Cantor Confusion
in [Math]
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From: Virgil on 25 Mar 2007 14:37 In article <1174839356.692786.303300(a)p15g2000hsd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 23 Mrz., 22:06, Virgil <vir...(a)comcast.net> wrote: > > In article <1174670803.869854.53...(a)o5g2000hsb.googlegroups.com>, > > > > > > Cantor, unless working every digit simultaneously, will never finish. > > > > Cantor's rule is not time dependent. > > The following rule is not *time* dependent either: > > 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. So for *every* > term number n you can determine, the number of transpositions required > to have it at the first position. The order in which digits are replaced in applying Cantor's rule makes no difference to the end result. . The order in which transpostions are applied does make a difference to the result. WM is wrong! Again! As usual! > > If Cantor can finish his replacement in no time, then the above rule > also finishes its task in no time. Except that Cantor's digit replacements can all be applied simultaneously with no difficulty , whereas transpositions into first position can nonly be done one at a time, as there is only one first place. > > > > > In fact we need not know how to proceed, but it is sufficient to know > > > hat the sum remains 2 what ever we do. > > > > But to get there, WM insists on patitioning that which has no parts. > > Nodes are like points in that respect, they have no parts. > > In other respects they have. Which part of a node links to its left child? And does that leave the rest of the node unlinked to that child? > > > > > However, you may proceed as follows (but only after having read my > > > book) > > > > That lets us off that hook. NO one need read WM's propagandizing. > > > > > > > Because there is no "all paths". > > > > > > Pray, provide a proof (within set theory, where we are arguing). > > > > > I am arguing within the tree. > > > > Which tree? > > That one which has all nodes but only finite paths. Since a path is, by definition maximal, with respect to length or number of nodes, and finite chains of nodes are not maximal in that respect in an infinite tree, whatever is limited to finite paths cannot be an infinite tree. > > > In any CIBT or T(oo) there are sets of all paths, and those > > sets are uncountable. > > There is not at all an infinite path detectable in any CIBT. There are certainly no finite paths detectable in any CIBT, as no such alleged path could be of maximal length, so WM is claiming to have infinite trees with no paths in them whatsoever. > > > > That WM has imagined some other tree is irrelevant. > > > > > There is never an uncountable number of > > > separated paths. > > > > As each path separates from every other at some node, there are > > uncountably many pairwise separable paths, which is all that counts. > > > > > You say all paths were uncountable and separable. > > > Conclusion. there is no "all paths" within the tree. > > > > By "separable" all we mean is that any two can be separated, > > Everything that can be separated is separated in the infinite tree. But those finite non-paths don't count. In a CIBT, if there are to be any paths at all, they too must be infinite, as finite chains of parent to one child linked nodes, however long, are not long enough to be paths. Any chain which is a subchain of any longer chain is immediately not maximal in length and therefore not a path, and in a CIBT, /every/ finite chain is a proper subchain of other chains. WM's irrational passion to make CIBTs behave unnaturally, blinds him to the facts.
From: Dik T. Winter on 26 Mar 2007 10:58 In article <1174670803.869854.53820(a)o5g2000hsb.googlegroups.com> mueckenh(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"? > 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? The problem with this is that there is a dependency in the order of transpositions, so they can not be performed simultaneously. > > > > Yes, I do. What now? Every two paths separate from each other at > > > > some specific node, through all nodes go uncountably any paths, > > > > there is no level where all paths do separate. > > > > > > Because there is no "all paths". > > > > Pray, provide a proof (within set theory, where we are arguing). > > 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. > > > 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. > > > I think we have cleared our positions sufficiently: We agree in: At no > > > level in the tree more than countably many groups of paths (including > > > single paths) can be distinguished. And outside of he tree there are > > > no paths. > > > > > > 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. > > 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. -- 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 26 Mar 2007 11:01 In article <1174671186.397336.238820(a)y66g2000hsf.googlegroups.com> mueckenh(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. > > > 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 set of all negative unit fractions is infinite but neither any > > > member nor the whole set does cover zero. > > > > What is the relevance? > > > To inform you about the difference between numbers and length of > paths. Well, you wanted us to belief that the length of paths would be the element of something, so I think you were a bit confused. -- 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 26 Mar 2007 11:02 In article <1174671358.437949.73880(a)e1g2000hsg.googlegroups.com> mueckenh(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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 28 Mar 2007 13:58
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? 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 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. > > > 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? > 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. Ever tried to count the students attending a lesson or the peas in a cup full of peas? > > > > > > Yes, I do. What now? Every two paths separate from each other at > > > > > some specific node, through all nodes go uncountably any paths, > > > > > there is no level where all paths do separate. > > > > > > > > Because there is no "all paths". > > > > > > Pray, provide a proof (within set theory, where we are arguing). > > > > 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. > > > > > 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. > > > > > I think we have cleared our positions sufficiently: We agree in: At no > > > > level in the tree more than countably many groups of paths (including > > > > single paths) can be distinguished. And outside of he tree there are > > > > no paths. > > > > > > > > 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. > > > > 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. There is no node in finite distance from the root node where all have separated. 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. Regards, WM |