Cantor Confusion
in [Math]
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From: WM on 9 May 2007 11:16 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. Bye, WM
From: WM on 9 May 2007 11:23 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? > > 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. Is in the union 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)? Regards, WM
From: WM on 9 May 2007 11:26 On 9 Mai, 16:16, William Hughes <wpihug...(a)hotmail.com> wrote: Time does not play a role in set theory but to branch off is a temporal act. Regards, WM
From: WM on 9 May 2007 11:27 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...? What then is it? Regards, WM
From: William Hughes on 9 May 2007 11:30
On May 9, 11:26 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 16:16, William Hughes <wpihug...(a)hotmail.com> wrote: > > Time does not play a role in set theory but to branch off is a > temporal act. > > Regards, WM Remember in Wolkenmuekenheim, whether something is part of set theory depends on whether or not this would inconvenience Muekenhiem. - William Hughes |