Cantor Confusion
in [Math]
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From: mueckenh on 28 Mar 2007 14:03 On 26 Mrz., 17:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174671186.397336.238...(a)y66g2000hsf.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 23 Mrz., 16:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1174655745.265882.25...(a)e65g2000hsc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > > The union of all *finite* paths gives the same infinite set of nodes > > > > as the union of all finite trees. But there is no infinite path > > > > involved / required. > > > > > > 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. > > > > > The union of all finite paths is an infinite union of finite paths, > > > > but not an infinite path. > > > > > > It is not even a path. It is the infinite tree. > > > > The union of all paths with only zeros 0., 0.0, 0.00, ... does not > > contain and not possess an infinite path. And it does not require an > > infinite path to pass every finite level. > > Again that word "contain". The union of all those paths *is* an infinite > path. The union of all finite paths is an infinite union of finite paths. Why should it be an infinite path? The union of all finite natural numbers is an infinite uion (with cardinality aleph_0) but it is not an infinite natural number. Regards, WM
From: mueckenh on 28 Mar 2007 14:06 On 26 Mrz., 17:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174671358.437949.73...(a)e1g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 23 Mrz., 17:11, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > Now we see that their number is countable, because there are never > > > > more than countably many paths-bundles including such with only one > > > > path isolated from one another. > > > > > > The last part, again, makes no sense. If we exclude path-bundles that > > > contain a single path, there are indeed countably many of such > > > path-bundles, because they can be clearly identified with terminating > > > paths. But that does say *nothing* about the number of single paths. > > > > If there were any single path, then it had to exist within the tree. > > The tree with its infinitely many levels contains every path which can > > exist, including the infinite paths. > > It still says *nothing* about the number of single paths. An uncountable number of paths existing separated from each other would form a level with uncountable man nodes. But we know that there cannot be such a level, including all infinity of the complete tree. Regards, WM
From: Virgil on 28 Mar 2007 18:04 In article <1175104726.767746.12720(a)n76g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 26 Mrz., 16:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174670803.869854.53...(a)o5g2000hsb.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 23 Mrz., 17:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > You need not to start. Reverse all terms of the series > > > > > simultaneously. > > > > > > > > Oh. How do I do that? What do I interchange the first one (1 with)? > > > > > > You are not so squeamish when exchanging every digit of Cantor's > > > diagonal. > > > > But that is not precisely the same. You want me to reverse a series that > > does not terminate. I ask you how I do that, and you refrain to answer. > > > > > Cantor, unless working every digit simultaneously, will never finish. > > > In fact we need not know how to proceed, but it is sufficient to know > > > hat the sum remains 2 what ever we do. > > > > 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? It does not seem to be defined in any reference available to me. In fact, there is not even any definition of any infinite sequence having a last member. Do you have definitions for any to this stuff? If so, please enlighten us. > If any infinite series ever had a value, then the > sum of this sequence is 2. > Proof: Sum the first n terms from the right hand side. There is no first term or first n terms. > > > The problem with this is that there is a > > dependency in the order of transpositions, so they can not be performed > > simultaneously. > > Cantor's work cannot be done simultaneously. The bits in the "anti-diagonal" for any list can all be done independently, in any order, therefore simultaneously. Transpostions cannot be done independently unless the pairs of positions they transpose are mutually exclusive and disjoint. For non-disjoint pairs of positions, order of transposition makes a difference. WM's ignorance destroys his arguments again. > > 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. As paths are subsets of sets of nodes, maximal chains with respect to connection by edges, it is only the cardinality of the power set of the node set which limits the cardinality of the path set. WM's "separation" arguments only hold for finite trees, as in an infinite tree no path is ever "separated" from all other paths at any node. A very > simple logical conclusion, in principle. > > > > > > > This > > > > > > still does not say anything about the number of non-terminating > > > > > > paths. > > > > > > > > > > That is where we disagree. All paths belong to groups of paths. > > > > > Even > > > > > single paths belong to groups and are groups, if isolated. > > > > > > > > Yes, so what? > > > > > > 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. Every path separates from any other single path at some node, but does not ever simultaneously separate from every other path at any one node. > > > > > You nevertheless believe in the uncountable while I do not. > > > > > > > > Well, within set theory it can be proven. > > > > > > Within the tree it can be disproven. > > > > Until now you did not succeed, because your proofs either contain a denial > > of the axiom of infinity or other logical flaws. > > Wrong. As WM is not a mathematician, his stated judgment of what is mathematically right or wrong is irrelevant until supported by a valid mathematical proof, which so far WM has shown himself incapable of producing even for statements which are mathematically right. > > > > > > But you do not believe in > > > > set theory. On the other hand, you have *not* proven that set theory > > > > is inconsistent. That you do not believe in the uncountable is *not* > > > > an argument against set theory. > > > > > > That you do believe in set theory is not a crime. But that you do > > > believe in uncountaly many separations without uncountably many > > > separations, that is hard to believe. > > > > I do not. You think I do, but that is due to a lack of logical reasoning. > > The lack of logical reason is as follows. You say: > For every pair of path, there is a node, where they separate. There is > no node where all have separated. > You should say, however: > For every pair of path, there is a node in finite distance from the > root node, where they separate. As every node is already at a finite "distance" from the root node, your addition is redundant. > There is no node in finite distance > from the root node where all have separated. Again, the "in finite distance from the root node" is redundant. > And you should recognize > that the binary tree is infinite. We have long recognized that, but WM seems to be a bit vague about it. > 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. WM's concatenation of irrelevancies and fallacies, does not disprove the uncountability of the set of paths in a complete infinite binary tree. And as there have been presented valid proofs of that unocuntability, WM is SOL. > > Regards, WM
From: Virgil on 28 Mar 2007 18:16 In article <1175105032.482322.292690(a)d57g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 26 Mrz., 17:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174671186.397336.238...(a)y66g2000hsf.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 23 Mrz., 16:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1174655745.265882.25...(a)e65g2000hsc.googlegroups.com> > > > > mueck...(a)rz.fh-augsburg.de writes: > > > > > > > > The union of all *finite* paths gives the same infinite set of > > > > > nodes > > > > > as the union of all finite trees. But there is no infinite path > > > > > involved / required. > > > > > > > > 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. You noticed! But what we do not understand is why you expected anything otherwise. > > > > > > > The union of all finite paths is an infinite union of finite paths, > > > > > but not an infinite path. > > > > > > > > It is not even a path. It is the infinite tree. > > > > > > The union of all paths with only zeros 0., 0.0, 0.00, ... does not > > > contain and not possess an infinite path. And it does not require an > > > infinite path to pass every finite level. > > > > Again that word "contain". The union of all those paths *is* an infinite > > path. > > The union of all finite paths is an infinite union of finite paths. > Why should it be an infinite path? Because it is a set of nodes which is maximal with respect to being a chain of nodes in that tree. > The union of all finite natural numbers is an infinite uion (with > cardinality aleph_0) but it is not an infinite natural number. Every natural is a set of naturals with their union, N being a set of naturals which is not a natural. Every path is a maximal directed chain of nodes in its tree and the union of nested sequences of such paths is also a maximal directed chain of nodes in the union tree, so is, by definition, a path in that tree. That works equally well for WM's infinite unions. Also don't forget those bijections between the set of paths of a CIBT and the power set of N.
From: Dik T. Winter on 28 Mar 2007 22:03
In article <1175104726.767746.12720(a)n76g2000hsh.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 26 Mrz., 16:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > But that is not precisely the same. You want me to reverse a series that > > does not terminate. I ask you how I do that, and you refrain to answer. > > > > > Cantor, unless working every digit simultaneously, will never finish. > > > In fact we need not know how to proceed, but it is sufficient to know > > > hat the sum remains 2 what ever we do. > > > > 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. > 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. > > > However, you may proceed as follows (but only after having read my > > > book): By transpositions (as we know them from Cantor's work) bring > > > the n-th term (with n = 2, 3, 4, ...) into the first position and > > > after that bring the n+1-th term into first position. If you are > > > really fast, then you reverse the whole sequence of terms (because you > > > can determine, for *every* term number n, the number of transpositions > > > required to have it at the first position). > > > > 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. 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}? In that case the result is not a sequence and you have to redo quite a bit of mathematics before you can draw any conclusions about it. > > The problem with this is that there is a > > dependency in the order of transpositions, so they can not be performed > > simultaneously. > > Cantor's work cannot be done simultaneously. You are in error. In > order to exchange a_n you have to find n. That is done by counting. > You cannot point to n unless you have pointed to n-1. The *exchanges* can be done simultaneously. This can *not* be done in your process. > 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. Or do you think you can do your exchanges all at once? > > > 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. > > > 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. > > > > Well, within set theory it can be proven. > > > > > > Within the tree it can be disproven. > > > > Until now you did not succeed, because your proofs either contain a denial > > of the axiom of infinity or other logical flaws. > > Wrong. Right. > > > That you do believe in set theory is not a crime. But that you do > > > believe in uncountaly many separations without uncountably many > > > separations, that is hard to believe. > > > > I do not. You think I do, but that is due to a lack of logical reasoning. > > The lack of logical reason is as follows. You say: > For every pair of path, there is a node, where they separate. There is > no node where all have separated. I do *not* state that because that would be nonsensical. If you change "node" in the last part to "level" I would agree. There is *no* node where more than two paths do separate. > 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. > 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |