Cantor Confusion
in [Math]
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From: WM on 10 May 2007 06:24 On 9 Mai, 22:07, William Hughes <wpihug...(a)hotmail.com> wrote: > 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 [more] than the number of sets unioned. Did you want to write that? You unveil your secret! Regards, WM
From: WM on 10 May 2007 06:32 On 9 Mai, 22:43, William Hughes <wpihug...(a)hotmail.com> wrote: > >...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. No, it has not. p'' is, by definition, that path which is with p also at the next node, wherever you rest. > > > 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. I asked you already quite some time ago: Give me two levels for which two different paths are required. If you can do so, then I will believe your claim. > > 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, I have not been taken in by your elephant. Give me two levels for which two different paths are required. Or refrain from claiming this nonsense. Regards, WM
From: WM on 10 May 2007 06:34 On 9 Mai, 23:36, Virgil <vir...(a)comcast.net> wrote: > There has been no discussion here for quite some time, only WM > reiterating his misunderstandings. You seem to be quite interested to take part. Did you mistake your contribution about "infinite definitions" for my? Regards, WM
From: WM on 10 May 2007 06:39 On 9 Mai, 23:42, Virgil <vir...(a)comcast.net> wrote: > > I could hardly have learned less. Of course, "infinite definitions" are always majoring anything else. > > > You need it taught by anyone at all with any more mathematical > > > competence than you have yourself. Only the ignorant laugh at their own > > > ignorance. > > > Then try to stop it. Do you know the definition of a binary relation > > by H&J? > > > 2.1 Definition A set R is a binary relation if all elements of R > > are ordered pairs, i.e., if for any z Î R there exist x and y such > > that z = (x, y). > > > 2.3 Definition Let R be a binary relation. > > (a) The set of all x which are in relation R with some y is called the > > domain of R and denoted by dom R = {x | there exists y such that xRy}. > > dom R is the set of all first coordinates of ordered pairs in R. > > (b) The set of all y such that, for some x, x is in relation R with y > > is called the range of R, denoted by ran R. So ran R = {y | there > > exists x such that xRy}. > > > Can you find the words domain and range in this text? > > I find 'domain' and 'range' to be defined in terms of a set of ordered > pairs, not the reverse, as WM has been trying to imply I did not try to imply anything as you try to imply, but I stated clearly that domain an range belong to a function. And here you see that they belong to a binary relation too. At Oxford and Harvard, to my knowledge, functions are formulas (or rules or whatever) with two sets, just as in Germany. Regards, WM
From: William Hughes on 10 May 2007 06:44
On May 10, 6:22 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 21:52, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > 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. > > You argued that there is an infinite number of branchings (numbers), > each of which takes place at a finite position (each of which is > finite). > > > > > >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. > > Nonsense. Branchings continue to create one single separation per > node. They continue and continue and continue... Correct. and since there is no last branching, an infinite path is not the result of a branching. The question is "is an infinite path the result of a branching?" Look! Over there! A pink elephant! The question is "how many infinite paths are there?" - William Hughes |