Cantor Confusion
in [Math]
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From: mueckenh on 7 Apr 2007 13:43 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. Reards, WM
From: mueckenh on 7 Apr 2007 13:58 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. > > > > > > 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"? By definition the final sequence is the result of the sequence of interchanges. > > > > > > 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? 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. > > > > 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. Wrong. It has been proven always only for a finite part of the list. > This kind of proof is pretty abundantly available > in mathematics. That is the reason for the abundance of errors in modern 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 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.) > > > > > 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. 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? > > > > > > > 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. Yes, counting to 500 is required, should this expression not be nonsense. > > > > > > > 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. If a stone would share your opinion, my chances to convince it would not be better than to convince you. "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." What is the difference between every finite ode and every finite segment of nodes with regard to the bijection n <--> {1,2,3,...,n}??? Happy Easter! Regards, WM
From: G. Frege on 7 Apr 2007 14:31 On 7 Apr 2007 10:43:12 -0700, mueckenh(a)rz.fh-augsburg.de wrote: > > Without nature (or reality, as we say today) there would not be any > natural number. > Well, since mathematical objects are fictional anyway (imho), a more reasonable claim (from this point of view) would be: "Without nature (or reality, as we say today) no one would _talk_ about natural numbers." :-) F. -- E-mail: info<at>simple-line<dot>de
From: G. Frege on 7 Apr 2007 14:34 On 7 Apr 2007 10:41:55 -0700, mueckenh(a)rz.fh-augsburg.de wrote: > > Do you claim that all paths exist as separated entities? If so: how or > where do they exist? Can you explain what you mean by existence of > real numbers at all? > Sounds like PHILOSOPHY to me. Not like MATHEMATICS (proper), sorry. F. -- E-mail: info<at>simple-line<dot>de
From: Virgil on 7 Apr 2007 14:51
In article <1175967715.832820.167690(a)q75g2000hsh.googlegroups.com>, mueckenh(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? > 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. > 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. |