Cantor Confusion
in [Math]
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From: Dik T. Winter on 5 May 2007 22:33 In article <1178191761.058003.238240(a)c35g2000hsg.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 1 Mai, 04:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > That is wrong in many cases. That's probably why mathematics is wrong. > > > Example: Each natural number is a single number. All natural numbers > > > are not a single number. > > > > All natural numbers *are* a single number. You are confusing "all" with > > "the set of all". > > And are all men a single man too, in your opinion? What meaning of single are you using here? 1. Not married 2. Unaccompanied by others 3. Consisting of or having only one part... 4. Consisting of a separate unique whole: "individual" <every single citizen> 5. (not relevant) 6. (not relevant) 7. having no equal or like 8. (not relevant) So my answer is yes if you are using meanings (4) or (7). > > Your conclusion is invalid. Indeed, each node is passed by a mutlitude > > of paths. But the path "0.10101010..." is nevertheless separated from > > all other paths. > > No. Only from those few you can ask for. And those are (in your opinion) only finitely many. Or can I ask for infinitely many paths? > > > That is irrelevant, because there is no node passed by less than > > > infinitely many paths. > > > > So what? For each two different paths it is possible to state a node where > > they do separate. For each path there is no node where it separates from > > all other paths. There is no contradiction. > > No. Only from those few you can ask for. Opinion again. > > > But it means that p is never single. If it is never single, then it is > > > at least double, i.e., there is at least one path p' with it. > > > > As long as you remain in the finite part. > > You always remain there. In particular Cantor's diagonal proof always > remains there. There is a subtle difference. With the diagonal proof we always remain in the finite. With the paths we can also always remain in the finite because any two different paths have a node where they differ *in the finite*. So for each node there is a path p' that has it in common with p. But also for each p' != p there is a node in p that is not in p'. > > > I am thinking in the terms of Cantor's diagonal p. "Up to all nodes" > > > we must be able to state that p is different from every other number > > > of its bunch of paths. But that is wrong, because p is never single. > > > > As I have no idea what you mean with "up to all nodes", it still makes no > > sense. > > "Up to all nodes" means the sequence of nodes has been worked > completely. But infinite paths do not have a final node, so what do you mean? > > > > We can name all natural numbers, just because that means that we can > > > > name each natural number. > > > > > > Wrong. If you could name all, then no one would remain unnamed. > > > > Just semantics. In mathematics when something is true for each element > > of a set it is true for all elements of a set. > > All elements of the set of theorems are a theorem? Yes. Pray note that theorems imply previous definitions and possibly axioms. So what is a set of theorems in the context of some axioms may not be a theorem in the context of another set of axioms. So the theorem that 'i' forms the algebraic closure of the 'numbers' is only true within the context of particular definitions and axioms. It is *not* true in the context of the p-adics numbers. > > > > It is your simplified finitistic view where that > > > > means that there is a larger number. > > > > > > It is the meanig of "completed". > > > > Your meaning, perhaps. > > "completed" means nothing remains. Perhaps. But as "completed" as such is not a mathematical term, I wonder what you mathematically *do* mean with it. > > > Deplorably this mathematics leads to the result that there are more > > > splitting results than splittings in the binary tree. > > > > As your opinion of "splitting results" is worthless, this makes no sense. > > First try to properly define how bunches split. Until now, with your > > definitions, I have not seen any splitting of bunches at all. > > A bunch terminate by splitting into two bunches, i.e., by separating > these two bunches. The number of separated bunches is and always > remains countable. See my other article on this. > > > > And for each two unequal paths p and p' there is an n such that > > > > K(p, n) != K(p', n). > > > > > > Has no relevance at all. For every node we know that p is not alone. > > > > Yes. So what? > > In the present state of affairs we know that at every node of path p > which we look at, path p is not unique. And we know, that we can test > and test for different paths p', but we cannot carry out more than > countably many tests (our list is limted). You can do countably many tests? I thought you could only do finitely many tests. But again, in mathematics it is *not* necessary to test each and every individual. Otherwise you could not even prove that sum{i = 1..n} i = (n + 1) * n / 2 for all n, but only for finitely many n. > > > 0.000... is not unique in the tree. The number of unique numbers is > > > countable. > > > > Define unique, and we can talk. With current definitions, the number is > > uncountable. > > A path is unique when we can prove that it is different from every > other path, not only from the few paths we can mention and test, but > from all the uncountably many paths we can never mention. But we can mention and test only finitely many paths. So you now say that there are only finitely many paths? And that sum{i = 1..n} i = (n + 1) * n / 2 holds only for those n for which it has been tested? > In the present state of affairs we know that at every node we look at, > the path p is not unique. Yes. But to mathematicians that is not a problem. Stronger, try to factorise largish numbers without the use of numbers that are not representable by a finite number of digits. All those numbers that are used in methods like the Number Field Sieve are unique paths within your trees, although you are not able to find a distinguishing feature at some finite point in your trees. -- 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 5 May 2007 22:40 In article <1178366406.361131.321710(a)y5g2000hsa.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: .... > Theorem 2: The set of finitely defined numbers cannot be put into a > list. > > Proof: Assume such a list exists. This means, such a list has been > finitely defined. Wrong. Existence does *not* mean finitely definable. Have a look at Turings work. There *does* exist a (finitely deinable) list of Turing machines that purport to generate numbers. So the list is certainly countable. However, the subset of that list that actually *does* generate numbers is not finitely definable. -- 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 5 May 2007 22:54 In article <virgil-8C7341.12472203052007(a)comcast.dca.giganews.com> Virgil <virgil(a)comcast.net> writes: > In article <1178191761.058003.238240(a)c35g2000hsg.googlegroups.com>, > WM <mueckenh(a)rz.fh-augsburg.de> wrote: .... > > > All natural numbers *are* a single number. You are confusing "all" with > > > "the set of all". > > > > And are all men a single man too, in your opinion? > > Dik's English often reflects the idiom of his native tongue, Dutch, I > believe, but that does not hide his idea that all natural numbers are > individually single numbers, just as all men are individually single men. Dutch, indeed. But I have read thousands (yes, indeed) English books since I was 13. My writing may reflect my native tongue, my reading is pretty well (I think). And what Wolfgang states is just his usual balderdash. Harping on many different meanings in colloquial English while trying to conduct a mathematical discussion. When he had looked up a dictionary he would have found the several meanings of "single" in English. And most of them would not translate to a "single" term in either Dutch or German. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: MoeBlee on 6 May 2007 05:34 On May 5, 4:54 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 3 Mai, 18:13, MoeBlee <jazzm...(a)hotmail.com> wrote: > > There are no implicit assumptions in formal Z set theory. > > What you say proves that you never have thought about that topic deep > enough. One implicit assumption is that, in Hessenberg's proof > ofCantor'stheorem, the set of all natural numbers which are not mapped > on subsets of P(N) which do contain them, does exist together with the > mapping pescription (you know that a function consists of formula, > domain and range, I hope. Such a set need exists as little as the set > which the largest natural number is mapped upon. What you say is inane. (1) If a presentation of a proof uses any principles or premises whatsoever other than first order logic applied to the Z axioms, then that presentation is not that of a Z set theory proof. So, to reiterate, since you are so very obtuse, if you find that some presentation or another uses some premises or principles other than first order logic applied to the Z axioms, then that presentation is not of a Z set theory proof. (2) Usual textbook presentations of the theorem that no set maps onto its power set use nothing but first order logic applied to the axioms of Z set theory. (3) In ordinary set theory, a function is a certain kind of set of ordered pairs and, in ordinary set theory, a function is NOT a triple of a formula, domain, and range. That you even think a set theoretic function must have a FORMULA (!) as a component shows, again, your lack of understanding even the basics of set theory. (4) I don't know about Hessenberg's presentations, but in ordinary presentations, even though we don't need to prove by contradiction, it is sometimes easy to set the proof up that way. Then from the reductio assumption of a function mapping N onto PN, as to the set of elements of N that are not mapped to a member of PN, we PROVE that set exists by a simple use of the axiom schema of separation. That you even question that shows that you do not understand even such basic things about set theory as the axiom schema of separation. (5) As to the function from N onto PN, again its existence is either a reductio assumption, or, by modus tollens or some other principle that is included in first order logic, we prove such a function does not exist. > You are wrong to assume that I did not study set theory. I infer it by the ignorance you display. > But the > formalism veils the problem it is built upon. Whatever you think that means, my point stands: Proofs of theorems of formal Z set theory make no use of anything other than first order logic applied to the axioms. Your claim about 'implicit assumptions' is pure bull. > > in formal Z set theory, with no implicit assumptions, that the set of > > real numers is uncountable. > > Even if this was clear, It's clear if you'd read a good textbook such as Enderton's or Suppes's. > then the result > 2-1-1+2-1-1+2-1-1+-... <= 2 > would show that set theory is self contradictive. You've not given any set theoretic definition of "2-1-1+2-1-1+2-1-1+-... " If you want to talk about an infinite sumation, then you need to be able to eliminate the '...' and put it in the form: Sum(from n to oo) F(n). You'll see that once you force yourself to move on from UNDEFINED expressions (such as "2-1-1+2-1-1+2-1-1+-... ") to actually defined notation of set theory in which an infinite summation is the limit of a sequence of sums each of which is based on a function F, then the contradictions you claim to have found are in fact not any contradiction in set theory itself. That is, you ought not conflate your undefined expressions with any rendering of a term or formula of set theory. Set theory does not have to answer to your fingerpainting with symbols. MoeBlee
From: WM on 6 May 2007 09:39
On 3 Mai, 20:18, Virgil <vir...(a)comcast.net> wrote: > In article <1178189738.719781.210...(a)y80g2000hsf.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 30 Apr., 20:04, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > On Apr 30, 5:35 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > On 27 Apr., 22:06, Virgil <vir...(a)comcast.net> wrote: > > > > Why then do you believe that Cantor's diagonal proof is true? > > > > You'd have to define 'true proof'. Meanwhile, with just some basic > > > knowledge of predicate calculus and set theory, one can see that > > > Cantor's diagonal argument is formalizable into a proof in first order > > > logic from the axioms of Z set theory. > > > There is an implicit assumption entering which is wrong. > > The difference of 1 between the digits of two numbers is insufficient > > to distinguish these numbers in the limit n --> oo. > > What WM is describing is not a difference between two fixed numbers but > between two sequences of numbers whose differences converge to 0, the > limiting difference is indeed zero but that is totally irrelevant, not for the sequence of digts of the diagonal number and any other entry of the list. > > In base b, numbers of form m/b^n have two expansions. But excluding > these, a difference of 1 or more in any digit position differentiates > between two different reals. It does only differentiate the reals at a finite index. But at a finite index it differentiates as well between 0.999... and 1.000.... So, why do you want to exclude these but not those? Regards, WM |