An uncountable countable set
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From: Tony Orlow on 14 Oct 2006 22:34 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? > > 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!!!! There is the additional constraint that, before removing any ball, ten have been inserted. This makes it impossible for the number of balls to ever decrease, except by that one, before increasing, and for it ever to "become" empty.
From: Virgil on 14 Oct 2006 22:48 In article <45319809(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <4531023e(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > > > >> Where is there defined in the problem any mention of a discontinuity in > >> the process? There isn't. > > > > The discontinuities follow from the statement of the problem. The > > "number of balls in the urn" has a discontinuity each time it changes. > > The formula for the number of balls after each iteration is linear. The formula for the number of balls as a function of time is non-linear, and time is the relevant variable here. > It > does not change direction. While the set may be discrete, the formula > mapping the set from the naturals is continuous in the reals. A function from the naturals is never continuous according to any standard definition of continuity. What alleged definition of continuity does TO pretend to be using? > Either you > see the significance of this, or you don't, I suppose. No one sensible sees the significance of TO's nonsense claims. > > > > The notion that an infinite sequence cannot take occur within a finite > > time interval has been dead since Zeno. > > > > You miss the point, as usual. One can look at this in time, in which > case the iterations all become squashed together at noon and > indistinguishable. Or, one can look at this in iterations, in which case > the sum clearly diverges. The Zeno machine is a deliberate attempt to > produce a paradox. Ho hum. Since we do not need any Zeno machine to do our analyses, it is only TO's Zeno machine which is paradoxical. > > > > >> 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. > > > > In TO's world, only TO can say what is or is not there, but in ZF and > > NBG, there is a least infinite ordinal. > > ZF and NBG don't handle sequences or their sums, but only unordered > sets, so they really have nothing to say on the matter. Where does TO get the peculiar idea that, say, NBG's set of finite ordinals is not ordered and cannot include sequences?
From: Virgil on 14 Oct 2006 22:58 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.
From: Virgil on 15 Oct 2006 00:09 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. And as the problem is posed in terms of times, it is quite artificial to eliminate time from one's analysis. > > > > >> The balls at noon are not distinguishable nor specific. They are non-existent. > > > > 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? They were one by one removed. How > does 9n become 0 when n=oo? It doesn't. and n doesn't become oo.
From: Virgil on 15 Oct 2006 00:16
In article <45319d93(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > 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. That "constraint" is irrelevant, as it by the clock that the insertions and removals are determined, not merely by the insertion or removal of other balls. And how can earlier insertions make for fewer balls at any time as TO claims it does. > > 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. To argue that adding balls earlier but removing them as before leaves fewer of them is even sillier. > > > > >>> 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". Correct in this context means an analysis in accord with the constraints of the original problem. Which TO's analysis is not. |