An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Tony Orlow on 16 Oct 2006 14:35 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?
From: Tony Orlow on 16 Oct 2006 14:38 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. > > 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. "One by one" and all removed, but no last one. Vigilogic at its worst. > > How >> does 9n become 0 when n=oo? It doesn't. > > and n doesn't become oo. Then noon doesn't happen, and the vase never empties.
From: Tony Orlow on 16 Oct 2006 14:40 Virgil wrote: > 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. Already explained, but not explained by you why the salient feature I mention is irrelevant. > >> 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. No, I already explained that. Apparently you didn't pay attention. >>>>> 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. When you claim that stated features of the process are "irrelevant" by edict, you're not one to speak.
From: Randy Poe on 16 Oct 2006 14:43 Tony Orlow 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, No, it's not. > so you cannot claim to be working with pure unquestionable > logic. You must choose which logical construction of the two is valid, > if either. When expressed in terms of iterations, you can say things about the number of balls in the vase at all times which are strictly less than zero (noon). There are no conclusions to be drawn about f(0). Information about f(0) must be derived by other means. Taking the limit of f(t) as t->0 tells you that limit, but does not tell you f(0) unless you know a priori (i.e. you ASSUME) f is continuous. Since f is provably not continuous, that would be a foolish assumption contrary to the other assumptions in the problem. - Randy
From: Tony Orlow on 16 Oct 2006 14:44
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? So, trichotomy is not a feature of time? Interesting. And I always thought that, given two moments, or points in time, a and b, either a<b, b<a, or a=b. But I guess there's a fourth option. How do you express that option? >>> 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 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. > > 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. > > 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. > > > 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. See? |