Cantor Confusion
in [Math]
|
From: mueckenh on 30 Mar 2007 09:35 On 30 Mrz., 05:14, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > No, the sum is undefined. If you think it is defined, *prove* it and > > > > > *prove* that the sum is 2. What is the sum of the "sequence": > > > > > "..., 1/4, 1/2, 1"? > > > > > > > > Well, what is it? If any infinite series ever had a value, then the > > > > sum of this sequence is 2. > > > > > > But it is not a sequence according to mathematical definitions. > > > > It is a sequence according to mathematical definitions, if you read it > > from the left hand side. Oh, that should read "from the right hand side"! But you read it as it was meant. > > So there is a first element, being 1, a second element, being 1/2, the only > difference is that you apply right to left reading. > > > Further you can determine a unique limit value by lim{n-->oo} (1/2^n > > + ... + 1/8 + 1/4 + 1/2 + 1}. > > Yes, because you can revert finite sequences without consequence. You > do not even need convergence for that. And if the series is absolutely converging, then you can exchange all terms you like. The result is independent of the order. > But in mathematics sequences > are defined as having a first element. On the other hand, I wonder how > you prove that the series of interchanges on the initial sequence lead > to your final "sequence". It is the same as Cantor's "proof" that he gets ready. > > > > > Proof: Sum the first n terms from the right hand side and find that > > > > the difference of their sum and 2 is not more than 1/2^(n-1). Further > > > > find that the sum is always less than 2. Voila. > > > > > > Ok, what node contributes '1' to that sum? I ask this because you want it > > > to apply to nodes contributing things to a path. > > > > You know, this node cannot be determined. Therefore I use the limit > > (which does exist): > > lim{n-->oo} (1/2^n + ... + 1/8 + 1/4 + 1/2 + 1} = 2 > > Yes, you use limits to show something which you can not determine. The > strange thing is that the first node contributes nothing, as does the > second nodes, as do all nodes in a finite distance from the root. And > as all nodes in an infinite path are a finite distance from the root. > Nevertheless you maintain that all nodes together contribute 2 because > there is no node that contributes one, there is no node that contributes > 1/2, etc. So there is a sequence of no nodes that contribute 2. And in > some mysterious way you conclude that that sequence of no nodes is the > same as the sequence of nodes. Take the geometric series. It contains exactly the same terms as my reverted series. > > > > > > Yes, and when are we done? > > > > > > > > When is Cantor done? > > > > > > Well, at each step in the reversal process you have a sequence with a first > > > element and no last element. I do not know of a way to define what the > > > result is after infinitely many steps. > > > > Do you know of a way how to finish Cantor's diagonal? > > Is there any need to finish it at all? Yes, if you want to conclude that it is different from any other list entry, then it must be finished. Otherwise you only know that it differs from some entries. > Apparently you see a need to finish > it. For the proof it is only needed to show that there *is* a real number > that is not on the list. The algorithm that describes that real number is > sufficient. You cannot say: "there is a real number that is not in the complete list" unless you have searched the complete list. > > > > What is the first element after > > > infinitely many steps? And how do you define that at all? Or is your > > > definition simply that it is {..., 3-rd term, 2-nd term, 1-st term}? > > > > My definition is that every finite part of the sequence can be > > reversed. "Every finite" part of a countable set means "all" - this is > > just like in Cantors diagonal. > > You are wrong here. And what you state is *not* a definition. Pray start > to distinguish "definition" from "theorem". That distinction depends on the axioms chosen. It is not absolute. > "Every finite part" means > just that: "every finite part", it does not apply to "infinite parts", > which is "all". On the other hand with Cantor we have "every finite element" > and that means "all elements", because there are no "infinite elements". > > > > > Ever tried to count the students attending a lesson or the peas in a > > > > cup full of peas? > > > > > > Why should I? To perform the n-th exchange in Cantor's process you do > > > *not* have to do the first n-1 preceding exchanges first. > > > > How would you know which element the n-th element is. > > By putting n in the mapping given. The mapping given includes and requires the counting. (That reminds me of an anti-nuclear-power-station fighter who argued: We don't need power stations, our electricity is supplied by the electrical outlet at home.) > > > > Therefore we know that only countably many are there. > > A proof, please, for once. 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. Regards, WM
From: mueckenh on 30 Mar 2007 09:38 On 30 Mrz., 05:23, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1175178082.736299.205...(a)l77g2000hsb.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 29 Mrz., 04:14, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1175105032.482322.292...(a)d57g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > ... > > > > > > > Not in the uniting, no. But the union *does* contain infinite > > > > > > > paths. Just like the union of all finite initial segments of the > > > > > > > natural numbers does have as a subset an infinite initial segment > > > > > > > of the natural numbers. > > > > > > > > > > > > Just like the set of all negative unit fractions does cover zero? > > > > > > > > > > What does *this* mean? The set of all negative unit fractions does > > > > > contain infinite subsets. *That* is what we are talking about. > > > > > > > > But these infinite subsets do not cover zero. > > > > > > And what is the relevance? We are talking about infinite subsets, not > > > about what they "cover", whatever that may mean. > > > > Infinite sets of paths need not contain or be infinite paths. > > Again "contain". But again you are pretty careless. Infinite sets of > finite paths do not contain infinite paths. All elements are finite paths. > Nor are they an infinite path because the are a set of paths. It is when > you *unite* infinite sets of paths that you *can* get infinite paths. 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? Regards, WM
From: Virgil on 30 Mar 2007 15:07 In article <1175261756.864645.326740(a)p77g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 30 Mrz., 05:14, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > Yes, because you can revert finite sequences without consequence. You > > do not even need convergence for that. > > And if the series is absolutely converging, then you can exchange all > terms you like. The result is independent of the order. But only if the result is an infinite sequence with terms indexable by, and therefore ordered by, the naturals. > > > But in mathematics sequences > > are defined as having a first element. On the other hand, I wonder how > > you prove that the series of interchanges on the initial sequence lead > > to your final "sequence". > > It is the same as Cantor's "proof" that he gets ready. The transposition of the first and second terms of a sequence followed by the transposition of the first and third produces a different result than the transposition of the first and third followed by the transposition of the first and second. abc -> bac -> cab versus abc -> cba -> bca Thus the order of application of transpositions makes a difference. The replacement of members of a sequence by a rule depending only on the value and not position of the member being replaced is independent of the order of operations. let the rule be to replace any lower case letter by its upper case equivalent. abc-> Abc -> ABc -> ABC is the same as abc -> abC -> AbC -> ABC even though the operations were differently ordered. So the Cantor rule for building an antidiagonal for a list of binary sequences can be applied independently to different digits whereas WM's sequence of transpositions depends on the order of application, at lest if any two transpositions involve the same index position. So that WM's theories are, as usual, based on assumptions contrary to fact. > > You cannot say: "there is a real number that is not in the complete > list" unless you have searched the complete list. In a sequence of binary strings, if one can say of some given string that it differs from each member of the sequence in at least one index position, then one can say that the given string is not in the sequence. The Cantor rule shows that there exists just such a string. > > You are wrong here. And what you state is *not* a definition. Pray start > > to distinguish "definition" from "theorem". > > > That distinction depends on the axioms chosen. It is not absolute. Then give us a system of axioms in which what you allege becomes only a definition and not a theorem. We have given you systems in which it is false. > > > > > > > > Why should I? To perform the n-th exchange in Cantor's process you do > > > > *not* have to do the first n-1 preceding exchanges first. > > > > > > How would you know which element the n-th element is. > > > > By putting n in the mapping given. > > The mapping given includes and requires the counting. But does not require that one do the exchange for any m < n before doing the exchange for n. For each n, we can count to n and do the exchange for that n independently of having done, or not done, the same for any other n. Why WM insists that we have to replace the digit in position n before working on position n+1, is not apparent, as it is not the case. On the other hand, the same independence of order of application does NOT hold for two transpositions where the same index is involved in both. > > > Therefore we know that only countably many are there. > > > > A proof, please, for once. > > Every separation takes place at a separation point. Does WM claim that P(N), the power set of N, is countable? Does WM dispute the bijection between paths and members of P(N)? Since both or these are eminently provable in ZF and NBG, and no doubt most other set theories, WM has a hard row to hoe. No separation > takes place at any other point. There are only countably many > separation points. And there is only one initial separate path. > > Regards, WM
From: Virgil on 30 Mar 2007 15:19 In article <1175261931.220013.38850(a)e65g2000hsc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de 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? Because a path as a set of nodes in an infinite tree, according to any reasonable definition of paths, can be, and must be, infinite in that infinite tree, but naturals cannot be infinite anywhere. Does WM deny that a path, as a set of nodes in a given tree, must be maximal in that tree in the sense that inserting any other node into such a set of nodes makes it no longer a path in that tree? WM must be denying it, as that is the only way his claims make any sense, but that means what he is calling paths are not the same as what everyone else is calling paths. Note that what may be a path in a finite tree need not be a path in any extension tree {an extention having the same root nodes and including all of the original nodes, but some or all of the leaf nodes of the original no longer being leaf nodes).
From: mueckenh on 1 Apr 2007 04:04
On 30 Mrz., 21:07, Virgil <vir...(a)comcast.net> wrote: > In article <1175261756.864645.326...(a)p77g2000hsh.googlegroups.com>, > > mueck...(a)rz.fh-augsburg.de wrote: > > On 30 Mrz., 05:14, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > Yes, because you can revert finite sequences without consequence. You > > > do not even need convergence for that. > > > And if the series is absolutely converging, then you can exchange all > > terms you like. The result is independent of the order. > > But only if the result is an infinite sequence with terms indexable by, > and therefore ordered by, the naturals. Every countable sequence can be indexed the naturals. In my application here is no problem. > > > > > > But in mathematics sequences > > > are defined as having a first element. On the other hand, I wonder how > > > you prove that the series of interchanges on the initial sequence lead > > > to your final "sequence". > > > It is the same as Cantor's "proof" that he gets ready. > > The transposition of the first and second terms of a sequence followed > by the transposition of the first and third produces a different result > than the transposition of the first and third followed by the > transposition of the first and second. > > abc -> bac -> cab versus abc -> cba -> bca > > Thus the order of application of transpositions makes a difference. That is not a problem at all. We can work like the cleaning service of Hilbert hotel: For the first sequence of transposition use half an hour, for the second sequence use quarter an hour and so on. > > The replacement of members of a sequence by a rule depending only on the > value and not position of the member being replaced is independent of > the order of operations. let the rule be to replace any lower case > letter by its upper case equivalent. > > abc-> Abc -> ABc -> ABC is the same as abc -> abC -> AbC -> ABC > even though the operations were differently ordered. > > So the Cantor rule for building an antidiagonal for a list of binary > sequences can be applied independently to different digits Nevertheless it cannot be applied to the n-th digit unless the positions 1 to n-1 are known. Because otherwise position n is unknown. This may be masked, for small n, by your knowledge acquired at school but it will become manifest at larger n. Regards, WM |