An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Tony Orlow on 14 Oct 2006 22:09 Virgil wrote: > In article <453102b5(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> 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 > lim(t->0: balls(t)) does not exist, > but balls(0) does exist and balls(0)=0 > Functions can exist at points at which their limits do not. > There are even functions with domain R which are discontinuous at every > rational argument but continuous at every irrational one. That sounds vaguely interesting. Can you give an example?
From: Tony Orlow on 14 Oct 2006 22:23 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: > > <snip> > >>>>>>> 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? >> > > You just implied it; when you claimed that at t>=0, n>=oo; where I > presume that by "n", you refer to the statement in the problem: > > "At time t = -1/n, where n is a natural number, we add balls labelled > (10*(n-1)+1) through 10*n inclusive, and remove the ball labelled n". > > That statement obviously does not refer to removals or additions of > balls at time t = 0, becuase there is no natural number n such that > -1/n = 0. > > Do you agree with this conclusion? > Of course. That's what I was saying. Your statement concerning n does not cover noon, because noon=f(oo), and oo is outside your range. So, you really don't have any claim with regard to what happens at noon. Its beyond your purview. >>> 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. > > This contradicts what you said above regarding whether we can determine > if a particular ball is in the vase at some time t; you wrote: > >>>>>> For any finite time before noon, when iterations of the problem are >>>>>> temporally distinguishable, yes, but at noon, no. How does that contradict that? Any specific finitely indexed ball has a specific finite time before noon at which it is inserted and another at which it is removed. At any of those times, there are a growing number of balls as t approaches noon. The set-theoretic claim is that, even though nothing happens AT noon, nevertheless BY noon the vase is empty, even though BEFORE noon there are potentially infinitely many balls in the vase. Ahem. When does this occur, if at all? > > Do you now agree that we can conclude from the problem statement that, > for each ball labelled with a natural number, that ball is not in the > vase at noon? Given any finitely numbered ball, we can calculate its entry and exit times. However, we can also say that when it exits, there are more balls in the vase than when it entered. If you had any upper bound to your set of naturals, you'd see your logic makes no sense, but there is none. Omega is a phantom. > >> The balls at noon are not distinguishable nor specific. > > Since the only statement in the problem regarding putting balls in the > vase is a statement about putting balls which are labelled with a > specific and unique natural number in the vase, I don't see how you > justify this conclusion /purely in terms of the given problem/. Before noon, there are balls. At noon, there are not. What happened? How does 9n become 0 when n=oo? It doesn't. > > At any rate, for each natural number n, the number of balls in the vase > which are "indistinguishable and not specific" at time t = -1/n is 0. None of which times are noon. > > By your own logic, why can't we conclude that therefore the number of > balls in the vase which are "indistinguishable and not specific" at t=0 > is also 0 (i.e., lim n->oo 0 = 0)? > It's a point of condensation, where an infinite number of iterations happen in a "moment". It's deliberate obfuscation of the problem. S'not necessary. >>> 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. > > You're a bit feisty today. There are no dumb questions; only dumb > answers. > > Cheers - Chas > There are lots of dumb questions. They waste a lot of time. The uncountable countable thread, like Finnigan's Wake.
From: Virgil on 14 Oct 2006 22:30 In article <4531961f(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > stephen(a)nomail.com wrote: > > Tony Orlow <tony(a)lightlink.com> wrote: > >> stephen(a)nomail.com wrote: > >>> Tony Orlow <tony(a)lightlink.com> wrote: > >>> > >>>> Why even mention the gedanken at all then? > >>> I am not the one who brought it up. I am not even sure > >>> why people think it has anything to do with set theory. > > > >> It doesn't. It's a distinct SEQUENCE of events, not a set without order. > > > > Yes, it is a distinct sequence of events. Each ball is involved > > in exactly two events. Ball #n is added at time -(1/2)^(floor(n/10) > > Indeed, and each iteration of the sequence is involved with 10-1 balls. > Which logic seems to have more of a flaw, when they conflict? > > >> Set theory doesn't apply. It's just another example of set theorists > >> trying to claim that everything falls under set theory. This experimant > >> obviously does not. Set theory is incapable of handling the concept of > >> sequence in a well-defined way over such a set. On the other hand analysis allows representation of the problem in terms of a real function. Let let t = 0 at noon. Let A_n(t) = 0 while ball n is not in the vase and equal 1 when it is in the vase or in transition. Note that A_n(0) = 0 for all n. Let B(t) = sum_{n in N} A_n(t). Then at any time t, B(t) >= the number of balls in the vase at time t. But then B(0) = sum_{n in N} A_n(0) = sum_{n in N} 0 = 0. and similarly for each t > 0, > > > >> No, set theory confuses the issue with its concentration on omega. The above does not use any omega. > > The concept that everything that went in also came out is based, as WM > would put it, on the concept of a completed infinity. The condensation > point at "noon" is a Zeno machine, design for the express purpose of > producing a paradox. But not a contradiction. > > > > You have never presented any sort of refutation for that > > argument, and instead are just responding with lame > > insults. > > > > Stephen > > > > > > What are you on? I have presented more than one argument showing that > what you claim is false. TO has produced a bunch of arguments, but none showing that the number of balls at noon is not zero. > > If only one ball can be removed at any time, and the vase becomes empty, > then there has to have been a single ball removed which produced that > result. If there has to be a last ball out, then there has to be a last ball in, since each ball in is followed by that ball coming out. > > The problem is precisely characterized as an infinite series which > diverges. There is no discontinuity at noon, as expressed in the problem > itself. If the values at points approaching noon diverge then there is no way that there can NOT be a discontinuity. > That's a transfinitological obfuscation. That is a fact of functions. If a function is to be continuous at a point then its values at all points of all sequences in its domain converging to that point must also converge, and must converge to the value of the function at that point. Anything else produces at least a discontinuity. It pays to learn a little about mathematiccs before you try to denigrate it. As it is, TO is only demonstrating his ignorance.
From: Virgil on 14 Oct 2006 22:31 In article <4531966e(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> David Marcus wrote: > >>> Han de Bruijn wrote: > >>>> Virgil wrote: > >>>> > >>>>> In article <66a8$452f4298$82a1e228$30886(a)news2.tudelft.nl>, > >>>>> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > >>>>> > >>>>>> Randy Poe wrote about the Balls in a Vase problem: > >>>>>> > >>>>>>> Tony Orlow wrote: > >>>>>>>> Specifically, that for every ball removed, 10 are inserted. > >>>>>>> All of which are eventually removed. Every single one. > >>>>>> All of which are eventually inserted. Every single one. > >>>>> None are reinserted after being removed but,each is removed after > >>>>> having > >>>>> been inserted, so that leaves them all outside the vase at noon. > >>>> Huh! Then reverse the process: first remove 1, then insert 10. It must > >>>> be no problem in your "counter intuitive" mathematics to start with -1 > >>>> balls in that vase. > >>> Consider this situation: Start with an empty vase. Add a ball at time 5. > >>> Remove it at time 6. > >>> > >>> How you would translate that into Mathematics? > >>> > >> What happens between times 5 and 6? > > > > Nothing. > > > >> Are there other balls involved? > > > > No. > > > > Hi David! > > Well, thanks for that elucidating example. It's all blindingly obvious > now. :) And it is TO who is blind.
From: Tony Orlow on 14 Oct 2006 22:31
Virgil wrote: > In article <45310688(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >>> In article <452fbf0e(a)news2.lightlink.com>, >>> Tony Orlow <tony(a)lightlink.com> wrote: > >>> So let us leave them coupled but merely change the coupling so that the >>> nth ball is inserted, say , 1/2^n minutes before it is removed. Both the >>> insertions and the removals are still all completed before noon, and it >>> is obvious that the vase is empty at noon. >>> >>> >> >> Then you are inserting balls one at a time, and removing them as you >> insert the next. What does that have to do with the original problem? > > The only necessary constraint on insertions of balls into the vase and > removals of balls from the vase is that each ball that is to be removed > must be inserted before it can be removed, and, subject only to that > constraint, the set of balls remaining in the vase at the end of all > removals is independent of both the times of insertion and of the times > of removal. WRONG!! :) There is the additional constraint that ten other balls (or nine, for the first) must be inserted before it can be removed. > > To argue otherwise is to misrepresent the problem. > To argue that the adding of ten balls can be coupled with the removal of one and get an eventual result of zero is just plain silly. > >>> When infinitely many are inserted and all of them removed, what is >>> obvious to TO is false to logic. >> Your take on logic is very, shall we say, provincial. > > You may say what you like, however it remains correct. Define "correct". |