Cantor Confusion
in [Math]
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From: mueckenh on 30 Apr 2007 08:35 On 27 Apr., 22:06, Virgil <vir...(a)comcast.net> wrote: > > 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. Why then do you believe that Cantor's diagonal proof is true? Regards, WM
From: mueckenh on 30 Apr 2007 08:40 On 27 Apr., 22:20, Virgil <vir...(a)comcast.net> wrote: > In article <1177671404.952318.309...(a)r35g2000prh.googlegroups.com>, > > > > > > mueck...(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? > > 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, There cannot be more splitting results than splittings. That is true in any case which is object to sensible thinking. If it is not true in infinity, then infinity is not subject to sensible thinking. Regards, WM
From: William Hughes on 30 Apr 2007 08:55 On Apr 30, 8:32 am, mueck...(a)rz.fh-augsburg.de wrote: > On 27 Apr., 15:37, William Hughes <wpihug...(a)hotmail.com> wrote: > > > 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. > > Why do you think to need to show that? You are claiming that path p is not separated from all other paths. This is equivalent to claiming that there is a least one path p' (and a path in a non changing thing), which is not separated from p. All you have shown is that for any n there is a path p'(n) which doesn't get separated from p until level n. This is not the same as showing that there is a path that never gets separated from p. Look over there! A pink elephant! This is exactly the same a showing there is a path that never gets separated from p. <snip> > If any infinite path p exists, then the bunch {p} does exist No, {p} is not a path bunch. A path bunch is (outside of Wolkenmuekenheim of course, inside of Wolkenmuekenheim, there are no fixed statments and no fixed definitions) the set of paths that enter a node. Since there is no last node in an infinite path, an infinite path cannot be identified with the set of paths that leave a node. - William Hughes
From: William Hughes on 30 Apr 2007 09:02 On Apr 30, 8:40 am, mueck...(a)rz.fh-augsburg.de wrote: > There cannot be more splitting results than splittings. An infinite path is not a splitting result. (Please join Tony and Albrecht, chant 100 times before breakfast, some paths do not have a last node) -William Hughes
From: mueckenh on 30 Apr 2007 09:17
On 30 Apr., 00:25, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1177672072.574980.291...(a)n15g2000prd.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 25 Apr., 04:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > 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). > > Yes. So what? Also for all p and p' != p there is an n such that > K(p, n) != K(p', n). If you look from the one side, then the result is different from the result you get from the other side. Why do you think that only your side is the correct one? At *every* node in the tree there is a bunch of paths existing. Hence your point of view is completely irrelevant. There is no specific path p in the tree. Otherwise the existence of p would imply the existence of {p}. > > > > > 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. > > No. The bunch 0.0 terminates and the bunches 0.00 and 0.01 continue. But only till the next node. Then they are split off. > > > > That is a matter of convention of notation. In any case a node > > increases the number of separated path bunches by 1. > > No. You are confusing bunches with edges. The number of paths does > *not* increase at a node. The number of paths remain the same, but > some go right and others go left. Definition: A bunch of paths is a nonempty set of paths which run together through a common set of nodes, as long as they do so. > > > 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. > > But that is wrong. The correct statement is "one edge comes in and two > edges come out". And indeed, the number of edges is countable. As I > already stated, there *is* not a single bunch or path that comes in at > a node. All bunches and paths that come out of a node also do come > in it. A bunch of paths is a set of nodes which have a node in common. Definition: A bunch of paths is a nonempty set of paths which run together through a common set of nodes, as long as they do so. > > > > > 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. > > But there *is* not 1 path bunch that comes in. In every node 1 path bunch comes in and one more paths bunch goes out. Regards, WM |