An uncountable countable set
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From: mueckenh on 23 Jul 2006 13:15 Dik T. Winter schrieb: > In article <1153478871.303029.5360(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > 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... > > 0.111... is not a finite unary number. Rather, I would say it is not a > unary number at all. 0.111.. is not a terminating rational number. Rather I would say it is not (a representation of) a rational number at all. Regards, WM
From: mueckenh on 23 Jul 2006 13:27 Dik T. Winter schrieb: > [reformatting to standard discussion style] > > In article <1153479321.840071.258130(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > 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? > > > > 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." > > You actually said that it made no sense. But I will try to find the > correct quote. Not necessary. I confirm: It makes no sense because omega makes no sense. But may also be that I argued with respect to the finite number of protons in the universe, which prohibits that natural numbers = indexes with 10^100 irregular digits can be written down or noted in any form. I am not sure, because I could not find the original quote. > > 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. > > So, I can perform the 102349176-th split without knowledge about previous > splits? Or was that not the 102349176-th split? When you do your splitting > you are performing a sequential process. Further splits depend on earlier > splits. There are no splits to perform in order to get the mapping between edges and paths. The knowledge about which path will represent which number is irrelevant for this purpose. We count only the number of paths. > > > Dik T. Winter wrote: > > > Four your count of edges and > > > paths you need limits, but those limits do not exist. > > > > Oh, the limit of 1 + 1/2 + 1/4 + ... = 2 is not existing? Strange. > > Is that the count of edges or the count of paths? It is the count of edges per path. The complete shares of two edges are mapped on every path. This is a high shareholder value. But the shares are scattered. > > > 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. > > You are confused. The count of paths have no limit. But if you wish > to interprete the count of paths as a real number, be my guest, and > tell me how to represent the count of paths as a real number. But > whatever, if you were ever to complete your picture, both the number > of edges and the number of paths would become uncountable (you did > confuse me a bit by first talking about nodes and now about edges). > It is the number of nodes that remains countable. The number of edges > after adding the n-th splitting is 2n+1, the number of paths is 2n, and > the number of nodes is n. ? In the tree the number of edges is the number of nodes (except for the first one) because every node (except the first one) is the end of an edge. > > > Again, you need something like a limit here, because adding nodes is a > > > sequential process. > > > > Adding lines in Cantor's list is not a sequential process? > > Perhaps, but the list is given by some (unidentified process). Without > the given list, there is even no start on the determination of the > diagonal. The proof clearly states "given a list", so the assumption > is there that there is a list. And the tree is given too by some unidentified process. > > Nope, the assumption is that you are able to give a list. And the proof > shows that you are not able to. The proof accepts that I am able to give a *countable* list, i.e., a list which is complete with repect to its enumeration by natural numbers n e N. And my tree requires that I am able to give a tree with *countably* many levels, complete with respect to N too. > > > But adding levels > > of the binary tree is a sequential process. Are you really in earnest > > about that question? > > Yes. I state that adding levels (or individual nodes, but I grant that > all nodes of the same level can be added simultaneously), is a > sequential process. The definition of the list may or may not be a > sequential process, but that is irrelevant. Given a list, the definition > of the diagonal number is not sequential. 1) Every counting and every identification of the n-th digit is a sequential process. 2) Given the tree, the calculation of edges per path however is *not* sequential. > > > Take the nodes as digits. There is no question that every path can be > > interpreted as the representation of a real number. > > Oh, yes. And every edge can be interpreted as such (with quite a bit of > duplication). And every node can be interpreted as a rational number > with a finite binary representation. (Obviously I ignore the dual > representations here.) Every edge can be interpreted as a binary digit, 0 if it goes to the left, and 1, if it goes to the right, for instance. > > When > something happens for natural n, and we can define a limit when n > grows without bound, it is common to state that that limit is the infinite > case in informal discourse. Just this is the mess in set theory. Infinity is not present in magnitudes of numbers. It is most important to distinguish this. > > > Therefore I say sometimes each natural n can index all digits m =< n > > 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 > > Yes, for each finite number n An covers all of A1 to An. No dispute about > that (and I never disputed that). In shorthand: > For each n, An = (*+){i = 1 .. n} Ai. > True, without a problem. The problem is, what is: > (*+){i = 1 .. oo} Ai > you have *never* defined that one. You claim it is An for some n, but as > you do not define it, and give no proof... oo is not a finite number, > so we need a definition to reasonably argue about it. oo is not a finite number, but all numbers Ai of the sum (*+){i = 1 ... oo} Ai are finite numbers. We h
From: mueckenh on 23 Jul 2006 13:34 Dik T. Winter schrieb: > In article <1153482158.663876.226910(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > 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. > > Perhaps. In that case his statement was clearly wrong. But nobody seems to have observed that up to now. > > > 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. > > As I said before. You are arguing agains the axiom of infinity. But your > inconsitency proofs show nothing, because in part of your proofs you always > use the negation of that axion. I do not use the negation of the axiom, but accept that every digit of 0.111... can be indexed by a finite natural. Hence every sequence of digits of 0.111... can be covered by a natural. (There is no digit by which 0.111... could be distinguished from every natural.) But 0.111... is the union of all of its finite sequences, i.e., of the naturals. Regards, WM
From: mueckenh on 23 Jul 2006 13:46 Dik T. Winter schrieb: > In article <1153482677.919976.211910(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > What has your statement in parentheses to do with the statement I gave? > > > There is no argument at all about K being different from all natural > > > numbers. It is just a plain statement: > > > For all p, there is an n such that An[p] = K[p] > > > you claim that is false, and again without direct proof. Do you disagree > > > with the statement: > > > For all p, Ap[p] = K[p]? > > > I know that Ap[p] = 1 = K[p]. Now chose in the above n = p and you get > > > the new statement. Why would this second statement be false? > > > > You said "for all p". If p is a natural number, then the segment K[p] > > K[p] is a single digit. > So I have misunderstood you all the time. > > is the unary representation of a natural number and, hence, is covered > > by a natural number. Remember: > > > > For each natural (i.e. finite) number p we know: to index and to cover > > are equivalent. > > p covers K[p] <==> p indexes p. > > > > Would K have only natural indexes p, then it would be covered by at > > least one natural number. > > Again, a conclusion without proof. Not if you agree that indexing n includes covering all m =< n too. Because then indexing every n leads to indexing all m smaller or equal than every n. I argue that this is proven by the construction of the finite unary numbers for each finite unary number. If you disagree, then please give an example of a *finite unary number* which indexes some n but does not cover every m =< n. >Consider the digits of K indexed by > natural numbers p. Further consider the statements: > For all p, Ap[p] = K[p] > and > For all p, Ap[p+1] != K[p+1] > Which of these is false, and why? None of these equations is false. But you misinterpret the last one. You show that for every digit p+1 of K there is a unary number Ap not indexing (and, therefore, not covering) it. Why should that be important? Already A1 does not index K[2]. The question is, whether all p exist and whether forall p there is an Ap. All p exist by the axiom of infinity. For each p there exists a natural which indexes it and covers all smaller positions. But K does not consist of more than all p. But here is a decisive test: Find a p in K which is not covered by a unary number (and with it all k < p). If you are unable, you must confess that K is completely covered by at least one natural number though you might be unable to name it. Regards, WM
From: mueckenh on 23 Jul 2006 14:14
> > Dik T. Winter wrote: > > > 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. > > But you particularly chose 0.111... because it was the representation of > a rational numbers. Pray be consistent. I used it only to prove that it is not in the list. Now we have finite unary numbers in the list. For those numbers we cannot find a *+ sum larger han any. > > > > 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. > > > > You imply that it is possible to take n = p, but that is impossible for > > p being not a natural number like aleph_0. > > But p *is* a natural number! A restatement: > For each natural p there is an n such that An[p] = K[p] > and > For each natural p there is an n such that An[p] != K[p] > what is wrong about those statements? Every even number is surpassed by an odd number. Every odd number is surpassed by an even number. Which set contains larger numbers? Arguing in the infinite does not yield consistent results --- or it yield the result you intend. > > > > > 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. > > > > 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? > > This includes what the meaning is when there are infinitely many symbols > in a column. It does not give meaning to what happens when there are > infinitely many columns. Each column is treated separately. Therefore the *+sum does not depend on the number of columns. > > I don't have to know anything. This process is defined. > > Yes you are showing that. > > > The axiom of infinity guarantees the existence of all transpositions. > > Yup. > > > It is even defined by a sequential process: IF the set M contains x > > THEN it contains x U {x}. Unless you advanced to x you cannot advance > > to x U {x}. Further the sequential process is already implied in every > > counting. If you can exhaust M, then you can exhaust the set of my > > transpositions. > > Yup. But the axiom of infinity does not guarantee the you can exhaust M > in the sense you mention. What constraints are to be observed? Please note, the set of transpositions together with a given well-order determines everything in my case. There is no difference to Cantor's list. > > > Or put it in other words: If the set M does exist, then > > the set of all my transpositions does exist too. > > Probably, but completely irrelevant. When we use your interpretation of > Cantor's statement, he was wrong. Yes, and that is true for all his statements! One can never exhaust N, whether it may be existing by the mercy of an axiom or not. > > > > (1, 2)(2, 3) != (2, 3)(1, 2) (still assuming that the numbers are > > > indices). So how can you state that they work at once? > > > > They are defined "at once" and lead to a well-defined result if applied > > to a given well-ordered set. > > Mathematics happens in a timeless environment. > > And I just showed you a set of two transpositions that can be applied to > a well-ordered set and do not lead to a well-defined result. That does not matter in my case because the sequence of the transpositions is determined. > > > There is an order dependency in Cantor's list too. The n-th digit > > prevails the digit umber n+1. Mixing them up destroys the proof. > > Eh? You are again confusing what is given (the list) with what is > calculated. When the list is given I can calculate the n-th digit > without needing any previous difits. How do you know what i s given in the n-th line unless you count up to n? > > > > You have to define in what *sense* the sequence of transpositions > > > converge. I think it is possible to give such a definition, others > > > think not, but > > > > > let that go, that was not my main critique. > > > > It is easy to see that a subset which is in order by size can never > > shrink. Once the element m < n precedes n by order it can never loose > > this property. > > I think you do not read what I write. I only stressed that other objection (not uttered by you) are invalid. > > > > My main critique was that you > > > have to show that a well-ordered set remains well-ordered in the limit. > > > There is not yet any proof forwarded, and a quote from Cantor that does > > > not contain a proof is not a proof. > > > > I have not to prove anything concerning a limit. There is not the > > faintest possibility that a well-order could be destroyed by a > > transposition which is enumerated by a natural number. > > Indeed, I never argued with that, so I wonder why you come up with that. You argue with infinity, but infinity is not present for an enumerated set. > > > Only such > > transpositions do occur. If the set does not remain well ordered "in > > the limit" then there is no limit. > > 1/n is never 0 when n is enumerated by natural numbers. Nevertheless, > lim{n -> oo} 1/n = 0 > distroying a property. So you are now actually stating that that limit > does not exist. In fact, it does not exist other than by definition: epsilon can be made arbitrarily small only for a = 0. But that does not imply that 0 is ever reached. In the same way omega is never reached by counting lines in Cantor's list. Hence, we are never sure whether the diagonal number is not present there. You may answer: We know for every natural n that b_n =/= a_n. We know b_n = 4 if a_n = 5. But that is false, because you must count to n in order to find out whether a_n = 5. And you cannot count to every n, though you may count to each n. That is the deliberate error of Cantor's: He mixed up each and every (or all). "Each" is possible, "every" is not. Regards, WM |