Cantor Confusion
in [Math]
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From: Virgil on 6 May 2007 16:26 In article <1178461785.380972.277030(a)e65g2000hsc.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 6 Mai, 03:53, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1178190762.081101.159...(a)h2g2000hsg.googlegroups.com> WM > > <mueck...(a)rz.fh-augsburg.de> writes: > > > Wrong. If there is always a path p' with p, > > > > What does this statement mean? > > Even in the *infinite* tree a path cannot be distinguished from all > other paths. That means, a real number which cannot be described by a > finite formula (and most of them cannot) does not exist. What axiom requires that a number must be defined by a finite formula in order to exist? > > > > > Hence Cantor's diagonal proof fails. > > > > Which proof? > > Cantor's diagonal proof fails, because the diagonal number is never > distinguished from all other real numbers (if uncountably many real > numbers exist). Cantor's "diagonal" proof only involves binary strings, not numbers, and a very simple rule distinguishes the "diagonal" form every member of the list. > > > > > > > > Bunches terminate by splitting into other bunches. If they would not > > > split, then they would not terminate. > > > > This is nonsense. Bunches *all* start at the root by your definition. > > If a bunch splits in two bunches at a node that would mean that those > > two new bunches start at the node, and they do not. > > All start at the root node, but they terminate by splitting. Does WM claim that none of the paths in a bunch extend beyond the "termination" of a bunch? And, as yet, WM has given no mathematically satisfactory definition of "bunch" nor of "termination" of a bunch, nor any of his other attempts to confuse by non-definition.
From: Virgil on 6 May 2007 16:33 In article <1178463181.745792.46180(a)y80g2000hsf.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 6 Mai, 04:40, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1178366406.361131.321...(a)y5g2000hsa.googlegroups.com> WM > > <mueck...(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. > > Marthematical existence does mean definable. But not necesssarily "finitely" so. > What else should it mean? > Definitions can only be finite. That may be one of WM's axioms, but does not appear in ZF or NBD. > > > Have a look at > > Turings work. There *does* exist a (finitely definable) 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. > > Fine. That means, the set is countable but there is no bijection with > N definable. Why then do you require a definable bijection between > paths and nodes in the tree? We don't! In fact, we reject any such thing in any tree with more than one node. There is, however, a bijection between the power set of the set of non-leaf levels (in any tree in which all paths are of the same finite length or are all endless) and the set of paths in that tree.
From: Virgil on 6 May 2007 17:02 In article <1178463202.150771.47030(a)y80g2000hsf.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. So one can ask for, and get, a set with infinitely many elements but it will not have infinitely many elements? That does not happen is a well regulated logical system. > > It is proven by Cantor that there are uncountably many real numbers. > It is also clear that there are only countably many questions, no? Define question! > > > > > > > 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. If it does not fail for infinite binary strings, and it doesn't, then it does not fail for numbers either. > > > 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. It must be true for every pair of paths whatsoever or else one does not actually have a complete infinite binary tree at all. > > > > > > "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". Not in ZF or NBG. What axiom system are you proposing in which things are otherwise? > > > > > > > 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 ZF and NBG there are infinite sets which are "completed" in the sense that nothing remains other than that which is already in them. So in ZF and NBG, your "completed" is irrelevant. If you want a system that behaves as you want it to, you will have to build it yourself, as our systems just don't work your way. > > > > > > > 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. As ZF and NBG are not time-bound, that is irrelevant in them. WM is quite free to include time in his axiom system, whenever he gets to finishing it, but it does not appear in ZF or NBG. > > > 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. When we get if for ALL, that means for every member of the set of all naturals, but not for the set itself, any more that a set of horses has to be a horse. > > > > > > > 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. Then WM is saying that it does NOT hold for some members of N. In ZF and NBG, when one gets around to defining the set of natural numbers N, one of N's properties is induction: (1) the first (0 or 1, depending on one's taste) is a member (2) for each member , x there is another called succ(x) such that (a) succ(x) is not the first, or is it x (b) for x and y in N, Succ(x) = Succ(y) iff x = y (3) If K is a subset of n such that (a) the first member of N is a member of K and (b) whenever x is a member of K so is succ(x) then K = N Thus in ZF and NBG, the kind of anomaly where sum{i = 1..n} i = (n + 1) * n / 2 is only true for finitely many n in N cannot happen, because one can show it true for n = 1 and also show that if true for n = x it is also true for n = succ(c), thus it must be true for all n in N.
From: Virgil on 6 May 2007 17:14 In article <1178467505.278081.271080(a)o5g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 6 Mai, 11:34, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > What you say is inane. > > What you say is insane. What Wm says is insane, at least in ZF and NBG. And he has yet to produce any system in which it is not 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. A function may be a instantiated as a procedure, but in ZF, it must be a set, as everything in ZF is a set. > > 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. It is perfectly clear to mathematicians. If it is not to WM, that is not the fault of mathematics. > > > > > 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. AS WM's assertions about set theory are quite often false in ZF and NBG, and he does not provide us with any system in which we can find them true, we are driven to conclude that WM's knowledge of set theory is considerably less than what he claims it is. > > > > > > > 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. We have, and it gives us in ZF an infinite tree in which the paths biject with the power set of N. This tree is so different from any tree WM visualizes, that WM cannot be using any recognizable set theory.
From: William Hughes on 6 May 2007 18:59
On May 6, 9:41 am, WM <mueck...(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. The fact that: p' is never single is true even though p can be separated from every p' and if p' separates from p it never rejoins p. has nothing to do with a last path Whoops! this is wrong. Look! Over there! A pink elephant! The fact that p' is never single is true has nothing to do with a last path - William Hughes |