An uncountable countable set
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From: Virgil on 31 Oct 2006 16:13 In article <45478e6b(a)news2.lightlink.com>, 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: > >>>>> 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. > > > > Given that you cannot answer how many Turing machines are, or > > how many sets of prime numbers there are, or pretty much > > any set that is not a member of P(N), I do not see how you > > can claim it works for sets in general. > > > > <snip> > > > > > >>>>> 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, > > > > I am not talking about the original problem. How many times > > do you have to be told that? > > > > <snip> > > > > > >>> 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.... > > > > Who are the one applying physical reasoning to it. > > > >> 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. > > > > I never said that was a formulation of the problem. It is however > > an accurate description of the set of balls that are added to > > the vase, and the set of balls removed from the vase. But that > > was not the point of the question. > > > >>> 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. > > > > More pointless insults. > > > >>> 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.... > > > > More pointless insults. You seem to be incapable of discussing > > this rationally. > > > >>> 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? > > > > What iterations? I am not talking about iterations. I am > > just posing a problem about sets that should be totally uncontroversial. > > > >> 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. > > > > No, the words "balls" and "vase" are the distraction. If you want > > to avoid distractions, phrase the question strictly in mathematical > > terms. > > > >>>>>>>>> Given that for every positive integer -1/(2^(floor(n/10
From: Virgil on 31 Oct 2006 16:16 In article <45479064(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > If that's what a partial ordering vs. a total ordering is, Bigulosity is > a partial ordering on sets, not total ordering. Different sets can have > the same Bigulosity. Since TO is apparently totally ignorant, not merely partially ignorant, of the difference, how does he even know what he means?
From: Lester Zick on 31 Oct 2006 17:13 On 30 Oct 2006 19:31:56 -0800, "Ross A. Finlayson" <raf(a)tiki-lounge.com> wrote: >Lester Zick wrote: > >> Lo mismo. >> >> ~v~~ > >Hi, > >Lester, you suggest to eschew axioms, or that axioms are unjustified. Hi Ross, what I say is that axioms are undemonstrated assumptions of truth. Axioms can be justified only through appeals to plausibility and dialectical sophistry. But there is no proof axioms are anything more than assumptions. >I agree. I proffer the null axiom theory, it has no non-logical >axioms, where logical axioms are the truth tables. Uh no, Ross, this isn't going to work. You can't just posit "truth" tables where the concept of truth itself is ambiguous. This is nothing more than axiomatic assumptions of truth by another name. >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. Well it really doesn't matter where you post it. There is no "theory of truth" which winds up with undemonstrated assumptions whether you call them axioms or not. Just getting rid of the word doesn't help. Either you can demonstrate "truth" in universal terms whether logical, physical, or mathematical or you're barking up the wrong tree. ~v~~
From: Lester Zick on 31 Oct 2006 17:24 On Tue, 31 Oct 2006 10:30:08 -0500, Tony Orlow <tony(a)lightlink.com> wrote: [. . .] Tony, I'm going to post several replies to this one post because I've come up with a couple of ideas which may (or may not) appeal to you. First off why not change your approach in the following way. It seems to me that you could arrange all the naturals on the x axis. Then instead of trying to cram in all the transcendentals on the same axis, try putting transcendental infinites on the ordinal y axis instead. However if you try this approach you may find that you need another mutually orthogonal z axis to accommodate another class of infinites. I don't know if this is going to work completely or not. But I think it holds considerably more promise than trying to accommodate it all on one more or less circular x axis alone. In any event this is the end of this particular suggestion. I hope it helps and sheds some light on what I think is going on in mechanical terms. In any event I'll get back to your original message now plus what I think will turn out to be definitive mechanical arguments on the subject of transcendentals and conventional linear analysis of the reals. ~v~~
From: Lester Zick on 31 Oct 2006 17:46
On Tue, 31 Oct 2006 10:30:08 -0500, Tony Orlow <tony(a)lightlink.com> wrote: >Lester Zick wrote: >> On Mon, 30 Oct 2006 11:18:19 -0500, Tony Orlow <tony(a)lightlink.com> >> wrote: >> >>> Ross A. Finlayson wrote: >>>> Lester Zick wrote: >>>>> Oops. My bad. Hit the wrong button on the duplicate reply. - LZ >>>>> >>>> De nada. >>> Hi Ross, Lester. Nice to see the two of you conversing. >> >> Just haven't had much to say previously on a technical level. >> >>> Mind if I >>> sprinkle some thoughts about? >> >> Not at all. Just keep it clean. >> > >I'll sweep up when I'm done. ;) Got it. >>>>>> ... >>>>>> Various considerations of the natural integers have there being a point >>>>>> at infinity. That can be useful in a form of nonstandard analysis, to >>>>>> say that, for example, an infinite sum with a limit actually equals in >>>>>> evaluation that limit. Consider for example something along the lines >>>>>> of >>>>>> >>>>>> 1 >>>>>> s = ----- >>>>>> 2^n >>>>>> >>>>>> for n from zero to infinity. The limit exists and is two and then >>>>>> consider replacing the 2 in the denominator with s. When you can >>>>>> actually say that the sum equals that limit, you get the same >>>>>> expression for s. >>>>>> >>>>>> Otherwise, s could never equal 2. >>>>> Okay. I take the point, Ross. But what rule is there requiring 00 to >>>>> be part of the same set as finites raised to a power of infinity? I >>>>> think the power of infinity could be defined using the "number of >>>>> infinitesimals" which is reciprocally defined with differentiation. >>> I think you both derive your concept of infinitesimals largely from >>> differentiation as dx->0 and 1/dx->oo. Would that be a fair assessment? >> >> Probably although I explicitly rely on the Newtonian notion and >> haven't tried to reason through the actual processes involved apart >> from noting definite integration and differentiation are reversible >> and mutually reciprocal processes. >> > >So, you're coming from a purely physical and calculus standpoint, but it >still seems to come to the same conclusion, that adding an infinite >number of infinitesimals can yield a finite value, no? Yes although I would prefer the term "combining" to "adding" since we still don't know exactly what kind of addition we'd be talking about. Basically I'm talking about Newtonian definite integration where you talk about "adding". >>> I think Ross is also deriving the same dx through subdivision, but I >>> would caution that A n>=0 1/n>1/2^n. In my book, for a set of size N, >>> using an alphabet of size S, we require strings of length L so as to >>> have enough strings to enumerate the set of size N, such that N=S^L. A >>> two-symbol alphabet such as binary mirrors the structure given by power >>> set, where there are two logical values, producing 2^L strings of length >>> L. Complete languages like digital systems, with alphabets of size S, >>> mirror the power set in structure also, where there are S possible >>> logical values, rather than just two, in the logical system. So, what am >>> I rambling about? Oh yeah, it's true, dividing the unit into n segments >>> is not equivalent to dividing it in half n times. They're two different, >>> not not incompatible, notions. >> >> Well here, Tony, you're frankly losing me. As noted above I just have >> very little interest in technical issues of this sort. I can't even >> make out what you're trying to say with "A n>=0 1/n>1/2^n" and at this >> stage of my life I'm really not inclined to try. > >Well, sorry about that. I thought you were into logic. Logic yes symbolism no. > That's a symbolic >logic statement. I've gotten into the habit of using them around here to >avoid confusion. The 'A' is supposed to be the upside down 'A', the >universal quantifier, "for all". So, I am saying that, for all n greater >than or equal to 0, 1/n is greater than 1/2^n, as n is less than 2^n. Still lost in space, Tony. I might puzzle it out with a little work. >Furthermore if as it >> seems you're trying to draw some computer analogy to general >> mathematics and science I'd consider the effort totally misdirected. >> Show me any infinite or infinitesimal inside a machine and I might >> reconsider. >> > >:) Are you familiar with 2's complement signed binary? No. I was only a professional systems and assembler programmer for thirteen years for two different major Fortune 500 companies. > It's in your >computer. Zero is 000...000 (how every many bits there are). To >increment, find the rightmost 0, and invert every bit from there >rightward. To decrement, find the rightmost 1 and invert every bit from >there rightward. To find the additive inverse, invert all bits and add >1. (there's a nice little explanation of why it works that way) So, >111...111 is -1, predecessor to 1, right? Well, values are generally >taken, if we have n bits, to go from -2^(n-1) to 2(n-1)-1. Why one extra >negative number, 100...000? Well, it's not a negative number. It is its >own additive inverse, just like 0. When you invert the bits you get >011...111, and when you add 1, you're back to 100...000. It's not a >positive or a negative. It's the element on the opposite end of the >integer ring from +/-0. It's +/-oo. > >So, you might reconsider. :) Sure. Just as soon as you can actually show me the infinitesimals and infinites inside the machine and instead of just talking about them as if on a good day they might be there. >>>>>> There are other reasons to consider the infinite naturals as containing >>>>>> an infinite element, that N E N. >>>>> N E N? >>> The nth element is n. If there are n elements, n is a member of the set. >>> The number of elements up to and including n, starting at 1, is always >>> n. The set of the first n naturals starting at 1 always has a maximum >>> element of n. So, N is not just the size of the set, it's also a member. >>> That sounded too much like Sy Sterling, from the Hair Club for Men..... ;) >> >> Okay. I agree with what you say here certainly with respect to the >> naturals. >> > >Not Sy Sperling |