An uncountable countable set
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From: Tony Orlow on 12 Sep 2006 11:35 David R Tribble wrote: > David R Tribble wrote: >>> 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. > > Tony Orlow wrote: >>> 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. > > Mike Kelly wrote: >> Bullshit. There is no bijection between the naturals and the set of all >> binary strings. > > Finite binary strings (only). Tony thinks that the set of all possible > finite bitstrings is the powerset of the naturals (the finite bit > positions or indices), so there are more (finite) bitstrings than > bit positions. > > Yet he admits that there is a bijection between the two. He > concludes that this demonstrates a flaw or inconsistency in set > theory. > > In a similar vein, he thinks he has a bijection between the naturals > (as binary strings) and all the subsets of N. Likewise, he thinks that > this proves that set theory is flawed. > > Obviously he's confused between powerset cardinalities (|P(S)| = 2^|S|) > and binary bitstrings (bit n = 2^n), probably because of the similarity > in notation. > You don't have it quite right. While there are countably infinite bit strings in the power set of the naturals which don't correspond to any standard finite natural, I contend that these strings cannot represent infinite values, given that all bit positions are finite, and that they qualify as finite naturals, since each has a successor and precedessor, generated the same way as for standard finite naturals. So, when these strings are included as finite naturals, indeed, that set of strings bijects with the power set of the naturals. Tony
From: Tony Orlow on 12 Sep 2006 11:40 David R Tribble wrote: > 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. > > David R Tribble wrote: >>> 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. > > Tony Orlow wrote: >> Not for the hypernaturals. > > Since I didn't say anything about hypernaturals, I take it that you > agree with what I actually did say. > > > David R Tribble wrote: >>> 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.) > > Tony Orlow wrote: >> Yes, I am including infinite values on the number line, since it's >> "infinitely long". > > Yet another thing you have to define or prove. Where do these > "infinite values" appear on your real number line? Further from 0 than any finite number. > > Are these infinite values connected (in the mathematical sense) to > the finite values on the line? If not, how is it a "line"? > Yes, they are on the line, with all finite values between them and 0. > > Tony Orlow wrote: >>> 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. > > David R Tribble wrote: >>> 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.) > > Tony Orlow wrote: >> 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. > > Without proving the second part about denseness. So we can > conclude that you're assuming that there are only a finite number > of reals in any interval (as I suspected). No, Big'un is an infinite unit. Given denseness derived from the axiom of internal infinity, ExeR EzeR x<z -> EyeR x<y<z, there are an infinite number of reals in the interval. > > Unless you have a definition or axiom about denseness that > is consistent with your theorem above? (The standard accepted > definition of "dense" is not, of course.) > I don't understand why you assume Big'un is finite, when it's an infinite unit. > > David R Tribble wrote: >>> Of course, if it's false in your system, your system cannot be >>> compatible with standard arithmetic. You see why, don't you? > > Tony Orlow wrote: >> Of course. > > I don't think you do. > You're wrong.
From: Tony Orlow on 12 Sep 2006 11:43 David R Tribble wrote: > 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. >>> >>> 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. > > David R Tribble wrote: >>> 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. > > Tony Orlow wrote: >>> In your concoction, the naturals are power set to nothing. > > David R Tribble wrote: >>> 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. > > Tony Orlow wrote: >> That's my point. You can biject the naturals with the bit posiitons, and >> also with the strings, > > That part is correct. It proves that they are the same set. > >> but the set of strings is the power set of the bit positions. > > That part is wrong. The bijection proves they are the same set, > so one cannot possibly be the power set of the other. > >> Therefore you have a bijection between a set and its >> power set, and no cardinality which is appropriate for the set of bit >> positions. > > You are obviously missing something. > What I left out was that the power set includes countably infinite subsets of N, and such countably infinite bit strings are not considered standard naturals. However, they represent finite whole values and all have successors and predecessors, and so are naturals in that nonstandard sense, as long as all bit positions are finite. Tony
From: Tony Orlow on 12 Sep 2006 11:53 David R Tribble wrote: > Tony Orlow wrote: >>> ....11111 binary (all bit positions finite) > > Mike Kelly wrote: >>> 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. > > Tony Orlow wrote: >>> 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. :) > > Mike Kelly wrote: >>> Does it not bother you that nobody else agrees with, or even >>> understands, your proof? > > Tony Orlow wrote: >> 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. > > As I posted earlier, it leads to the conclusion that 0 > 0. > True, considering the number circle breaks the transitive nature of >, except that it's only one way of looking at the number line. In computer science, ....1111 is either -1 or the greatest possible value for the register, depending on whether one is using signed or unsigned integers. When we look at the number circle, we are assuming that one can never make the transition from the largest positive to the largest negative, since there is no end to reach. However, the number line can be viewed as the number circle in the limit, as the number of bits increases without bound. Whether ...1111 is viewed as the largest positive finite integer or the smallest negative finite integer, its successor is either the smallest infinite or 0. In computer science, the carry is lost, and the successor is ...000. That's an overflowed for unsigned integers, but not for signed integers. SO, it depends on the interpretation of the bit string. > >> 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. > > Then since an infinite string of non-zero bits can only yield a > finite value, you can never have an infinite value, can you? > Where do you get your infinite values for your "infinite cases" > then? > From the dense set of reals in the unit interval. That's my unit infinity. Once we allow that value to exist, we can apply N=S^L to see that we need log2(Big'un) bits to represent that value. Any such finite formula on Big'un will produce a specific infinity or infinitesimal.
From: Tony Orlow on 12 Sep 2006 12:10
David R Tribble wrote: > Tony Orlow wrote: >>> ....11111 binary (all bit positions finite) > > Mike Kelly wrote: >>> That isn't a natural number, Tony. > > Tony Orlow wrote: >> 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. 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. Therefore the value must be finite. > > Breathtaking. You have the annoying habit of ignoring the > completely obvious when it suits you. If ...111 is a binary number > with an infinite number of digits, why is it a finite value? > Shouldn't an infinite value have an infinite number of digits? > Yes, it should, and Wolfgang's and my position is that N is unbounded but finite, given that it only contains finite values, and only one per unit of value range. Since the value range is finite, given that no two naturals are infinitely different, and given that only a finite number of elements can be fit sparsely in a finite range, the set is finite, even though it's unbounded and infinite by the Dedekind definition. At which bit position can the string achieve an infinite value? None that exists in that string. > >> 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, ... > > If ...111 is a finite value, then there must be another finite value > that is larger than it. You claim that this is 0, but that gives you > a contradiction: 0 = ...111+1 > ...111 > 0, or 0 > 0. > (We can now use this to prove that any finite number is larger than > itself: k = 0+k = (...111+1)+k > ...111+k > 0+k = k, so k > k.) > > This should be a rather big clue that something's wrong with your > concept, and that it does not, in fact, "work as a natural". > If 0 is to be the successor to ....1111, then we are working in a signed system, and ...1111=-1. In that case, we cannot express the largest finite, since it would be a string of all 1's with a 0 in the most significant bit, which does not exist. Likewise, we cannot expess the infinite string of 0's with a 1 in the most significant bit, the largest negative. We can only count up and down from 0, with all standard positive integers having an unending string of leftward 0's at some point, and all standard negative integers having an unending leftward string of 1's at some point. In this system, one cannot say whether a repeating string of 1's and 0's, such as ...011011011 is negative or positive, unless you stipulate that it "terminate" at the end of a full cycle of repetition. 0 doesn't have to be successor to ...1111, and it can't be, if we are working with unsigned numbers. In that case ...1111 has no successor which can be expressed. > >> and in fact, this is the way signed >> integers work in your very computer. > > Computers only deal with fixed-length finite numbers, > modulo 2^B for some word width B. That does not provide a > very useful model for the entire set of naturals. > It can when considered in the limit. > >> So, while ...111 may not be considered a standard natural, I see no >> reason why it should not be considered, say, an extended natural. > > Which presumably requires some extended Peano axioms in order > to exist? > No, it just requires a loosening of the requirement that we only look at the minimal set satisfying those axioms. I don't see that it contradicts them as they stand. Extensions of the naturals such as this also fit into that model. Tony |