Cantor Confusion
in [Math]
|
From: WM on 9 May 2007 14:57 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. 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. It has not a single node more than the union of these paths. Therefore this path is the result of aleph_0 branching-offs. Regards, WM
From: Virgil on 9 May 2007 14:58 In article <1178721265.006218.171950(a)e51g2000hsg.googlegroups.com>, WM <mueckenh(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. 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. The original tree has a set of paths bijecting with the set of subsets of N, so of cardinality greater than that of N.
From: WM on 9 May 2007 15:04 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. > > 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. But there must be such a path. Why not call it 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, because p is not single at any node. As long as we are not forced to give a numerical value, let's call it p''. There are more than enough real numbers which must be represented by paths which can never be given. Regards, WM
From: Virgil on 9 May 2007 15:05 In article <1178723452.432944.32310(a)e51g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 15:50, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1178628554.931048.21...(a)y80g2000hsf.googlegroups.com> WM > > <mueck...(a)rz.fh-augsburg.de> writes: > > > On 7 Mai, 22:36, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > > > (1) Who is the author of that book? I'd like to look it up to see just > > > > what is written there. > > > > > > Karel Hrbacek and Thomas Jech: "Introduction to Set Theory" > > > Marcel Dekker Inc., New York, 1984, 2nd edition. > > > > They do not *use* the word "formula". > > No, they don't, but I use it. Nevertheless it is synonymous to what > they use. Not what they use in any formal definition. > > > But that is not the definition, their formal definition is: > > "A binary relation F is called a function (or mapping, correspondence), if > > aFb_1 and aFb_2 imply b_1 = b_2 for any a, b_1, and b_2." > > And how do you think the a's and b's are selected? And from what sets > do you think > are they selected? And why do you think H&J give that informal > statement? Just in order to confuse the students? It is ridiculous to > follow this discussion. I can assure you, if one of my students would > not know that a function is a formula (or rule or whatever) together > with a domain where it is defined and a range, then he or she would > not pass the exame. And this is the same in the better math courses in > Germany. A listing of elements of a set defines that set without reliance on anything that can be called a "rule" of "formula". {<1,7>, <2,11>, <3,3>, <5,3> } defines a function without any 'formula' or 'rule' needed to define it.
From: Virgil on 9 May 2007 15:06
In article <1178723800.220312.183120(a)w5g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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 It is YOUR void responses to our postings which are the problem here. |