Cantor Confusion
in [Math]
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From: Virgil on 5 May 2007 15:04 In article <1178366311.849575.234890(a)w5g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 3 Mai, 19:57, Virgil <vir...(a)comcast.net> wrote: > > > > > > What axiom system does WM claim allows him to prove otherwise? > > > > > 2-1-1+2-1-1+2-1-1+-... is never much more than 2. > > > > Irrelevant to the existence of bijections between the power set of N and > > the set of all paths of a CIBT. Which bijections prove WM wrong. > > They prove > > 2-1-1+2-1-1+2-1-1+-... =< 2 > > wrong, i.e., they prove 2-1-1 =/= 0. They prove WM's claims irrelevant to the validity of ZF and NBG. In ZF and NBG all assumptions over and above the validity of formal logic are explicit. None of WM's implicit assumptions are, or can be, required to hold in such systems. In particular, there exists in ZF the set N of all naturals, and, P(N), the power set of all naturals which is of larger cardinality than the set of all naturals, and a CIBT whose set of paths bijects with P(N). These prove beyond mathematical doubt that, in ZF and NBG, CIBTs exist and have uncountably many paths. ZF specifically allows all of this, and no explicit assumptions in ZF prohibit any of this, and no implicit assumptions are allowed in ZF. Similarly for NBG. That WM wants to have a system in which things go his way is obvious. It equally obvious that he is not competent to construct such a system in which every assumption has been made explicit, as mathematical axiom systems require.
From: Virgil on 5 May 2007 15:19 In article <1178366406.361131.321710(a)y5g2000hsa.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 3 Mai, 19:47, Virgil <vir...(a)comcast.net> wrote: > > > > Cantor's proof says that all members of the set R are more than all > > > members of the set N. > > > > In the sense that there is an injection from N to R but none from R to > > N, right! > > > > > This is a wrong conclusion. > > > > It is a validly proved conclusion in ZF and NBG, and WM has not come up > > with any axiom system in which he can prove it false. > > > > And as far as mathematics is concerned, that settles the issue. > > Definition: A set M is countable, if there is a one-to-one > correspondence with the set N of natural numbers. > There is a one-to-one correspondence of M and N if and only if all > elements m_n of M can be enumerated, i.e., if and only if a list can > be defined: > > 1) m_1 > 2) m_2 > 3) m_3 > ... > n) m_n > ... > > Theorem 1: The set of finitely definable numbers is countable and can > be put into a list. WM has not defined "finitely", nor "finitely defineable" nor "set" nor "list". If he is trying to use definitions from ZF or NBG then he must use all of ZF or all of NBG, both of which require the power set of N to exist and can prove it to be of greater cardinality than N. > > Proof: All finite definitions are finite words over a finite alphabet. > All finite words can be ordered lexicographically. This is the finite > definition of a list. > > Theorem 2: The set of finitely defined numbers cannot be put into a > list. > > Proof: Assume such a list exists. This means, such a list has been > finitely defined. Such a list of infinitely many assignments can be "finitely defined" by giving a rule of assignment which applies equally well to infinitely many cases. The function which tales a natural and returns its square is finitely defined and provides a list of squared naturals. So that WM's argument that such list cannot exist is demonstrably false. Besides which, there is nothing in ZF or NBG which requires anything to be "finitely defined", whatever than may mean.
From: Dik T. Winter on 5 May 2007 21:07 In article <1178106864.347366.36570(a)c35g2000hsg.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 1 Mai, 04:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > Dear Dik, > you are invited: > > http://groups.google.nl/group/AOTI/browse_thread/thread/da0a93f42305b2c8 There are no articles in this discussiongroup, they are either expired or removed... And, I do not discuss through google.groups. -- 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 5 May 2007 21:15 In article <virgil-4EF71A.12152102052007(a)comcast.dca.giganews.com> Virgil <virgil(a)comcast.net> writes: > In article <1178106864.347366.36570(a)c35g2000hsg.googlegroups.com>, > WM <mueckenh(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. Ah, it is about that. That is also in his book, and I have mentioned it in the review. I think it is what I call number 8 in his chapter 9. In his book he does *not* prove that it is an equivalence relation (transitivity is not proven, only shown by a single example), moreover, it only works for ordered sets. So if not all sets can be ordered it is not an equivalence relation on sets. -- 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 5 May 2007 21:53
In article <1178190762.081101.159970(a)h2g2000hsg.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > 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, What does this statement mean? > 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. Why not? For every node of p there is a p' such that K(p, n) = K(p', n) and for every path p' different from p there is an n such that K(p, n) != K(p', n). There is no contradiction. > Hence Cantor's diagonal proof fails. Which proof? > > > 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. Now a semi-finitistic view. The path that goes only to the left at every node is different from all other paths, because there is no other path that goes only to the left at every 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. This is nonsense. Bunches *all* start at the root by your definition. If a bunch splits in two bunches at a node that would mean that those two new bunches start at the node, and they do not. > > > > 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. You mean edges, not bunches. > > > 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'. So you retract your statement that "a bunch of paths is a set of nodes which have a node in common"? To what does the "which" refer? You *did* mean "a bunch of paths is a set of paths which have a node in common". > > > > 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. What are 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, What are 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. What are separated bunches? > It is the number of separations which counts! Before you say something like this you should first define your terms. When are two bunches separated? When I analise carefully, I find the following: (1) A path bunch is the maximal set of paths that have some set of nodes in common. (2) Two path bunches are separated if they have no path in common. (3) From (1) we can biject the path bunches with (ordered) sets of nodes that start at the root and where nodes are connected by edges. So a path bunch is either a finite path or an infinite path. (4) Two bunches are separated if there is a node in one of them (after the bijection) that is not in the other and the other way around. (5) A bunch comes in at a node when that node is (after the bijection) in the set of nodes, and it is the last node. (6) A bunch comes out at a node when that node is (after the bijection) in the set of nodes, and it is *not* the last node. When I use this I find that I can biject the bunches that are finite paths with the edges in the tree (and they are countable), because for all such bunches there is a terminating edge. I am still at a loss with those bunches that are an infinite path (actually a single path). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |