Cantor Confusion
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From: G. Frege on 21 Jan 2007 14:09 On Sun, 21 Jan 2007 14:00:26 -0500, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > > [WM's] arguments can't be wrong because they are obvious! > Indeed, obviously wrong! :-) F. -- E-mail: info<at>simple-line<dot>de
From: G. Frege on 21 Jan 2007 14:32 On Sun, 21 Jan 2007 19:08:02 +0000 (UTC), stephen(a)nomail.com wrote: >> >> [WM's] arguments can't be wrong because they are obvious! >> > And Han claims to understand them, which is the definition > of mathematical truth. :) > Another case of "congeniality"! :-) F. -- E-mail: info<at>simple-line<dot>de
From: David Marcus on 21 Jan 2007 14:40 mueckenh(a)rz.fh-augsburg.de wrote: > It is interesting: > William knows that T1 and T2 exist, but T1 is different. > Virgil is also accepting T1 but, like William, sees only finite trees > therein. > Dik knows that T1 contains infinite trees, but doubts that its paths > are the union of the paths of the finite trees. > You doubt the existence of T1 (seeing that otherwise set theory is > contradiced?). > > Couldn't you get to a consensus? It would spare me a lot of work. Perhaps it would help if you look at the following and see if it is what you are saying. If it isn't, please note any corrections. An edge is an ordered pair of nodes. A tree T is an ordered pair (M,E) where M is a set of nodes, E is a set of edges. Assume that any tree we mention is a complete binary tree. Define a path P in T to be a function P: X -> M where X is either {0,...,n} or N, P(0) is the root node, (P(i),P(i+1)) is in E for all i such that P(i) and P(i+1) are both defined, and if the domain is {0,...,n}, then P(n) is a leaf node. Denote by Q(T) the set of paths in T. Define a nested sequence of trees to be a sequence {T_i} where T_i = (M_i,E_i), M_i is a subset of M_{i+1}, and E_i is a subset of E_{i+1}. Define the union of these trees to be T* = (M*,E*) where M* = U{M_i} and E* = U{E_i}. Claim: |Q(T*)| <= |U{Q(T_i)}|. -- David Marcus
From: David Marcus on 21 Jan 2007 14:42 G. Frege wrote: > On Sun, 21 Jan 2007 14:00:26 -0500, David Marcus > <DavidMarcus(a)alumdotmit.edu> wrote: > > > [WM's] arguments can't be wrong because they are obvious! > > Indeed, obviously wrong! :-) Obviously! -- David Marcus
From: mueckenh on 21 Jan 2007 15:48
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). > > > 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? But it is not important, because we use the union U V over all V. 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). > 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. 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. Regards, WM |