An uncountable countable set
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From: David Marcus on 31 Oct 2006 02:27 imaginatorium(a)despammed.com wrote: > Virgil wrote: > > In article <1162268163.368326.64650(a)m73g2000cwd.googlegroups.com>, > > imaginatorium(a)despammed.com wrote: > > > > > Virgil wrote: > > > > In article <45462ba0(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: > > > > > > <snipola> > > > > > > > > >> 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. > > > > > > > > It is the only definition of set size that is known to produce a valid > > > > partial ordering on sets. > > > > > > Huh? I thought cardinality produced a valid *total* ordering on sets. > > > > The cardinalities are totally ordered, but the sets are not. > > A total order on sets would require that when neither of two sets was > > "larger than" the other that they must be the same set, not merely the > > same size. > > Oh, right. But - and I'm not quite sure how to say this, but the > cardinalities _are_ totally ordered; for any two sets A and B, c(A) < > c(B), or c(A) = c(B), or c(A) > c(B). If you "reduce" the sets by the > cardinality equivalence relation, they are totally ordered. Or, if you use the equivalence relation (bijectivity) instead of equality in the definition of "total order", then cardinality satisfies the total-order properties. -- David Marcus
From: David Marcus on 31 Oct 2006 02:29 Ross A. Finlayson wrote: > Say what? > > I'm trying to figure out dividing R into (-oo, 0] and [0, oo), to > actually have the mereological boundary at zero, not to the left or > right, for symmetry about the origin. > > This is "sci.math", not "Tony and Virgil rip on each other." If you > can't sustain a reasonable argument then you're not. With Tony's > semi-glossolalia on the one side and Virgil's vicious character stabs > bait-switched with feigned academic innocence, so he feels OK, on the > other, I prefer Tony, because I'm not here for the mainstream except > for the end, of the Hilbert program. > > Tony, you make some sense, please endeavor to do so in general. Have a > point, remain honest. Mathematics actually demands conformance. > > Orthodoxy I can get out of a book, and meanness I don't really stand in > person, toad. Stuff it, jive turkey. > > That witch's brew talk is a weak Halloween reference. > > Here's something that's frightening, to some: Gorin and Kukushkin > divide by zero and evaluate Gamma of negative integers to integrate the > Devil's "Scare"-case. > > P(P(V)): set? > > Infinity: less than nothing. Non-archimedean? Not bad for poetry. -- David Marcus
From: Tony Orlow on 31 Oct 2006 09:51 David R Tribble wrote: > David R Tribble wrote: >>> Every member of N has a finite successor. Can you prove that your >>> "infinite naturals" are members of N? > > Virgil wrote: >>> The property of not being an infinite natural holds for the first >>> natural, and holds for the successor of each non-infinite natural, so >>> that it must hold for ALL naturals. > > Tony Orlow wrote: >> It holds for all finite naturals, but if there are an infinite number of >> naturals generating using increment, then there are naturals which are >> the result of infinite increments, which must have infinite value. > > Can you show us one of those infinite naturals? > > And while you're at it, show us the finite natural that is its > predecessor. Of course, you first have to define what you mean > by "infinite increments". > I meant an infinite number of increments, each being a successive difference of +1 in measure. Here is an infinite natural, Big'un: 100...000. That's in binary, and there are log2(Big'un) 0's after the 1, so the most significant bit is at log2(Big'un). Its predecessor, 0111...111, with the 0 also at log2(Big'un), is also infinite, as are countably, or evenly uncountably many, in the chain of predecession. Just as I have been talking about "finite neighborhoods around limit points" int he T-riffic, such as log2(Big'un), Robinson defines the '~' operator as defining a countable zone within the uncountable sequence such at a~b means that there are a finite number of elements between a and b. Within the uncountable sequence, there are an uncountable number of such neighborhoods, I believe. I don't think there is any countable regression to countable neighborhoods from the continuum. Within the countable neighborhood we can specify the equivalent of any finite number. Between countable neighborhoods, The T-riffics use repeating infinite strings, corresponding to rational portions of infinite values. So, while the T-riffics can have the range of an uncountable set of naturals, it can only address a countable set using only finite strings. But, that's the case with any number system. Tony
From: Tony Orlow on 31 Oct 2006 10:01 David R Tribble wrote: > David R Tribble wrote: >>> Each ball n is placed into the vase at time 2^int(n/10), and then later >>> removed at time n. This happens for every ball before noon. So every >>> ball is inserted and then later removed from the vase before noon. >>> >>> At any given time n before noon, ten balls are added to the vase and >>> then ball n (which was added to the vase in a previous step) is >>> removed. Your entire confusion results from assuming a "last" time >>> prior to noon, but there is no such time. > > Tony Orlow wrote: >> At no time prior to noon are all balls removed. Nor are any removed at >> noon. It cannot be empty, then. > > The problem states that every ball (every ball) is added to the vase > and then later removed from the vase. It states the specific times of those events, which imply that there are always more balls added than removed at any time. > > We conclude from this that every ball is removed (eventually). Yes, you conclude an end to the unending set by compressing events at a point in time so they cannot be distinguished and the difference between in and out is hidden. Whoopedy doo. It's a parlor trick. > You conclude that at no time are all balls removed. There is no finite t<0 when all balls have been removed. Agree? There are no balls removed at t=0. Agree? If t1<t2, and a ball is in the vase at t1 but not at t2, then it had to have been removed at some point t such that t1<t<=t2. Agree? > > Obviously you think that there are balls left in the vase that never > got removed. In fact, you say that there are an infinitude of balls > left in the vase. Yet somehow you cannot name a single one of them. > I can, as soon as you tell me how many you inserted to begin with. Multiply that by 9/10 and you have an answer. Except that you can't, because what you're doing is not math, but Zeno-esque logic trick.
From: Tony Orlow on 31 Oct 2006 10:30
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. ;) >>>>> ... >>>>> 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? >> 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. 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. 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? 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. :) >>>>> 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? I think it was "Sperling", now... Anyway, so do WM and Han and others, it seems. That's Ross' N E N, unless of course Ross corrects me on that. Ross? >>>>> In these arguments here, if there is no infinite value for n, then the >>>>> process never completes. >>>> Well infinitesimal subdivision certainly never completes. >>>> >>> Les, Lester, there is some consideration that it does. Is not the >>> differential intuitively the atomic subdivision of one? >>> >>> In an interesting way, variously on what you consider interesting, the >>> notion of subatomic particles in physics is a similar one as the >>> consideration of sub-iota reals in mathematics. >>> >> Lester hates when I talk mathematical TOE. > > Only because it's not true, Tony. People call things "theories" when > they're at best nothing but undemonstrable rank speculation and at > worst when they're nothing but analytical academic formalisms and what > they're really after is just the cachet of scientific truth. > How do y |