Cantor Confusion
in [Math]
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From: WM on 3 May 2007 07:12 On 1 Mai, 03:40, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1177939040.660107.315...(a)y5g2000hsa.googlegroups.com> mueck...(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. Wrong. If there is always a path p' with p, then for every node of p we have a path p' such that K(p, n) = K(p', n). Hence p is never different from every other path. Hence Cantor's diagonal proof fails. > > > 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. Because p can never be distinguished from all other paths in the tree, but only from those few (countably many) paths p' which you can ask for. > > > > > > 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. Bunches terminate by splitting into other bunches. If they would not split, then they would not 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. They terminate because they split. > > > > > 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? Let p in P. P is a set of paths p'. p is a set of nodes. Let K be a node of p. It is common to the paths of the bunch P, if E K in p: A p' in P: k in p'. > > > 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. But not at separated bunches. > 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. But not as separated path bunches, > The only path bunch that does go in > and does not come out is 0.0. Correct. And at this node the bunches 0.00 and 0.01 go out as separated bunches, containing the other bunches you mentioned which separate later. It is the number of separations which counts! Regards, WM
From: WM on 3 May 2007 07:14 On 2 Mai, 20:15, Virgil <vir...(a)comcast.net> wrote: > In article <1178106864.347366.36...(a)c35g2000hsg.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > >http://groups.google.nl/group/AOTI/browse_thread/thread/da0a93f42305b2c8 > > WM's "intercession" of two sets is merely having each set of two sets > dense in their suitably ordered union, and does not eliminate the very > real differences between sets incapable of being bijected. Very real? Real is reality, but no tansfinite beliefs. Regards, WM
From: William Hughes on 3 May 2007 07:18 On May 3, 7:12 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > Wrong. If there is always a path p' with p, then for every node of p > we have a path p' such that K(p, n) = K(p', n). Look over there! A pink elephant! p' never changes. >Hence p is never > different from every other path. since there is a single fixed path, p', which is not the same as p. - William Hughes
From: WM on 3 May 2007 07:29 On 1 Mai, 04:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1177941733.864803.29...(a)h2g2000hsg.googlegroups.com> mueck...(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". And are all men a single man too, in your opinion? > > > > 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. No. Only from those few you can ask for. > > > > > 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. No. Only from those few you can ask for. > > > > > 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. You always remain there. In particular Cantor's diagonal proof always remains there. > > > > > 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. "Up to all nodes" means the sequence of nodes has been worked completely. > > > > > 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. All elements of the set of theorems are a theorem? > > > In a completed, i.e., finite set, > > Ah, again contradicting the axiom of infinity. Correct: contradicting, not refuting 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. "completed" means nothing remains. > > > > 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. A bunch terminate by splitting into two bunches, i.e., by separating these two bunches. The number of separated bunches is and always remains countable. > > > > 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? In the present state of affairs we know that at every node of path p which we look at, path p is not unique. And we know, that we can test and test for different paths p', but we cannot carry out more than countably many tests (our list is limted). > > > > 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. A path is unique when we can prove that it is different from every other path, not only from the few paths we can mention and test, but from all the uncountably many paths we can never mention. In the present state of affairs we know that at every node we look at, the path p is not unique. Regards, WM
From: WM on 3 May 2007 08:50
On 3 Mai, 13:18, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 3, 7:12 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > Wrong. If there is always a path p' with p, then for every node of p > > we have a path p' such that K(p, n) = K(p', n). > > Look over there! A pink elephant! > > p' never changes. > > >Hence p is never > > different from every other path. > > since there is a single fixed path, p', which is not > the same as p. At every node K(p, n) of p there is a single fixede path p', which has been with p for all nodes K(p, m < n). Yes, without pink elephants, we can find at every node K(p, n) such a path p'. So p is never single. If not being completely stubborn, we know or recognize that. Regards, WM |