Cantor Confusion
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From: Virgil on 25 Jan 2007 16:00 In article <1169735624.784496.98310(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 24 Jan., 21:42, Virgil <vir...(a)comcast.net> wrote: > > > T = T(oo). > > > Only if one also requires that it have every path that can be > > constructed from those nodes and edges. > > So a countable set can get an uncountable set by your *requirement*? > > > > 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 trees as I defined them this is wrong. There is but one cut tree > with 3 nodes, namely > > 0. > /\ > 0 1 That is because in naming those nodes, you incorporated the information normally included in the set of edges, so WM has implicitly declared a set of edges. From the set of 3 randomly names nodes, we can construct 6 trees. Note that if the names of the nodes are randomized, WM does not have a tree at all, so he need, even if only implicitly, the set of edges, > > Please stay to my definition when discussion my proof. Or indicate > that you are discussion something else which mght be of interest to > somebody else. Since your definition hides necessary information, it is not a fair tree. One must also have the information provided by a set of edges in order to determine a tree if there is to be more than one node. > > > > > > 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. > > but only by such who deny the law of cause and effect in mathematics. Cause and effect does not hold in mathematics, and is even a bit vague sometimes in quantum physics. Logical consequences follow from premises in mathematics, but that is not at all the same thing as "cause and effect". > Trying to establish some fuzzy set theory? Trying to make mathematics into a purely experimental science? > > > > But in either case, we have no problem with the existence of the > > sequence itself. > > Only because you have no problem with thought craft, telekinesis and > that stuff. I do not work with telekenesis, but I do work with thought, and I hope with a bit of craft. > > > > Conclusion: By finite induction, for all n in N, the pseudo-union of > > binary trees of max path length up to n is a finite binary tree. > > > Of course. Every natural number is a finite number. Why do you think > > induction would not reach every finite number? > > > One never gets to any infinite pseudo-unions in such standard inductive > > arguments. > > Do you know why this is so? There are only finite numbers. They do not > lead to an infinte number of numbers. That one cannot get from point A to point B by one route does not mean one cannot get there at all. > > One never gets to any infinite union by standard inductive arguments. > You have said it. You know i5t. You know why it is so and why it cannot > be else. You only resist to admit that it is so. I quire agree that one cannot get from finite unions to infinite unions by finite induction. Why WM should have supposed that I thought otherwise is not at all clear. As the source of many of his other false assumptions is not clear. On the other hand, that route being barred does not mean that one cannot get from finite unions to infinite unions by other methods. See ZF.
From: Virgil on 25 Jan 2007 17:09 In article <1169736308.284550.236030(a)v45g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 25 Jan., 13:39, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > 1) If we identify path-lengths with natural numbers, then there is no > > > infinite path in any of the finite trees, because there is no infinite > > > natural number. Therefore, also the union of all finite trees cannot > > > contain an infinite path. > > > Clearly the tha path-length of an infinite path cannot be a > > natural number. > > The path length is the number of nodes of the path. As long as the > nodes are enumerated by natural numbers, the length (distance from the > root node) is a natural number. That assumes that every path has "last" node. Which need not be the case. And in an infinite binary tree cannot be the case. WM has lost his way again! > > > > S can be and is larger than L. S is not countable. > > That again is an unjustified and provably wrong assertion. WM 'proves' things only by assuming them, so his judgment is untrustworthy concerning what is provable. > The sequences belong to the countable union of the finite trees. > *Everything* in this union is countable. WM's opinions on it are certainly unaccountable! > > Please spare such nonsense in future. Only if you do first! > > In a union of trees containing only finite paths, there cannot be a > finite path. Depends on how many trees. For a finite set do trees it must be true, but for an infinite set of trees, depending on how "union" of trees is defined, it can be, and often is, false.
From: Virgil on 25 Jan 2007 17:10 In article <1169736539.298735.176670(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Sorry. S contains for instance the paths 0.111... and 0,010101... the > bits (nodes) of which can be indexed by natural numbers. > If it contains all possible such paths, it is uncountable.
From: Virgil on 25 Jan 2007 17:33 In article <1169737999.047174.247580(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 25 Jan., 15:34, Franziska Neugebauer > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 24 Jan., 21:05, Franziska Neugebauer > > > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > > [...] > > >> Using Virgil's trees we have the result that the induced std-tree > > >> (M(P), E(P)) may be the same if different sets of paths P are given. > > > > > Then it is not what I am talking about.So what _are_ you then talking > > > about? > > > > > If T(oo) = T concerning the set of nodes and edges,Axiom of > > > Extensionality: Two sets a and b are equal if > > > > A c (c e a <-> c e b) > > > > There is no room in equality for "concerning" this or that. > > Tell it Virgil. But you are in error. The set {1, 0, 1} is identical to > the set {0, 1, 1}, concerning the elements. Concerning the order of > elements, both sets differ. The sets do not differ, but unless WM is confusing the notation of sets { ...} with ordered sets (...), neither of his "ordered" sets are ordered. > > > > > then they are identical with respect to paths too. There is no further > > > parameter to fix. If you do not accept this identity, then further > > > discussion is not meaningful. WM's declarations of certainty without logical foundations are not what anyone else would call meaningful discussion anyway! > > >The set of finite trees V* = { T(n) | n e N } is not a finite set. > > Induction over its members does not prove a property of V*. > > Then it does not consist of only finite levels. Which shows how little WM knows about induction. Induction can possibly prove that all the members of V* have some property, but can prove nothing about V* itself. > > > This is quite the same as in the case of the infinite set > > N = { n | n e N }. Induction over n for some property p(n) does not > > prove p(N). Maybe you forgot that. > > No. All I require is that a property holds (or not) for all the natural > numbers or all the levels or nodes or edges of trees which can be > enumerated by natural numbers. Then you best conclusion is that it holds for all natural numbers but not that it holds for N. At least using the principle of induction. To prove that that property, whatever it may be, holds for N as well requires something other than induction. > > > > >> when you claim that > > >> induction proves existence of U V* where in fact is does only prove > > >> existence of UV(n) A n e N. > > I am fully satisfied with the latter. I call it U T(n) for A n e N = > T(oo). Nothings remains to be desired. Can you spot a difference to > your "UV*"? Perhaps not a node but a path? Let it be. I am not > interested in fairy tales. Since you produce so many fairy tails, we thought they were your entire purpose in life. > > > > I must admit this is probably not simple enough at applied university > > level but nonetheless valid. > > Better fitting to scientology or matheology? We are not discussing what you teach, but only what you are claiming here. > > > > > As I said above: If you conclude from one and the same tree that there > > > are different paths, > > > There exist different sets of paths (Virgil's trees) which *induce* > > the same standard-tree (M(P), E(P)). > > There exist different systems of paths which may induce the same tree. > A a simple finite example: > 0.1 and 0.11 and 0.111 are paths which unioned result in the path > 0.111. Of course you can also get to this same result when starting > from the paths 0.1 and 0.111 or from the paths 0.11 and 0.111. That is > clear. But once the union tree T(oo) is constructed, the set of path > contained in it is uniquely defined. > > You always express very carefully that Virgls approach "leads to the > same tree". Please state explicitly wether you are willing to assert > that this same tree, T(oo), after construction, does contain different > sets of paths, induced by the mental constitution of the observer? Depends on how one defines a tree. If one goes by standard definitions, as found in, say, Wikipedia or Mathworld, they only consider finite trees, so one needs a special definition to cover infinite trees. If one defines a tree as being entirely defined by its nodes and edges, and allows a path to be any chain of edges starting at the root node and being as "long" as possible, then a maximal infinite binary tree, whether constructed as a limit or not, contains uncountably many paths. If, on the other hand, one defines an infinite tree by its set of paths, and having at least one endless path, then one can have an infinite tree with only countably many paths, or only finitely many, for that matter.
From: Virgil on 25 Jan 2007 17:46
In article <1169740229.047316.94890(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > I do not use the union of sets of paths but the union of nodes. Wrong! WM requires incorporation of the information usually included in the set of edges in addition to a mere set of unrelated objects as nodes. In WM's trees the root node is always identifiable as the root without any set of edges, and the parent node child nod connection is always apparent without ant set of edges, so that WM requires all the information obtainable from the set of edges without the set itself. But without a set of edges to provide that information, where does WM get it from? His crystal ball? > > Correct! The set of paths in the union is a subset of the union of all > sets of paths. It is the set of all sequences of nested paths, one path from each tree in the union. Which for an infinite sequence trees gives infinite sequences of paths, giving uncountably many possible such infinite sequences. |