An uncountable countable set
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From: Virgil on 13 Jul 2006 15:40 In article <1152791376.515947.207400(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152718980.956031.285220(a)75g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > > > > > > > > > > You are using "to index" simultaneously for two different meanings. The > > > > first is "to index a specific position" the second is undefined. What > > > > does "0.1111 is indexed by the list number 4 = 0.1111" exaclty mean? > > > > > > The unary number n indexes every position 1,2,3,...,n. > > > > Actually a unary numeral, or any other numeral for a natural number, can > > only index one position. The MEMBERS of a natural, however expressed, > > can index up to that natural. The members of N can index all natural > > positions. > > Why then is 0.111... not in the sequence 0.1, 0.11, 0.111, ...? Because ordinals are not members of themselves. > 0.111... cannot be indexed by the members of any natural. Does 0.111... > have more 1's than can be indexed by any natural? Or what prevents it > from being indexed by any of the naturals? What cannot be indexed by any member of N is N itself, but N itself can be "indexed" by the members of Succ(N) = N \union {N}. Each ordinal has a successor, even limit ordinals like N, and any ordinal can be indexed by the members of its successor.
From: Dik T. Winter on 13 Jul 2006 11:02 In article <1152786578.943557.52630(a)s13g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > Cantor himself generalized his proof (in the same paper) to all > well-defined sets like the real numbers (= Linearcontinuum) and > functions which only can take the values 0 and 1. Yes, that is easily transformed into a proof that the set of subsets of the reals has grater cardinality than the reals themselves. > (Therefore I said one > could safely interpret his list as a list of binary numbers. Zermelo > only remarked the problem with double representation of binary > fractions.) Zermelo *transformed* his first proof to a proof with binary numbers. In Cantor's proof you can *not* safely interpret his list as a list of binary numbers, because in his list there are no two elements that are equal, while if you interprete them as binary numbers, there are. > Cantor concluded: dassF die Maechtigkeit wohldefinierter > Mannigfaltigkeiten kein Maximum haben oder, was dasselbe ist, dass > jeder gegebenen Mannigfaltigkeit L eine andere M an die Seite gestellt > werden kann, welche von staerkerer Maechtigkeit ist als L. For > non-German-readers: He recognized that cardinalities do not have a > maximum. Indeed. That was the second part of the proof which can easily be transformed from a proof that the powerset of the reals has greater cardinality than the reals themselves to the proof that for any set S, |P(S)| > |S|. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 13 Jul 2006 11:13 In article <1152787358.727132.194480(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > I translate (not needed for Franziska, but needed for others): > > When in a well-ordered set any two elements m and m' change place in > > the common order, the order-type will not change, so also the ordinal > > number will not change. From that it follows that those transformatio= > ns > > of a well-ordered set leave the ordinal number unchanged, that can be > > written as a finite or infinite sequence of transpositions of two > > elements, that is, all such changes that emerge through permutations > > of elements. > > > > I see a problem here: > > > > It is not clear what the meaning is of the words "transformations" > > ("Umformungen") and "permutations" ("Permutation"). I think, it is clear > > that it does not include *any* re-ordering, although many of them can be > > written as an infinite sequence of transpositions, while some those *do* > > change the ordinal number. > > These words are uninteresting. Transpositions or exchanges of two > elements are clearly defined. Infinitely many are admitted by Cantor. This is wrong. Not any sequence of transpositions is permitted. What *is* permitted is *transformations that can be written as a sequence of transpositions*. Because if you admit *any* sequence of transpositions you may destroy the order-type. > > The reordering of the naturals (0, 1, 2, ...) to (1, 2, 3, ..., 0) can > > be written as an infinite sequence of transpositions, but it changes the > > ordinal number. But is it a permutation? > > Unimportant for Cantor's statement. It *is* important. Because if that falls under Cantor's definition of "transformation" and "permutation", Cantor's statement is trivially false. > There is no transposition which > could change the ordinal number. Only the infinite process does what > cannot be done by any finite set of transpositions. This shows that > infinite processes and infinity is doubtful. My example with |Q shows > that infinity is impossible. Eh? You assume that Cantor's conclusion is true for your sequence of transpositions. I state that Cantor's conclusion is false for your sequence of statements. > > On the other hand, there are many infinite sequences of transpositions > > that do not change the ordinal number. Consider a well-ordered set of > > type w * w. Interchange the first two elements of each maximal subset of > > order type w. > > > > Anyhow, either WM's interpretation is wrong, and so his conclusion is > > wrong, or WM's interpretation is right and Cantor's statement is wrong, > > and so (again) WM's conclusion is wrong. > > > > I think the interpretation by WM is wrong. > > I know your interpretation is wrong. What part of my interpretation is wrong? > But that does not at all affect my proof, because, as Virgil, > emphasized frequently, there is no transposition which changes the > well-order to normal order while maintaining the initial enumeration (= > well-order). Would infinitely many transpositions be possible, this > would necessarily happen. No. There would still not be a single transposition at which well-order changes to non well-order. As in lim{n -> oo} 1/n, there is not a single natural at which 1/n becomes 0. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 13 Jul 2006 11:15 In article <1152787731.201543.289190(a)35g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > Pray tell me how. Somewhere else you stated that the representation > > 0.111... did not exist. What are you arguing? Either the representation > > 0.111... does exist or not. And if it does exist the definitions of the > > mathematical operations are not sufficient to give a meaning to it. > > > > Anyhow, how do you show that 1.000... - 0.999... = 0 with the definitions > > you are using? > > In "usual" mathematics we have 10 * 0.999... = 9.999... and from that > we get easily 0.999... = 1. What "usual" mathematics are you using? > I argue that 0.111... does not exist. When I will have succeded, also > 0.999... will be abolished. But that has not yet been generally > accepted. OK? I argue that 0.111... does exist as it is defined as sum{n = 1, oo} 10^(-n). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 13 Jul 2006 11:23
In article <1152788625.612288.116290(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > The axiom says "there is an infinite set". It does not say that > > > 0.111... does belong to that set, in particular because all numbers > > > which in fact do belong to the set are different from 0.111... . > > > > You are using the word set with two different meanings in the same sentence. > > The set of naturals shold be the same set as the set of indices of > 0.111... . But it cannot. But it is. > > *And* you do not answer my primary comment. There is a specific definition > > for a notation like 0.111... Without such a definition it is just a string > > of eight symbols. > > There is a specific definition that the set of all naturals has a > cardinal number larger than any element of the set. I would think there is a proof of such. > This definiion is > wrong. I show this by the fact that 0.111... is not in the set of unary > reprsetations of all natural numbers. aleph_0, the number of 1's in > 0.111... is *not* the cardinal number of |N. Proof, please. But you still refrain from answering my primary comment. > > Again, not an answer to a specific question. I wrote: > > > For all p there is an n such that An[p] = K[p]; > > and you wrote that is wrong. What is wrong about that statement? > > Te assumption that all digits of 0.111... can be inexed by natural > numbers while 0.111... has the property to to be outside of the list. Still waiting for a proof. > > > But as K is not in the list > > > of natural numbers, it must differ from the numbers in the list. And > > > this difference cannot be accomplished other than by K having more > > > digits than any number in the list. Again, this is impossible. Hence we > > > have a contradiction. > > > > Pray *prove* why that is impossible. You always only state it but provide > > not prove. Each An has finitely many digits. K has infinitely many digits. > > If you disagree with the second you disagree with the axiom of infinity, > > but that is philosophical. > > K cannot have infinitely many digits if all can be indexed by finite > numbers. Infinite is larger than any finite number. Therefore every > list number taken will not be capably of indexing K. > 0.111... - 0.111...1 = 0.000...0111... You mean covering here. And you are right, K is not in the list. I have never argued that it is. Still, what is the problem? > > > > The last is true. But it is also true that for all p there is an An. > > > > Or please exhibit a p for which it is not wrong. > > > > > > This equation > > > 0.111... - 0.111...1 = 0.000...0111... > > > says that for any An = 0.111...1 there is a p = n+1 which cannot be > > > indexed. > > > > This makes absolutely no ense to me. And again is no answer to the > > question I posed. > > Whatever p you choose, there is always a place in 0.111... which cannot > be indexed. Therefore there is a place which cannot be indexed in > principle. What do you mean with "cannot be indexed"? > > Again you fail to answer a question, Your implication was: > > > > > If you think the sentence "all positions of K = 0.111...are indexed by > > > > > list numbers" is not equivalent to the sentence "K > > > > > is in the list", then you seem to imply that more than one list number > > > > > is required to index the digits of K. > > Care to explain it? > > You know that one list number is not capable of indexing 0.111... , bu > you assume that 0.111... can be indexed completely. You use the word "indexed" in a strange way. You mean "covered". I never assume that 0.111... can be "covered" by some An. Nevertheless, *each digit position can be indexed by some An". > > Well, to index can be stated as to cover. But not the other way around. > > Therefore never more than one number is required for indexing purposes. Covering? Well, in the finite. What holds in the finite does not necessarily hold in the infinite. > > > > > 0.111... *- (0.1 + 0.11 + 0.111 + ...) = 0.000... > > > > > > > > I can make no sense of the second term. What do you *mean* with > > > > (0.1 + 0.11 + 0.111 + ...) > > > > > > The sum of all list numbers by digit, as shown in the following > > > example: > > > > > > 0.1 > > > +0.11 > > > +0.111 > > > _____ > > > =0.321 > > > > Ok. What about "+ ..."? > > Definition: 1+1+1+... := 1 That makes absolutely no sense. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |