An uncountable countable set
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From: Mike Kelly on 7 Sep 2006 14:47 Tony Orlow wrote: > Mike Kelly wrote: > > Tony Orlow wrote: > >> Mike Kelly wrote: > >>> Tony Orlow 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. > >>>> > >>>>>> 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) > >>> That isn't a natural number, Tony. > >>> > >> Are you sure? Pay close attention. > >> > >> For any finite bit position n, it and all predecessors can only sum to a > >> finite bit string value of 2^(n+1)-1. > > > > OK. Any string 111....1111 with a finite number of 1s represents a > > natural in binary. > > > >> Since there are only finite bit positions in the string, it can never achieve any infinite value at any position in the unending string of bits. > > > > OK. Any string 111....1111 with a finite number of 1s represents a > > natural in binary. > > > >> Therefore the value must be finite. > > > > Why? You're supposed to be *demonstrating* that the string represents a > > value, but you're *assuming* it instead. > > > > You've shown that any finite bit string of all 1s represents a finite > > natural number. And concluded from this that an infinite string of 1s > > represents a finite natural number. Why? Total non sequitur. > > At which bit does this string attain an infinite value? It doesn't. I never said it does so please STOP asking that question. I say that 111..... doesn't represent a (natural) value at all. > All bits are > finite, and have a finite number of predecessors. There is none where > the string can ever become infinite in value, or even extent. Yeah, finite strings are finite and represent naturals, got it. Why does that imply that an infinite string represents a natural? > >> Furthermore, since any such number does have a predecessor and > >> successor, in this case ....1110 and ...0000, respectively, it fits in > >> the successorship model. The only concept this breaks is that 0 is now a > >> successor as well, creating an infinite ring of successorship. Other > >> than that, it works as a natural, and in fact, this is the way signed > >> integers work in your very computer. > >> > >> So, while ...111 may not be considered a standard natural, I see no > >> reason why it should not be considered, say, an extended natural. > > > > And you think this is a better model of "natural numbers" than standard > > ones? You think it accords better with the intuitive picture people > > have of what "counting numbers" are? > > > > Yes, in the infinite realm, it does. It's precisely the binary signed > integer in the limit as the number of bits approaches oo. I laughed out loud here. You think anybody but you on the PLANET would think that this is relevant to the intuitive, naive idea of what "numbers" are? -- mike.
From: Mike Kelly on 7 Sep 2006 14:49 Tony Orlow wrote: > 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. Bullshit. There is no bijection between the naturals and the set of all binary strings. -- mike.
From: Mike Kelly on 7 Sep 2006 14:52 Tony Orlow wrote: > 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, you're chanelling Finlayson. Are you aware that your post is A) Incoherent babble B) Completely unrelated to the post it responds to ? -- mike.
From: stephen on 7 Sep 2006 15:13 Mike Kelly <mk4284(a)bris.ac.uk> wrote: > Tony Orlow wrote: >> Mike Kelly wrote: >> > >> > 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. 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. 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. > 1) A finite string of 1s represents a (finite) natural number. > 2) An infinite string of 1s represents a (finite) natural number. > 1) doesn't imply 2). Tony's argument seems to be a upside-down inside-out version of one of Zeno's paradoxes. ....111111 (in binary) = 1 + 2 + 4 + 8 + ... All the numbers on the right are finite, so the sum must be finite. Apparently a sum can only be infinite if one of the terms in the sum is infinite. An infinite sum of finite terms cannot possibly be infinite in Tony's mind. Typically people have problem with an infinite sum of finite terms being finite, aka Zeno's paradox, which is counter intuitive at first glance. However insisting that an infinite sum of finite terms must be finite is just weird. Stephen
From: Virgil on 7 Sep 2006 15:42
In article <450018b5(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <44fe1669(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > >> I have yet to see any valid > >> counterexample to my rules regarding inductive proof in the infinite > >> case. Where an equality between expressions is proven for all n, it is > >> valid for infinite n. > > > > What standard rule of induction or standard definition does TO claim > > justifies these claims? > > > > None contradict it, except for transfinitology. If one defines, as is usual, the naturals as being the smallest (by inclusion) of inductive sets, then axiom of infinity and axiom schema of separation prohibit it. > > > > > > >> Where an inequality is proven for all n greater > >> than some finite m, and the difference upon which the inequality is > >> based does not have a limit of 0 as n->oo, it holds also in the infinite > >> case. > > > > What standard rule of induction or standard definition does TO claim > > justifies these claims? > > None contradict it. There is nothing in the Peano axioms which precludes > infinite natural values. If one defines, as is usual, the naturals as being the smallest (by inclusion) of inductive sets, then the Peano axioms specifically exclude any but "finite" naturals. > > > > > If neither can be justified by any standard rule, then TO must present > > his system in its entirety to get it in, as it is in conflict with every > > standard system. > > It is a suggested rule, a simple extension to standard induction, which > conflicts with transfinitology, but not with anything else. That is TO's opinion, and we have all seen how little his opinions are worth on matters of what the axiom systems allow and do not allow. |