) = 0. > > > > But by definition, step0(0) = 1 > > > > Therefore 0 = 1. > > A function with such a declared discontinuity has two limits at that > point, depending on the direction of approach. So, what else is new? Ah. Is a "declared discontinuity" somehow significantly different from a simple discontinuity? I mean, is there such a thing as an "undeclared discontinuity" to which different rules apply? (I've no idea: this is not normal mathematical terminology you see.) > That proves nothing. It illustrates that for a function f(), the value of f(0) is not necessarily equal to lim(x->0) f(x). Which is of rather crucial importance in the current problem. > What causes a discontinuity at noon? I'll tell you. The von Neumann > limit ordinals. That's schlock. Um, that's foaming at the mouth. Von Neumann limit ordinals have nothing to do with it - the very simplest notion of the natural numbers being an unending sequence - Wolf Kirchmeir's six? eight?-year old grandson's understanding - is absolutely all that is needed. You can't grasp the notion of an unending sequence, which is why you are in such a total tangle. According to your "view" then, there is no discontinuity at noon - is that right? The number of balls identified by natural numbers increases without limit, and despite the fact that there is no ball not removed before noon, at ten past an unlimited number of them are somehow still lurking in the vase? Look, I know my "sliver" corresponds to a slightly different sequence, but it's simpler. Consider the sliver between y=-2/x and y=-1/x, for x<0. Consider it "hatched" with horizontal lines on integral values of y. Think of every one of the horizontal bars as representing a time some ball spends in a vase. You seem to agree that the sliver goes ever upward, ever closer but not actually reaching the y-axis. If we were to travel upward, we would see each line corresponding to a ball's stay in the vase - always in then out halfway towards the y-axis; and importantly, this viewing journey would never end. But if we were to travel along the x-axis towards the origin, looking upwards (this is maths, not physics; we pretend we could view the sliver however far away), we would notice that the number of balls was increasing without limit. Then we would reach the origin. Looking up we would see the sliver to the left of the y-axis. You agreed at one point that the sliver is entirely in the neg-x/pos-y quadrant, so obviously there is no sliver to the right of the y-axis. But in your view (do I understand?) somehow there would just be some more sliver that had crept around the "top"? Or what? Do enlighten us... Meanwhile, it would greatly help if you also considered some different slivers. How about the one between y=-2/(x+1) and y=-1/x ... (sliver-2: height diverges) This is exactly the same as the first, except that the upper hyperbola lobe has been shifted left by one. This sliver does not become indefinitely narrow - the width as we go up tends to exactly 1. This is the case where every ball is inserted one minute (are we in minutes? makes no difference...) earlier, so by 11:59 all the balls are in the vase. In this case how much of the sliver leaks into the x>0 quadrant? Or: y=-1/x + 1 and y = -1/x ... (sliver-3: height constant = 1) Or: y = -1/x - x and y = -1/x ... (sliver-4: height tends to 0) For sliver-3 ther |