Cantor Confusion
in [Math]
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From: mueckenh on 30 Apr 2007 10:02 On 30 Apr., 00:45, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1177674968.281638.41...(a)u32g2000prd.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > 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". > > And in mathematics if something is proven for "each", it is proven for "all. That is wrong in many cases. That's probably why mathematics is wrong. Example: Each natural number is a single number. All natural numbers are not a single number. > Meaning that if something is proven for "each" element of a set, it is > proven for "all" elements of a set. And because the set of all natural > numbers exists by the axiom of infinity, this also holds for natural > numbers. If something is proven for each natural number, it is proven > for all natural numbers (which does *not* mean that it is proven for > the set of all natural numbers, because that is not a natural number). > That means, all the nodes in the tree are passed by path bunches of many paths. There is never a single path 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). > > Yes, I never did contradict this. On the other hand, for each p and p' != p > there is an n such that K(p, n) != K(p', n). That is irrelevant, because there is no node passed by less than infinitely many paths. > > > 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. > > And indeed, for "all" nodes given a path p that goes through that node, > there is a path p' that also goes through that node. But that does > *not* mean that p and p' have the same set of nodes. But it means that p is never single. If it is never single, then it is at least double, i.e., there is at least one path p' with it. > > > Why do you think, "each" mean "all" in case of Cantor's diagonal but > > not in case of the tree? > > It does, but you have to properly interprete it. You are thinking along > the lines of "up to all nodes", but that makes no sense. I am thinking in the terms of Cantor's diagonal p. "Up to all nodes" we must be able to state that p is different from every other number of its bunch of paths. But that is wrong, because p is never single. > > > > 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. > > We can name all natural numbers, just because that means that we can > name each natural number. Wrong. If you could name all, then no one would remain unnamed. (Otherwise you had not named all.) In a completed, i.e., finite set, there must be a largest one. (A supremum can only exist in case you have every number you know but not in case you have all numbers as a completed set.) > > > All means "completed". Completed means, there is nothing to be added. > > All means indeed completed. The set of all natural numbers is completed > by the axiom of infinity. But it is not completed by the existence of all natural numbers. > > > In case of a set the elements of which obey trichotomy, there is a > > largest one. > > Prove it. The law of trichotomy only states that given two numbers it > is possible to determine that the first is larger, the second is large, > or they are equal. Given all numbers, it is possible to determine the largest. > It is your simplified finitistic view where that > means that there is a larger number. It is the meanig of "completed". > But that is not a mathematical > discussion. Within mathematics, it is possible to determine which of > the three possibilities is true (although it is likely not possible > with the amount of resources we have available). For instance, it is > known that the functions pi(x) and Li(x) cross each other infinitely > many times, but even the first cross-over point will not be known ever. Deplorably this mathematics leads to the result that there are more splitting results than splittings in the binary tree. > > And for each two unequal paths p and p' there is an n such that > K(p, n) != K(p', n). Has no relevance at all. For every node we know that p is not alone. > > > 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. > > As far as mathematics is concerned, 0.000... is unique (as long as we use > the standard definition for that notation). 0.000... is not unique in the tree. The number of unique numbers is countable. Regards, WM
From: mueckenh on 30 Apr 2007 10:10 On 30 Apr., 14:55, William Hughes <wpihug...(a)hotmail.com> wrote: > 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. Wrong. It is sufficient to show that 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), because a path has only nodes indexed by natural numbers. There is no part of p surpassing "forall m in N". Why do you think otherwise? > > > If any infinite path p exists, then the bunch {p} does exist > > No, {p} is not a path bunch. Please be precise. {p} is a path bunch. What you are arguing is that the path bunch {p} cannot exist. > A path bunch is 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. If {p} is impossible even in the infinite tree, then the nonexistence of {p} implies the nonexistence of p. Regards, WM
From: mueckenh on 30 Apr 2007 10:13 On 30 Apr., 15:02, William Hughes <wpihug...(a)hotmail.com> wrote: > 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. The splitting of a path bunch into two path bunches is a splitting result. The number of splitting results is the number of splittings. Regards, WM
From: William Hughes on 30 Apr 2007 10:17 On Apr 30, 10:02 am, mueck...(a)rz.fh-augsburg.de wrote: > 0.000... is not unique in the tree. In Volkenmuekenheim this is true. In Volkenmeukenheim the phrase "is not unique in the tree" means exactly what Muekenheim wants neither more nor less. Outside of Volkenmuekenheim the path 0.000... is contained in the tree and it is the only path that contains only 0's so it is unique. - William Hughes
From: mueckenh on 30 Apr 2007 10:34
On 30 Apr., 16:17, William Hughes <wpihug...(a)hotmail.com> wrote: > > 0.000... is not unique in the tree. > the path 0.000... is contained in the > tree and it is the only path that contains only 0's so it is unique. Of course p = 0.000... is the only path which contains only 0's. But the tree is not able to separate it from any other path, because at every node there is another path p' ijn a bunch with p. In short: It is not sufficient for p to have only 0's because the levels with natural indexes cannot separate p from any other path. Regrds, WM |