Cantor Confusion
in [Math]
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From: Virgil on 10 May 2007 14:21 In article <1178792057.501828.120870(a)p77g2000hsh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. What WM declares obvious is provably false in ZF to NBG, as the set of paths is clearly equinumerous with the set of all subsets of the set of levels, and the set of all levels is equinumerous with the set of 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. Not even that. In a finite tree, the number of paths is the same as the number of leaf nodes, something which obviously does not carry over to infinite trees. > The infinite paths are nothing but unions of finite paths. > These unions do not create any new node. Depends on how one defines paths and how one defines union as to whether a union of paths is a path. And even then, only unions of certain related paths could again be paths. > > > > 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. On the contrary, in a CIBT, every branching at any node separates two sets of paths, each equinumerous with the set of paths of the entire original tree.
From: Virgil on 10 May 2007 14:26 In article <1178792572.147577.287520(a)o5g2000hsb.googlegroups.com>, WM <mueckenh(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), Branchings are not single numbers but sets of numbers, unless one cuts things off with leaf nodes at the branching. > > > > >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... But what the separate branchings at each node separates in a CIBT are sets of paths each equinumerous with the set of paths of the entire tree.
From: William Hughes on 10 May 2007 14:32 On May 10, 12:52 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 10 Mai, 04:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1178722674.540579.289...(a)e51g2000hsg.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > On 9 Mai, 04:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > 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. > > > > Obviously not. For each node there is a co-path p' which has been with > > > p all the way long. > > > Where is the contradiction? > > If you say > 1) for every path p', there is a node where p' leaves p > and > 2) there is no node where every path p' has left p > then you must be able to name a node K(n) such that for nodes K(m < n) > at least two paths p' and p'' are required to accompany p. > Because, if for all nodes which you can enumerate, one path p' is > sufficient to accompany p, why should anybody share your belieflthat > for all paths you can enumerate, not one path p' is sufficient to > accompany p? > In Wolkenmuekenheim, the number of paths you can enumerate is always finite, so for all paths you can enumerate, one path p' is suff icient to accompany p. Of course the set "all paths you can enumerate" can change. Look! Over there! A pink elephant! p' never has to change. - William Hughes
From: Virgil on 10 May 2007 14:39 In article <1178793140.441281.239800(a)l77g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. Then p'' = p, as in any tree, finite or infinite, if every node of each of two paths is a node of the other, they are but one path. A path is completely determined by the set of nodes it passes through. > > > > > > 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, WM's claim above is ambiguous. To say that at any non-leaf node of a complete tree there is more than one path through that node, that would be true. But to say that for one path there is a different path through every node of the first path is false. It is not clear which one WEm is trying to say. > > 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! It is just WM's DTs lurking.
From: Virgil on 10 May 2007 14:45
In article <1178793587.372588.308230(a)q75g2000hsh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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 � 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. At british and American universities, functions are first of all sets of ordered pairs. While some functions may be defined using 'formulas or rules or whatever', those 'formulas or rules or whatever' are not the functions themselves, the functions 'are' the sets of ordered pairs. |