Cantor Confusion
in [Math]
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From: mueckenh on 27 Apr 2007 06:40 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). If something is valid for all elements, then it does not change if the number of elements is infinite. 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. Remember: All paths are sets of nodes. Not all sets of nodes are paths. We cannot distingush more than countably many paths. Regards, WM
From: mueckenh on 27 Apr 2007 06:56 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. 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. Wrong is the conclusion (5,6) > > If None are, then WM is wrong. > > > Further, if > > they were correct, I show a contradiction in set theory. > > 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. Regards, WM
From: mueckenh on 27 Apr 2007 07:07 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. The tree shows us that every path is simultaneously a path bunch, 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 others are not available. 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 < n^2 : K(p',m) = K(p,m). > > > There are bunches of paths like the bunch 0,1 which contains all paths > > from 0,1000... to 0.111... > > A searated path is a bunch of paths with only one element like > > 0,111... > > > > In order to get two separateed bunches of paths, there is one bunch > > required which contains all these paths and one node to split the > > incomin bunch into wo bunches. > > As far as I understand from your book, there is no splitting of bunches. > There are only nodes where bunches terminate. A bunch like 0,0 is split into two bunches 0.00 and 0.01 by the due node. > According to the book a > bunch of paths is a set of paths with a common initial segment (i.e. > set of nodes). So at each node at least three bunches are involved > (actually more). That is a matter of convention of notation. In any case a node increases the number of separated path bunches by 1. > There is the bunch that terminates with the node in > question, and there are the two bunches that go left and right. All > three are incoming bunches of the node, and only the latter two are > outgoing bunches. 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. > > > The infinite tree contains all separated paths. Therefore there are as > > many nodes (minus one) required in the tree. > > A proof is still required. Simply think: 1 pathbunch goes in, X pathbunches are asserted to be in the tree. Every node creates one mor pathbunch. X-1 nodes are required. > > > Should no separated paths exist, then they would also not exist in > > Cantor's list as entries or as diagonal number. > > Makes no sense. Cantor's list? I agree. There is no isolated sequence in the list. Regards, WM
From: William Hughes on 27 Apr 2007 07:56 On Apr 27, 6:40 am, mueck...(a)rz.fh-augsburg.de wrote: On April 17 Muekenenheim writes M: Each "point of separation" is a single node On April 18 Muekenheim writes M: a set of separation points can separate a path from all other paths. The obvious conclusion is that A path can be separated from all other paths by a set of nodes However, Muekenheim does not want to admit this. After trying to evade, he decides on direct self contradiction 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) Muekenheim outdoes Humpty Dumpty A word means exactly what I want it to mean, and its meaning can change whenever I want. - William Hughes
From: mueckenh on 27 Apr 2007 07:56
On 25 Apr., 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > Cantor's diagonal can be defined up to any line n of the list. But > > > > this says *nothing* about an infinite sequence of digits. > > > > > > It is defined for *any* line of the list. > > > > That does not define it for *all* lines. > > It does. If it were, than the number of digits were complete. Otherwise there can be no judgement concerning "all" but at most concerning "each". But completeness of digits means presence of all, up to the last one (which does not exist). If you do not agree to my arguing, then you also must not agree on the idea that single paths exist in the tree. In the tree we have, if K(p,n) denotes the n-th node of path p, 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). Up to "each" node of p, there is a path p' having the same set of nodes up to level n. If "each" means "all" in this case, then for path p there is a path p' =/= p which has the same set of nodes as p, i.e., which is p' = p. Why do you think, "each" mean "all" in case of Cantor's diagonal but not in case of the tree? > > > Here is an example. (You know that all rational numbers obey the laws > > of trichotomy.) > > > > Cantor's (first diagonalization) method allows to enumerate every > > rational number. So each rational has its place in the sequence. But > > this sequence does not enumerate *all rationals". Otherwise there must > > be the largest rational less than 1. > > Wrong. That must only be true if there is a largest natural number. Of course. If all natural numbers are available, then there must be a largest, because the naturals obey trichotomy. Please do not mistake all for each. You can name each natural number (in fact you cannot, but let's assume, you could), but you cannot name all natural numbers. All means "completed". Completed means, there is nothing to be added. In case of a set the elements of which obey trichotomy, there is a largest one. > Consider the following mapping: > Let's have 'r' a rational number > 0. > Rewrite 'r' as a continued fraction [r0, r1, ..., rn+1], where the > ri are natural numbers, except for r0, which can be 0. It is easily > shown that such a representation is unique. Next map that number to: > 2^r0.(1 + 2^r1.(1 + ... (1 + 2^rn)...)). So the continued fraction is > mapped to an integer which has in its binary expansion a specific spacing > of the 1 bits. From this a reverse mapping is obvious. > Pray tell which rational does not map to an integer, and which integer does > not map to a unique rational. And also tell me which integer maps to the > largest rational smaller than 1. Your mapping seems to be a unique one but this is only due to your point of view. The other possible point of view which is not less justified yields: Every path is simultaneously a path bunch, 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 others are not available. 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 < 10^n : K(p',m) = K(p,m). This also covers decimal expansions and continued fractions and shows that there is no uniqueness, not even for the representation 0.000... of the unique number 0. > > > The absence of this rational is as natural as the absence of the set > > {N,k,f} in Hessenberg's proof. > > That is only because you put the cart before the horse. You do not understand > the conditions. That is because barbers are not members of the subsets of natural numbers? Regards, WM |