Cantor Confusion
in [Math]
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From: Virgil on 30 Apr 2007 14:24 In article <1177942395.372951.189000(a)p77g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Which subset of the set of all levels does not represent the unique path branching left from just those levels and branching right from all other levels? Failure to find such a subset shows that WM's position is untenable.
From: Virgil on 30 Apr 2007 14:25 In article <1177942663.514897.21460(a)o5g2000hsb.googlegroups.com>, William Hughes <wpihughes(a)hotmail.com> wrote: > 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. So that Volkenmeukenheim is a Humpty Dumpty land? > > 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: Virgil on 30 Apr 2007 14:29 In article <1177943670.511746.99050(a)h2g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. In a finite tree, every path is determined by a finite set of nodes and can be separated from all others by a finite set of nodes. In an infinite tree, it takes an infinite set of nodes to determine a path and an infinite set to separate it from all others. Why can WM not accept that? Which subset of the set of all levels does not represent the unique path branching left from just those levels and branching right from all other levels? Failure to find such a subset shows that WM's position is untenable.
From: Dik T. Winter on 30 Apr 2007 22:02 In article <1177941733.864803.29600(a)h2g2000hsg.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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: .... > > > 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. All natural numbers *are* a single number. You are confusing "all" with "the set of all". > > 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. Your conclusion is invalid. Indeed, each node is passed by a mutlitude of paths. But the path "0.10101010..." is nevertheless separated from all other paths. > > > 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. So what? For each two different paths it is possible to state a node where they do separate. For each path there is no node where it separates from all other paths. There is no contradiction. > > > 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. As long as you remain in the finite part. > > > 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. As I have no idea what you mean with "up to all nodes", it still makes no sense. > > > 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. Just semantics. In mathematics when something is true for each element of a set it is true for all elements of a set. > In a completed, i.e., finite set, Ah, again contradicting the axiom of infinity. > (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.) Makes no sense. Do you mean that within the reals sup{x < 1} does not exist? > > > 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. What *is* "the existence of all natural numbers"? Back to philosphy again? > > > 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. Prove it. > > It is your simplified finitistic view where that > > means that there is a larger number. > > It is the meanig of "completed". Your meaning, perhaps. > > 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. As your opinion of "splitting results" is worthless, this makes no sense. First try to properly define how bunches split. Until now, with your definitions, I have not seen any splitting of bunches at all. > > 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. Yes. So what? > > > 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. Define unique, and we can talk. With current definitions, the number is uncountable. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 30 Apr 2007 21:40
In article <1177939040.660107.315400(a)y5g2000hsa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 30 Apr., 00:25, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > 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? But there are not two different sides. Both statements are not in contradiction with each other. > 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}. If you mean with {p} the set containing the path p, that is trivially existing. Why you think that set does not exist escapes me. > > > > 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. A severe terminology problem. Bunches do not split, they continue until they terminate. > > > 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. Right. So there are infinitely many bunches of paths starting at the root node. And so bunches of paths do not split, they either terminate, or they continue. > > > 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. That is trivially wrong. In common with what? > 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. Yes, I assume such a definition. > > > 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. Trivially wrong. The path bunches 0.000, 0.001, 0.010, 0.011 all come in at the node 0.0. Or is it now your opinion that they do not start at the root but somewhere else? And I would state that from the node 0.0, amongst others all four path bunches go out. The only path bunch that does go in and does not come out is 0.0. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |