An uncountable countable set
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From: mueckenh on 21 Jul 2006 06:47 Franziska Neugebauer schrieb: > Dear Wolfgang, > > I would appreciate if you not cut away that parts of my previous posts > you are trying to argue against. It may also be helpful if you read > before cutting. Dear Franziska, I cut what I regard as unnecessary and not helpful in order to make reading more comfortable. Your counting starting from zero is superfluous and does not support your arguing. > > 2 The *-sum of the sequence does not necessarily "inherit" the > properties from their individual sequence members. If this > "inherentance" is kind of a "general principle" you should refer to a > recognized source. > > But there is even a finite counterexample: > > 101 > 011 > 110 > ---- *-sum > 111 > > The property "does not contain at least one 0" is not inherited by > the *-sum. So the "inheritance principle" is not so obvious. But it is obvious for *all* *finite* *unary* number. And only finite unary numbers are involved in the second list below: The *- sum of 0.1 0.11 0.111 .... 0.111... is 0.111... i.e. the result is a number of the list. The *- sum of 0.1 0.11 0.111 .... is 0.111... The result is allegedly not in the list. This is a contradiction, because by tertium non datur either the first case or the second must be true (if aleph_0 is a number with respect to trichotomy with natural numbers). Regards, WM
From: mueckenh on 21 Jul 2006 06:52 Franziska Neugebauer schrieb: > What is false unter the abovementioned presumption omega = { 0, 1, > 2, ... }? > I am not willing to discuss this strange counting. 0.1 means 1 and the first (1) digit after the point is a 1. 0.11 means 2 and the first (1) and the second (2) digit after the point are 1's. > > When it is not important "what I mean", why do you discuss with me? This > is rather insulting. This remark concerned only the rather strange set-theoretic counting. Other opinions of yours are often welcome, as you know. Regards, WM
From: mueckenh on 21 Jul 2006 06:55 Dik T. Winter wrote: You again refrain from answering questions. The definition of 0.999... is as follows: 0.999... = lim{n -> oo} sum{k = 1 .. n} 10^(-k) You said that definition is silly. What is silly about it? _______________ WM: Sorry, could you help me with "silly". My find function shows with "silly" only the sentence: "It is only in modern set theory that such silly things as the rejection of the binary tree occur." _______________ Dik T. Winter wrote: what indices in the limit above are undefined? ________________________- WM: Those which cannot be enumerated by natural numbers. ________________ Dik T. Winter wrote: Because, and I state it again, there is a difference between sequential processes and simultaneous processes. ________________ WM: I know that in the infinite binary tree *every* split is realized by the presence of two edges and would be impossible without them. | o /\ Nothing can be more simultaneous than this knowledge which is present without considering any sequential process and which concerns the whole tree. The tree "in its length" is nothing else than a Cantor-List. The only difference is that the tree guarantees the presence of every "diagonal number". _______________ Dik T. Winter wrote: Four your count of edges and paths you need limits, but those limits do not exist. _________________ WM: Oh, the limit of 1 + 1/2 + 1/4 + ... = 2 is not existing? Strange. There is no question that every path can be interpreted as the representation of a real number. And now "those limits do not exist"? Very strange. _____________ Dik T. Winter wrote: Again, you need something like a limit here, because adding nodes is a sequential process. __________________ WM: Adding lines in Cantor's list is not a sequential process? They are given by the "higher Being which rules matheology". But adding levels of the binary tree is a sequential process. Are you really in earnest about that question? By the way: Notes are not added, they are there. Take the nodes as digits. There is no question that every path can be interpreted as the representation of a real number. ______________________ Dik T. Winter wrote: Also, it is true for each natural that it cannot index all 1's of 0.1111. That means that for all integers it is impossible to index all 1's of 0.1111. On the other hand, with the *set* if integers it is possible to index all 1's of 0.1111, and also of 0.111... . That is, when with "to index" we mean "point to an individual element" as in K[p]. The reason is that each natural can index one and only one digit. On the other hand, you on occasion use index to mean something completely different. For that I could use the word "cover" where A covers B if in all positions where B has a 1, A also has a 1. Now in the finite case we find that the digits of 0.1111 are covered by all naturals greater than or equal to 4. But the digits of 0.111... are not covered with any natural. ________________________ WM: Why then do you expect that all digits are indexed "in the infinite case"? The corresponding definition is of the same quality as "there are 10 natural numbers between 0 and 1". There are only finite numbers. ________________________ Dik T. Winter wrote: But nobody has stated that that is possible, it is only you who *claims* that that should be possible, without giving any reasonable proof of it. ________________________ WM: There is no difference! If a natural indexes the digit n, then it covers all digits m =< n. This is guaranteed by the construction of the list of unary representations - for *all finite numbers*!!! How can you believe you could make someone believe this situation would change? Therefore I say sometimes each natural n can index all digits m =< which it covers. Whether you call this to cover or to index is unimportant. Both definitions imply each othe as long as *finite numbers* are concerned _____________________________ > It is true for each natural, that it cannot index all 1's of 0.111... . Dik T. Winter wrote: Again, you mean cover here. ________________________ WM: There is no difference between covering the digits 1 to n or to index the n-th digit, because of the construction of the naturals. AND WE HAVE ONLY NATURALS. ___________________________ Dik T. Winter wrote: I would like to ask you to refrain from such confusing use of words. And that is indeed true. But that does *not* mean that For all p there is an n such that An[p] = K[p] is false. That is true, take n = p. _____________________ WM: You imply that it is possible to take n = p, but that is impossible for p being not a natural number like aleph_0. __________________________________ > > If stepwise exhaustion is impossible, your reordering could be completed. > > But you can with your *+ operation give the proof in one sweep, but you > > failed to answer to my question: "what happens at infinity?". > > 1) There is o infinity. Ever Transposition has a natural number. Dik T. Winter wrote: Sorry, that was loosely speaking. What happens if you add *all* of them together? You still fail to tell what happens in that case. (It is quite easy to tell it, and you need a limit, but you are unwilling, or perhaps unable to tell it. ___________________ WM: I defined: The +* sum is 1 if the + sum is 1 or larger. This includes the infinite case, or would you insist that infinitely many 1's are less than one 1? ___________________ Dik T. Winter wrote: When I ask you what the meaning of *+ ... is you do not give a reasonable answer.) Answer: for each digit position, when there is an Ap with a one in that digit position, the identical digit position in K is also a 1. And that answers the question. The result is: for all p K[p] = 1 which we notate as K = 0.111..., because: for all p there is an n such that An[p] = 1. _____________________ WM: Yes. And as there is no difference between indexing digit number n and covering all digits m =< n, there is, according to your arguing, a natural number which covers all digits of K. ____________________ > > You can exhaust an infinite set if you take out all elements at once. > > You may think that would be possible, but it is not. You always need > the belief expressed by the three "..." Dik T. Winter wrote:
From: Franziska Neugebauer on 21 Jul 2006 07:33 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > >> Dear Wolfgang, >> >> I would appreciate if you not cut away that parts of my previous >> posts you are trying to argue against. It may also be helpful if you >> read before cutting. > > Dear Franziska, I cut what I regard as unnecessary and not helpful in > order to make reading more comfortable. lame excuse > Your counting starting from zero is superfluous and does not support > your arguing. In set theory one usually counts from zero. As pointed out it is a quite simple task to renumber the naturals if you want to insist on counting from 1. >> 2 The *-sum of the sequence does not necessarily "inherit" the >> properties from their individual sequence members. If this >> "inherentance" is kind of a "general principle" you should refer to a >> recognized source. >> >> But there is even a finite counterexample: >> >> 101 >> 011 >> 110 >> ---- *-sum >> 111 >> >> The property "does not contain at least one 0" is not inherited by >> the *-sum. So the "inheritance principle" is not so obvious. > > But it is obvious for *all* *finite* *unary* number. Can't see that. If it is so obvious then please show (prove!) it. I consider my 3-item list a valid counterexample to an "inheritance principle". > And only finite unary numbers are involved in the second list below: > > The *- sum of The following notation I will call "A". > 0.1 > 0.11 > 0.111 > ... > 0.111... > > is > > 0.111... > > i.e. the result is a number of the list. You want to put 0.111... into the sequence. On which position j e N? What does the sequence-'...' in your context (0.1; 0.11; 0.111; ...; 0.111...) ^^^ mean? > The *- sum of > I will call this "B". > 0.1 > 0.11 > 0.111 > ... > > is > > 0.111... > > The result is allegedly not in the list. We did not put it into the list, so it's not there. > This is a contradiction, because by tertium non datur either the first > case or the second must be true (if aleph_0 is a number with respect > to trichotomy with natural numbers). First of all you produced two notations A def= (0.1; 0.11; 0.111; ...; 0.111...) and B def= (0.1; 0.11; 0.111; ...) Notation A is (not yet) defined since the first occurence of '...' has (not yet) a defined meaning. Fomally different is notation B which simply denotes the original "list". *Assumed* that A and B denote different sequences the proposition 0.111... is in *A* & 0.111... is not in *B* is not a contradiction and does not violate the tertium non datur principle. A contradiction is a proposition of the form p & ~p Hence you have (not yet) shown a contradiction as has been pointed out all too often. F. N. -- xyz
From: mueckenh on 21 Jul 2006 07:42
Dik T. Winter schrieb: > What your > example shows is that an infinite sequence of transpositions can destroy > well-orderedness. What is the difference? The only conclusion is that > Cantor's statement is incorrect or that the reading of Cantor's statement > is incorrect. Take your pick. Your hinting at any doubt about the clear meaning of Cantor's statement is outrageous. >I have found already a few instances where > Cantor's statements are in conflict with modern set theory. There are many. See for instance for a very simple one: "On Cantor's proof of continuity-preserving manifolds" on http://www.fh-augsburg.de/~mueckenh/ > And I am not > surprised. He found the building blocks but was not entirely sure about > the way to go. His articles were research in progress, as it happens. And now it happes that all his way turns out as a deadlock. > > And now for a debunking of a myth. Nowhere (at least I could not find > any place) has Cantor used the diagonal argument to show that the reals > are not countable. His proof about the reals shows that a complete, > densely ordered set is not countable. That one does not use the diagonal > argument at all. So far we agreed recently. > His diagonal proof shows that the set of infinite > sequences of two symbols is not countable, and as an extension, that the > powerset of a set has cardinality strictly larger than the cardinality > of the original set. > > So all your arguments about 0.999... = 1.000... are *not* directed against > Cantor. Wrong. My arguing is directed against the complete existence of the set of all naturals, the set of all digits of a real number, the actual infinity, the first transfinite number. That all is purest Cantor. Regards, WM |