An uncountable countable set
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From: David R Tribble on 8 Sep 2006 23:12 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. > 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: David R Tribble on 8 Sep 2006 23:23 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.
From: David R Tribble on 8 Sep 2006 23:35 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? Are these infinite values connected (in the mathematical sense) to the finite values on the line? If not, how is it a "line"? 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). 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.) 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.
From: David R Tribble on 9 Sep 2006 00:17 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.
From: mueckenh on 9 Sep 2006 17:11
Dik T. Winter schrieb: > In article <1157574332.235885.113030(a)d34g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > Talking about 0.111... defined as: for all natural p digit p is 1, there > are no other digtits. And the list is the list of natural numbers. > > > > > Your "each" means in symbols of logic: "A = (for) all". > > > > The number is nothing than all of its digit positions. > > > > Therefore your statement is a self contradiction. > > > > > > Where? In logical terms (A meaning "for all" and E meaning "there is"): > > > (1) A{p = digit position} E{q = list item} {such that q indexes p} > > > > That is your definition. But what we can safely say is only: > > (1') A{p = digit position of list item} E{q = list item} {such that > > q indexes p} > > Why can we only say that? The definition of 0.111... is such that (1) > holds. Then it would be in the list. > If it does not hold you should be able to give an index position > such that it is false. It is not in the list, although we cannot give an index position which is responsible for that fact. All possible indexes are in the list. > > > > (2) A{p = digit position} E{q = list item} {such that q covers p} > > > > If your definition could be satisfied, the construction of the list > > would imply this, yes. What we can safely say, however, is only: > > (2') A{p = digit position of list item} E{q = list item} {such that > > q covers p} > > The same here. But apparently you think my definition of 0.111... can not > be satisfied. Why not? Because it is not in the list which, by definition, contains all numbers which can be indexed and which can index. > > > Now we may ask: Is it possible that a list item indexes or covers other > > digits than those which are indexed or covered by list items? The > > answer is: no. > > You are repeating yourself again, and I have already agreed to that. Why > then repeat again and again? Because you assert that all digits of 0.111... can be indexed, which is wrong. Indexing and covering "by all list numbers" is equivalent. Both is true or both is false. > > > And we may further ask: How can 0.111... be distinguished from all list > > items? The answer is the only one possible: By digit positions occupied > > by 1's which are neither indexed nor covered by list items. Why? > > This is false. 0.111... is distinguished from all list items in that it > does not terminate. "Not to terminate" is not a property which can serve to distinguish a number from others, because we can never observe the end (because it does not exist) neither the non-end (because it is nothing but a negation of an unobservable property). What can serve to distinguish two numbers in unary representation is a 1 at a digit position. All numbers which can be distinguished by such an 1 are in the infinite list. Numbers are not processes. > Consider them as decimal fractions. The list consists > of the numbers (1-10^(-n))/9, 0.111... is 1/9. There is no n in the list (and in N) which yields 1/9. Therefore 1/9 = 0.111... cannot completely be indexed. > > > because 0.111... does not have any other component. > > It does not have any other component, but also it does not terminate, > in contrast to all numbers on the list. This cannot occur other than by more digits 1 than available by list numbers. > > > So we may finally ask: Is it possible that all digit positions of > > 0.111... are indexed or covered by list items? > > The answer is obvious: no. > > Every digit position can be indexed and covered. Otherwise state which > digit position can not be indexed, use my definition: > Talking about 0.111... defined as: for every natural p digit p is 1, > there are no other digtits. > but there is no natural that covers all digits. > > And if you think that that definition is wrong, please *prove* that. It has been proven by showing that 0.111... does not belong to the set of numbes which can be indexed completely. Regards, WM |