Cantor Confusion
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From: mueckenh on 21 Jan 2007 16:16 Virgil schrieb: > > But if you, personally, think it is necessary, then simply use the > > axiom of infinity. There is a set of all natural numbers, a set of all > > levels, a set of all finite trees. > > But nothing that makes a union of finite trees into an infinite tree. Therefore the set of paths of the union tree T1 is countable. > > While a union of sets is necessarily a set, trees are not merely sets, > they have additional structure required, and the unioning of sets does > not automate any appropriate amalgamation of the additional structures. > > WM is relying on his false assumption that an infinite tree is totally > determined by its set of nodes and set of edges, That is the case. There is nothing else determining the tree and the paths in it. >even though he has been > presented with two different infinite trees with the same sets of nodes > and edges but differing sets of paths. That is a contradiction. It shows that R is not an uncountable set. Regards, WM
From: Virgil on 21 Jan 2007 16:17 In article <1169395778.466876.185410(a)a75g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1169332163.222236.89960(a)v45g2000cwv.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Andy Smith schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de writes > > > > > > > > > >Andy Smith schrieb: > > > > > > > > > >> I > > > > >> > > > > The union of all finite binary trees contains all levels > > > > >> > > > >which can be > > > > >> > > > > enumerated by natural numbers: > > > > >> > > > > > > > > >> > > > > 0 0. > > > > >> > > > > / \ > > > > >> > > > > 1 0 1 > > > > >> > > > > / \ / \ > > > > >> > > > > 2 0 1 0 1 > > > > >> > > > > ............................... > > > > >> > > > > > > > > >> > > > > >> Out of interest, aren't the set of all numbers defined by the union > > > > >> of > > > > >> all paths through a finite binary tree with N levels just all the > > > > >> numbers addressed by the first N bits? If so, why do you bother > > > > >> with > > > > >> the tree construction - does it have some special significance? > > > > > > > > > >The real numbers are represented as infinite paths in the "complete" > > > > >infinite tree. Some even twice. > > > > > > > > > >The union of all finite trees is an infinite tree. > > > > >Every finite tree contains only a finite set of paths. > > > > >The countable union of all paths of the finite trees is therefore the > > > > >countable union of all finite paths. > > > > >The countable union of all finite paths is in the union of all finite > > > > >trees. > > > > >The "complete" tree containing all paths is identical to the union of > > > > >al finite trees, with respect to nodes and edges. > > > > >Identical trees cannot contain different sets of paths. > > > > >Therefore, both trees contain the same set of paths. > > > > >Therefore the "complete" set of all path is countable. > > > > >Therefore the set of all real numbers is countable. > > > > >Therefore ZFC is inconsistent. > > > > > > > > I would have said that the set of all paths in a finite tree of depth N > > > > correspond 1:1 with the address range of N bits. > > > > > > You use N as a natural number? It is usual here to denote the set of > > > all natural numbers by N, a natual number by n. > > > > > > > > An infinite tree corresponds to a number encoded in a countably > > > > infinite > > > > set of bits. > > > > > > The nodes of the tree, yes. The edges of the tree too. The paths of the > > > tree, no. > > > > An infinite tree corresponds fairly closely to the set of all real > > numbers encodable in a countably infinite set of bits, except that some > > numbers have dual encodings. > > So every number is represented by at least one path in the infinite > tree? In the usual encoding, only those reals between 0 and 1 are represented. While it is possible to represent all reals by one tree, the method of doing so makes the correspondence obscure, whereas the correspondence between infinite paths and reals between 0 and 1 is relatively straightforward and obvious. > > > > > > > > > > Cantor's diagonalisation argument then applies. > > > > > > It does not apply to the paths of the tree with all nodes, because this > > > tree contains the representations of all real numbers of the interval > > > [0, 1]. > > > > That does not prevent it from applying, > > > Of course, you may apply it, but in vain, because every real number is > represented by at least one path in the infinite tree. Actually the diagonalization argument was first applied by Cantor to what was essentially a complete infinite binary tree, in the form of the set of all functions from the set of naturals to a set with two members, and only later modified to apply to real numbers. So that WM's claim ignores history as well as logic. > > > and, in fact, the > > diagonalization argument applies perfectly to the set of all infinite > > sequences of binary bits, which is where it was originally applied. > > So you can apply the argument *and* find a real number which is not > represented by a path in the infinite tree? However did WM manage to misread that into what I actually said? Cantor, using his original diagonal argument, and anyone else who wants to emulate him, can show that any /list/ of paths in a complete infinite binary tree is incomplete.
From: Franziska Neugebauer on 21 Jan 2007 16:21 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> >> By induction you only define (if at all) every _finite_ union >> >> >> >> T(1) U ... U T(n) >> > >> > That is completely sufficient as long as there is no upper bound, >> > but >> > n --> oo (potential infinity). >> >> Then I would like to have an explanation what >> >> T(1) U T(2) U ... >> >> shall mean in contrast to >> >> T(1) U T(2) U ... U T(n) n e N. > > There is no contrast but identity (for n --> oo). "M�ckenheim-Limes"? >> > But if you, personally, think it is necessary, then simply use the >> > axiom of infinity. >> >> The axiom of inifinity in relation with the axiom of union does not >> apply here since: >> >> ,----[ <45af6ca0$0$97262$892e7fe2(a)authen.yellow.readfreenews.net> ] >> | 7. Now let V* denote the set of all finite trees { T(i) | i e N }. >> | >> U V is only defined for V having card(V) e N. Since V* does not >> | meet this requirement > > How do you know? Because card(V*) = card(omega) by definition. > But it is not important, It IS important. > because we use the union U V over all V. Again: U { T(i) | i e N } is not defined. > V* is not in the union (as omega is not in the union over all natural > numbers or over all finite intial segments of omega). I want you to define T(1) U T(2) U ... >> we (you?) have to define what >> | >> | U V* >> | >> | shall mean. The obvious definition >> | >> | "U V* = T(max(D(V*))" >> | >> | fails due to the reason that max(D(V*)) = max(omega) is not >> | defined. >> | >> | The questions is: How do you define U V_omega? > > No. "No" is not an appropriate answer to a question introduces by "How". > The union of all natural numbers does not include omega. > The union of all segments does not include it. > The union of all finite trees does not include it. I am asking for a definition of "the union of all fintie trees". You have not yet given one. F. N. -- xyz
From: Franziska Neugebauer on 21 Jan 2007 16:29 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> > nevertheless the union of {1,...,n} and >> > {1,..., m} and the infinite union of all segments are defined. >> >> We are writing about trees. > > The trees have levels. For them we have the same as for the initial > segments given above. Again: Your notations T(1) U T(2) U ... and U {T(i) | i e N } are undefined. >> >> | >> >> | The questions is: How do you define U V_omega? > > V_omega is not in the union. What is V_omega? > The union is only over omega elements, but that is not a problem > because it is covered by the axiom of infinity. Unions are covered by the axiom of union. But only if real set-theoretical unions are involved. It is an equivocation (fallacy) to claim that your tree-union (which selects the deepest out of two trees as "union") is a union. It has been explained by Virgil, WH and me that a set-theoretical union of trees is hardly a tree. Hence you are writing on undefined notations. > Everything else is fine Indeed. I cannot spot any contradiction in a set theoretical treatment of binary trees. > and finite. >> > There is no question open. >> >> There is still an open question: How do *you* _define_ >> >> The infinite union of all finite trees? > > The finite trees have levels at finite positions. For the union of > trees we have the same with respect to levels as for the initial > segments given above. There is not the slightest difference. Again: You have not defined "infinite union of all finite trees". F. N. -- xyz
From: David Marcus on 21 Jan 2007 16:31
Franziska Neugebauer wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > > V* is not in the union (as omega is not in the union over all natural > > numbers or over all finite intial segments of omega). > > I want you to define > > T(1) U T(2) U ... Do you think WM can actually produce a definition? His knowledge of math is rather limited. > > The union of all natural numbers does not include omega. > > The union of all segments does not include it. > > The union of all finite trees does not include it. > > I am asking for a definition of "the union of all fintie trees". > You have not yet given one. -- David Marcus |