Cantor Confusion
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From: Virgil on 24 Jan 2007 15:17 In article <1169639256.391493.236600(a)s48g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 23 Jan., 13:35, Franziska Neugebauer > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > > > ** The std-union of the set of sets of paths of every finite binary > > std-tree is not identical to the set of paths of the infinite > > binary std-tree G. ** > > Summary > > 1) Every complete infinite binary tree T (containing all nodes and > edges) contains all paths. > 2) The union tree T(oo) of all finite trees is well defined (as I have > shown elsewhere) What Wm has constructed is not a union of trees. > and yields the complete infinite binary tree > containing all nodes and edges: T = T(oo). > 3) The union of all finite trees includes the union of all nodes and, > with it, the union of all such subsets which are paths (because every > path is a well defined subset of the set of nodes if the structure of > the tree is well defined). > 4) The set of paths in T(oo) is a subset of the countable set of finite > sets of all paths in the finite trees. Except that (1) it contains paths not in any finite tree, such as paths which evantually alternate directions of branching (2) It is not countable, as Cantor proved. . > 5) A countable union of countable sets is a countable set (according to > ZF with AC). The set of nodes of the total binary tree is countable and the set of edges is countable but the set of paths is provably not. > Going on, we can say: > > 6) T(oo) = T contains only finite paths. Then it does not contain all paths, not even the infinite paths of his own weeping willow trees. > 7) T(oo) = T contains all paths including all infinite paths. > ==> There are no infinite paths. (There are no irrational numbers.) That is an irrational conclusion based on an irrational argument. > > Nothing further remains to say. Except that WM claiming something is not proof and he has not provided valid proofs for his claims. All his attempts at proofs involve unjustified assumptions.
From: Franziska Neugebauer on 24 Jan 2007 15:22 G. Frege wrote: > On 24 Jan 2007 07:33:13 -0800, mueckenh(a)rz.fh-augsburg.de wrote: >> >> What is the union of {a, b, c} and {a, b, c, d, e, f, g} in your >> opinion? >> > {c, r, a, c, k, p o, t} ? No. It is {c, r, a, k, p o, t}. LOL F. N. -- xyz
From: Virgil on 24 Jan 2007 15:42 In article <1169639909.549632.314770(a)k78g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 1) Every complete infinite binary tree T (containing all nodes and > edges) contains all paths. Avery finite tree is a subtree of a finite tree in which all paths are of the same maximal length. So that any conjunction ( union in WM's world) of finite trees will be the same as a conjunction of such finite trees of all equal path lengths and so will result in a tree in which all paths have the same length. If that length is finite, then that supposedly infinite binary tree is finite, so that there must exist paths of infinite length in an infinite tree constructed in this way. > 2) The union tree T(oo) of all finite trees is well defined (as I have > shown elsewhere) It is not well defined in any mathematical sense, since "union" doesn't define it, but it is mathematically defineable. > and yields the complete infinite binary tree > containing all nodes and edges: T = T(oo). Only if one also requires that it have every path that can be constructed from those nodes and edges. > 3) The union of all finite trees includes the union of all nodes and, > with it, the union of all such subsets which are paths (because every > path is a well defined subset of the set of nodes if the structure of > the tree is well defined). Wrong! A set of nodes is not enough. For a set of only three nodes, there are at least 3 different binary trees which can be formed, 6 if one counts mirror images as distinct. For larger node sets the number of possible trees for that node set increases factorially. > 4) The set of paths in T(oo) is a subset of the countable set of finite > sets of all paths in the finite trees. Often claimed, but never proved by WM, and often disproved by others. > 5) A countable union of countable sets is a countable set (according to > ZF with AC). Quite so, but quite irrelevant to the cardinality of the set of all paths in the total infinite binary tree. > ==> The set of all path is countable. (==> The real numbers are > countable.) WM again tries to conclude what he does not and cannot prove. > > Going on, we can say: > > 6) T(oo) = T contains only finite paths. Then it is a finite tree with only finitely many nodes and edges. > 7) T(oo) = T contains all paths including all infinite paths. And 2 = 1 in WM's world. > ==> There are no infinite paths. (There are no irrational numbers.) > > Nothing further remains to say. WM has now gone as far from reason as it is possible for him to go. > > Regards, WM
From: Virgil on 24 Jan 2007 15:48 In article <1169640439.710405.16290(a)k78g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dave Seaman schrieb: > > > > > > N is an infinite set, but each member of N is finite. > > Translated: The set of all segments {1,2,3.,..,n} is infinite, but each > segment {1,2,3.,..,n} is finite. > > Now the number does not play a role. If there is no infinite segment, > then they all put together, always starting from 1 again) cannot be > actually infinite (finished infinity, reached infinity) but only > potentially infinite. > > Regards, WM WM must be defining "actually infinite" to mean endless but having an end anyway. To the rest of us, at least for sequences, either a given sequence has a last member and is thus a finite sequence or does not have a last member and is not a finite sequence (which we abbreviate by calling it infinite). But in either case, we have no problem with the existence of the sequence itself. Does WM claim that for such a sequence to be allowed to exist mathematically one must have some sort of "actual existence" different from ordinary existence of functions?
From: Virgil on 24 Jan 2007 16:13
In article <1169641853.359139.100270(a)j27g2000cwj.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > Summary > > 1) Every complete infinite binary tree T (containing all nodes and > edges) contains all paths. Then, unless it contains leaf nodes, it must contain endless paths. But every node that is a leaf node in one finite tree is not a leaf in a larger tree, so that can be no leaf nodes. > 2) The union tree T(oo) of all finite trees is well defined (as I have > shown elsewhere) and yields the complete infinite binary tree > containing all nodes and edges: T = T(oo). Except that "union" is not a legitimate name for the operation as WM defines it, agreed. One might note that it also contains all possible paths. > 3) The union of all finite trees includes the union of all nodes and, > with it, the union of all such subsets which are paths (because every > path is a well defined subset of the set of nodes if the structure of > the tree is well defined). Actually paths would be "subsets" of edges, as it is only sequences of nodes connected by edges that can be paths. > 4) The set of paths in T(oo) is a subset of the countable set of finite > sets of all paths in the finite trees. Trivially false. Since the "union" tree cannot contain any leaf nodes, no path can be finite. > 5) A countable union of countable sets is a countable set (according to > ZF with AC). True but irrelevant. > ==> The set of all path is countable. False, and proven false. > (==> The real numbers are > countable.) False, and proven false. > > Going on, we can say: > > 6) T(oo) = T contains only finite paths. False, and proven false. > 7) T(oo) = T contains all paths including all infinite paths. True but irrelevant. > ==> There are no infinite paths. (There are no irrational numbers.) False, and proven false. > > No. I *define* T(oo) is an ordered set of nodes, the order being > expressed by the edges. AS a set of nodes only, it is not a tree at all. One can get by with the set of edges only, but not with the set of nodes only. Try thinking before writing. > > > *Not* > > a tree. And a set of paths does not have nodes or edges as elements, it > > has paths as elements. > > The union of paths is a set of nodes as a path is set of nodes. The union of sets of paths is set of paths. A path is a sequence of edges in which the child node of one edge is the parent node of the next, not merely a set of nodes. There are lots of sets of nodes which are not paths. but every maximal sequence of connected edges is a path and every path is such a sequence. > > > And if you consider a path as a set of node, then > > T1 *still* does not have nodes as elements but sets of nodes. > > T(oo) is the union of the T(n) which are sets of nodes. Therefo T(oo) > is a set of nodes. If T(oo) is merely a set of nodes, then it is not a tree at all, as node by themselves do not form trees. > > The union of trees is defined by nodes! Not so, as nodes by themselves are not enough to determine edges, and the set of edges is necessary to determine which of many possible trees on a given set of nodes one has. > Paths are maximal subsets > following some prescription. Therefore in long trees there are no short > paths, but there are long paths only. "Tall" trees? |