Date: Jan 26, 2013 11:45 PM Author: Subject: Re: ZFC and God On Jan 26, 3:30 pm, Virgil <vir...@ligriv.com> wrote:

> In article

> <06a85bef-99c1-4104-862c-27351c153...@f6g2000yqm.googlegroups.com>,

>

>

>

>

>

>

>

>

>

> WM <mueck...@rz.fh-augsburg.de> wrote:

> > On 26 Jan., 16:06, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> > > WM <mueck...@rz.fh-augsburg.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 parent-child 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 parent-child 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 parent-child 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

>

> --

I know this is all very interesting to all of you but what does

Colonel Sanders have to do with God other than his restaurant? Thanks,

Martin