Cantor Confusion
in [Math]
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From: William Hughes on 30 Apr 2007 10:45 On Apr 30, 10:34 am, mueck...(a)rz.fh-augsburg.de wrote: > On 30 Apr., 16:17, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > 0.000... is not unique in the tree. > > the path 0.000... is contained in the > > tree and it is the only path that contains only 0's so it is unique. > > Of course p = 0.000... is the only path which contains only 0's So outside of Wolkenmuekenheim it is a unique path inside the tree. > But > the tree is not able to separate it from any other path, because at > every node there is another path p' ijn a bunch with p. Look over there! a pink elephant! At every node it is the same path. - William Hughes
From: William Hughes on 30 Apr 2007 10:58 On Apr 30, 10:10 am, mueck...(a)rz.fh-augsburg.de wrote: > On 30 Apr., 14:55, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On Apr 30, 8:32 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > On 27 Apr., 15:37, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On Apr 27, 8:16 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > forall K(p,n) with n in N thereis p' such that forall m in N with m =< > > > > > n : K(p',m) = K(p,m). > > > > > p' may be different for different n's, so I should write p'(n). > > > > I need to show that I can find a single, non changing, > > > > p' that hold for every n. > > > > Why do you think to need to show that? > > > You are claiming that path p is not separated from all other paths. > > This is equivalent to claiming that > > there is a least one path p' (and a path in a non changing thing), > > which is not separated from p. > > Wrong. It is sufficient to show that > forall K(p,n) with n in N thereis p' such that forall m in N with m > =<n : K(p',m) = K(p,m), > because a path has only nodes indexed by natural numbers. There is no > part of p surpassing "forall m in N". Why do you think otherwise? > > > > > > If any infinite path p exists, then the bunch {p} does exist > > > No, {p} is not a path bunch. > > Please be precise. {p} is a path bunch. Only in Wolkenmuekenheim, where on alternate Tuesdays, and whenever Muekenheim finds it convenient, a path bunch is a set of paths. Outside of Wolkenmuekenheim, not every set of paths is a path bunch. In particular, {p} is not a path bunch. So showing that there are only a countable number of path bunches does not show there are only a countable number of paths. <snip> > If {p} is impossible even in the infinite tree Outside of Volkenmuekenheim, saying that {p} is not a path bunch does not mean that {p} is impossible. There are more things in heaven and earth than just path bunches. - William Hughes
From: William Hughes on 30 Apr 2007 11:08 On Apr 30, 10:13 am, mueck...(a)rz.fh-augsburg.de wrote: > On 30 Apr., 15:02, William Hughes <wpihug...(a)hotmail.com> wrote: > > > On Apr 30, 8:40 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > There cannot be more splitting results than splittings. > > > An infinite path is not a splitting result. > > The splitting of a path bunch into two path bunches is a splitting > result. So a splitting result is two path bunches. Only in Wolkenmuekenheim is two path bunches an infinite path. - William Hughes
From: Virgil on 30 Apr 2007 13:28 In article <1177936379.128093.174340(a)n76g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Anyhow. The number of path bunches surpassing the number of nodes is > zero. If any infinite path p exists, then the bunch {p} does exist > too, and {p} is a path bunch. If, on the other hand, {p} does not > exist, then it is ridiculous to try to count the number of paths. So that WM is essentially saying that there is no such thing as an infinite binary tree. The problem is that he has no axiom system in which he can prove his conclusion whereas we have several in which we can prove that such trees do exist and have uncountably many paths in them. The ZF system and the NBG system, for example, both not only allow, they require that CIBTs exist and that the set of paths in them are uncountable. What axiom system does WM claim allows him to prove otherwise?
From: Virgil on 30 Apr 2007 13:33
In article <1177936511.069913.212850(a)e65g2000hsc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 27 Apr., 22:06, Virgil <vir...(a)comcast.net> wrote: > > > > If something is valid for all elements, then it does not change if the > > > number of elements is infinite. > > > > What is true of elements one at a time need not be true for infinite > > sets of them. > I.e., properties of members of a set need not be inherited by the sets themselves. > Why then do you believe that Cantor's diagonal proof is true? Cantor's proof does not say anything is true for a set itself, only for each member of a set, so it is not the same thing. |