An uncountable countable set
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From: Virgil on 13 Oct 2006 22:28 In article <45304682(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <452fc7c2(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> 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? > > > > At no time BEFORE NOON have all the balls been removed. > > Since in the original statement of the problem, one was given the > > precise time before noon for removal of every ball that was to be > > inserted, by noon it does not matter a whit how long the ball was the > > vase. > > Hi Virgil. > Wow! It's really nice to see you! I can't believe you actually stooped > so low as to respond to one of my little comments. Thank you, Sir. I > really appreciate what little attention I get. :) > > So, what were you saying? > > Oh yeah, that all balls were in the vase before noon. But that's not true. I did not say precisely that "all the balls were in the vase before noon", as that would imply falsely that might all have been in the vase at one time. It is TO's inability to make these small but critical distinctions that keeps leading him down the primrose path to personal perdition.
From: Virgil on 13 Oct 2006 22:33 In article <45304714(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David Marcus wrote: > > 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? > > > I gave you the translation, to the last iteration of which you did not > respond. > > > > "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)?" > > > > Think in terms on n, rather than t, and you'll slap yourself awake. David makes more sense asleep that TO does awake. Since the transfer of balls into and out of the vase is marked to occur at specific times, it makes sense to consider the status of the vase, e.g., the number of balls in it, to be a function of time.
From: Tony Orlow on 14 Oct 2006 10:50 imaginatorium(a)despammed.com wrote: > 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.) You have defined your step function with an explicit discontinuity. There is no explicit discontinuity in the gedanken under discussion. The discontinuity is introduced with the application of omega. > >> 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. Where is there defined in the problem any mention of a discontinuity in the process? There isn't. > >> 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. I am in no tangle. That you cannot see that your conclusion is based on the notion of omega as the first limit ordinal and first discontinuity in the ordinals is disappointing. If the sequence never ends, then noon never comes. > > 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? The process at noon is not well defined, since the distinction between iterations disappears. How do you know there are countably many iterations, and not some uncountably number? You don't. You base your argument on all iterations being finite, but there is no least upper bound to the finites, because there is no least infinite. > > 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. So, where is "noon" in your graph? > > 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 t
From: Tony Orlow on 14 Oct 2006 11:30 Randy Poe wrote: > Tony Orlow wrote: >> 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. >>> >> I know that at no time > > Crucial phrase missing here: "at no time BEFORE noon" > >> have all the balls previous inserted been >> removed, but only 1/9th of them, since 1 is removed for every 10 >> inserted. > > You have correctly described the situation at every one > of the infinite values of 1 < t < 0. > >> What is the flaw in that logic? > > That you somehow think f(x), x<0 forces a value of f(0). > > - Randy > lim(t->0: balls(t))<>0
From: Tony Orlow on 14 Oct 2006 11:33
cbrown(a)cbrownsystems.com wrote: > Tony Orlow wrote: >> cbrown(a)cbrownsystems.com wrote: >>> Tony Orlow wrote: >>>> cbrown(a)cbrownsystems.com wrote: >>>>> Tony Orlow wrote: >>>>>> cbrown(a)cbrownsystems.com wrote: >>>>>>> Tony Orlow 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. >>>>>>>> The original statement contrasted two situations which both matched this >>>>>>>> scenario. The difference between them was the label on the ball removed >>>>>>>> at each iteration, and yet, that's not relevant to how many balls are in >>>>>>>> the vase at, or before, noon. >>>>>>> Do you think that the numbering of the balls is not relevant to >>>>>>> determining the answer to the question "Is there a ball labelled 15 in >>>>>>> the vase at 1/20 second before midnight?" >>>>>>> >>>>>>> Cheers - Chas >>>>>>> >>>>>> If it's a question specifically about the labels, as that is, then it's >>>>>> relevant. It's not relevant to the number of balls in the vase at any >>>>>> time, as long as the sequence of inserting 10 and removing 1 is the same. >>>>>> >>>>> Putting aside the question of /how/ (limit? sum of binary functions?) >>>>> one determines the /number/ of balls in the vase at time t for a >>>>> moment... >>>>> >>>>> Do you then agree that there is some explicit relationship described in >>>>> the problem between what time it is, and whether any particular >>>>> labelled ball, for example the ball labelled 15, is in the vase at that >>>>> time? >>>> For any finite time before noon, when iterations of the problem are >>>> temporally distinguishable, yes, but at noon, no. >>>> >>> I don't understand why you think this would be the case. >>> >>> Why do you think the relationship holds for t < 0? >>> >>> Why you do think it does not hold for t >= 0? >>> >>> Cheers - Chas >>> >> Because for t>=0, n>=oo. > > Actually, for t>=0, there is /no/ natural number n such that t = -1/n. > Similarly, for t = -1/pi, there is no natural number n such that t = > -1/n. Yeah, no idding. Who said oo was a natural number? > > But what do either of those statements have to do with whether or not > ball 15 is in the vase at t=0? Nothing to do with ball 15. That has a specific time of removal. Every specific ball does. The balls at noon are not distinguishable nor specific. > > Do you believe that we cannot state whether ball 15 is in the vase at > 1/pi seconds before midnight, because there is no step associated with > 1/pi? > > Cheers - Chas > That's a dumb question, made to make me look dumb. It backfired. |