Cantor Confusion
in [Math]
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From: WM on 6 May 2007 10:53 On 6 Mai, 04:33, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > 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? No. You ma yask for a set of infinitely many paths, but not 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. It is proven by Cantor that there are uncountably many real numbers. It is also clear that there are only countably many questions, no? > > > > > 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. That's why it fails for numbers with infinitely many digits. > 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'. No. Only for those p' you can ask for. There must be others, because there is no node which belongs only to p alone. > > > > > 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? I mean that the speech of "all or complete or finished" is in clear contradiction with "infinite". > > > > > 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. Sure. > > > > 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. Please read carefully. We cannot do more than countably many. And in finite time, we cannot do more than finitely many. > 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. Of course this is only true for finitely many n. For infinitely many natural numbers, we get aleph_0. > > > > > 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? It holds for finitely many natural numbers. The natural numbers are a countable set. > > > 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. Have you ever tried to handle a natural number with 10^100 digits which are not given by an abbreviation which can be defined by less than 10^100 bits? Or do you doubt the existence of those natural numbers? Regards, WM
From: WM on 6 May 2007 12:05 On 6 Mai, 11:34, MoeBlee <jazzm...(a)hotmail.com> wrote: > > What you say is inane. What you say is insane. > (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. Function, as understood in mathematics, is a procedure, a rule, assigning to any object a from the domain of the function a unique object b, the value of the function at a. A function, therefore, represents a special type of relation, a relation where every object a from the domain is related to precisely one object in the range, namely, to the value of the function at a. Please do not believe that I will repeat for another 1000 postings your controversy with other great mathematicians about the properties of a function. > > (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. Best by avoiding any formula and any clear idea what you do. > > You are wrong to assume that I did not study set theory. > > I infer it by the ignorance you display. Obviously you are not very good in concluding from given facts on hidden facts. > > > 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. Then use your axioms to understand the binary tree. > > You've not given any set theoretic definition of > "2-1-1+2-1-1+2-1-1+-... " Everybody who has studied a bit of set theory should know 2-1-1+2-1-1+2-1-1+-... < 2+1+1+2+1+1+2+1+1+... = alep_0 So whatever misunderstanding you may want to apply, 2-1-1+2-1-1+2-1-1+-... < 3 < aleph_0. > > If you want to talk about an infinite summation, then you need to be > able to eliminate the '...' and put it in the form: Sum(from n to oo) > F(n). Sum [n in N] (2-1-1) < 3. That is but another way of writing 2-1-1+2-1-1+2-1-1+-... < 3. If you have problems with infinite sums or products then you should look up set theory, for instance Koenigs theorem. Regards, WM
From: William Hughes on 6 May 2007 14:16 On May 6, 10:53 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > Fine. That means, the set is countable but there is no bijection with > N definable. No it means that while a bijection exists it is not finitely definable. Outside of Wolkenmuekenheim this is a result. Inside of Wolkenmuekenheim it is not a resut because Muekenheim goes "WAAH WAAH WAAH", I can't hear you" if anyone tries to explain it to him. - William Hughes Why then do you require a definable bijection between > paths and nodes in the tree? > > Regards, WM
From: Virgil on 6 May 2007 16:10 In article <1178458775.741379.141710(a)n59g2000hsh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. Since the "diagonal" differs from each listed member at some finite place value, what happens at later place values is irrelevant to whether they are equal or different. While there may be some subsequence of the original which converges to the diagonal, no member of that subsequence EQUALS that diagonal, which is all that is required. > > > > 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? The rule by which the diagonal is constructed makes it differ from both representations when those representations have equal values. In any case, for the set of all binary sequences (not numbers) no two are the same, so the issue does not arise.
From: Virgil on 6 May 2007 16:13
In article <1178458882.422814.275130(a)h2g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 5 Mai, 19:09, William Hughes <wpihug...(a)hotmail.com> wrote: > > On May 5, 7:42 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 3 Mai, 16:05, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > At every node K(p, n) of p there is a single fixede path p', which > > > > > has been with p for all nodes K(p, m < n). Yes, without pink > > > > > elephants, we can find at every node K(p, n) such a path p' > > > > > > Look over there! A pink elephant! > > > > > > p' never changes when n changes. > > > > > > >So p is never single. > > > > > > However, it is not necessary for p to be single for p to be separated > > > > from > > > > a path p'. It is possible for p to be separated from every p' without > > > > p ever being single. > > > > > That is obviously nonsense for linear sets which, after having been > > > separated, never meet again. > > > > Thats right, after the last path p' has separated p has to be single. > > Whoops, how does one show there is a last path. > > That has nothing to do with a last path. > > If in the infinite binary tree a path does not get single, then it > will never exist as a single. For every subset of the set of levels in a CIBT there is a SINGLE path branching left from just those levels and right from all others. How much more "single" does a path have to get? |