Cantor Confusion
in [Math]
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From: MoeBlee on 10 May 2007 00:01 On May 9, 7:46 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > I have got the same interface as here. I will answer there. This > thread has become fairly long. Some credit for that goes to me. Some > credit goes to MoeBlee for his understanding of function. I have a correct understanding of the set theoretical definition of 'function' as well as I am correct that that definition is also used in many textbooks of mathematics on subjects other than set theory, and I defended that. As to the length of the thread, it has become as long as it has because several people have wished to express themselves frequently on various subjects that have come up, and after a certain point, mainly about an argument you have about trees. And, just to note (in case it is of any issue whatsoever), with regard to the stated topic of this thread, your argument about trees is not any more on-topic than my discussions about the definition of 'function'. That may not have been needed to say, but I say it JUST IN CASE there is to be taken any insinuation whatsoever that the tangents I have participated in are any less legitimate subject matter than the huge tangent which is the discussion you've been having with your interlocuters about your tree argument. MoeBlee
From: WM on 10 May 2007 06:04 On 9 Mai, 20:31, Virgil <vir...(a)comcast.net> wrote: > > 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. > > The union of infinitely many path bunches is a single path? As the union of infinitely many initial segments of N is N. > > > > > All nodes of the tree are last nodes of finite paths. > > The union over all finite paths is the set of all infinite paths. This > > set has not more elements than were unified. > > Then it is NOT the same 'tree' as a CIBT in ZF or NBG. > > In any CIBT of ZF or NBG, the set of paths is easily shown to > equinumerous with P(N). > > And WM cannot refute this bijection. And Virgil cannot refute that this is a contradiction. Regards, WM
From: WM on 10 May 2007 06:07 On 9 Mai, 20:47, Virgil <vir...(a)comcast.net> wrote: > In fact, to establish the difference between the diagonal and any one > other only requires finitely many digits, which WM knows but chooses to > ignore. This is also true for the numbers 1.000... and 0.999... Why must this case be excluded from the proof? Regards, WM
From: WM on 10 May 2007 06:14 On 9 Mai, 20:58, Virgil <vir...(a)comcast.net> wrote: > In article <1178721265.006218.171...(a)e51g2000hsg.googlegroups.com>, > > 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. > > That assumes, contrary to fact, that within each "branching off" there > are at most countably many paths branching in either direction, but this > has not been established. > > In fact the set of paths in a CIBT through any node is equinummerous > with the set of all paths. Yes, but this set is obviously not larger than the set of all nodes. > > Just take that node as a new root snipping off everything in each path > preceding the new root, and you now have a tree isomorphic in every tree > property to the original, and with as many paths as the original. Fine. The number of all finite paths is equinumerous to the number of all nodes. The infinite paths are nothing but unions of finite paths. These unions do not create any new node. > > The original tree has a set of paths bijecting with the set of subsets > of N, so of cardinality greater than that of N. Any separation by one node yields not more than one further path. This continues infinitely, but does not change the inequality 2-1-1+2--1+2-1-1+-... < 3. It simply continues and continues and continues ... Regards, WM
From: WM on 10 May 2007 06:22
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... , and for any countable number of elements 2-1-1 we have 2-1-1+2-1-1+2-1-1+-... < 3 > > Look! Over There! A pink elephant. > 2-1-1+2-1-1+2-1-1+-... > 3 Uuups Regards, WM |