Cantor Confusion
in [Math]
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From: G. Frege on 4 Apr 2007 13:10 On Wed, 4 Apr 2007 13:48:01 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: >>> >>> [...] natural numbers are finite by definition. >>> >> Not by definition, but by nature. [WM] >> > Natural numbers have nothing to do with nature. From the definition it > immediately follows that all natural numbers are finite. > But in contrary to ordinary Mathematics M�ckenmatics is a /natural science/; WM's natural numbers are part of nature! (Don't ask!) :-) F. -- E-mail: info<at>simple-line<dot>de
From: Virgil on 4 Apr 2007 14:52 In article <1175678971.644490.125770(a)d57g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 2 Apr., 14:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > 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. That in no way proves that a path which must pass through infinitely many levels must be "separated" from all other paths in any one, or any finite number, of such levels. > > > 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. But WM does not know how to apply them. > 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. Separated from what? For every node except the root node in a complete infinite binary tree, the uncountably many paths passing through that node are "separated" from the uncountably many not through that node. > > > > > > 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. Only in WM's version of a set theory. Every standard set theory easily avoids that problem by avoiding WM's invisible axioms. > > > > > 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. Since what we understand by logic conforms to what logicians undersatnd is logic a good deal better than what WM understands by logic, WM will now have to prove all the logicians wrong as well as all mathematicians.
From: Virgil on 4 Apr 2007 14:57 In article <1175680624.978978.36490(a)y80g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. In standard set theories, finiteness and infiniteness of sets are not distinguished from each other prior to having been defined. If WM wants a system in which everything is defined outside of the system, he must work outside of mathematics, as that is not the way mathematics works.
From: Virgil on 4 Apr 2007 15:13 In article <1175681058.802334.28720(a)e65g2000hsc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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 you cannot add them unless every term has an index, and WM's "permutation" leaves at least one term without any index. > > > > > > 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. Then Wm does not understand either Cantor or himself. > > 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. Actually, one can, like Cantor did, prove that for any given listing of binary strings there is a /general rule/ which produces a string not in that list based on the fact that two such binary strings are different as a whole if they are different at ay index position. WM keeps trying vainly to invalidate that argument, but never succeeds. > > 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. It only requires one bit per listed string, which is easy to accomplish. > > > > 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. Wm may so limit himself, but he cannot limit others to his own incapacities. > > The same holds for paths. Every finite node of a path means the whole > path because there are no "infinite nodes". But there are infinite paths in every compete infinite binary tree. > > > > 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.) N is created by axiom. So it is WM who is deplorable. > > > > > > 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. A tree with only one path has only one node. A complete infinite binary tree has as many paths as there are subsets of N.
From: mueckenh on 7 Apr 2007 13:41
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? Can you explain what you mean by existence of real numbers at all? How are they related to separated paths? Are the numbers represented by paths which are completely within the infinitely many levels of the tree or not? Regards, WM |