An uncountable countable set
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From: Virgil on 9 Jul 2006 19:29 In article <1152479971.443896.173440(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > Not so. Cantor's rule works for each n independently of all other > > members of N, so all can be done simultaneously by a single rule. > > > > Yours does not allow simultaneity because of the non commutativity of > > permutations. > > Nonsense! That "mmueckenh" says so confirms me in my belief that I am correct. >I do not need and do not want to commute any transpositions. You do if you want to be able to avoid having to do them all in sequence. Cantor: simultaneous, gets them all done in one step. "Mueckenh": sequential, never gets them all done.
From: Virgil on 9 Jul 2006 19:46 In article <1152480309.668400.137270(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > > > No, but it may depend on the ordering of the calculations. So there is > > an inherent difference between calculations that are independent of each > > other and calculations where some calculation can not be done before > > all previous calculations are done. So when calculating the digits of > > the diagonal number it does not matter what digit you calculate first. > > In your re-ordering sequence it makes a strong difference what re-ordering > > you do first. > > And that is all you have to object against my proof? We have a fixed > scheme of transpositions. Since there is no fixed well-ordering of the rationals, there can be no fixed sequencing of transpositions to re-order" them. Any attempt must be conditional on what one finds in the initial well-ordering. > But there is no limit process. Then the process is doomed. The only way to "complete" a countably infinity sequence of steps , one at a time, is if there IS some limit process. > > Yes, that is what I wrote in some of my articles (that you have read). > > The limit process is *not* needed to distinguish the diagonal number from > > all other numbers of the list. It is needed to show that the diagonal > > number is a real number. > > But that is not interesting. Whyever not? > It is easy to see that the diagonal is a > sequence of the same sort as are the list entries. Whether they are > real numbers is uninteresting. Interesting is that such sequences are > uncountable. It may be interesting to "Mueckenh", but it is also false. > > But in order to prove that the diagonal differs from every entry, there > a limit is required but not available. Not by me. I can easily do it without any limit process at all. > The missing limit has not been remedied but has only been put aside. My > objection remains: But that there is no satisfactory limit > consideration becomes clear from the following: We know that 0.999... = > 1.000... This leads to the result that a change of 1 in the limit where > the digit number goes to oo does not have the effect which would be > required in order to distinguish the diagonal number from the list > numbers. Cantor's rules avoid that particular pitfall by never using either 0 or 9 in diagonal constructions.
From: Virgil on 9 Jul 2006 19:52 In article <1152480817.482681.144170(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > The set ordered by size and by natural indices is the limit of my > > > transpositions. > > > > 1. What is a limit of transpositions? > > 2. How do you prove its existence and uniqueness? > > How does Cantor prove the existence and uniqueness of the well-ordered > set of rationals or of the diagonal number of his list? Cantor "proves" the existence of a well-ordering of the rationals by proving existence of a bijection from N to them. This induces a well ordering on the rationals. > > The *only* criterion about countable infinite sets is whether one can > determine *precisely* at which natural number something happens: After > how many steps in a > well-order of |Q the fraction 4711/235537 will appear, for instance, or Given an arbitrary bijection, f: N --> Q, "mueckenh" claims to be able to tell us the value of n such that f(n) = 4711/235537, even though that will differ from one such bijection to another. Another occasion of his just not thinking.
From: Virgil on 9 Jul 2006 20:11 In article <1152481328.191617.303380(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > >> After each of your countably many transpositions, there is still a > > >> countably infinite subset of the rationals left which are not in > > >> natural order. That is a necesssary consequence of the entire set > > >> being originally well ordered. > > > > > > After each of his countably many exchanges there is still a > > > countably infinite subset of the list numbers left which are not yet > > > exchanged. > > Excuse me. There is not a single exchange carried out. All aere > simultaneously defined. More is not required. You have to show (1) that they can be carried out and (2) that after they are carried out the result will be what you claim it should be. You can do neither. > Have you ever tried to > well-order the set of rationals for more than, say, 20 elements. One can do it for all of Q by establishing a rule for what rational follows any given rational. This works simultaneoulsy for all of Q, and has often been done. > You are right. I responded only to Virgil's very naive belief that > there something really should be carried out. Of course it is > sufficient to give the principle by a short expression like this > > (1,2) > (2,3) > (1,2) > ... > > to intelligent people, and all is instantaneously ordered by magnitude. The effect of transpositions (1 2)(2 3)(1 2) applied successively to the first 3 elements of any list of 3 or more is simply to reverse the order of those first 3. This means that unless one has the exact reverse of the order by magnitude to start with, the result will not be ordered by magnitude. Until "mueckenh" demonstrates that he can at least put a finite lists into order by magnitude by a sequence of traspositions of successive elements, one may safely ignore his attempts to do so with infinite lists.
From: Dik T. Winter on 9 Jul 2006 20:19
In article <1152450584.362072.151080(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > You start with the list of natural numbers in unary notation: > > A0 = 0. > > A1 = 0.1 > > A2 = 0.11 > > A3 = 0.111 > > ... > > calculate the diagonal number K: > > K = 0.111... > > and claim that K equals An for some value of n. > > I stated: either it does or it does not. We claim it is not. > > > > Note the following: > > > > It is not by pure accident that I chose 0.111... . This number has also > > > > a meaning as decimal representation of 1/9. So we can do the > > > > calculations. > > > > Indeed. Let's do the calculations, and let each An also have the meaning as > > decimal representation, so > > An = (1 - 10^(-n))/9 > > and now you claim that for some n > > (1 - 10^(-n))/9 > > eqals K = 1/9. > > > > If that were true, there must be some n such that 10^(-n) equals 0. > > Indeed, if you consider 1.000... - 0.999... = 0 That is true because of the *definition* of that notation. Without a special definition that notation has no meaning. (There do not exists things like infinite sums in arithmetic.) So the definitions: 1.000... =(def) 1 + sum{i = 1..oo} 0/10^i and 0.999... =(def) 0 + sum{i = 1..oo} 9/10^i where sum{i = 1..oo} is defined as lim{n -> oo} sum{i = 1..n}. Note, again, the limit implied in the notation. And, at no place do we leave the finite real. > Indeed, if you consider 1.000... - 0.999... = 0 and if you claim that > all positions are enumerated by natural numbers, then there must be > natural number n with 10^(-n) = 0. Why? Can you *prove* that? Because that would mean that there is a natural number that is not finite, but there is no such number. > > A0 = 0.11111111... > > A1 = 0.01111111... > > A2 = 0.00111111... > > and so > > K = 0.000000... > > Now we see: > > For all p there is an n0 such that for all n >= n0 An[p] = K[p]. > > and your claim is that from that it follows that > > There is an n such that for all p An[p] = K[p], > > But as An = 1/(9 . 10^n), this would imply that K (the limit of the > > sequence) must be in the list and for some n, An must be 0. > > > > So, are you now arguing that the limit of a sequence is an element of that > > sequence? Or what else? And if so, for what natural number (eh, > > finite...) is 1/(9 . 10^n) equal to 0? > > Either K is in the list (which would contradict analysis in the same > way as my original example), then there is an n with 10^(-n) = 0. Yes, we claim K is not in the list. > Or K > is not in the list. Then there must be a position which cannot be > enumerated by natural numbers (the unary representations of them are > now the zeros behind the decimal point). Again, why? This is merely an assertion. There is no proof (and I have not seen a proof yet). I think you are assering here that from: > > For all p there is an n0 such that for all n >= n0 An[p] = K[p] it follows that: > > There is an n such that for all p An[p] = K[p] What logical reasoning do you use to come from the first to the second? I asked already twice before, and have not yet seen an answer. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |