Cantor Confusion
in [Math]
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From: William Hughes on 16 Apr 2007 08:27 On Apr 16, 7:53 am, mueck...(a)rz.fh-augsburg.de wrote: > > If every finite node of a path means the whole path (you never > contradicted *that*), then "every finite part of the path" means "the > whole path". > So what? i finite parts of the path have property X ii the whole path is the union of the finite parts of the path This does not imply that iii the whole path has property X. Which bit of "you can't use i and ii to prove iii" do you fail to understand? - William Hughes
From: mueckenh on 16 Apr 2007 09:53 On 16 Apr., 14:27, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Apr 16, 7:53 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > If every finite node of a path means the whole path (you never > > contradicted *that*), then "every finite part of the path" means "the > > whole path". > > So what? > > i finite parts of the path have property X > > ii the whole path is the union of the finite parts of the path > > This does not imply that > > iii the whole path has property X. Look for the answer of Dik T. Winter: "the whole path is not the union of the finite parts of the path" Guess why? Regards, WM
From: William Hughes on 16 Apr 2007 10:31 On Apr 16, 9:53 am, mueck...(a)rz.fh-augsburg.de wrote: > On 16 Apr., 14:27, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Apr 16, 7:53 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > If every finite node of a path means the whole path (you never > > > contradicted *that*), then "every finite part of the path" means "the > > > whole path". > > > So what? > > > i finite parts of the path have property X > > > ii the whole path is the union of the finite parts of the path > > > This does not imply that > > > iii the whole path has property X. > > Look for the answer of Dik T. Winter: "the whole path is not the union > of the finite parts of the path" > > Guess why? > > Regards, WM Only someone with no ethics at all would use quotation marks to indicate a paraphase. Dik Winter did not say the whole path is not the union of the finite parts of the path Nothing that he did say contradicts this. The "special part" is a figment of your imagination. - William Hughes
From: Virgil on 16 Apr 2007 14:23 In article <1176723918.670619.273750(a)q75g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > > But not all real numbers coexist in the tree? Only a countable number > > > of them is admitted simultaneously? > > > > Why? > > There cannot be more separated paths in the whole tree than are points > of separation in the whole tree, (unless there is more than one > separation per point of separation which, however, can be excluded by > the construction of the tree). That presumes, falsely, that one node is necessary or sufficient to separate one path from all others in a CIBT. It takes infinitely many nodes in a CIBT to simultaneously separate any one path from all other paths. All any one node can do is to separate an uncountable set of paths fro another uncountable set of paths.
From: Virgil on 16 Apr 2007 14:29
In article <1176724396.229576.101420(a)p77g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 16 Apr., 02:43, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1176300393.408149.165...(a)w1g2000hsg.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > If all paths exist simultaneously, then there must exist uncountably > > > many in the infinite tree. > > > > But there are. > > Without the chance that uncountably many are separated in the whole > tree? Uncountably many paths are separated from another uncountably many at every node in that infinite tree. To separate any one path from all others takes a set of countably many nodes. > > > > > > > So every finite part of the path does > > > > *not* > > > > mean all parts of the path. > > > > > > You did not talk about such strange parts before. And this explanation > > > does not fit my original statement: > > > "Every finite node of a path means the whole path because there are no > > > infinite nodes". > > > > Yes, I never contradicted *that*. > > > > > Nevertheless, we can agree upon these things if we > > > clarify their meaning. Does "Every finite part of the path" mean "the > > > whole path"? Does "Every finite part of the path" include or cover > > > also that special part you just invented? > > > > And now you switch again from "finite node" to "finite part of the path". > > As "every finite part of the path" would mean to me (without further > > definition) a set of parts of paths, I would state: no, because a path > > is not a set of parts of paths but (by your own definition) a set of nodes. > > And even as a set of parts of paths, "every finite part of the path" does > > *not* contain the path itself as an element (if the path is not finite). > > What you do mean with the word "cover" escapes me. > > Every node n is in bijection with the set of nodes from the first to > n. The latter is a finite part of a path. > > If every finite node of a path means the whole path (you never > contradicted *that*), then "every finite part of the path" means "the > whole path". Why bother to contradict that which only a nut would ever claim? One might argue that all finite parts together form a path, but never that every part is the whole. |