Virgil
Posts:
9,012
Registered:
1/6/11


Re: ZFC and God
Posted:
Jan 26, 2013 6:30 PM


In article <06a85bef99c14104862c27351c153ccb@f6g2000yqm.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 26 Jan., 16:06, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > > WM <mueck...@rz.fhaugsburg.de> writes: > > > On 26 Jan., 02:50, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > > > >> I asked how you define terminating decimal representation. How is > > >> that meaningless? > > > > > Sorry, where did you ask? > > > > You've snipped the question three times, in the thread directly > > preceding this post. > > Please excuse me, but there are very many text that I have to read. > Sometimes I overlook something. Nevertheless, I answered it: > > > > The latter is not quite correct, because a terminating decimal > > > representation has nothing behind its last digit d_n, neither zeros > > > nor any other digits. (But of course, we can expand every terminating > > > decimal by a finite set of further decimals d_j = 0 for every j with n > > > <j <m, m in N.) > > > > I don't know why you want to avoid using the usual convention that > > 0.1 = 0.1000...., but okay. It makes no difference. > > Sorry, it makes a difference. 0.1000... is an infinite path in the > Binary Tree.
Since 0.1 = 0,1000... is presented as a decimal, it is not at all the same as the binary 0.1000..., and while the binary 0,1000... can be identified with a path in a Complete Infinite Binary Tree, no such idenification is relevant here.
> But in the original question I distingusihed the Binary > Tree constructed by all finite initial segments of infinite paths and > the Binary Tree constructed by (all) infinite paths. Of course it > makes not a difference with respect to nodes. But, according to the > belief of matheologians, it makes a big difference with respect to > paths.
If, as is usual, one defines a path in any binary tree as a maximal sequence of parentchild linked nodes in the node set of that tree, then a Complete Infinite Binary Tree cannot have any finite paths, as no finite set of nodes can be a maximal sequence of parentchild linked nodes. > > > > Let's state the definition explicitly then: > > > > Let x be a real number in [0,1]. We say that x has a terminating > > decimal representation iff there is a natural number k and a > > function f:{1,...,k} > {0,...,9} such that > > > > x = sum_i=1^k f(i) * 10^i. > > > > Right? > > Right. > > > > Now, let {t_i} be a list of all the finite decimal representations of > > reals, that is, each t_i is a finite decimal representation, and every > > finite decimal representation is in the list. For each t_i, let k_i > > be the "length" of t_i. > > > > And we define a sequence d_j so that > > > > d_j = 7 if j > k or t_j(j) != 7 > > d_j = 6 if j <= k and t_j(j) = 7. > > > > As before, we can notice the following facts: > > > > d_j is defined for every j in N. > > d_j = 7 or d_j = 6 for every j in N. > > > > Clearly, d_j is *NOT* a finite sequence. Moreover, since the sequence > > d_j does not end in trailing 0s or 9s, the real number d defined by > > > > d = sum_i=1^oo d_i & 10^i > > > > has no finite decimal representation. > > > > Now, please tell me what is unclear about these obvious facts? > > It is unclear why you apparently are unable to understand, that we are > working in the set of terminating decimals. Therefore the diagonal > cannot be actually infinite, although there is no last digit.
The WM must be working in Wolkenmuekenheim again, as a sequence with no last term is not finite, and outside of Wolkenmuekenheim "not finite" and "infinite" mean the same thing. > > Can't you understand that the Binary Tree constructed by all > terminating paths has no last level and is nevertheless not actually > infinite because it does not contain any actually infinite path like > that of 1/9, 1/7, 1/3, and many, many more?
Then it is not a Complete Infinite Binary Tree. > > You have to distinguish these both cases, because the most important > argument of matheologians is the following: The tree constructed by > means of all finite paths does not contain the actually infinite > paths.
Outside of Wolkenmuekenheim , the definition of a path is any such binary tree is that is it a maximal parentchild connected set of nodes. This means that every path must start at the root node and every finite path must end in a terminal node having no children.
But in a Complete Infinite Binary Tree every node has exactly two child nodes so such terminal nodes are not possible is such a tree.
> Therefore I restrict the discussion to all finite paths.
NONE of which can exist n any COMPLETE Infinite Binary Tree.
> And > obviously the diagonal cannot be longer, because its digits consist > only of digits of the finite paths.
Only in Wolkenmuekenheim.
Outside of Wolkenmuekenheim, all paths in all Complete Infinite Binary Trees are necessarily infinite sets of nodes.
It the diagonal nevertheless > appears infinite to you, then this fact only shows that the most > important argument is wrong. > > Regards, WM 

