Cantor Confusion
in [Math]
|
From: WM on 9 May 2007 11:32 On 9 Mai, 15:36, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 9, 9:21 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 9 Mai, 00:56, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > On May 8, 5:26 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > On 7 Mai, 21:32, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > On May 7, 3:19 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > That is not the point here. The point is that there are subsets of > > > > > > countable sets which do not allow for a bijection with N. > > > > > > No. > > > > > Why "no"??? > > > > Because you made the statement " there are subsets of > > > countable sets which do not allow for a bijection with N." > > > > The "No" was meant to indicate that this statement is false. > > > What is true is that there are subsets of a finitely definable > > > set which are not finitely definable. The difference is... > > > > WAHH! WAHH! WAHH! I'm not listening. > > > > > > There are subsets of a finitely definable set which are not > > > > > finitely definitable. > > > > > And they do allow for a bijection with N? > > > > Yes, but not a finitely definable one. The difference is... > > > > WAHH! WAHH! WAHH! I'm not listening. > > Gosh, I am sure that this line read > > WAHH! WAHH! WAHH! I'm not listening. > > when I wrote it. I must be wrong. Only a total slimeball would write that. If you repeat that once again only, our discussion will have been finished. (This is not a temporal notice.) Regards, WM
From: William Hughes on 9 May 2007 11:39 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? > > > > > > > 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. > > > > All nodes of the tree are last nodes of finite paths. > > > The union over all finite paths is the set of all infinite paths. > > > Not without a pink elephant. > An infinite path is set of nodes. A set of infinite paths is a set of sets of nodes Look! Over there! A pink elephant A set of infinite paths is a set of nodes > Is in the union a set of nodes > a node too much (i.e., a node which does not belong to > an infinite path)? > Or is in the union a node too less (i.e., a node which does not belong > to a finite path but to an infinite path)? > - William Hughes
From: William Hughes on 9 May 2007 12:04 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. > What then is it? > 0,010101... is an infinite path; a union of an unbounded number of finite paths. An infinite path is not a finite path. A result of branching off is a path bundle. According to Professor Muekenheim, a path bundle is a finite path. A result of branching off is not an infinite path. - William Hughes
From: Carsten Schultz on 9 May 2007 12:18 WM schrieb: > On 9 Mai, 16:05, Carsten Schultz <cars...(a)codimi.de> wrote: > >> I see, > > You cannot see. You only believe to do it, even when I showed you that > you can't by a clear example (only in order to help you to recognize > that, of course, - alas in vain). But I will not take the trouble to > translate your proof of failure here. And I will not further respond > to your void postings. Remain as you are. I still would advise you to look up the definition of a relation. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: William Hughes on 9 May 2007 12:44
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 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 <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. Look! Over There! A pink elephant! > every path p there is another path *existing in the tree* which is not > different from p. - William Hughes |