Cantor Confusion
in [Math]
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From: WM on 6 May 2007 09:41 On 5 Mai, 19:09, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 5, 7:42 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 3 Mai, 16:05, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > 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' > > > > Look over there! A pink elephant! > > > > p' never changes when n changes. > > > > >So p is never single. > > > > However, it is not necessary for p to be single for p to be separated > > > from > > > a path p'. It is possible for p to be separated from every p' without > > > p ever being single. > > > That is obviously nonsense for linear sets which, after having been > > separated, never meet again. > > Thats right, after the last path p' has separated p has to be single. > Whoops, how does one show there is a last path. That has nothing to do with a last path. If in the infinite binary tree a path does not get single, then it will never exist as a single. Regards, WM
From: WM on 6 May 2007 09:47 On 6 Mai, 03:07, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178106864.347366.36...(a)c35g2000hsg.googlegroups.com> WM <mueck...(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... Excuse me, here are the valid URLs: Binay Tree: http://groups.google.nl/group/AOTI/browse_frm/thread/1f7faf620bb90808/# Intercession http://groups.google.nl/group/AOTI/browse_frm/thread/6386cb0c4c23e619/# > > And, I do not discuss through google.groups. Sometimes it is preferable to have a calm atmosphere. Regards, WM
From: WM on 6 May 2007 09:49 On 6 Mai, 03:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <virgil-4EF71A.12152102052...(a)comcast.dca.giganews.com> Virgil <vir...(a)comcast.net> writes: > > > 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. > > 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), for real numbers and their subsets it is simple, but would not fit to the frame. For other sets, much has to be investigated. > moreover, > it only works for ordered sets. So if not all sets can be ordered it is > not an equivalence relation on sets. Just these points are to be discussed. Regards, WM
From: WM on 6 May 2007 10:29 On 6 Mai, 03:53, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178190762.081101.159...(a)h2g2000hsg.googlegroups.com> WM <mueck...(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? Even in the *infinite* tree a path cannot be distinguished from all other paths. That means, a real number which cannot be described by a finite formula (and most of them cannot) does not exist. > > > 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? Cantor's diagonal proof fails, because the diagonal number is never distinguished from all other real numbers (if uncountably many real numbers exist). > > > > > 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. All start at the root node, but they terminate by splitting. > > > > They terminate because they split. > > You mean edges, not bunches. No. Look at the discussion about that: http://groups.google.nl/group/AOTI/browse_frm/thread/1f7faf620bb90808 > > What are separated bunches? Let us continue this discussion there. > > > 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. yes. > (2) Two path bunches are separated if they have no path in common. yes. > (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. yes. > (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. yes. > (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). Do they occupy more nodes than all the nodes occupied by finite bunches. Look at a formalizaton by Franziska Neugebauer which I replied to in: http://groups.google.nl/group/AOTI/browse_frm/thread/1f7faf620bb90808 Regards, WM
From: WM on 6 May 2007 10:53
On 6 Mai, 04:40, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178366406.361131.321...(a)y5g2000hsa.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > ... > > 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. > > Wrong. Existence does *not* mean finitely definable. Marthematical existence does mean definable. What else should it mean? Definitions can only be finite. > Have a look at > Turings work. There *does* exist a (finitely definable) list of Turing > machines that purport to generate numbers. So the list is certainly > countable. However, the subset of that list that actually *does* > generate numbers is not finitely definable. Fine. That means, the set is countable but there is no bijection with N definable. Why then do you require a definable bijection between paths and nodes in the tree? Regards, WM |