An uncountable countable set
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From: Tony Orlow on 7 Sep 2006 13:27 Dik T. Winter wrote: > In article <44f1b114(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > Dik T. Winter wrote: > > > In article <44ef36ae$1(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > ... > > > > Such that one can specify which finite number of iterations will get > > > > one within a specific finite range of accuracy, gven a specific > > > > method of approximation? It's a limit concept, really, yes? > > > > > > Computer scientist you are? How wrong you are. The very first definition > > > is that a number is computable if there is a Turing machine so you can give > > > it a specific integer n and it will calculate all digits from the first > > > to the n-th. There is no limit concept involved at all. All algebraic > > > numbers are computable, as are a host of non-algebraic numbers (e, pi). > > > > Dik, you didn't even try to think about that one. Sorry. It was a lame > > answer. > > Oh. > > > When you say you can calculate to the nth digit, > > But you stated "that you can specify which finite number of iterations" etc. > That is something different. A number is computable when there is a Turing > machine such that, when given an arbitrary number n, it will calculate the > 1-st through n-th digit. There is nothing about a specification of a > finite number of iterations. Well, if you set up a Turing machine such that it will take an infinite number of operations to get to any finite digit, then you have probably not designed it well.
From: Tony Orlow on 7 Sep 2006 13:42 David R Tribble wrote: > Virgil wrote: >>> Then show that the set of all natural numbers does not have cardinal number >>> aleph-0 and ordinal number w. A proof please. > > Tony Orlow wrote: >>> I have already shown how the set of bit positions in the binary naturals >>> has no cardinal or ordinal that can be assigned to it. > > Virgil wrote: >>> TO has claimed it, but not proved it. >>> TO claims a lot but never proves any of it. >>> Since the set of digit positions in any positional notation for the >>> members of N is indexed by N, the cardinality of the set of bit >>> positions equals the cardinality of its index set. > > Tony Orlow wrote: >> Since the bit strings are the power set of the set of bit positions, >> each natural being a unique subset of 1 positions, the set of naturals >> is power set to the set of bit positions. > > That sounds reasonable at first glance. But further study shows > that there is a direct one-to-one correspondence between the bitstrings > and the naturals. So the logical conclusion is that they are the same > size. If you associate all the (finite) bitstrings with naturals, they > are in fact the same set. > > So there can't be an in-between cardinality because the sets > have the same cardinality. > > >> In your concoction, the naturals are power set to nothing. > > Exactly. To be so, there would have to be an infinite set smaller > than N that could not be bijected with N. But there is no such set. > > Specifically, like I said above, the set of bit positions bijects with > the set of bitstrings. For every natural, there is a bit index, and > vice versa. For every natural, there is a bitstring, and vice versa. > So every bit index is a natural, and every bitstring is a natural. > Hence they are exactly the same set. > That's my point. You can biject the naturals with the bit posiitons, and also with the strings, but the set of strings is the power set of the bit positions. Therefore you have a bijection between a set and its power set, and no cardinality which is appropriate for the set of bit positions.
From: Tony Orlow on 7 Sep 2006 13:45 David R Tribble wrote: > Tony Orlow wrote: >>> In fact, mapping the naturals in [1,Big'un] to the reals in [Lil'un,1] >>> using the mapping function f(x)=x/Big'un yields Ross' Finlayson numbers, >>> and is perfectly consistent with IFR. Not only do we obviously have >>> Big'un^2 reals on the line because we have the sum of Big'un unit >>> intervals each containing Big'un reals, ... > > David R Tribble wrote: >>> That implies that there are BigUn naturals in the real number line, >>> so that |N| = |R|. You state this as though it's a fact, but what is >>> your proof? > > Tony Orlow wrote: >> What? No. I mapped the reals in [Lil'un,1] (Lil'un is successor to 0, in >> the Finlayson system. He calls it iota, which is finite) to the naturals >> in [1,Big'un] (the real line). The reals in the unit interval are the >> image of the naturals on the entire line. > > Which implies that there are the same number of naturals in the > real number line (or in N) as there are reals in (0,1]. That's > provably false. Not for the hypernaturals. > > Either that or you're omitting an awful lot of reals in (0,1] to get > your mapping to the naturals to work. Or you're using some alternate > version of the "real number line" that is not dense in the reals. > (Based on your previous alleged well-ordering of the reals, and > your acceptance of Ross's iota ordering, that could very well be > the case.) > Yes, I am including infinite values on the number line, since it's "infinitely long". > >> You are mistakenly applying >> the standard cardinalistic fact that the number or reals in [0,1] is the >> same as the number of all reals. That is false in my system. > > Since it's rather easily proved true in standard theory, you need > to provide that missing proof of yours showing that it's false in > your system. For instance, you have to demonstrate that there > are more reals in (0,2] than in (0,1] while showing that both > intervals are dense. Good luck with that. (Hint: assume that > there are only a finite number of reals in any interval.) Given the axiomatic statement that there are Big'un reals in every unit interval, that 2*Big'un>Big'un, and that (0,1] U (1,2] is (0,2], that's done. > > Of course, if it's false in your system, your system cannot be > compatible with standard arithmetic. You see why, don't you? > Of course.
From: imaginatorium on 7 Sep 2006 13:50 Tony Orlow wrote: > Mike Kelly wrote: > > Tony Orlow wrote: > >> Virgil wrote: > >>> In article <44fe2642(a)news2.lightlink.com>, > >>> Tony Orlow <tony(a)lightlink.com> wrote: > >>> > >>>> Virgil wrote: > >>>>> In article <44fd9eba(a)news2.lightlink.com>, > >>>>> Tony Orlow <tony(a)lightlink.com> wrote: > >>>>> > >>>>>> Dik T. Winter wrote: > >>>>>>> In article <44ef3da9(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> > >>>>>>> writes: > >>>>>>> Your axiom uses things that are not defined. What is the *meaning* of > >>>>>>> "x<z"? > >>>>>> Geometrically it means that x is left of z on the number line. > >>>>> And for someone standing on the other side of the number line would x be > >>>>> on the right of z? > >>>>> > >>>>> And does the line stay horizontal as one moves around earth? Which way > >>>>> is larger if the line ever goes vertical. And how does the "larger" work > >>>>> at antipodes? > >>>>> > >>>> Silly questions. > >>> In response to a silly definition. > >>>>>> It means > >>>>>> A y (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y). That says about all it > >>>>>> needs to, wouldn't you say? > >>>>> Not hardly. > >>>>> A y (y = x) v (y = z) v (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y) > >>>>> is a bit better but still insufficient. > >>>> True, I should have specified y<>x and y<>z. I guess it's usually done > >>>> using <= for this reason, eh? > >>>> > >>>>>>> > > That is not a definition, because it makes no sense. "The set of > >>>>>>> > > naturals > >>>>>>> > > is as large as every natural"? > >>>>>>> > > >>>>>>> > It is not larger than all naturals > >>>>>>> > >>>>>>> That is something completely different again. > >>>>>> It's not LARGER than every finite. > >>>>> Which natural(s) is it "not larger" than", in the sense of not being a > >>>>> proper superset of that natural or having that natural as a member? > >>>> ....11111 binary (all bit positions finite) > >>> Unless that string has only finitely many bit positions as well as only > >>> finite bit positions, it is not a natural at all, as it is then neither > >>> the first natural nor the successor of any natural, and every natural > >>> has to be one or the other. > >> It is the successor to ....11110. Duh. I've already proven that this is > >> a finite value, given that all bit positions are finite, and that > >> therefore no place in that string can achieve an infinite value, and > >> that any such number has predecessor and successor. The cute thing is > >> that the successor to ...1111 is 0, and that ...1111 is essentially -1. :) > > > > Does it not bother you that nobody else agrees with, or even > > understands, your proof? > > > > I find it disappointing, but not surprising, that you don't understand > such a simple proof, since it's contradictory to your education. Hmm, presumably also "contradictory to [the] education" of all mathematics professors too. Well, never mind: I'm sure we underestimate the degree to which you are unable to spot the implausibility of your working assumption, which must I suppose be that you are simply unimaginably brighter than, um, well everyone who's studied maths properly. > I do > find it annoying that you feel the right to disagree with it without > understanding it. If you feel there is a problem with the proof, please > state the logical error I made. Your logical errors have been pointed out endlessly. You are impervious to reason. Though to tell the truth the problem here is much simpler. We suspect you have some private (actually rather incoherent) notion of what the I-word and the F-word mean. You have never been able to explain, except circularly. > If the string is all finite bits, and > none of them ever can possibly achieve an infinite value, then how can > the string have an infinite value? There's nowhere in the string where > that can occur. It's that simple. Grok it. OK, grok this. Call a subset of the pofnats "endless" if when enumerated in natural order, one never gets to an end. Let us say that a set "achieves" endlessness when it has no end, i.e. is endless. Consider the initial segment Pn of the pofnats {1, 2, 3, ... n-2, n-1, n} where n is some pofnat (um, greater than 5). Plainly the set Pn is not endless, since it ends at n. Now consider the set X including all pofnats. For every pofnat m in the set X, n is a finite value (that's the 'f' in "pofnat"). Not a single one of the pofnats m in the set allows a set that ends at it to achieve endlessless. So how can the set X be endless? There's nowhere in the set X where endlessness can occur. It's that simple. Grok it. I can't see any essential difference between this argument and yours. (I.e. both appear to me to be obviously non-sequiturs with wrong conclusions.) What do you think? If your argument is valid and mine is invalid, can you show some principled distinction between them? ("Principled" excludes the "distinction" that you believe, however fervently, that one conclusion is right and the other wrong.) Or do you think my argument is correct too? So the entire endless set of pofnats is in fact not endless? Brian Chandler http://imaginatorium.org
From: Tony Orlow on 7 Sep 2006 13:52
stephen(a)nomail.com wrote: > Tony Orlow <tony(a)lightlink.com> wrote: >> stephen(a)nomail.com wrote: >>> Mike Kelly <mk4284(a)bris.ac.uk> wrote: >>> >>>> stephen(a)nomail.com wrote: >>>>> imaginatorium(a)despammed.com wrote: >>>>> >>>>>> Tony Orlow wrote: >>>>>> given that all bit positions are finite, and that >>>>>>> therefore no place in that string can achieve an infinite value, and >>>>>>> that any such number has predecessor and successor. The cute thing is >>>>>>> that the successor to ...1111 is 0, and that ...1111 is essentially -1. :) >>>>>> Hmm. So -1 is "essentially" a lot larger than 1, for example, whereas >>>>>> add 5 to both sides and you get the other thing. Well, that's faintly >>>>>> amusing ice pose. >>>>> Given that Tony apparently thinks that if you keep adding 1's, you >>>>> eventually get back to zero, I cannot understand why he was >>>>> so upset by the balls and vase problem. >>>>> >>>>> Stephen >>>> I was actually thinking of that problem in the car a week ago. Surely >>>> even if there are "infinite integers" to go on the balls, those balls >>>> get removed infinitesimally before Noon? The nth ball gets removed at >>>> the nth iteration, at time -(1/2) ^ n. Surely Tony would argue that >>>> this is valid in the infinite case. Oh well. >>> Tony's objection was that because you are always adding balls, >>> you can never get 0. But apparently his own math says that >>> if you keep adding 1, you eventually get, ..111111111111, and >>> if you add 1 to that, you get 0. So in Tony's world, if >>> you just add 1 ball at a time, and never remove any balls, >>> you can end up with 0 balls at noon. >>> >>> Stephen >>> > >> It's nice to see you are enjoying making fun of the number circle. >> People used to make fun of the Round Earth Theory too. What shape is our >> universe, and why? Have you ever asked yourself that question? I thought >> not. > > I'm making fun of you, not the number circle. Once again > you are ignoring an argument and responding with adolescent > philosophical maunderings. > > Here is a simple question for you: if you keep adding balls > to a pile, and never remove any, and you do this an "infinite > number of times", is it possible to end up with zero balls? > > Your math seems to say "yes". Do you agree with your math? > Or does your answer depend on what shape of the Universe > happens to be that day? > > Stephen The universe is always expanding at the speed of light. At any given moment, it wraps back upon itself, but given that we cannot travel as fast as it expands, we cannot circumnavigate it. We cannot make the full circle, and never can, given the relationship between space and time. While a snapshot of the continuum may lead us to the conclusion that adding forever gets us nothing, it only ever gets us farther away from our origin, as that continuum expands. Tony |