Cantor Confusion
in [Math]
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From: Dik T. Winter on 25 May 2007 22:28 In article <1179932826.710774.250710(a)k79g2000hse.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 22 Mai, 04:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179750646.698997.275...(a)b40g2000prd.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > On 21 Mai, 04:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > No. They do not state it is a sequence. > > > > > > They write 1 + 2 + 3 + .... That is the sum of a sequence, i.e., a > > > series. > > > > Yes, you like to use their informal statements as formal statements. > > 1) Even informal statements are not always wrong. No, in general they are even correct. > 2)This equation is a formal statement. It is not. > 3)Is their theorem formal enough? 1.3 Theorem Let L be an > infinite cardinal, let K_A (A < L) be nonzero cardinal numbers, and > let > K = sup {K_A | (A < L)}. Then Sum {A < L} K_A = L * sup {K_A | (A < > L)} And *where* in that formal statement is there a sequence? Just as I said: 1 + 2 + 3 + ... is an informal notation for: sum{i in N} or sum{i < aleph_0} no sequence at all. Just summation over a set. > > > > But definitions *can* lead to > > > > another result. > > > > > > For instance? > > > > Definition: sum{i in N} i = 0. > > Wrong definition. What is wrong about it? > > Yes, so you can *not* state that the set of natural numbers can be > > summed up. > > If I can state that the union of a countable collection of countable > sets is countable, then I can state that the set of natural numbers > can be summed up. When I used the countable union of countable sets > nobody opposed up to now. Indeed. That has been proven. But if you want to add the set of natural numbers, using addition for natural numbers, you have a lot more to do than just state it. In that book they add sets of *cardinal* numbers, and give a definition which leads to a result that is a *cardinal* number. Assuming you can replace "cardinal" by "natural" in that proof leads to wrong conclusions. > > > Their notation shows the sum of the sequence of natural numbers > > > > > > 1 + 2 + 3 + ... + n + ... (n in N) > > > > > > They write, it is easy to see that this sum is equal to aleph_0. > > > > And they also write that they are doing cardinal arithmetic. > All natural numbers are cardinal numbers! Yup. But not all cardinal numbers are natural numbers. > > But there are other results possible. Consider my definition: > > sum{i in N} = 0 > > now try to prove that that is not possible. > > Wrong definition. > Proof. 1 > 0. There is no negative natural number. Every sum of > natural numbers is >= 1 > 0. Yes. This does *not* prove my definition wrong. Because in arithmetic on natural numbers every *finite* sum is a natural number. Arithmetic on natural numbers tells us nothing about *infinite* sums. > > > And n fact, > > > if there were all the natural numbers actually existing, then an > > > infinite one must be among them. > > > > A common assertion by you, unproven. Pray give a proof, assuming the > > axiom of infinity, and with using the negation of that axiom. > > 1+2+3+...+X = X(X+1)/2 > > for every X in N. Proof by induction left as an exercise. Yes. So not for infinite X. Unless you first prove that there *is* an infinite X in N. You cannot use this as proof that there is indeed an infinite X. -- 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 28 May 2007 22:21 In article <1179933246.747971.247250(a)q66g2000hsg.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 22 Mai, 04:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179750984.210277.71...(a)n15g2000prd.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: .... > > Yup, you are still living some hundred years ago. Mathematics has gone > > forward since that time. > > Unfortunately today "finite" is meaning "infinite" only when "forward" > is meaning "backwards". O. But you do not even know what has changed since Cantor's time. > > > My > > > argument should only show that there are countable sets (like the > > > paths in the tree) which cannot be put in a bijection with N. > > > > They can, but that requires an infinite definition. > > That means they cannot. By "can" I understand: "It can be done in real > world on a sheet of paper, a blackboard or in a brain." I do not > understand "cannot" by "can", even if this is not up to date. Yes, again your own personal meaning of "can". If there is an injection from A to B and an injection from B to A, there is an injection between the two. The proof of that dates from Cantor's time, so I think you should know about that. Whether the actual bijection can be put in finite terms is, with respect to mathematics, irrelevant. And, at least in my brain, such a bijection is conceivable. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: WM on 3 Jun 2007 16:39 On 26 Mai, 03:52, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1179864527.794567.66...(a)p77g2000hsh.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > But from > > > your book: > > > "also eine sogenannte normale irrationale Zahl, die keine erkennbares > > > Muster der Ziffernfolge aufweist" > > > > Note the last sentence! > > Yes, I note it. And I note also that the first part of that sentence is > nonsense, and that is what I do remark on. Whether or not pi is normal > has *no* relationship with the remainder. Therefore I stated the remainder! > > > > > > In the first place, a normal number can have a well defined sequence of > > > digits (as the Champerowne numbers show). > > > > Of course, but then we could compute the numerals. > > Yes, we could indeed. Therefore I stated the remainder! > > > > And, in the second place, there > > > are numbers without a well defined sequence of digits that are *not* > > > normal. > > > > And there are red cars, which are not normal cars. > > > > > (Normal here meaning normal to a particular base.) The two are not > > > equivalent. So even if no rule can be given for the digits of pi, that > > > does *not* mean that it is normal. > > > > But if pi is normal with no recognizable pattern, then no rule can be > > given. (If I say A ==> B then I do not imply B ==> A.) > > Normality of pi has nothing to do with it. If pi is normal with no recognizable pattern, then no rule can be given. > If pi is not normal and with no > recognizable pattern, then also no rule can be given. But I said: "If pi is normal with no recognizable pattern, then no rule can be given." And this statement is true. I used a normal number as an example. I did not mention the reversed statement, in particular I did not state that normality is necessary. Can't you follow a logic conclusion any longer? > To wit: > normal => no recognizable pattern > and > no recognizable pattern => normal > are *both* false. I did not use either of them. I did not use "=>". I used a normal irrational number (as a suggestive example) "and" I required that no recognizable pattern is visible. Regards, WM
From: WM on 3 Jun 2007 16:46 On 26 Mai, 04:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1179931587.894431.146...(a)u30g2000hsc.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 22 Mai, 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1179747854.168331.52...(a)x35g2000prf.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > > On 21 Mai, 04:37, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > In article <1179654356.461792.242...(a)n59g2000hsh.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > > > > Its is more. You cannot answer the question whether the numbers P = > > > > > > [pi*10^10^100] and P' = P with the last digit replaced by 3 > > > > > > nsatisfy P < P'. > > > > > > > > > > Yes, so what? Your distinction is just terminology, and not more than > > > > > that. > > > > > > > > It is by far more than a difference in terminology if one will never > > > > or always be able to answer a question. > > > > > > Makes absolutely no sense. You have some implicit notion of number in mind > > > that I do not have. > > > > Trichotomy. > > But there is trichotomy. We know that either P < P' or that P = P' or that > P > P'. But we do not have the means to determine which of the three holds. We know nothing about that number. > > > > You are obviously confused. For an injection from paths to nodes you need > > > a catalogue of all paths. > > > > The tree *is* a catalogue of path, namely of *all* existing paths > > which represent existing real numbers of the interval [0, 1]. > > And I thought that a catalogue in your sense numbered the paths from 1 > onwards. Apparently you used a different meaning, again. A catalogue is not a list. "Catalogue" is not normed as far as I know. It is simply a collection o all paths. > > > Each of > > these paths p can only be distinguished from another path p' when both > > have been separated from each other. > > > > You know that each node separates one more path, the number of > > separated paths minus number of nodes is 2-1-1+2-1-1+2-1-1 +-... > > No. Each node splits a set of paths in two sets of paths. When you > come in at a node with a set of K paths, you come out with two sets of > K/2 paths (this is a bit informal). If K is infinite, so is K/2. And > if K is uncountable, so is K/2. So all this splitting does show nothing. > And it is not the case that each node separates one more path. If that > were the case you should be able to indicate what path is separated by > the root node. Every path which can be identified must be separated from all other paths. For this sake there must be as many nodes as separated paths. > > > Therefore, there must be as many separation points, or nodes, as > > separated paths. > > Wrong. Absolutely correct. But I you wilfully adhere to some believe in ghost paths, it is no longer useful to maintain this discussion. Regards, WM
From: RLG on 3 Jun 2007 19:36
"WM" <mueckenh(a)rz.fh-augsburg.de> wrote in message news:1180900794.289843.91710(a)p77g2000hsh.googlegroups.com... > On 27 Mai, 04:08, "RLG" <J...(a)Goldofo.com> wrote: >> "Carsten Schultz" <cars...(a)codimi.de> wrote in message >> > > The infinite equilateral triangle IET > > 0.1 > 0.11 > 0.111 > > can be considered as a (very special) Cantor list. Indexing all digits > by natural numbers, we get the following version of the EIT: > > 0.1 > 0.12 > 0.123 > ... > > We can set up a bijection between the initial segments of the diagonal > and the entries of the list > > 0.1 <--> 0.1 > 0.12 <--> 0.12 > 0.123 < --> 0.123 > ... > > It is easy to see that this bijection proves the following statement: > There is no actually infinite diagonal unless there is an infinite > entry too. > > Conclusion: As there is no infinite entry, i.e., there is no infinite > natural number, it is false to claim the existence of an actually > infinite diagonal or, in general, to claim the existence of an > actually infinite set. There is no actual infinity. If I have understood you correctly, you seem to be suggesting that the naturals should be considered as a proper class instead of a set. This seems a bit unnatural to me but why are you emphasizing this diagonal? Are you also trying to argue that Cantor's theorem is false? R |