Cantor Confusion
in [Math]
|
From: mueckenh on 27 Apr 2007 08:16 On 27 Apr., 13:56, William Hughes <wpihug...(a)hotmail.com> wrote: > On Apr 27, 6:40 am, mueck...(a)rz.fh-augsburg.de wrote: > > On April 17 Mueckenheim writes > > M: Each "point of separation" is a single node > > On April 18 Mueckenheim writes > > M: a set of separation points can separate a path from all other > paths. You quote this assertion, casually forgetting that it is only valid in case the path does exist. Given the path exists in the tree, separated from all other paths, then there must be a set of separation points. Meanwhile we have found out that this is not the case. > > M: A path cannot be separated from all other paths at all > > (justifying this with the usual Volkenmuekenheim one, two, look > over there a pink elephant, three) Do you have some arguments too? Then you may analyzed this conclusion: forall K(p,n) with n in N thereis p' such that forall m in N with m =< n : K(p',m) = K(p,m). Regards, WM
From: William Hughes on 27 Apr 2007 09:37 On Apr 27, 8:16 am, mueck...(a)rz.fh-augsburg.de wrote: > forall K(p,n) with n in N thereis p' such that forall m in N with m =< > n : K(p',m) = K(p,m). p' may be different for different n's, so I should write p'(n). I need to show that I can find a single, non changing, p' that hold for every n. For every s in N there is a p'(s) so that p'(s) works for every n<=s Look, over there, a pink Elephant! There is a p' that works for every n in N. - William Hughes
From: Virgil on 27 Apr 2007 16:06 In article <1177670420.844384.220400(a)n15g2000prd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 24 Apr., 18:20, William Hughes <wpihug...(a)hotmail.com> wrote: > > > A path can be separated from all other paths by a set of nodes > > A path *is* a set of nodes. A path cannot be separated from all other > paths at all because down to level L(n) > every node belonging to path p is also a node of path p' =/= p. This > holds for every n in N - and other nodes are not available in p. If we > denote the n-th node belonging to path p by K(p,n) then we have: > forall K(p,n) with n in N thereis p' such that forall m in N with m =< > 2n : K(p',m) = K(p,m). There is no one node that separates an infinite set of nodes from all other infinite sets of nodes, but that does not mean that all infinite sets of nodes are the same. > > If something is valid for all elements, then it does not change if the > number of elements is infinite. What is true of elements one at a time need not be true for infinite sets of them. Any two distinct subsets of P(N) can be distinguished by a single element that is in one but not the other, but there is no finite set of members of N which will do the same for all members of P(N). So that WM's dreams about infinite trees do not match the reality of such trees, > Remember:The infinite product is 0 if > one factor is 0. > > > There are uncountably many sets of nodes > > That may be. But there are only countably many sets of nodes which are > paths. There are countably many levels in a CIBT, and each different subset of the set of levels generates a separate path in that tree, the path that branches left at just the members of that set of levels. So how can WM claim that the number of paths is less than the cardinality of the power set of a countable set? Answer: He can't do it validly, so he fakes it > Remember: All paths are sets of nodes. Not all sets of nodes > are paths. Remember: Every subset of the set of levels "is" a path! And there are uncountably many such subsets of the set of levels. > > We cannot distingush more than countably many paths. Speak only for yourself, WM! > > Regards, WM
From: Virgil on 27 Apr 2007 16:20 In article <1177671404.952318.309270(a)r35g2000prh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 24 Apr., 20:55, Virgil <vir...(a)comcast.net> wrote: > > These uncountability proofs are wrong. > > > > Saying that they are wrong does not make them wrong. You must show that > > some step in their proof is wrong in order to establish that they even > > can be wrong, much less that they are wrong > > > > Here it is again: > > (1) The set of levels in a CIBT is indexed by N. > > (2) For each subset of N there is a unique path which branches left > > at all the levels indexed by that set and right elsewhere. > > (3) this establishes a bijection between the set of all paths in the > > CIBT and the set of all subsets of N > > (4) Card(set of all subsets of N) = Card(P(N)) > Card(N) > > (5) Card(set of all paths in a CIBT) > Card(N). > > (6) The set of all paths in a CIBT is uncountable, > > meaning of cardinality greater than that of N. > > > > Which step is wrong, WM? > > Wrong is the assumption (2) that all paths can be distinguished. There > are no unique paths but only unseparated bunches. If there are no unique paths, there is no tree at all. Does WM deny that there is a countably infinite set of levels? Does WM deny uncountably many subsets to that set of levels? Does WM deny that any two different sets of levels can be distinguished? Does WM deny one different path for every subset of the set of levels? So that > > Consider the path p with nodes K(p,n) for all n in N. Then: > forall K(p,n) with n in N thereis p' such that forall m in N with m < > 2n : K(p',m) = K(p,m). > > Wrong is further statement (4), established only by some faulty > proofs. What is faulty about the set of levels, L, being infinite? What is wrong with P(L) being uncountable? What is wrong with every distinct member of P(L) determining a different path, the path branching left at just the levels in that set and right elsewhere? > Wrong is the conclusion (5,6) Unless something is wrong with card(set of paths) = Card(P(L)), it is WM's objections that are wrong! > > > > Your alleged rebuttal does not occur in any set theory recognized by > > mathematics , nor in any set theory which does not presume the contrary > > a priori, nor in any set theory all of whose presumptions (axioms) have > > been clearly stated so that one can determine unambiguously what does or > > does not follow from those axioms. > > > > > This cannot > > > be rebutted by a different proof. > > > > Each such attempted proof has been validly rebutted by showing one or > > more flaws in that proof, frequently by different critics. > > The number of critics is unimportant, in particular if they are biased > and unable to think in logic terms like: > If there are X separated paths in the tree, then there must e X-1 > nodes separating them in the tree. That may hold for finite trees, but does not invalidate the L based proof of the uncountability of the set of paths in a XIBT.
From: Virgil on 27 Apr 2007 16:30
In article <1177672072.574980.291080(a)n15g2000prd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 25 Apr., 04:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > That depends on the notation. The "bunch of" makes it somewhat > > > clumsy. > > > > This does not make a difference. When you remove the stuff between > > parenthesis it makes no sense. There is *no* node that separates a specific > > path from another one. > > Small wonder. There is no specific path in the tree. Then WM is looking at different trees that everyone else. In a CIBT, there are infinitely many levels, and for every subset of that set of levels, L, there is a unique path which branches left at those levels and no others. So clearly there is the /specific/ path corresponding to L and the /specific/ path corresponding to {} as two specific paths. And I challenge WM to find any subset of L for which there is not a /specific/ corresponding path. > > The tree shows us that every path is simultaneously a path bunch, > because down to level L(n) Consider the set of all levels, L, as limiting things to L(n) is only relevant for finite trees. > No matter how these are defined, there is the simple mathematical > equation: > One bunch comes in, two bunches go out, the difference is due to the > node splitting the incoming bunch. As in CIBTs each of the bunches going out has as many paths as the entire tree, consideration of bunches is irrelevant. |