Cantor Confusion
in [Math]
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From: Virgil on 29 Mar 2007 15:52 In article <1175177285.625501.231650(a)b75g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > It is a sequence according to mathematical definitions, if you read it > from the left hand side. But since nodes are, like points, that which has no part, anything requiring nodes to have parts is fallacious. For example, which "part" of a node has the left child and which part the right child? Or wold you, like Solomon, command the children to be split. > Further you can determine a unique limit value by lim{n-->oo} (1/2^n > + ... + 1/8 + 1/4 + 1/2 + 1}. > > > > Ok, what node contributes '1' to that sum? I ask this because you want it > > to apply to nodes contributing things to a path. > > You know, this node cannot be determined. Then the whole thing is a chimera. > Do you know of a way how to finish Cantor's diagonal? When the rule is stated, the diagonal is finished. > > > Or do you think > > you can do your exchanges all at once? > > I do not think it, I know it. It is ridiculous to claim you could do > infinitely many exchanges simultaneously. If the order in which operations are to be performed /does/ affect the result, one cannot perform them simultaneoulsy, as in WM's rule for permuting positions of terms in an infinite sequence. If the order in which operations are to be performed does not affect the result, one can perform them simultaneoulsy, as in Cantor's rule for determining the digits of the "anti-diagonal". That WM cannot see the difference is a measure of his mathematical incompetence. > > > > > > > I am arguing within the tree. There is never an uncountable number > > > > > of > > > > > separated paths. You say all paths were uncountable and separable. > > > > > Conclusion. there is no "all paths" within the tree. > > > > > > > > You have to first *prove* your statement that there is not an > > > > uncountable > > > > number of separated paths. Until now your proofs depend crucially on > > > > the > > > > assertion that there is not an uncountable number of separated paths, > > > > and > > > > so are circular. > > > > > > Wrong. The proof depends on the fact that the set of nodes is > > > undisputedly countable and only nodes are points of separation. A very > > > simple logical conclusion, in principle. > > > > So, please, prove it, using simple logic. > > 1 + 1 = 2. Good so? Which proves nothing, not even that 1 + 1 = 2. > > > > > > > Therefore, if they existed isolated and were so many as you say, > > > > > they > > > > > would populate a level with uncountably many nodes. > > > > > > > > Why? Each two paths are isolated from each other (by your > > > > definition), so > > > > all paths are isolated from all other paths. *But* there is *no* > > > > level > > > > where a single path is isolated from all other paths. > > > > > > There is no *finite* level where all paths are separated. But the tree > > > is *infinite* and it contains all levels where something could happen. > > > A path which is not separated form another one within the tree is > > > never separated. > > > > Yes, you are not contradicting what I state. Each path is separated from > > each other path. > > Therefore we know that only countably many are there. Is that a royal "we"or an editorial "we"? It is certainly not plural. > > > > Nothing happens after any finite distance. There is *no* level where > > all paths are separated from each other. Why you think that there should > > be such a level escapes me. > > It is very simple. Every digit of a real number is indexed by a finite > natural number. Yet the number of such real numbers exceeds the number of natural numbers. The set of paths in an infinite tree is like the set of binary infinite sequences, and both are uncountable. There are numerous proofs of both of these facts, and WM has not ever been able to fault any of these proofs in ZFC or NBG, or in any system which does not presume a priori that no such things as uncountable sets can exist. So that WM is not arguing about any mathematical system, but only some private system of his own which he will not, or can not, specify.
From: Virgil on 29 Mar 2007 16:37 In article <1175178082.736299.205080(a)l77g2000hsb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 29 Mrz., 04:14, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > Infinite sets of paths need not contain or be infinite paths. If no infinite set of finite paths or infinite set of nodes forms a path, just what are the paths of a CIBT? A directed chain of nodes in tree is a set of nodes in which there is a first node, and each node other than that first has a parent node in the chain and each node in the chain has at most one child node in the chain. A path in a tree is chain that is maximal as a chain it that tree, i.e., any superset of that set of nodes is n longer a chain in that tree. In finite trees, every path is a directed chain directed from the root node to some leaf node having no children in that tree. In an infinite tree, every path is an endless chain including the root node but not having any end node at all. If WM's infinite trees contain no such paths then they are pathless trees. > > > > > > The union of all finite paths is an infinite union of finite paths. > > > > Again shifting position. You were talking about the union of a specific > > set of finite paths. The union of *all* finite paths is equal to the > > union of *all* finite trees, which is a set of nodes that is not even a > > path. > > > > > Why should it be an infinite path? > > > > The union of the paths 0., 0.0, 0.00, ..., is an infinite path by your own > > definition of infinite path, namely a path that does not terminate. > > If the union of finite paths contains, or is, an infinite path, then > the union of finite natural numbers contains, or is, an infinite > natural number. The union of the set of von Neumann finite naturals /is/ the set of finite naturals, but is not a natural at all. In the same way, the union of a nested infinite sequence of finite paths is not a finite path, but can be a path in a suitable tree > > > > But I think you fail to see the distinction. Natural numbers are finite. > > That is their definition. Paths can be infinite. That is their > > definition. > > But finite paths cannot be infinite. Aha! Perhaps WM has finally learned something!
From: Virgil on 29 Mar 2007 16:55 In article <1175178657.434003.303550(a)d57g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 29 Mrz., 04:23, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1175105211.612045.221...(a)l77g2000hsb.googlegroups.com> > > > 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. If WM is presenting that collection of garbage as a proof of anything, it only proves his ineptitude at proofs. > > > > > 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. Its wrongness is so non-obvious to those who know what is actually going on that it is WM who is wrong. > The necessity of as much separation points as > separated paths is not restricted to the finite tree. Nonsense. It is required > in any case. Otherwise there must be paths with no connection to the > root node. But those constructs are not paths. > > > 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. For any two, or, indeed, any finite set of paths, in an infinite binary tree there is a node "separating" any one path from all the others, and a finite level at which each of those paths passes through a different node. This is not the case for infinite sets of paths, which situation cannot occur in finite trees. Since WM's mind is so limited that he cannot understand anything except finite trees, it is not surprising that he tends to think properties only holding for finite cases carry over to the infinite. > > 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. > Obviously bad logic. The bad logic is on WM's part. We say that there are uncountably many paths in the CIBT. It is only WM who worries about separating them in some irrelevant way. If we define the tree-distance between two paths as 1/2^n where n is the level of the first nodes not common to both, then, just as with real numbers, every two paths are a positive distance apart, though for any path one can find a sequence of paths "converging" to it, in the sense that the distances become arbitrarily small. So that the infinite paths form a dense set, like the reals. > But you say, it is good logic. So let it be. > Antilogic cannot be disproved by logic. As WM does not have logic, nor does he have proofs, he loses all around.
From: Dik T. Winter on 29 Mar 2007 23:14 In article <1175177285.625501.231650(a)b75g2000hsg.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 29 Mrz., 04:03, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1175104726.767746.12...(a)n76g2000hsh.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > On 26 Mrz., 16:58, "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. 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. 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". > > > Proof: Sum the first n terms from the right hand side and find that > > > the difference of their sum and 2 is not more than 1/2^(n-1). Further > > > find that the sum is always less than 2. Voila. > > > > Ok, what node contributes '1' to that sum? I ask this because you want it > > to apply to nodes contributing things to a path. > > 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. > > > > Yes, and when are we done? > > > > > > When is Cantor done? > > > > Well, at each step in the reversal process you have a sequence with a first > > element and no last element. I do not know of a way to define what the > > result is after infinitely many steps. > > Do you know of a way how to finish Cantor's diagonal? Is there any need to finish it at all? 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. > > 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". "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". > > > Ever tried to count the students attending a lesson or the peas in a > > > cup full of peas? > > > > 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 woud you know which element the n-th element is. By putting n in the mapping given. > > Or do you think > > you can do your exchanges all at once? > > I do not think it, I know it. It is ridiculous to claim you could do > infinitely many exchanges simultaneously. These two sentences are in contradiction with each other. > > > > You have to first *prove* your statement that there is not an > > > > uncountable number of separated paths. Until now your proofs > > > > depend crucially on the assertion that there is not an uncountable > > > > number of separated paths, and so are circular. > > > > > > Wrong. The proof depends on the fact that the set of nodes is > > > undisputedly countable and only nodes are points of separation. A very > > > simple logical conclusion, in principle. > > > > So, please, prove it, using simple logic. > > 1 + 1 = 2. Good so? A singular lack of logic. That is not a proof of your statement. Do you know how to construct proofs? > > > There is no *finite* level where all paths are separated. But the tree > > > is *infinite* and it contains all levels where something could happen. > > > A path which is not separated form another one within the tree is > > > never separated. > > > > Yes, you are not contradicting what I state. Each path is separated from > > each other path. > > Therefore we know that only countably many are there. A proof, please, for once. > > > You should say, however: > > > For every pair of path, there is a node in finite distance from the > > > root node, where they separate. There is no node in finite distance > > > from the root node where all have separated. > > > > The same problem here. But if you correct again the "no node" to "no > > level" I would agree, and I have stated such already quite some time. > > So we both agree. Yes, on this statement, *when* it is adapted. > > > And you should recognize > > > that the binary tree is infinite. Therefore, everything that happens > > > even *after* any finite distance from the root node, nevertheless > > > happens in the tree - unless it does not happen at all. > > > > Nothing happens after any finite distance. There is *no* level where > > all paths are separated from each other. Why you think that there should > > be such a level escapes me. > > It is very simple. Every digit of a real number is indexed by a finite > natural number. Every due level is in the tree. Yes, so what? It still escapes me. -- 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 29 Mar 2007 23:23
In article <1175178082.736299.205080(a)l77g2000hsb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 29 Mrz., 04:14, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1175105032.482322.292...(a)d57g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: .... > > > > > > Not in the uniting, no. But the union *does* contain infinite > > > > > > paths. Just like the union of all finite initial segments of the > > > > > > natural numbers does have as a subset an infinite initial segment > > > > > > of the natural numbers. > > > > > > > > > > Just like the set of all negative unit fractions does cover zero? > > > > > > > > What does *this* mean? The set of all negative unit fractions does > > > > contain infinite subsets. *That* is what we are talking about. > > > > > > But these infinite subsets do not cover zero. > > > > 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. > > The union of the paths 0., 0.0, 0.00, ..., is an infinite path by your own > > definition of infinite path, namely a path that does not terminate. > > If the union of finite paths contains, or is, an infinite path, then > the union of finite natural numbers contains, or is, an infinite > natural number. Why? As by definition natural numbers are finite, that is false. > > > The union of all finite natural numbers is an infinite union (with > > > cardinality aleph_0) but it is not an infinite natural number. > > > > Indeed. Because by definition natural numbers are finite. > > > > But I think you fail to see the distinction. Natural numbers are finite. > > That is their definition. Paths can be infinite. That is their > > definition. > > But finite paths cannot be infinite. That is their definition. Yes, so what? The union of finite paths can be an infinite path, because *paths* can be infinite by your own definition. The union of finite natural number can not be an infinite natural number because by definition all natural numbers are finite. So their union is something else. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |