Cantor Confusion
in [Math]
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From: Virgil on 1 Apr 2007 15:42 In article <1175414674.648660.163490(a)d57g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 30 Mrz., 21:07, Virgil <vir...(a)comcast.net> wrote: > > > > 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. If, for example, the nth transposition exchanges the current occupants of positions n and n+1, what is the final position of the object originally in first position? If it does not have a final position, then what you have constructed is not a permutation of the members of the list. > > > > 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. It can be applied, as shown above, before anything is /applied/ to prior positions, which is all that is needful for the validity of the Cantor proof. WM's pseudo-permutation is, unlike Cantor's rule, dependent on order of application.
From: Dik T. Winter on 2 Apr 2007 08:34 In article <1175178657.434003.303550(a)d57g2000hsg.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 29 Mrz., 04:23, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > It still says *nothing* about the number of single paths. > > > > > > An uncountable number of paths existing separated from each other > > > would form a level with uncountable man nodes. > > > > Pray, for once, show a *proof* of this statement. > > It is a result of set theorey that uncountably many is greater than > counatbly many. If you have a greater set, then there must be more > elements than in a smaller set. If n elements exist in a set, then > they must exist simultaneously. That means there must be some domain > where this happens. Yes. The domain is the set of paths. But you state there must be a *level* where it happens. Pray show a for once a *proof* of that statement. > > > But we know that there > > > cannot be such a level, including all infinity of the complete tree. > > > > Indeed, there is not. Nevertheless there are uncountably many infinite > > paths. > > That is obviously wrong. The necessity of as much separation points as > separated paths is not restricted to the finite tree. It is required > in any case. Otherwise there must be paths with no connection to the > root node. But those constructs are not paths. Show a *proof* of that necessity for infinite trees. > > You do not believe it, but you fail to prove it. It is just your > > insistence that if all paths do separate from each other that there must > > be a level where all paths are separated from each other. > > In particular, there must be all separation points in the tree. > > You say: There are all uncountably many separated paths in the tree. > But there are not all uncountably many points of separation in the > tree. Right. And that is provable. > Obviously bad logic. But you say, it is good logic. So let it be. > Antilogic cannot be disproved by logic. You are indeed not able to disprove it. Simply because you do not understand the logic. -- 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 2 Apr 2007 08:43 In article <1175261756.864645.326740(a)p77g2000hsh.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. 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. > > 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. It is not. > > > 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. What is that in relevance to my remark? > > > 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. No. You can *prove* that whatever entry you take the diagonal is different. And, this is independent of the entry you take. So the diagonal differs from *all* 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. You can. If you can prove there is a number that is different from each member of the list, as is done. > > > > 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. Nevertheless, you are wrong. > > "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". You have no remark to this, which *shows* you are wrong? > > > > 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. Why? How do you know? If the mapping is (for instance) f: N -> R, f(n) = sqrt(n), I see no counting involved at all. > > > 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. Yes. So what? -- 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 2 Apr 2007 08:46 In article <1175261931.220013.38850(a)e65g2000hsc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 30 Mrz., 05:23, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > 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? Careless again. You get an infinite number, but not an infinite natural number. The reason is simply that natural numbers are finite by definition. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 4 Apr 2007 05:29
On 2 Apr., 14:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1175178657.434003.303...(a)d57g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > It is a result of set theorey that uncountably many is greater than > > counatbly many. If you have a greater set, then there must be more > > elements than in a smaller set. If n elements exist in a set, then > > they must exist simultaneously. That means there must be some domain > > where this happens. > > Yes. The domain is the set of paths. But you state there must be a *level* > where it happens. Pray show a for once a *proof* of that statement. That is simple. There are no parts of paths outside of any leve. Paths exist only where levels are. > > > > > > > But we know that there > > > > cannot be such a level, including all infinity of the complete tree. > > > > > > Indeed, there is not. Nevertheless there are uncountably many infinite > > > paths. > > > > That is obviously wrong. The necessity of as much separation points as > > separated paths is not restricted to the finite tree. It is required > > in any case. Otherwise there must be paths with no connection to the > > root node. But those constructs are not paths. > > Show a *proof* of that necessity for infinite trees. Whether infinite or not: The laws of logic remain valid. Even in the infinite sequence 121212... theer is no point where 1 gets larger than 2. Such laws show that there is no separated path without a separation point. > > > > You do not believe it, but you fail to prove it. It is just your > > > insistence that if all paths do separate from each other that there must > > > be a level where all paths are separated from each other. > > > > In particular, there must be all separation points in the tree. > > > > You say: There are all uncountably many separated paths in the tree. > > But there are not all uncountably many points of separation in the > > tree. > > Right. And that is provable. But only by another proof. Therefore there is an inconsistency in set theory. > > > Obviously bad logic. But you say, it is good logic. So let it be. > > Antilogic cannot be disproved by logic. > > You are indeed not able to disprove it. Simply because you do not > understand the logic. Not what you pretend to be understood by logic. Regards, WM |