Cantor Confusion
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From: mueckenh on 1 Nov 2006 13:29 Dik T. Winter schrieb: > > Therefore it is impossible to exchange omega letters in a diagonal. > > Wrong. For each element of the list a digit is calculated in the diagonal. > As there are infinitely many (omega) elements in the list, there are > infinitely many (omega) digits in the diagonal. omega is the supremum, not the maximum. It does not contribute a diagonal digit. An infinite diagonal requires not only an infinite length but also an infinite width of the matrix. Therefore your absurd infinite number of finite lines does not help you. Here we have the same facts as in our old problem 0.1 0.11 0.111 .... you remember? Without an infinite number in the list there is no infinite diagonal defined. > > > The > > diaogonal cannot be roader than the list. The length of the diagonal is > > the minimum of width and length. This knowledge is prior to your > > axioms. > > Width and length are equal. Fine. But the width is finite by definition. (We do not put finite segments together, but we have only finite seqments.) > > > > All finite sequences are countable. They yield another finite sequence. > > > > Hence they are uncountable. Contradiction. > > > > > > The statement "they yield another finite sequence" is wrong in the > > > context of the axiom of infinity. > > > > A matrix with width A and length B has a diagonal which has min(A,B) > > elements. If your axiom contradicts this, then the axiom contradicts > > mathematics and should be abolished. > > That is not contradicted. Width and length are equal. This amounts to say that there are infinite natural numbers or that the diagonal is longer than any line. Impossible. The maximum of a set of finite numbers which has no maximum is simply not present. It is *not* an infinite number which is larger than any finite number, because a maximum must belong to the set. And a supremum not belonging to the elements of the set does not yield a diagonal digit. > > > > > Forget the Turing machines. The diagonal of the list of all finite > > > > sequences (words) of a finite alphabet is a finite sequence because the > > > > diagonal cannot have more places than the words in the list. > > > > > > Forget that arguing with negation of the axiom of infinity. I am talking > > > in the context of that axiom. Due to that axiom there is a set of > > > natural numbers (all finite), that is itself not finite. > > > > And if you can conclude that in this context every straight line > > crosses itself 17 times, then you will also take that as a fact? > > You, if that follows from some axiom, it would really be possible, unless > the added axiom leads to an inconsistencey. But I think there might be > surfaces where that is even valid. Let us stick to Euclidean geometry. But that is unimportant. I see it is impossible to convince you of the existence of reality. Regards, WM
From: MoeBlee on 1 Nov 2006 13:31 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > All entries of the list have a finite number of letters. An infinite > > > sequence is larger than any finite sequence. The diagonal of a list > > > cannot have more letters than the lines. > > > > > > According to your logic the list can have infinitely many lines. But > > > even if that was correct it would not facilitate an infinte diagonal. > > > > > > The number of diagonal elements is the minimum of columns and lines. > > > > 0 > > 1 2 > > 3 4 5 > > 6 7 8 9 > > ............... > > ................. > > ................... > > > > infinitely downward for an infinite list of finite lists. > > > > The diagonal is 0 2 5 9 14 ... infinitely across. > > > > All entries in the infinite list are finite lists. > > better say finite sequences or numbers or entries 'sequence' and 'list' are synonymous here. I'm pefectly happy to use just 'sequence'. Doing so does not at all harm my argument. > > The infinite list is > > longer than any finite list. > > The entries surpass every finite entry. Nevertheless you call all of > them finite. I don't know what you're trying to say. Even using just the word 'sequence', my point is correct. We have an infinite sequence S of finite sequences. Being an infinite sequence, the length of S is longer than the length of any finite sequence. > > The diagonal of the list is infinite. > > That is your assertion. But obviously the diagonal elements are > simultaneously elements of the entries. No, we trivially PROVE the diagonal sequnece is infinite. Also, since you have pointed to more specific language - 'sequence' rather than 'list' - and you are now using the word 'element', I'll state my terms here with reasonable precision short of putting this in the formal language: A sequence is a function such that the domain of the function is an ordinal. A finite sequence has a natural number as its domain. A denumerable (countably infinite) sequence has omega as its domain. An uncountable sequence has an uncountable ordinal as its domain. The entries in a sequence are members of the range of the sequence. Each entry is indexed by a member of the domain. The elements of the sequence are ordered pairs of the form <x y> where x is a member of the domain (which is an ordinal) and y is a member of the range of the sequence (so the y's are the entries). dom(S) = the domain of S. length(S) = dom(S). In my example S is the denumerable sequence recursively defined as follows: S(0) = {<0 0>} S(n+1) = the unique finite sequence f such that length(f) = length(S(n))+1 and such that f(0)= S(n)(max(dom(S(n)))) + 1 So S is a denumerable sequence such that each entry of S is a finite sequence. The diagonal of S = the unique denumerable sequence D such that, for all n in omega, D(n) = S(n)(n). D is a denumerable sequence. The first four members of S are: {<0 0>} {<0 1> <1 2>} {<0 3> <1 4> <2 5>} {<0 6> <1 7> <2 8> <3 9>} For visual simplicity, we strip the parentheses, angle brackets, and indexes, leaving just the entries of each member (this is informal, just for visualization, and no part of my argument depends on this): 0 1 2 3 4 5 6 7 8 9 To represent the denumerable sequence, we add ellipses (this is informal, just for visualization, and no part of my argument depends on this): 0 1 2 3 4 5 6 7 8 9 .............. ................. .................... The first four members of the diagonal are (this is informal, just for visualization, and no part of my argument depends on this): <0 0> <1 2> <2 5> <3 9> For visual simplicity, we strip the angle brackets, and indexes, leaving just the entries of each member (this is informal, just for visualization, and no part of my argument depends on this): 0 2 5 9 To represent the denumerable sequence, we add ellipses (this is informal, just for visualization, and no part of my argument depends on this): 0 2 5 9 ... > > And that be formalized easily in set theory. And I did that: diagonal of S = the unique denumerable sequence D such that, for all n in omega, D(n) = S(n)(n). > That may be, therefore it is no wonder that set theory yields > selfcontradictions. You just claimed that set theory "yields selfcontradictions". I suppose by "yield" you mean that set theory contradicts your own personal mathematizing, since the last time you claimed to have shown a contradiction of set theory, it turned out that you had not shown a sentence P in the language of a particular set theory such that both P and ~P are theorems of that theory. > The diagonal elements are simultaneously elements of the entries. > Therefore the diagonal elements cannot sum up to a number which is > larger than any natural number unless also the elements of list entries > sum up to a number which is larger than any natural. In my example, I said nothing about summing up. And I said nothing about anything in S being larger than any natural number. > Or put it so: Every segment of the diagonal is covered by an entry. Which 'entries'? > There is no segment which is not covered. What is the initial segment {<0 2>}, of the diagonal, covered by? And what does it matter? > If all entries are finite, Yes, all entries of S are finite sequences. > then the diagonal cannot be infinite (if infinite omega is larger than > any finite n). In this post I PROVED that the diagonal of S is infinite. The diagonal of S is an infinite set. It is an infinite sequence. It has an infinite domain. It is an infinite set of ordered paris. (And, by the way, it has an infinite range.) And you've not shown that that contradicts any theorem of any Z set theory. That it may contradict your own confused word jumbles is not of concern to me. MoeBlee
From: mueckenh on 1 Nov 2006 13:35 Dik T. Winter schrieb: > In article <1162300936.776151.45540(a)e3g2000cwe.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > Attention: 0.111... has only finite initial segments - and nothing > > > > more. Only those can be indexed. > > > > > > You keep stating that, without proof. That it has only finite initial > > > segments, I agree. Not the remainder. > > > > It is you who denies proof. Give an example of a digit which does not > > belong to a finite sequence. If you cannot do so, then every digit > > belongs to a finite sequence. > > I do *not* deny that. But there are infinitely many digits, each being > part of a finite segment. That is but a statement without value. If you do not deny my assertion, then it is not necessary and not useful to assume any infinity, because every number we can name belongs to a finite set. > > > The sequence 0.111... consists of every > > digit but not of more. Therefore, there is not need and no use of > > talking about infinite sequences. > > How can those infinitely many digits form a finite sequence? There are not infinitely many digits. That is only your illusion. > > > > > You need no transfinity to show that lim [n-->oo] 1/2^n = 0 and that > > > > lim [n-->oo] (1 - 1/2^n) / (1 - 1/2) = 2. Because that was known before > > > > transfinity was introduced. > > > > > > But I never argued that. > > > > But it was argued in connection with my infinite binary tree. > > No, something quite different was argued there. Namely that the limit *also* > is the number of edges in the infinite tree (or somesuch) requires transfinite > induction. But is does not. > > > > > > You need transfinity when you want to show that something that holds > > > > > in the finite case also is valid in the infinite case. Induction > > > > > will not show that 0.111... is rational, > > > > > > And indeed. Induction will *not* show that 0.111... is rational. > > > > Induction will show everything that *can* be shown. Up to every finite > > position 0.111...1 is rational. More is not possible. Irrational > > numbers don't exist, but that is another topic. > > Yes, you again deny that the infinite exists. Again denying the axiom of > infinity. While still maintaining that you are arguing within the > axiom of infinity. With the axiom of infinity it exists, and irrational > numbers exist, 0.111... does exist, and induction can not show that > 0.111... is rational. > > > Anyhow, in the binary tree there is no transfinity required. > > There is. Namely to show that the number of edges in the infinite tree > is equal to the limit of the number of edges in the finite subtrees. There is no omega required in the definition of the real numbers, but transfinite induction? There is transfinite induction required to see that the set of digits of a real number has cardinality aleph_0. There is transfinite induction required to see that LIM 1/n = 0 and that 1/9 = 0.111... ? Where is the transfinite step in Cantor's list, showing that the limit of the digits of the diagonal number is a real number? And that this real number differs from every list entry also in this limit? Just this latter assumption being wrong. Regards, WM In article <1162301133.288356.234...(a)m73g2000cwd.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1162139168.281239.183...(a)b28g2000cwb.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > You need not tell that to me. You should tell that to Wolfgang > > > > Mueckenheim who insists that sqrt(2) does not exist because it is > > > > impossible to know all the decimals in its decimal expansion. > > > > > > It does not exist as a number. It has no b-adic representation. It > > > exists as a geometric entity. It is an idea. > > > > It exists as the continued fraction [1,2,2,2,2,...]. It also exists > > as 1 in the base sqrt(2) notation. > > And it exists in many other disguising. But it does not exist as a > number which can be put in trichotomy with all other numbers. Why not? Anyhow, I presume that you use linear algebra on occasion and eigenvalues of matrices. How can you use things that do not exist? You should know it, in particular if talking about transfinity and transfinite induction. But to answer your question: I can use the idea. I can use the undisputed fact that the square of this idea is 2. Regards, WM > But whatever, did you read Hardy and Wright? Yes, also Zuckerman, Niven and Weil, about 15 years ago. Sorry, that is not a valid HTML document and will not display properly. It is mhtml. Regards, WM > An existence proof does *not* show that it is realisable. A proof showing that something is not realizable is the contradiction of an existence proof, because the existence does not have a room other than in the mind which is provably incapable of housing it. Dik: I think it is excluded in your finitistic world. And, I think, that indeed the universe is not able to provide the computer power to actually observe such a sign change. To do so would mean to actually calculate pi(x) for x about 10^316. The best current method has a time complexity of O(x/(ln x)^3) and a storage complexity of O(x^(1/2)/(ln x)). WM Please calculate that in your infinitistic world - if you can. Regards, WM
From: MoeBlee on 1 Nov 2006 13:47 mueckenh(a)rz.fh-augsburg.de wrote: > Sorry, to say, but you always mix up these two very different things. > It should be comprehensible that potential infinity is possible without > the axiom of infinity. But that is not what set theory requires. > Therefore it is correct to say that in set theory theory there is no > infinity present or detectable without the axiom of infinity. "Detactable", informally, okay. Yes, without the axiom of infinity, in set theory (throughout all these discussions, by 'set theory' I mean, in any given instance, some given Z or NBG variant) we cannot prove there exists an infinite set. So, in that sense, we cannot "detect" the existence of an infinite set. But we still are not permitted to claim that infinite sets do not exist. We can only say that we cannot ever detect whether they exist or not. You use these words such as 'is present', 'detecable', etc., which are your words for your OWN way of comprehending. But set theory is NOT in those terms that you use to try to comprehend set theory. Instead of your own METAPHORICAL language, let's look at the formulas (or English renderings of actual formulas) of set theory: Without the axiom of infinity, we cannot prove either of these two formulas ('finite' and 'infinite' I define, respectively as 'equinumerous with a natural number', 'not equinumerous with a natural number'): Ex x is infinite ~Ex x is infinite So without the axiom of infinity (but with no added axioms to the ZFC axioms), we cannot prove: Ax x is finite To prove that formula, we need to adopt an axiom that entails it. Just dropping the axiom of infinity does NOT entitle us to conclude Ax x is finite. Would you please just say whether you understand this point. MoeBlee
From: MoeBlee on 1 Nov 2006 13:53
MoeBlee wrote: > "Detactable", > 'detecable' I meant 'detectable' of course. MoeBlee |