An uncountable countable set
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From: mueckenh on 16 Jul 2006 07:59 Dik T. Winter schrieb: > In article <1152978585.858028.263870(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > > > In usual mathematics 0.999... is assumed to exist. > > > > > > It is not assumed. There is a definition floating around giving a meaning > > > to that notation. > > > Without the definition that notation makes no sense. > > > > And with that definition it makes no sense either, as the definition is > > of he same kind as the squared circle. > > You do not believe in limits? Otherwise, why does the definition not make > sense? Quote the definition for precisely that notation: > 0.999... = lim{n -> oo} sum{k = 1..n} 9.10^(-k) > what part of that definition makes no sense? "n --> oo" because there is no natural number oo. n cannot become oo whether there are infinitely many natural numbers or not. Each is finite. Thus, for n e |N, the result is always different from 1.000... .. It is not that I fight against the identification of 0.999... with 1 or against the identification of the limit of the series of Gregory-Leibniz with pi/4 for computational purposes. But the assertion that the digits of these limits could be used to construct a diagonal number is simply nonsense. Regards, WM
From: mueckenh on 16 Jul 2006 08:10 David Hartley schrieb: > I cannot comment on the German text, but Dik's reading of the English is > clearly correct, WM's wrong. To put it even more clearly > > Question: Which transformations preserve ordinal number? > > Answer: Those which are realised by a finite or infinite set of > transpositions. > > "Transformation" is not defined here. But "such transformations which do not change the well-order" are defined here. And only those are interesting. In German we read "Umordnung" which even better would be translated by "changing the order". And this can be done by transposing two elements infinitely often. By the way, why should Cantor have written that text, if it did not express something? > For that matter, neither is the > action of an infinite set of transpositions, nor is there any proof. If > you (WM) want to use this proposition in your argument, you must fill > all these gaps. That kind of arguing shows why set theory cannot be contradicted. Regards, WM
From: mueckenh on 16 Jul 2006 08:15 Dik T. Winter schrieb: > In article <1153039371.156834.8960(a)s13g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Virgil schrieb: > ... > > > Every ordinal has successors, whether in ZF or NBG or most other set > > > theories, and to index any of them it is sufficient to use at the > > > members of any successor. > > > > But the successor of 4 does not consist of the ingredients of 4 only. > > Eh? > The successor of 4 is 5, and 5 = 0.11111 which cannot be indexed by 4 = 0.1111. 5 in unay representation consists of one more 1 than 4. Hence, 0.111... as the succesor of all naturals must consist of more 1's, than any natural, if it is to be a number. > > So the successor of all natural numbers cannot consist of only all > > natural numbers only. > > As stated this makes no sense. But the successor of every natural number > is a natural number. The successor of all naturals is not a natural and, therefore, must be larger (because it is not less). Regards, WM
From: David Hartley on 16 Jul 2006 08:18 In message <1153051034.733810.190150(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de writes > >David Hartley schrieb: > > >> >> Can you provide a proof of that statement? Because you state just above: >> >> "Cantor's conclusion is correct, with no doubt, if infinite sets like >> >> the set of my transpositions do exist." Which contradicts what you state >> >> here. >> > >> >No. Cantor obviously was correct, if infinite sets would exist. But >> >they do not. So he was wrong. And you are wrong, as I have shown above >> >by Cantor's text which you misinterpreted grossly. >> >> Cantor's proposition (as you interpret it) leads directly to a >> contradiction in standard set theory. So it is disproved. To show that >> set theory is inconsistent you must also show that it can be proved. You >> cannot do this by arguing over quotations from Cantor. He does not >> appear to have offered a proof and he was not infallible. You claim it >> is obvious, so prove it yourself, with no more quotations, (except >> references to standard theorems}. > >Obvious is that no transposition of two elements of a well-ordered set >can destroy the well-order. >Obvious is that, if infinite sets do exist and can be exhausted, the >set of my transpositions does exist and can be exhausted. >That's all needed. >Belief in anything else is matheology. Obvious is, you haven't the faintest idea of how to prove your claims. -- David Hartley
From: mueckenh on 16 Jul 2006 08:25
Dik T. Winter schrieb: > > > there is one natural number missing: 0.0... > > > > That is not a natural number. > > Depends on whether you are a follower of Bourbaki and Halmos, or of others. > My learning was with Bourbaki as a basis. 0 was invented less than 2000 years ago. Natural numbers need not be invented. However, Bourbaki and Halmos tried this trick in order to prevent set theory from too easily been demasked as inconsistent. > > > The Addition > > *+ was defined by "if one 1 is present the result is 1, else it is 0". > > This covers also the infinite case. > > It does not. Define: > A1 = 0.10... > A2 = 0.110... > etc. Now I can do: > SUM{i = 1 .. n} A_i > where SUM means repeated application of *+. There is nothing in your > definition that covers the infinite case. If any case includes at least one 1 then the *+ sum is 1. Hence there is no necessity at all to distinguish cases. > > > > > What is the result? 0.111... > > > > > > How do you get there? What rule are you using to calculate that > > > "infinite sum"? But even if we allow that sum, we know that that > > > is not the unary representation of a natural number. > > > > But it is a unary representation of a whole number, namely of the > > number of digits of a real number. Nevertheless, in decimal > > number of digits of a real number. Nevertheless, in decimal > > representation 0.111... does exist, but not in the list. The addition > > of the list numbers gives the largest number of the list. > > In the finite case. As you have not provided a way to do the infinite > sum, I have no idea what happens in that case. Define: If any case includes at least one 1 the the *+ sum is 1. > > > Whatever this > > may be, it is not 0.111... . Proof: *Every* list number ends up with at > > least one zero. Therefore there must be at least one column with only > > zeros. > > Wrong. Provide a proof of that statement if you think it is correct. What is wrong? The statement that every number in the list ends by 0? > > > > Makes no sense to me. I did state: > > > For all p there is an n such that An[p] = K[p]. > > > You said that was wrong. I asked you why that was wrong, and you come > > > up with some nonsense. I state: select n = p. Isn't it a fact that > > > For all n, An[n] = K[n]? > > > > All n which can be indexed are in the list. K is not in the list. > > This is not an answer to my question. What is wrong with > For all n, An[n] = K[n]? > Pray, is there an n such that An[n] != K[n]? If so, what n is it? It is not such an n. But if K would consist of such sequences only, then it was in the list. > > > All segments K[n] are in the list. K is not in the list. What > > distinguishes K from all its segments if not at least one more 1at a > > position which cannot be indexed? K has nothing else to offer. > > What K offers is that it has no zeroes in its representation. And all > its digit positions are natural numbers. Then all its digit positions are in sequences in the list, because all natural numbers are in the list. This would mean K is in the list. Regards, WM |