An uncountable countable set
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From: Tony Orlow on 24 Oct 2006 00:51 David R Tribble wrote: > Tony Orlow wrote: >> You have agreed with everything so far. At every point before noon balls >> remain. You claim nothing changes at noon. Is there something between >> noon and "before noon", when those balls disappeared? If not, then they >> must still be in there. > > Of course there is a "something" between "before noon" and "noon" where > each ball disappears. At step n, time 2^-n min before noon, ball n is > removed. This happens for every ball, since there is a step n for > every ball. The balls are removed, one by one, one at each step, > before noon. > As each ball n is removed, how many remain? Can any be removed and leave an empty vase?
From: Virgil on 24 Oct 2006 01:20 In article <453d953e(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > cbrown(a)cbrownsystems.com wrote: > > Tony Orlow wrote: > >> cbrown(a)cbrownsystems.com wrote: > Isn't time considered, by you and others, to be a crucial element of the > argument? But, if the set-theoretical argument rests on time, > doesn't it matter whether you preserve the order of events in time? The only critical time dependency is that each ball to be inserted shall be inserted at a time before noon and also removed at a later time but still before noon. And any set of insertions and deletions satisfying these conditions must leave the vase at noon as it was before the first insertion. > Are > you allowed to change the problem with imagined discontinuities and time > vortexes, rearranging events in order to get whatever answer you want? All we want is that each ball be inserted before noon and removed before noon, as the statement of the problem requires. Given only that, the vase is empty at noon of anything of any balls which were inserted. > Sure, you can "prove" that kind of stuff from "the axioms", but not > while preserving the stated order of events. As the only ordering needed is that each ball inserted into the vase before noon also be removed before noon, and we have that, only TO and his ilk need to create "time vortices" or any such nonsense. > > >> That's the salient fact here. > > > > Regardless of its saliency, we can at least agree that it follows > > logically from (1)-(8). > > Not without changing the experiment. The only relevant issue is "Is each ball which is inserted into the vase before noon also removed from the vase before noon according to the rules set up in the problem?" My answer is "Yes". TO's answer apparently is "No". >You agree that it's reasonable to > conclude that every time a ball is removed 10 are added. You then > suggest that, if we rearrange events by making them happen at different > times than originally scheduled, we can argue that the numbers in and > out are equal. The point is, you can't do that, and at the same time > claim to be discussing the original problem. Can you? The question is "According to the rules set up in the problem, is each ball which is inserted into the vase before noon also removed from the vase before noon?" An affirmative answer confirms that the vase is empty at noon. A negative answer violates the conditions of the problem. > > > > >> You never remove as many as you > >> add, so you can't end up empty. > > > > That is not a conclusion we can draw from (1)..(8). > > Yes it is, given the times of insertions and removals. The limit is oo > at t=0. The only relevant question is "According to the rules set up in the problem, is each ball which is inserted into the vase before noon also removed from the vase before noon?" An affirmative answer confirms that the vase is empty at noon. A negative answer violates the conditions of the problem. > > Instead, this is > > the unstated assumption to which you return again and again: > > > > (Proposition T) If you never remove as many as you add, then there is a > > ball in the vase at time 0. > > If you never remove as many as you add then the vase can only become > fuller and is at no time any less full than it was at any time before. The question is "According to the rules set up in the problem, is each ball which is inserted into the vase before noon also removed from the vase before noon?" A negative answer violates the conditions of the problem. An affirmative answer confirms that the vase is empty at noon. > > > > If we simply /require/ (T) to be true, and damn the consequences, then > > of course it /is/ true, That is what we mean when we say "that argument > > is circular". > > You're not even making sense. TO is hardly in a position to carp about others not making sense. > That doesn't mean the state of the vase changes, that is, it becomes > empty. That would require the removal of balls, now, wouldn't it? Or, > perhaps just the labels.... The question is "According to the rules set up in the problem, is each ball which is inserted into the vase before noon also removed from the vase before noon?" An affirmative answer confirms that the vase is empty at noon. A negative answer violates the conditions of the problem. > No, there is an obfuscation caused by the Zeno Machine's reduction of > the Twilight Zone between finite and infinite to a single point in > "time" which actually doesn't exist. I mean, it's a really nice example > of what's wrong with trying to focus on that non-boundary, or pretend > that it exists. This should be considered a proof by contradiction that > there is something wrong with the set theoretical approach on this one, > much like Banach-Tarski. All these paradoxes point to lapses in > understanding. It's clearly a divergent series. The only critical issue is whether every ball the gets into the vase before noon is removed before noon. If this is the case, then, regardless of any other details, the vase is empty at noon. > >> > > > > Coungter-factual hypotheticals again. > > Not. The only factuals of relevants are that every ball in the vase at any time before noon is removed before noon. > > So, you agree, noon cannot occur in the experiment, of course.... Why should anyone do that? > > I guess I am "assuming" that lim(n->oo: 1/n)=0 and that lim(n->0: > 1/n)=oo. Sorry I kept that a secret. Irrelevant! > > > > > Secondly, until you say what /logical relationship/ there is between > > the statement "nothing happens at noon" and the conclusion "there > > exists a natural number n such that -1/n = 0", it's a non-sequituur. > > Every event occurs at time t=-1/n for n e N. There is no n e N such that > t=0=1/n. Is that really that complicated? Do you get it now? > > > > > And you cannot begin to establish that relationship until you specify, > > mathematically, what you mean by "nothing happens at noon". > > > > Ugh! There are no events which occur at time t=0. Events include > insertions and removals of balls. > > >>> Assuming that by "all events happen before noon", you mean "balls are > >>> only placed in or removed from the vase at times before noon"; then > >>> thes
From: Virgil on 24 Oct 2006 01:26 In article <453d99f8(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> David Marcus wrote: > >>> Tony Orlow wrote: > >>>> David Marcus wrote: > >>>>> Tony Orlow wrote: > >>>>>> Either something happens an noon, or it doesn't. Where do you stand on > >>>>>> the matter? > >>>>> What does "something happens" mean, please? I really don't know what > >>>>> you > >>>>> mean. > >>>> ??? Do you live in the universe, or in a static picture? When "something > >>>> happens" o an object, some property or condition of it "changes". That > >>>> occurs within some time period, which includes at least one moment. > >>>> There is no moment in this problem where the vase is emptying, > >>>> therefore, that never "occurs". If you are going to insist that time is > >>>> a crucial element of this problem, then you should at least be familiar > >>>> with the fact that it's a continuum, and that events occurs within > >>>> intervals of that continuum. > >>> Thanks. That explains what "something happens" means. Now, please > >>> explain what "emptying" means. > >> "Empty" means not having balls. To become empty means there is a change > >> of state in the vase ("something happens" to the vase), from having > >> balls to not having balls. > >> > >> Now, when does this moment, or interval, occur? > > > > A reasonable question. Before I answer it, let me ask you a question. > > Suppose I make the following definitions: > > > > 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, > > undefined if t = A_n or t = R_n. > > > > Let V(t) = sum{n=1}^infty B_n(t). > > > > Then V(-1) = 1 and V(0) = 0. If we consider V to be a function of time, > > at what time does it become zero? > > > > Just answer the question, and stop beating around the bush. The only relevant question is "According to the rules set up in the problem, is each ball which is inserted into the vase before noon also removed from the vase before noon?" An affirmative answer confirms that the vase is empty at noon. A negative answer directly violates the conditions of the problem. How does TO answer?
From: Virgil on 24 Oct 2006 01:27 In article <453d9a1d$1(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> 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 The only relevant question is "According to the rules set up in the problem, is each ball which is inserted into the vase before noon also removed from the vase before noon?" An affirmative answer confirms that the vase is empty at noon. A negative answer directly violates the conditions of the problem. How does TO answer?
From: Virgil on 24 Oct 2006 01:27
In article <453d9a7c$1(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> David Marcus wrote: > >>> Tony Orlow wrote: > >>>> Your examples of the circle and rectangle are good. Neither has a height > >>>> outside of its x range. The height of the circle is 0 at x=-1 and x=1, > >>>> because the circle actually exists there. To ask about its height at x=9 > >>>> is like asking how the air quality was on the 85th floor of the World > >>>> Trade Center yesterday. Similarly, it makes little sense to ask what > >>>> happens at noon. There is no vase at noon. > >>> Do you really mean to say that there is no vase at noon or do you mean > >>> to say that the vase is not empty at noon? > >> If noon exists at all, the vase is not empty. All finite naturals will > >> have been removed, but an infinite number of infinitely-numbered balls > >> will remain. > > > > "If noon exists at all"? How do we decide? > > > > We decide on the basis of whether 1/n=0. Is that possible for n in N? > Hmmmm......nope. The only relevant question is "According to the rules set up in the problem, is each ball which is inserted into the vase before noon also removed from the vase before noon?" An affirmative answer confirms that the vase is empty at noon. A negative answer directly violates the conditions of the problem. How does TO answer? |