Cantor Confusion
in [Math]
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From: mueckenh on 4 Apr 2007 05:57 On 2 Apr., 14:46, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1175261931.220013.38...(a)e65g2000hsc.googlegroups.com> mueck...(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. Not by definition, but by nature. Regards, WM
From: mueckenh on 4 Apr 2007 06:04 On 2 Apr., 14:43, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1175261756.864645.326...(a)p77g2000hsh.googlegroups.com> mueck...(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. You can mirror the series without change of value. You can decide to write from right to left. > > > > 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. It is. > > > > > 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? To show tha its sum is 2. > > > > > 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. But you cannot prove that for entries which you did not yet take. And that is the majority and remains so. > 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. > > > > 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. Only for finite segments. > > > > "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? The same holds for paths. Every finite node of a path means the whole path because there are no "infinite nodes". > > > > > > 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. That is deplorable. (N is created by counting.) > > > > > 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? Yes. That's it. Regards, WM
From: Dik T. Winter on 4 Apr 2007 09:47 In article <1175678971.644490.125770(a)d57g2000hsg.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. Irrelevant. Show for once a *proof* of that statement. > > > > 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. I have no idea what this means. But indeed the laws of logic remain valid. > 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. > > > 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. No. Because you have not yet proven that it is false. > > > 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. Oh. -- 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 4 Apr 2007 09:48 In article <1175680624.978978.36490(a)y80g2000hsf.googlegroups.com> mueckenh(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. -- 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 4 Apr 2007 10:00
In article <1175681058.802334.28720(a)e65g2000hsc.googlegroups.com> mueckenh(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... > > > > 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. > > It is. If so, *prove* it. How do you *prove* that your series of interchanges on the initial sequence leads to your final "sequence"? > > > > 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? > > To show tha its sum is 2. What is the relevance to my remark? I will quote again: "... 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." Now what? > > > 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. > > 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. This kind of proof is pretty abundantly available in mathematics. > > 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 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. > > Only for finite segments. What do you mean? I am talking about numbers, not about segments. > > > > "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? > > The same holds for paths. Every finite node of a path means the whole > path because there are no "infinite nodes". Yes, so what? Every finite node of a path means all the nodes of the path. Every finite part of the path does *not* mean all parts of the path. > > > > > 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. > > That is deplorable. (N is created by counting.) Oh. So you require counting to use sqrt(500)? Strange. > > > > > 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? > > Yes. That's it. No that is not a proof. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |