Cantor Confusion
in [Math]
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From: Dik T. Winter on 10 Apr 2007 21:56 In article <1175968689.240804.64510(a)e65g2000hsc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 4 Apr., 16:00, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > 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. That is not what I state. In your mirrored sequence you have a first element on the right. Well if you wish to go with the right-left writing people, do so. But that does not change the sequence at all, Only the way you look at it. > > > > 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"? > > By definition the final sequence is the result of the sequence of > interchanges. By *what* definition? Pray state the definition. And do not come up with the single definition: [lim{n->oo} (1 n)] {1, 2, 3, 4, ...} = {..., 4, 3, 2, 1} because in that case I come up with the single definition: [lim{n->oo} (1 n)] {1, 2, 3, 4, ...} = {2, 3, 4, 5, ...} Furthermore you should be able to prove that with your definition the definition does not change the sum. > > > > 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? > > For every path there is a node assumed to contribute 1. If it turns > out that this node is shared by another path, then take the next node > and continue to go on. And you never will terminate searching because there is no last node. Now what? > > > > 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. > > Wrong. It has been proven always only for a finite part of the list. You think so. In that case you should throw away *all* of mathematics, because there is (for instance) no direct proof that sqrt(n) is rational only if n is a square. When something is proven for an arbitrary element that means that is has been proven for all elements. > > This kind of proof is pretty abundantly available > > in mathematics. > > That is the reason for the abundance of errors in modern mathematics. Try to prove that sqrt(n) is rational if and only if n is a square, using your mathematics. > > > 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.) Again, your mantra. Proof, please. > > > 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. > > You don't see a split-brain inconsistency in your remarks? The path > *is* its (well-ordered) set of nodes - nothing else. > > Or is there a tail of a path which does not contain finite nodes? No. But there are parts of paths that do not contain the initial node, or whatever. There is, for instance, an infinite part of the path that starts at the second node. That is, obviously, part of the path, but, also obviously, not a finite part of the path. So every finite part of the path does *not* mean all parts of the path. > > > > 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. > > Yes, counting to 500 is required, should this expression not be > nonsense. Opinion, again. When do you start talking mathematics? > > > 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. I think you are the stone. > "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." Right. > What is the difference between every finite ode and every finite > segment of nodes with regard to the bijection > n <--> {1,2,3,...,n}??? What does map to {2,3,4}? I would think that is a finite segment of nodes. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 10 Apr 2007 22:18 In article <JGB7KK.F20(a)cwi.nl>, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: > In article <1175967715.832820.167690(a)q75g2000hsh.googlegroups.com> > mueckenh(a)rz.fh-augsburg.de writes: > > 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? If so: how or > > where do they exist? > > Philosophy? I would expect mathematics in this newsgroup. > > > Can you explain what you mean by existence of > > real numbers at all? > > Dedekind-cuts, Cauchy-sequences, whatever you want. > > > How are they related to separated paths? > > Each path is (in principle) a Cauchy-sequence. Actually, more like a binary sequence. > > > Are the > > numbers represented by paths which are completely within the > > infinitely many levels of the tree or not? > > This sentence is quite difficult to parse. Yes, each real number is > represented by a path which is completely within the infinitely many > levels of the tree.
From: mueckenh on 11 Apr 2007 09:41 On 11 Apr., 03:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1175967715.832820.167...(a)q75g2000hsh.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > 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? If so: how or > > where do they exist? > > Philosophy? I would expect mathematics in this newsgroup. I think that numbers belong to mathematics. > > > Can you explain what you mean by existence of > > real numbers at all? > > Dedekind-cuts, Cauchy-sequences, whatever you want. > > > How are they related to separated paths? > > Each path is (in principle) a Cauchy-sequence. To be or not to be, that is the question. If yes, then the tree must have a width of 2^aleph0. If not, then there are no irrational numbers - nowhere, neither in the tree nor else in mathematics. > > > Are the > > numbers represented by paths which are completely within the > > infinitely many levels of the tree or not? > > This sentence is quite difficult to pars. Yes, each real number is > represented by a path which is completely within the infinitely many > levels of the tree. But not all real numbers coexist in the tree? Only a countable number of them is admitted simultaneously? Regards, WM
From: mueckenh on 11 Apr 2007 10:06 On 11 Apr., 03:56, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1175968689.240804.64...(a)e65g2000hsc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 4 Apr., 16:00, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > 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. > > That is not what I state. In your mirrored sequence you have a first element > on the right. Well if you wish to go with the right-left writing people, > do so. But that does not change the sequence at all, Only the way you > look at it. In the geometric series there is no last element. Nevertheless you think the series has a limit value. In a geometric series with last but not first element, the things are different? Why is there not a limit value for lim[n --> oo] SUM [1/2^n + ... + 1/4 + 1/2 + 1]? > > By definition the final sequence is the result of the sequence of > > interchanges. > > By *what* definition? Pray state the definition. And do not come up with > the single definition: > [lim{n->oo} (1 n)] {1, 2, 3, 4, ...} = {..., 4, 3, 2, 1} > because in that case I come up with the single definition: > [lim{n->oo} (1 n)] {1, 2, 3, 4, ...} = {2, 3, 4, 5, ...} > Furthermore you should be able to prove that with your definition the > definition does not change the sum. Here is a mathematical proof: We know that the sum cannot be larger than 2. Proof: SUM [1/2^n + ... + 1/4 + 1/2 + 1] = 2 - 1/2^n, and 1/2^n >= 0 for every natural number n. We know that the sum is larger than 2 - eps for any positive real eps. Proof: Take any real eps > 0. Find n such that 1/2^n < eps. What do you require further? > > > For every path there is a node assumed to contribute 1. If it turns > > out that this node is shared by another path, then take the next node > > and continue to go on. > > And you never will terminate searching because there is no last node. Now > what? That is the same for searching the digits of pi. Nevertheless you say pi exists. > > > > > > 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. > > > > Wrong. It has been proven always only for a finite part of the list. > > You think so. In that case you should throw away *all* of mathematics, > because there is (for instance) no direct proof that sqrt(n) is rational > only if n is a square. When something is proven for an arbitrary element > that means that is has been proven for all elements. Yes, it has been proven for all elements (which exist). But it has not been proven for the sequence eventually constructed by these elements. So to say: It has not been proven for the diagonal number. > > > > 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 other paths in the binary tree.) > > Again, your mantra. Proof, please. If all paths exist simultaneously, then there must exist uncountably many in the infinite tree. > > > > > 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. > > > > You don't see a split-brain inconsistency in your remarks? The path > > *is* its (well-ordered) set of nodes - nothing else. > > > > Or is there a tail of a path which does not contain finite nodes? > > No. But there are parts of paths that do not contain the initial node, or > whatever. There is, for instance, an infinite part of the path that starts > at the second node. That is, obviously, part of the path, but, also obviously, > not a finite part of the path. So every finite part of the path does *not* > mean all parts of the path. You did not talk about such strange parts before. And this explanation does not fit my original statement: "Every finite node of a path means the whole path because there are no infinite nodes". Nevertheless, we can agree upon these things if we clarify their meaning. Does "Every finite part of the path" mean "the whole path"? Does "Every finite part of the path" include or cover also that special part you just invented? Regards, WM
From: mueckenh on 11 Apr 2007 10:10
On 11 Apr., 03:29, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1175967792.541552.222...(a)q75g2000hsh.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 4 Apr., 15:48, "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. > > And, so what? If I now define that the natural numbers are the ordinal > numbers, there is nothing wrong with that. It is just contrary to common > nomenclature. So, no, finiteness of the natural numbers is *not* because > of nature, but it is because of definition. You could define a triangle with four corners. It would be as meaningful and as possible. > I may put in your mind another > system where '0' is also called a natural number. Mathematicians use terms > as they see fit. It is the same with physicians. How red is a red quark? > And in what way is it red? Physicists will never look for a red quark because of wavelength problems. Regards, WM |