Cantor Confusion
in [Math]
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From: Dik T. Winter on 28 Mar 2007 22:14 In article <1175105032.482322.292690(a)d57g2000hsg.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 26 Mrz., 17:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > > 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. And what is the relevance? We are talking about infinite subsets, not about what they "cover", whatever that may mean. > > > 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. 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. > The union of all finite natural numbers is an infinite uion (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. -- 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 28 Mar 2007 22:23 In article <1175105211.612045.221910(a)l77g2000hsb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. Pray, for once, show a *proof* of this statement. > 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. 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. -- 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 29 Mar 2007 10:08 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: > ... > > > 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. It is a sequence according to mathematical definitions, if you read it from the left hand side. Further you can determine a unique limit value by lim{n-->oo} (1/2^n + ... + 1/8 + 1/4 + 1/2 + 1}. > > > 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 > > > > > 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. Do you know of a way how to finish Cantor's diagonal? > 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. > 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. What is the > > > > 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. How woud you know which element the n-th element is. > 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. > > > > > 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? > > > > > 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. > > > > 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. So we both agree. > > > 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. Regards, WM
From: mueckenh on 29 Mar 2007 10:21 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: > > > On 26 Mrz., 17:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > > > 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. > > 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. To show that, you can consider the negative unit fractions. The infinite set of these fractions does not cover the interval [-1, 0] but only the interval [-1, 0). Thi is a similar case which might instruct you. > > > > > 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. > > 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 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. Regards WM
From: mueckenh on 29 Mar 2007 10:30
On 29 Mrz., 04:23, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1175105211.612045.221...(a)l77g2000hsb.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > 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. > > 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. > > > 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. 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. > 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. 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. But you say, it is good logic. So let it be. Antilogic cannot be disproved by logic. Regards, WM |