An uncountable countable set
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From: stephen on 16 Oct 2006 21:32 MoeBlee <jazzmobe(a)hotmail.com> wrote: > stephen(a)nomail.com wrote: >> Tony has said that he does not like the axiom of infinity and >> the axiom of choice. Unlike many people who object to the >> axiom of infinity, Tony believes in infinite sets. > Okay, so that leads to the second part of my message. If he rejects the > axiom of infinity (does he actually?), then how does he get the > existence of infinite sets? Of course, this is meaningless for me to > ask, since he has no theory - just jumbles of undefined and circularly > defined verbiage. > MoeBlee Here is the message: http://groups.google.com/group/sci.math/msg/8a51619c2ae7f28f Tony thinks the axiom of infinity is "insufficient". So is the axiom of choice used in the argument that the vase is empty at noon? Stephen
From: Tony Orlow on 16 Oct 2006 21:41 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: >>>>>>>> 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. > > Was it? When you said: > >>>>>> Because for t>=0, n>=oo. > > you seemed to be saying that we could use the rule: > >>> "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". > > and the fact that "for t>=0, n>=oo" to prove that we cannot conclude > that ball 15 is not in the vase at t=0. Since "n>=oo" is never true, I > don't see how your logic applies. > > In fact, from the problem statement, it follows logically that ball 15 > is not in the vase at t=0. > > Now, I will grant this (alone) does not imply that you cannot /also/ > prove that ball 15 /is/ in the vase at t=0. > > If you can /also/ prove that, then you have proven that the problem is > contradictory - i.e., it contains assumptions that allow us to prove > that something is both true and false. > > But even if you /did/ have such a proof, that would not change the fact > that we can /also/ prove that ball 15 is not in the vase at t=0. > > Do you accept the above statements, or do you still claim that there is > /no/ valid proof that ball 15 is not in the vase at t=0? > 15 is a specific finite number for which we can state its times of entry and exit. At its time of exit, balls 16 through 150 reside in the vase. For every finite n in N, upon its removal, 9n balls remain. For every n e N, there is a finite nonzero number of balls in the vase. Every iteration in the sequence is indexed with an n in N. Therefore, nowhere in the sequence is there anything other than a finite nonzero number of balls in the vase. Now, where, specifically, in the fallacy in that argument? >> Your statement concerning n does >> not cover noon, because noon=f(oo), and oo is outside your range. > > You've lost me. Nothing happens at noon, if all sequential iterations are finite, given the time sequence. At all moments before noon, as has been conceded, there are a nonzero number of balls in the vase. > > What is f? What does it mean to say "noon = f(oo)"? How does this > disprove the assertion that ball 15 is not in the vase at t=0? > It means that every finite iteration occurs before noon, so the only ones that can happen AT noon are infinite. You have no infinite iterations, so noon does not occur, at at every moment BEFORE noon there is a nonzero number of balls in the vase. >> So, >> you really don't have any claim with regard to what happens at noon. Its >> beyond your purview. > > On the contrary, in a mathematical sense, a thing "happens" (i.e., can > be concluded from the problem statement) if, and /only/ if, it can be > logically deduced from assertions in the problem. Deduction depends on assumptions. Set theory's are phony in this case. At every moment before noon there are balls, and nothing happens at noon. Therefore, there are balls. > > Suppose we modify the original problem by appending "and, at each time > t = -1/n, where n is a natural number, we do not add a solid gold > statuette of Richard Nixon kissing Henry Kissinger to the vase". Then you would have to get your office decorations elsewhere. > > Do you conclude that therefore "it is beyond our purview" to state > "there is not a solid gold statuette of Richard Nixon kissing Henry > Kissinger in the vase at noon"? I'm sure there is somewhere in the back of your office. > > Assuming you see the absurdity of the above, why do you then claim > "because there is not natural number n such that -1/n >= 0, therefore > it is our beyond our purview to claim that ball 15 is not in the vase > at t>=0"? > I stated my argument. Refute it, stating specifically the logical error. >>>>> 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? > > Well, you claimed above that your statements have /nothing to do/ with > whether ball 15 is in the vase at t=0. You previously claimed that your > statements /do/ have something to do with whether the ball 15 is in the > vase
From: Tony Orlow on 16 Oct 2006 21:54 Virgil wrote: > In article <4533d0ff(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >>> In article <45319846(a)news2.lightlink.com>, >>> Tony Orlow <tony(a)lightlink.com> wrote: >>> >>>>> 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? >>> It is standard fare for anyone who knows any analysis. >>> >>> Let f: R --> R, be such that for each irrational x f(x) = 0, >>> and for each rational x whose expression in lowest terms is p/q, >>> let f(x) = 1/q. Then that function is provably continuous at each >>> irrational and provably discontinuous at each rational except 0. It is >>> provably the case that for each real a, lim_{x --> a} f(x) = 0, which >>> establishes the claim. >> Can you give an example of two irrational numbers, between which there >> are no rational numbers? > > Irrelevant to the example above. > > For the function,f, defined above one has > at every irrationals value of x, lim_Py -> x} f(y) = 0 = f(x) > but at every rational value, x = p/q in lowest terms, > lim_Py -> x} f(y) = 0 but f(p/q) = 1/q != 0 Uh, I suppose that establishes it in basic terms. Of course, that doesn't have anything to do with vase, but okay.
From: Tony Orlow on 16 Oct 2006 21:56 Virgil wrote: > In article <4533d18b(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >>> In article <45319b8c(a)news2.lightlink.com>, >>> Tony Orlow <tony(a)lightlink.com> wrote: >>> >>>> 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. >>> >>> When expressed as functions of time, rather than the number of >>> operations, there is no problem with having an empty vase at noon. >> When expressed in terms of iterations, the conclusion is quite the >> opposite, so you cannot claim to be working with pure unquestionable >> logic. You must choose which logical construction of the two is valid, >> if either. > > As the problem is stated in terms of the times at which events occur, > expressing things in terms of time is natural and indicated. > > When expressed in terms only of events, there is no longer any > requirement that there even be an event of completing all the > insertions-removals, or if there is, that it be close in time to any > other event. You cannot dismiss time and still have it. You can say, "What if we do this an infinite number of times?", or, "If we do this n times, how many balls are in the vase?" > >>>> Before noon, there are balls. At noon, there are not. What happened? >>> >>> They were one by one removed. >> "One by one" and all removed, but no last one. Vigilogic at its worst. > > It is apparently TOlogic, that if one moves from point A to point B one > must first cover half the distance then half the remaining distance, and > so on ad infinitum, so that one never reaches point B. > It is Virgilogic that one can go from one such midpoint to the next "one > by one" but still reach point B in finite time. Zeno's paradox is long since explained. This one is too.
From: Tony Orlow on 16 Oct 2006 22:04
Virgil wrote: > In article <4533d315(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >>> In article <45319e2d(a)news2.lightlink.com>, >>> Tony Orlow <tony(a)lightlink.com> wrote: >>> >>>> Virgil wrote: >>>>> In article <453108b5(a)news2.lightlink.com>, >>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>> >>>>>>> So how many balls are left in the vase at 1:00pm? >>>>>>> >>>>>> If you paid attention to the various subthreads, you'd know I just >>>>>> answered that. Where the insertions and removals are so decoupled, there >>>>>> is no problem. Where the removal of a ball is immediately preceded and >>>>>> succeeded by insertions of 10, the vase never empties. >>>>> It may that it never "empties", but at noon, and thereafter, it has >>>>> "become empty". >>>>> >>>> And when did that happen? >>> At the invisible transition from forenoon to noon. >> Oh, you mean the moment between all moments before noon and the moment >> of noon? > > > No! The set of all moments before noon is, in the real number model of > times, an open set whose least upper bound, noon, is not a member of > that set but such that there are no times between that set and its LUB, > noon. > So that as sets of times, the forenoon and what comes after, form a > partition of times into disjoint sets with every time being in or or the > other but none in both. > So, something happens at this partition, which is not actually a point in time, but a conceptual separation between this point and everything that came before it? Is this your quantum math again? Do we step out of time for less than a moment to make the balls disappear? > >>>>> 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. >>>>> >>>>> To argue otherwise is to misrepresent the problem. >>>> You already said that.....WRONG!!!! >>> What is wrong about it? >> This: >> >>>> There is the additional constraint that, before removing any ball, ten >>>> have been inserted. > >>> >>> Then let us put all the balls in at once before the first is removed and >>> then remove them according to the original time schedule. >> Great! You changed the problem and got a different conclusion. How >> very....like you. > > Does TO claim that putting balls in earlier but taking them out as in > the original will result in fewer balls at the end? If the two are separate events, sure. > If so, by what logic, and if not, what difference does it make? That the removal of balls in the second phase is unrelated to the addition of balls in the first stage. In the original problem, no ball is removed with the immediately preceding addition of ten. >>> Does TO claim that by putting balls in earlier there can be at ANY time >>> fewer balls in the vase that when putting them in later? >> Yes, I've already explained that. > > Not to anyone else's satisfaction. I'm sure to that of some. You're never satisfied except with yourself. >>> But when one puts them all in early enough, it becomes obvious that the >>> vase must be empty at noon. >> Yes, if all insertions occur before all removals. > > How does that change things? Is there any time at which putting balls in > earlier forces fewer balls to be in the vase? If so, at what time does > the number of balls from the earlier insertions become less that the > number of balls from the later insertions. > > As far as I can see, putting balls in earlier can only increase the > number of balls in the vase at some times, but not decrease the number > at any time. > > Why, did you put more balls in, when you added them earlier? You really don't understand the implications of the Zeno machine, do you? >>> So TO must argue that having more balls in the vase at all times before >>> noon results in less balls in the vase at noon. >>> >>> Now that is ...REALLY... WRONG!!! >>> >>> >> It is correct to say that decoupling the insertions from the removals >> such that each sequence is distinct time-wise and with its own point of >> condensation creates a different situation. > > As long as each ball is inserted at least as early as originally and no > ball is removed earlier that originally, if there is to be any set of > times at which there are fewer balls that originally then it must be > after the first ball is inserted in the new schedule, so the times are > bounded below and must have a GLB (greatest lower bound). > What is that GLB? > > Unless you can provide one, your claim fails. No it doesn't. |