An uncountable countable set
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From: Tony Orlow on 7 Sep 2006 12:37 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. 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.
From: Tony Orlow on 7 Sep 2006 12:39 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? 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. > >> 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.
From: Tony Orlow on 7 Sep 2006 12:41 Dik T. Winter wrote: > In article <4500215f(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > Mike Kelly wrote: > ... > > >> ....11111 binary (all bit positions finite) > > > > > > That isn't a natural number, Tony. > ... > > 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. > > Why not use the proper name such numbers already have? 2-adics. No reason, except that 2-adics could possibly have infinite bit positions. I wanted to make sure that detail was not lost, because it's crucial to the proof of finiteness.
From: Tony Orlow on 7 Sep 2006 12:46 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.
From: stephen on 7 Sep 2006 13:06
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 |