Cantor Confusion
in [Math]
|
From: WM on 4 Jun 2007 10:40 On 4 Jun., 03:23, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1180903197.893443.23...(a)g4g2000hsf.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > 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! > > But the first part is nonsense. No, it is a suggestive example. Most normal irrational numbers do not exhibit a regular pattern. Do you disagree? > > > > > > 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! > > Yes, and you should have left out the first part. Why? Do you think it is wrong? Do you think that usually normal irrational numbers are like Champerowne numbers? > > > > > > 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. > > Right. That is what I said. Should your German be too bad? > > > > 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? > > No, you stated: "presumed that pi is really a number without rules, hence > a so-called normal number that has no recognizable pattern in the sequence > of digits". The part about "normal number" is nonsense. In German: > "vorausgesetzt, daß pi wirklich eine regellose Zahl ist, also eine > sogenannte normale irrationale Zahl, die kein erkennbares Muster > der Ziffernfolge aufweist". > What is the reason you mention normality here?# Because an unnormal number would easily obey a rule, for instance it could show no 2,3,4,5,6,7,8, or 9 after a certain index, which would increase the possibility of predicting a digit by chance. > > > > 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. > > What normal irrational number *did* you use as an example? pi? Do you > have a proof that it is normal? If so you should publish it. I said: "vorausgesetzt", which means: "assumed", > How > should I interprete your "also eine sogenannte normale irrationale Zahl"? > This looks, to me, more like a definition of normal number than anything > else. It is a definition. "Regellos" implies "normal", because otherwise there would be a rule like "only few 1's after the digit number 10^100" or so. Is that the only "error" which you mean to have discovered? Or why are you so much chopping on it? Regards, WM
From: WM on 4 Jun 2007 13:43 On 26 Mai, 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1179932826.710774.250...(a)k79g2000hse.googlegroups.com> WM <mueck...(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? Can't you see simplest facts any longer? Sum {A < L} K_A for L = aleph_0 is the sum of all cardinals less than aleph_0. This sum can also be written by 1 + 2 + 3 + .... > Just as I said: No. Not just as you said. It is really funny to see how you try to veil your errors. > 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. Yes, summation over the ordered set of natural numbers. In general this is written as a series. Hracek and Jech even state the identity on p. 188: "It is not very difficult to evaluate infinite sums. For example, consider 1 + 2 + 3 + ... + n + ... (n in N). It is easy to see that this sum, SUM{n in N}n is equal to aleph_0." Can you read: 1 + 2 + 3 + ... + n + ... = this sum = SUM{n in N}n? But probably you will not acknowledge these authors, because they also use, as I did, the term "adic", which you opposed to as not being common use. Then you were as wrong as you are here. > > > > > > But definitions *can* lead to > > > > > another result. > > > > > > > > For instance? > > > > > > Definition: sum{i in N} i = 0. > > > > Wrong definition. > > What is wrong about it? Simply the fact that the equality is not an equality. > > > > 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. I do not replace anything but use the notation they use, namely the symbols 1, 2, 3, ... and n in N. By the way: Every finite cardinal is a natural. And only finite cardinals are added. > > > > > 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. That is completely irrelevant as long as only natural numbers are summed up. > > > > 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. Now I understand why you (and some others) are so unable to follow logic arguments, for instance concerning the tree. You think "infinite" is some holy grail? Even the infinite sum of natural numbers cannot be less than 1. > > > > > 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. Wrong again. Hrbacek and Jech proved that the right hand side is an infinite number, so X on the right is infinite. This does not imply the same symbol on the left being infinite? Regards, WM
From: WM on 4 Jun 2007 14:22 On 4 Jun., 01:36, "RLG" <J...(a)Goldofo.com> wrote: > "WM" <mueck...(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? It cannot be actually infinite unless there is also an actually infinite line. The height of the Cantor list is said to be actually infinite, the width, however, is not. By the diagonal height and width are bijected. Therefore both are actually infinite or both are not. It is simple to see. > Are > you also trying to argue that Cantor's theorem is false? It is false. Consider the problem with double representation of some rationals. For any finite index we have 1.000...0 =/= 0.999...9. So in Cantor's proof there is absolutely no problem for any 9 at a finite position. Only n the infinite the problem 1.000... = 0.999... occurs. This, however, is the same with any other limit. The real numbers in Cantor's list as well as the diagonal are limits, namely SUM{n=1...oo} a_n * 10^-n. and SUM{n=1...oo} b_n * 10^-n, respectively. There is no problem for any finite index n: (a_n - b_n) * 10^-n > 0. In the limit, however, LIM{n-->oo} (a_n - b_n) * 10^-n = 0. So in this limit the diagonal cannot be distinguished from an entry. But this limit is required. Cantor, in his oruginal paper, considered only a_n =/= b_n and held his sufficient for the diagonal to be different from any list entry. The case with double representation was discovered later (cp. the first diagonal proof of real numbers in sci. math. research http://groups.google.com/group/sci.math.research/browse_frm/thread/5d9926c33ee6e40f?scoring=d&hl=de ) I cannot understand, why the general problem with LIM{n-->oo} (a_n - b_n) * 10^-n = 0 has not been discovered yet. Regards, WM
From: WM on 4 Jun 2007 14:32 On 4 Jun., 01:17, Virgil <vir...(a)comcast.net> wrote: > In article <1180902384.062112.256...(a)m36g2000hse.googlegroups.com>, Corrected: Would you also reject the possibility that the initial segments of the diagonal (including the complete segment) can be bijected with themselves? If not, why is the bijection of the initial sements of the diagonal (including the complete diagonal) with the lines possible, but the bijection of the initial sements of the diagonal (including the complete diagonal) with the columns is not? Regards, WM
From: WM on 4 Jun 2007 14:40
On 4 Jun., 01:22, Virgil <vir...(a)comcast.net> wrote: > In article <1180903197.893443.23...(a)g4g2000hsf.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > If pi is normal with no recognizable pattern, then no rule can be > > given. > > False! It is just that the "rule" is not simple, but one can always say > what it takes work out the "next digit" of pi once one has any given > number of digits. There is a (relatively small natural) number n which needs the complete memory of the universe. If you would attempt to define it, you must kill yourself. And even if you would do so, there is another natural number m (less than 10^10^100) which cannot be defined at all. Therefore the method of BBP cannot be applied to find the digit of pi belonging to this index m. Why are mathematicians so short sighted? They believe in inaccessible cardinals but don't recognize that already some very small natural numbers are inaccessible, and Peano's complete induction fails at a very low level. Regards, WM |