Cantor Confusion
in [Math]
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From: Virgil on 7 Apr 2007 14:54 In article <1175967792.541552.222680(a)q75g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 4 Apr., 15:48, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1175680624.978978.36...(a)y80g2000hsf.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 2 Apr., 14:46, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > If you unite natural numbers, you cannot get an infinite number but > > > > > only an infinite set of natural numbers. Why is this different for > > > > > paths? > > > > > > > > Careless again. You get an infinite number, but not an infinite > > > > natural > > > > number. The reason is simply that natural numbers are finite by > > > > definition. > > > > > > Not by definition, but by nature. > > > > Natural numbers have nothing to do with nature. From the definition it > > immediately follows that all natural numbers are finite. > > Without nature (or reality, as we say today) there would not be any > natural number. Without nature, there would not be any people to argue about natural numbers, but without people's imaginations, there would be no place for those naturals to exist.
From: Virgil on 7 Apr 2007 15:21 In article <1175968689.240804.64510(a)e65g2000hsc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 4 Apr., 16:00, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1175681058.802334.28...(a)e65g2000hsc.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 2 Apr., 14:43, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > And if the series is absolutely converging, then you can exchange > > > > > all > > > > > terms you like. The result is independent of the order. > > > > > > > > Not arbitrarily. The result must also be a sequence. If I start with > > > > the > > > > sequence (1, 1/2, 1/4, 1/8, ...) I have a convergent series. If I > > > > apply in sequence the transpositions (1 2), (2 3), (3 4), ... I do not > > > > know what I get because it is not defined in mathematics. On the > > > > other > > > > hand, the only reasonable definition for a result I can come up with > > > > is > > > > the sequence (1/2, 1/4, 1/8, ...), because there is *no* place for the > > > > element "1". So the sum is changed. > > > > > > You can mirror the series without change of value. You can decide to > > > write from right to left. > > > > Yes. In that case the mirroring is just virtual and the first number is > > on the right. But that means that the '1' applies to the *first* node > > in the path. I do not think you want that... > > The mirror is not "just virtual" unless you claim that the infinite > paths exist only virtually. All mathematics is equally virtual. Finite trees and their finite paths are as virtual as infinite trees and their infinite paths, as they exist only in imaginations. > > > > If so, *prove* it. How do you *prove* that your series of interchanges on > > the initial sequence leads to your final "sequence"? > > By definition the final sequence is the result of the sequence of > interchanges. In which case, the final sequence is missing a term. > > > But you cannot prove that for entries which you did not yet take. And > > > that is the majority and remains so. > > > > You take them all at once, by taking an "arbitrary" entry. As the entry is > > arbitrary the proof goes for each and every entry, so in effect you have > > proven it for all entries. > > Wrong. It has been proven always only for a finite part of the list. If one proves something true for ALL naturall numbers, WM has just claimed that there are natural numbers to which that proof does not apply! Which members of the N does such a proof not cover, WM? > > > This kind of proof is pretty abundantly available > > in mathematics. > > That is the reason for the abundance of errors in modern mathematics. Except that WM cannot find any such errors in the math itself but must assume things not assumed in the original mathematics to find his alleged errors. If one assumes, as WM does, the equivalent of 1 + 1 =!= 2 as well as 1 + 1 = 2, then one is bound to find anomalies. But without adding assumptions which are false in the original system, WM cannot fault ZF, or ZFC, or NBG. They are each internally consistent as far as anyone can prove. > > > > > > And, this is independent of the entry you take. So the diagonal > > > > differs > > > > from *all* entries. > > > > > > Wrong. That holds only fnly for finite segments, not for infinite > > > segments of the diagonal. > > > > There is only one infinite segment of the diagonal, and that is the > > complete > > diagonal. So what are you talking about? > > You cannot apply Cantor's proof to the whole diagonal. In particular > because the diagonal does not exist. (It does not exist as a path > separated from all oter paths in the binary tree.) As there is no binary tree involved in Cantor's proof, WM is talking nonsense again. Or is it "still". Cantors's proof shows that given any list of functions from N to {0,1}, there must exist (without any need of contructing it) a function from N to {m,w} not in that list: Given F as the set of functions from N to {m,w}, and given any function g: N -> F, then h defined by h(n) = 1-g(n)(n) is a member of F but for all n e N, h =/= g(n). > > > > > Every separation takes place at a separation point. No separation > > > > > takes place at any other point. There are only countably many > > > > > separation points. And there is only one initial separate path. > > > > > > > > Yes. So what? > > > > > > Yes. That's it. > > > > No that is not a proof. > > If a stone would share your opinion, my chances to convince it would > not be better than to convince you. When one refuses to be persuaded by nonsense, those purveying that nonsense often descend to argumenta ad hominem.
From: mueckenh on 10 Apr 2007 14:32 On 7 Apr., 20:51, Virgil <vir...(a)comcast.net> wrote: > In article <1175967715.832820.167...(a)q75g2000hsh.googlegroups.com>, > > mueck...(a)rz.fh-augsburg.de wrote: > > On 4 Apr., 15:47, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > Such laws show that there is no separated path without a separation > > > > point. > > > > Indeed. But that does *not* prove your statement. As I have said time > > > and time again, for every two paths there is a single node where they > > > separate. On the other hand, there is not a single level where all the > > > paths are separated. > > > Do you claim that all paths exist as separated entities? > > Does WM deny that any two paths distinct are separated in the sense that > they have a last node in common? Do all paths exist separated from one another in the infinite binary tree or not? > > > If so: how or > > where do they exist? > > The paths of an infinite tree exists as subsets of the set of paths of > that tree, just as with finite trees,except that in infinite trees one > has infinite paths. > > > Can you explain what you mean by existence of > > real numbers at all? How are they related to separated paths? > > It is not so much real numbers, but functions from N to {L,R} that > correspond to paths, and any two different such functions will have > different values at some n in N. So you agree that in the infinite binary tree there are all those paths existing and separated from one another? > > > Are the > > numbers represented by paths which are completely within the > > infinitely many levels of the tree or not? > > Every real number in [0,1] is represented by at least one such path and > some by two paths. Why does no cross section (= level) of the tree contain uncountably many paths? Regards, WM
From: mueckenh on 10 Apr 2007 14:36 On 7 Apr., 20:54, Virgil <vir...(a)comcast.net> wrote: > In article <1175967792.541552.222...(a)q75g2000hsh.googlegroups.com>, > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > On 4 Apr., 15:48, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1175680624.978978.36...(a)y80g2000hsf.googlegroups.com> > > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 2 Apr., 14:46, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > ... > > > > > > If you unite natural numbers, you cannot get an infinite number but > > > > > > only an infinite set of natural numbers. Why is this different for > > > > > > paths? > > > > > > Careless again. You get an infinite number, but not an infinite > > > > > natural > > > > > number. The reason is simply that natural numbers are finite by > > > > > definition. > > > > > Not by definition, but by nature. > > > > Natural numbers have nothing to do with nature. From the definition it > > > immediately follows that all natural numbers are finite. > > > Without nature (or reality, as we say today) there would not be any > > natural number. > > Without nature, there would not be any people to argue about natural > numbers, but without people's imaginations, there would be no place for > those naturals to exist.- Zitierten Text ausblenden - The solar system would have eight planets with and without any people's imaginations. Perhaps nobody would call the numer "eíght". But it would be the very number which is also represented by XXXXXXXX or by oooooooo . Regrads, WM
From: mueckenh on 10 Apr 2007 14:42
On 7 Apr., 21:21, Virgil <vir...(a)comcast.net> wrote: > In article <1175968689.240804.64...(a)e65g2000hsc.googlegroups.com>, > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > On 4 Apr., 16:00, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1175681058.802334.28...(a)e65g2000hsc.googlegroups.com> > > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 2 Apr., 14:43, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > ... > > > > > > And if the series is absolutely converging, then you can exchange > > > > > > all > > > > > > terms you like. The result is independent of the order. > > > > > > Not arbitrarily. The result must also be a sequence. If I start with > > > > > the > > > > > sequence (1, 1/2, 1/4, 1/8, ...) I have a convergent series. If I > > > > > apply in sequence the transpositions (1 2), (2 3), (3 4), ... I do not > > > > > know what I get because it is not defined in mathematics. On the > > > > > other > > > > > hand, the only reasonable definition for a result I can come up with > > > > > is > > > > > the sequence (1/2, 1/4, 1/8, ...), because there is *no* place for the > > > > > element "1". So the sum is changed. > > > > > You can mirror the series without change of value. You can decide to > > > > write from right to left. > > > > Yes. In that case the mirroring is just virtual and the first number is > > > on the right. But that means that the '1' applies to the *first* node > > > in the path. I do not think you want that... > > > The mirror is not "just virtual" unless you claim that the infinite > > paths exist only virtually. > > All mathematics is equally virtual. Finite trees and their finite paths > are as virtual as infinite trees and their infinite paths, as they exist > only in imaginations. Wrong. The finite tree 0. /\ 0 1 does exist here, on your screen. > > > > > > If so, *prove* it. How do you *prove* that your series of interchanges on > > > the initial sequence leads to your final "sequence"? > > > By definition the final sequence is the result of the sequence of > > interchanges. > > In which case, the final sequence is missing a term. > > > > > But you cannot prove that for entries which you did not yet take. And > > > > that is the majority and remains so. > > > > You take them all at once, by taking an "arbitrary" entry. As the entry is > > > arbitrary the proof goes for each and every entry, so in effect you have > > > proven it for all entries. > > > Wrong. It has been proven always only for a finite part of the list. > > If one proves something true for ALL naturall numbers, WM has just > claimed that there are natural numbers to which that proof does not > apply! > > Which members of the N does such a proof not cover, WM? Cantor's proof does not cover the number behind the last one proved. Same as: The binary tree does not contain all paths, because otherwise it must contain a level with uncountably many nodes. Regards, WM |