An uncountable countable set
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From: Tony Orlow on 30 Oct 2006 22:16 stephen(a)nomail.com wrote: > 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: >>>>>> stephen(a)nomail.com wrote: >>>>> <snip> >>>>> >>>>>>> What does that have to do with the sets IN and OUT? IN and OUT are >>>>>>> the same set. You claimed I was losing the "formulaic relationship" >>>>>>> between the sets. So I still do not know what you meant by that >>>>>>> statement. Once again >>>>>>> IN = { n | -1/(2^(floor(n/10))) < 0 } >>>>>>> OUT = { n | -1/(2^n) < 0 } >>>>>>> >>>>>> I mean the formula relating the number In to the number OUT for any n. >>>>>> That is given by out(in) = in/10. >>>>> What number IN? There is one set named IN, and one set named OUT. >>>>> There is no number IN. I have no idea what you think out(in) is >>>>> supposed to be. OUT and IN are sets, not functions. >>>>> >>>> OH. So, sets don't have sizes which are numbers, at least at particular >>>> moments. I see.... >>> If that is what you meant, then you should have said that. >>> And technically speaking, sets do not have sizes which are numbers, >>> unless by "size" you mean cardinality, and by "number" you include >>> transfinite cardinals. > >> So, cardinality is the only definition of set size which you will >> consider.....your loss. > > If somebody presents another definition of set size, I will > consider it. You have not presented such a definition. > > I have presented an approach that works for the majority of infinite bijections, and explained some of the exceptions. IFR works for all numeric sets mapped from a common set. N=S^L works for all languages, including those that express the first set. Both work on a parameteric basis, using infinite case induction to finely order the values of formulas for a specific infinite n. Rare exceptions include the set 1/n for neN, whose inverse is itself, which IFR ends up saying has size 1, but that's because the natural indexes and fractional mapped reals only share one point in their range, 1. So, I think Bigulosity is worth considering. It subsumes Ross's EF, and asymptotic set density, and works perfectly for finite sets, so there should be no issue trying to generalize it, as a sound foundation. Most set definitions which aren't handled by Bigulosity are poorly formed, or can be restated in ways that make them easily handled. >>> In any case, it still does not make any sense. I am not sure >>> what |IN| is for any n. IN is a single set. There is only >>> one set, and it does not depend on n. In fact, there isn't >>> an n specified in the problem. Yes I used the letter n in >>> the set description, but that does not define an entity named 'n'. >>> > >> There most certainly is an 'n'. The problem describes a repeating >> process, each repetition of which is indexed with a successive n in n, >> and during each repetition of which ball n is removed. What do you mean >> there's no n??? > > I am talking about two sets that I have explicitly defined. > There is no 'n'. I have been very clear about what I am > talking about. I cannot help it if you cannot get over > your fixation with balls and vases. > Yes, from that perspective, they are the same set. But, there is an n in the original problem, which this characterization ignores. For each n in N, balls 10n-9 through 10n are added, and ball n is removed, at time t=-1/n. If your representation of the problem does not include all three of these facts with their proper relations, that's a good reason for getting an incorrect answer. Where is the n in your formulation? You have an expression for t, without mentioning t, but no realtion between t and n. So, you are missing a piece of your puzzle. >> There is no spoon. There is no God. There is no mind. There is no n. > >> No no no no. No. > > You really are not paying attention, are you? As I said earlier, > if you are not going to bother to read what I write, you should > not bother responding. > You should make it pertinent, and respond to the obvious holes in its relevancy. > So here is one last attempt. The balls and vase problem > is not a physical problem. No kidding. And I was going to set up a test bed in the basement.... Pretending that it is a physical > problem is pointless. It is phrased in such a way to suggest > a physical problem, so that it might confuse your physical > intuitions, but it is not a physical problem. Show me where your formulation relates t and n. Where is the schedule stated in that formulation? It's not. Try again. > > Even 'finite approximations' of it are not physical problems. > Consider the case where you start adding balls at 11pm. > Suppose you had 2000 balls. The last set of 10 balls will > be added at time at 10^-56 seconds before noon. But that > is less than Planck time, and is therefore impossible according > to known physics. How big are these balls? Lets suppose > they are 1cm in diameter, and moving them into the vase requires > the ball be moved 1cm. Once you get up to ball 350 or so, > you will have to move the balls faster than the speed of light. Blah blah blah. Thanks for the reality lesson. Now take a lesson in realism. > > This is simply not a physical problem. It is an abstract > problem. It is a mathematical problem. The "physical" part > is just a distraction. So is all your blabbering.... > > So if instead, someone had just posed this problem > Let > IN = { n | -1/2^(floor(n/10)) < 0 } > OUT = { n | -1/2^n < 0 } > > What is | IN - OUT | there would be controversy. > Note, there are not balls or vases, or times or anything > in this problem. Just two sets. It would help if you > bother to stop and think for a second before responding. > > Where are the iterations mentioned there? You're missing the crucial part of the experiment. By your logic, you could put them in in any order and remove them in any order, and when you say both processes are done, nothing's left, but that's BS. It ignores the sequence specified. This is just a distraction. >>>>>>> Given that for every positive integer -1/(2^(floor(n/10))) < 0 >>>>>>> and -1/(2^n) < 0, both sets are in fact the same set, namely N. >>>>>>> >>>>>>> Do you agree, or not
From: Ross A. Finlayson on 30 Oct 2006 22:18 Virgil wrote: > > Does TO wish to claim some balls are not removed before noon? > > Which one(s), TO? The ten more added for that one in the next timestep before noon, those. Remove those, for each add ten. For however many you removed, you must remove ten times as many. For however many you remove, you must add ten times as many. I don't really have a resolution to that. Standardly, there is no last ball removed from the vase, nor last added. For the "completion" of infinity to occur, it does. The count of balls added, in summary, to the vase, diverges as t-> noon, or let's say between zero and one inclusive t -> 1. If you don't really care, dump in all the balls at t = 0 and then just remove them, at some point all the balls were in the vase. That's similar in notion to the equivalency function, N/U EF, except instead of going to zero it goes to one. The difference between f(x) and f(x)+1 is positive and less than 1/n for finite integer n. It seems simple to describe, and there is ample historical verbiage to frame it in classical and modern terms. I figure someone must have presented it before in concept, EF or something along those lines, in my cursory researches I've never otherwise heard of it, except implicitly it's ubiquitous. Its integral has some interesting properties that lead to various analytical extensions in the polydimensional context. Points are defined by space, all of it, space is defined by points. Also it has some unique structural features in terms of ordering its range. If there's infinity in the real numbers, a question arises "is 1/oo zero?", and the answer is qualified, yes and no. Add it to three, the result is three. Add it infinitely many times to three, no less and no more, the result is four. That's too casual, but it's not an informal argument. Ross
From: Tony Orlow on 30 Oct 2006 22:19 stephen(a)nomail.com wrote: > Tony Orlow <tony(a)lightlink.com> wrote: >> stephen(a)nomail.com wrote: >>> Tony Orlow <tony(a)lightlink.com> wrote: >>>> imaginatorium(a)despammed.com wrote: >>>>> Tony Orlow wrote: >>> <snip> >>> >>>>>> The formulaic relationship is lost in that statement. When you state the >>>>>> relationship given any n, then the answer is obvious. >>>>> Do "state the relationship given any n"... I mean, what is it, exactly? >>>>> >>>> Uh, here it is again. in(n)=10n. out(n)=n. contains(n)=in(n)-out(n)=9n. >>>> lim(n->oo: contains(n))=oo. Basta cosi? >>> >>> What is in(n)? The sets I and everyone but you are talking about are >>> IN = { n | -1/2^(floor(n/10)) < 0 } >>> OUT = { n | -1/2^n < 0 } >>> Noone has ever mentioned or defined in(n) >>> >>> What is the definition of in(n)? Is is a set? >>> >>> Stephen >>> >> out(n) is the number of balls removed upon completion of iteration n, >> and is equal to n. > >> in(n) is the number of balls inserted upon completion of iteration n, >> and is equal to 10n. > > But there are no balls or iterations in the problem I posed. > So why do you keep talking about balls and iterations? > Are you really that incapable of participating in a discussion? > Are you incapable of responding to a particular formulation of the problem? Did you start the discussion? What makes you think you can steer it? >> contains(n) is the number of balls in the vase upon completion of >> iteration n, and is equal to in(n)-out(n)=9n. > >> n(t) is the number of iterations completed at time t, equal to floor(-1/t). > >> contains(t) is the number of balls in the vase at time t, and is equal >> to contains(n(t))=contains(floor(-1/t))=9*floor(-1/t). > >> Lim(t->-0: 9*floor(-1/t)))=oo. The sum diverges in the limit. > >> See how that all fits together? Its almost like physics, eh? > >> Tony > > The balls and vase problem has absolutley nothing to do > with physics. Try it out. Don't be an idiot. What makes you think my objection has anything to do with physics? It has to do wih properly representing the problem mathematically without ignoring pertinent details, like the times of insertion and removal for each ball. Your example without this information is irrelevant to the problem. So, what's wrong with my logic above, besides the fact that it's not what YOU are trying to spew? > > Stephen > > >
From: Ross A. Finlayson on 30 Oct 2006 22:31 Lester Zick wrote: > Lo mismo. > > ~v~~ Hi, Lester, you suggest to eschew axioms, or that axioms are unjustified. I agree. I proffer the null axiom theory, it has no non-logical axioms, where logical axioms are the truth tables. Then, with some introspection, all and only true statements are theorems. Goedel can't dam it, the tower of rain. I relate it to Yggdrasil it is very simple. It's, quite simple. There's quite a bit more exposition about it here and on sci.logic and physics. For, if it is, it's the T.o.E. There's only one theory with no axioms. More later. Warm regards, Ross
From: Randy Poe on 30 Oct 2006 22:47
Lester Zick wrote: > On 30 Oct 2006 10:46:23 -0800, "Randy Poe" <poespam-trap(a)yahoo.com> > wrote: > > > > >Lester Zick wrote: > >> On 30 Oct 2006 08:34:21 -0800, "Randy Poe" <poespam-trap(a)yahoo.com> > >> wrote: > >> > >> > > >> >Lester Zick wrote: > >> >> On 28 Oct 2006 12:54:51 -0700, "Randy Poe" <poespam-trap(a)yahoo.com> > >> >> wrote: > >> >"There is a least integer" and "there is a least real" > >> >are both false. > >> > >> They are? > > > >Yes. If you disagree, perhaps you can name me the minimum > >element of the sets Z and R. > > Or perhaps you can show us how it is "there is a least integer" and > "there is a least real" are both false since it's your contention not > mine. > > >> Perhaps you should take that up with theologians then. > > > >We are discussing mathematics. In the mathematical objects > >called "the set of integers" and "the set of reals" there is no > >least member. > > Well perhaps you can just prove that since it's your contention not > mine? Why, do you think that there's a least integer? What, around -1000? Proof: A least member x0 would have the property that x0 <= x for all other members x. Let x0 be any integer. x0-1 is also an integer, which is <x0. Thus x0 can't be a least member. Similar argument for x0 being any real. QED - Randy |