Cantor Confusion
in [Math]
|
From: WM on 9 May 2007 15:33 On 9 Mai, 20:25, Virgil <vir...(a)comcast.net> wrote: > In article <1178716861.173911.234...(a)y5g2000hsa.googlegroups.com>, > WM has no power to constrain mathematics to what he thinks it should be. > > That it is not what WM thinks it should be is by now apparent, but that > it is quite productive anyway is equally apparent Producing what? Yes, the working mathematicians, they are quite productive, but they do not need transfinite set theory. Transfinite set theory is the most useless science ever created. Regards, WM
From: William Hughes on 9 May 2007 15:52 On May 9, 2:57 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 18:04, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On May 9, 11:27 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 9 Mai, 16:46, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On May 9, 10:34 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > I gave a bijection between the set of nodes and the set of branching- > > > > > offs of paths bunches. As there can be not more results of branching- > > > > > offs than branching-offs, the task is done. > > > > > There are only a countable number of results of branching-off. > > > > > Whoops, an inifinite path is not a result of branching-off > > > > No? Is there no infinite branching off in 0,010101...? > > > No. There are an unbounded number of branchings, each of which takes > > place at a finite position. There is no branching which takes place at > > an infinite position. > > Correct. In other words: There is no infinite branching off in 0,010101..., so when I wrote "No?" I was being stupid. >The infinite path representing 1/3 is the union of all finite > paths of this kind: 0.0, 0.01, 0.010, ... > It has not a single branching more than the union of these paths. Therefore this path contains aleph_0 branchings This path is the result of the last branching-off Whoops. Look! Over There! A pink elephant. This path is the result of a branching off - William Hughes
From: William Hughes on 9 May 2007 16:07 On May 9, 2:49 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 17:39, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On May 9, 11:23 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 9 Mai, 16:10, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On May 9, 9:39 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > On 9 Mai, 01:10, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > > > > But why do you write down these paths: > > > > > > > > > > Let p'(1) be the path 0111... > > > > > > > > > Let p'(2) be the path 00111... > > > > > > > > > Let p'(3) be the path 000111... > > > > > > > > > > ? > > > > > > > > > Any set p(1) up to p(n) can be substitued by a single path. > > > > > > > > In fact? Why then did you write down so many useless paths? > > > > > > > > > Look! Over there! A pink elephant! > > > > > > > > > The set P' can be substitued by a single path. > > > > > > > > Oh, does the set P' consist of any paths which are not paths p(n) with > > > > > > > a finite argument n? > > > > > > > i: Any set of paths {p(1) ... p(n)) can be replaced by a single > > > > > > path. > > > > > > Why then did you write the unnecessary paths? > > > > Why then did you write the unnecessary paths? > > Why then did you write the unnecessary paths? > > > > > > > > > > > ii: The set P' is the union of sets {p(1) ... p(n)} > > > > > > and as such a set of nodes. > > > > > > In order to avoid the clumsy expression of path bunches, I call a path > > > > > bunch now a finite path. A finite path ends at some node. > > > > > > The infinite path {0.000...} is the union of all finite paths with > > > > > only 0's, namely {0.} U { 0.0} U {0.00} U ... > > > > > The union is one infinite path and, therefore, has not more paths than > > > > > were unified. > > > > > There are only a countable number of finite paths in an infinite path > > > > The infinite path {0.000...} is the union of all finite paths with > > > only 0's, namely {0.} U { 0.0} U {0.00} U ... > > > accepted? > > > Yes, but don't let the pink elephants distract you. > > The infinite path {0.000...} is a union of finite paths, > > or a *set* of nodes. > > The number of these sets is 1, i.e., less than the number of sets > unioned. The number of infinite paths equal to {0.0000....} is 1. Therefore, the number of infinite paths equal to {0.000...} is less than the number of sets unioned. Look! Over there! A pink elephant. The number of infinite paths is than the number of sets unioned. - William Hughes
From: William Hughes on 9 May 2007 16:43 On May 9, 3:04 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 18:44, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On May 9, 10:57 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 9 Mai, 04:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > In article <1178461785.380972.277...(a)e65g2000hsc.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > > > On 6 Mai, 03:53, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > ... > > > > > > > > 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. > > > > > Why not? For each other path there is a finite node where it goes a different > > > > way. > > > Call this other path, p'', and the node., M. > > p'' can be distinguished from p at node M > > Wrong. Then it would not be the path which is with p at the next node. Nor is it. Look above at the descrption of the "other path" For each other path there is a finite node where it goes a different way. The finite node is M and the other path is p'' Since p'' goes a different way from p at node M, p'' is not the path which is with p at the next node. > > > > However, For every node M, there is a path, p', for which > > > p' cannot be distinguished from p at node M. > > > Look! Over There! A pink elephant. > > > p'' cannot be distinguished from p at node M > > I did not say that I could name a path which does never branch off > from p. No, you cannot name a path which does not branch off from p. However, you claim that such a path must exist. Your existence proof is: Look! Over there! A pink elephant! >...there must be such a path. >Why not call it p''. because the name p'' has already been used for a path that does branch off from p. > > > <snip> > > > >The proof in the tree is, that for > > > every path p, and level M, there is another path *existing in the > > tree* > > which is not different from p at a level less than or equal > > to M. > > Yes. But why only see the one side of the medal? For every path p > there is another path *existing in the tree* which is not different > from p at any node, up to an arbitrary level M. This means that p is not single at any node. However, for different levels you may have to use different paths, and there is no one path that will work for all levels. Look! Over there! A pink elephant! For every path p there is another path *existing in the tree* which is not different from p at any node, - William Hughes
From: Virgil on 9 May 2007 17:36
In article <1178724769.155620.137410(a)e51g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > If you repeat that once again only, our discussion > will have been finished. (This is not a temporal notice.) There has been no discussion here for quite some time, only WM reiterating his misunderstandings. |