Cantor Confusion
in [Math]
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From: Virgil on 23 Mar 2007 16:14 In article <1174671186.397336.238820(a)y66g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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? No! Like the cardinality of that set is the same as the cardinality of N, namely aleph_0. Or does WM wish to require that the union of the set of negatives of unit fractions be the negative of a unit fraction? > > > > > > 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. If those paths are regarded as sets of nodes then their union is a set of nodes which is the corresponding path in the "union" tree. If those paths are not sets of some sort, then one cannot form the union of a family of them at all, as unions are only defined for families whose members are sets.
From: Virgil on 23 Mar 2007 16:31 In article <1174671358.437949.73880(a)e1g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 23 Mrz., 17:11, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174664333.103257.4...(a)y80g2000hsf.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 23 Mrz., 16:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > > Pray define "isolated". > > > > > > > > > > Two path are isolated at level k if they have digits m and n with m > > > > > =/ > > > > > = n at a level i <= k. > > > > > > > > So each two non-terminating paths are isolated from each other. And > > > > now > > > > what? > > > > > > 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. In a CIBT or T(oo), when one has a maximal set of nodes in which each node is parent to exactly one other node (as if it were a unary tree), such a set of nodes defines a path which will have no last node. Each such path has for each node a unique successor node, so we have an obvious bijection from such a path to N. For any path p, let L(p) be the set of node indices in N for which the successor is a left child. Then p -> L(p) defines a bijection from the set of all paths in CIBT to the power set of N, P(N), which has been proven uncountable. > The tree with its infinitely many levels contains every path which can > exist, including the infinite paths. It only contains infinite ones as paths, paths being necessarily maximal. Those that are only paths in finite subtrees, we can distinguish by calling them pre-paths, starting at the root, being connected by edgesm and having at most one child per node but having a terminal node, and not being maximal in an infinite tree or in a tree with longer paths.
From: mueckenh on 24 Mar 2007 13:20 On 23 Mrz., 20:47, Virgil <vir...(a)comcast.net> wrote: > In article <1174664147.264581.197...(a)n76g2000hsh.googlegroups.com>, > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > On 23 Mrz., 15:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1174654681.038185.299...(a)e65g2000hsc.googlegroups.com> > > > mueck...(a)rz.fh-augsburg.de writes: > > > How do you obtain that result? You have not shown a proof at all for it. > > > > > If we exchange infinitely many terms of the geometric series, its sum > > > > remains 2, because it is absolutely converging. > > > > > As the geometric series contains only the smallest possible infinity > > > > of terms, it should not cause problems to read it from behind. > > > > If there were a last one. As there is not a last one I have some > > > difficulty with it, because I do not know where to start. > > > You need not to start. Reverse all terms of the series simultaneously. > > As nodes, like points, are that which has no part, talking of parts of > nodes is nonsense. There was a time when subdividing the unity was considered impossible. Is Virgil so far behind? Regards, WM
From: Virgil on 24 Mar 2007 13:57 In article <1174756852.611987.163530(a)l77g2000hsb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 23 Mrz., 20:47, Virgil <vir...(a)comcast.net> wrote: > > In article <1174664147.264581.197...(a)n76g2000hsh.googlegroups.com>, > > > > > > > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 23 Mrz., 15:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1174654681.038185.299...(a)e65g2000hsc.googlegroups.com> > > > > mueck...(a)rz.fh-augsburg.de writes: > > > > How do you obtain that result? You have not shown a proof at all for > > > > it. > > > > > > > If we exchange infinitely many terms of the geometric series, its > > > > > sum > > > > > remains 2, because it is absolutely converging. > > > > > > > As the geometric series contains only the smallest possible infinity > > > > > of terms, it should not cause problems to read it from behind. > > > > > > If there were a last one. As there is not a last one I have some > > > > difficulty with it, because I do not know where to start. > > > > > You need not to start. Reverse all terms of the series simultaneously. > > > > As nodes, like points, are that which has no part, talking of parts of > > nodes is nonsense. > > There was a time when subdividing the unity was considered impossible. > Is Virgil so far behind? Subdividing the natural number one still is impossible. Subdividing the rational number one, or the real number one, is quite different. But nodes, like points, are that which have no part, trying to subdivide them is like trying to subdivide 0, any part is the whole.
From: mueckenh on 25 Mar 2007 12:15
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. If Cantor can finish his replacement in no time, then the above rule also finishes its task in no time. > > > 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. > > > 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. > 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. > > 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. Regards, WM |