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Replies: 306   Last Post: Aug 13, 2011 6:32 PM

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 mueckenh@rz.fh-augsburg.de Posts: 12,782 Registered: 1/29/05
Re: Balls and vase dyslexia
Posted: Jul 20, 2011 3:16 AM

On 19 Jul., 20:31, MoeBlee <modem...@gmail.com> wrote:
> On Jul 19, 12:04 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
>
>

> > On 19 Jul., 18:00, MoeBlee <modem...@gmail.com> wrote:
>
> > > for each a in w_1\w we have that
> > > P_a is a denumerable binary sequence.

>
> > Correct. You got it.
>
> > > But WM has not specified WHICH
> > > denumerable binary sequence P_a is for a in the domain of P. So P
> > > remains undefined.

>
> > That is necessarily so (because there are only countably many finite
> > definitions) and I cannot use an infinite definiiton. Could you help
> > me?

>
> Yes, I can help you. First, you seem to misunderstand in this sense:
> To define a function with a countable domain, it is not required to
> give a SEPARATE definitions for the value of assigned for each member
> of the domain.

Sorry, for separate functions separate definitions are required.
All paths in the Binary Tree are separate functions.

> > On the other hand, it is not very important, to specify P_a, because
> > for every a and every k we have: P_a is not a subset of B_k.

>
> Whatever the case about that, you simply have not defined P. It is
> utterly unstated by you even WHICH denumerable binary sequences are in
> P. Not even stated by you some property that determines whether or not
> a given denumerable binary sequence is in P.

Sorry if you cannot read definitions. There is every infinite string
of bits after the initial "0." What do you not understand?

As a simple execise try to find out what are the numbers of nodes that
belong to such path as 0.000... or 0.010101... or 0.111... and so on,
when enumerating the nodes according to lexical order, first from top
to bottom, second from left to right:

0
1, 2
3, 4, 5, 6
7, ...

>
> My guess is that you mean for P to be the set of ALL the denumerable
> binary sequences.

P_a is one of the paths. We could use P to denote the set of them. But
until now I don't have done so, IIRC.

> P = {f | f is a denumerable binary sequence}.
>
> Then P is not a function, but okay.
>
> But now if you say the "tree" is P, then what you are calling the
> "tree" is very different from what mathematicians take to be the
> complete binary tree.

You could understand the tree being constructed by all P_a.
>
> Your tree is the set of infinite PATHS for the complete binary tree.
> The set of PATHS of the tree is not the same as the tree. And the tree
> is not just the sef of nodes either. The tree is a set of nodes along
> with an ordering on those nodes.

Correct, I never denied that.
>
> OF COURSE, no infinite path is a subset of any B_k. But that doesn't
> prove that it's a contradiction to say "the set of nodes is countable
> but the set of paths is uncountable".

No, it does not.
My point is very different.

>
> Again, look at Virgil's formulation. The set of nodes is the set of
> natural numbers. The ordering is the progeny relation (that starts
> with the child relation). (I simplified in an earlier post, but for a
> tree, the ordering should be the progeny relation). And the set of
> paths is the set of countable sequences f of nodes such that for all
> n, we have f(n+1) is a child of f(n). The set of nodes is countable.
> The set of paths is uncountable. And you've not shown any

No, I have only shown that the construction of the Binary Tree does
not include the complete set of nodes of any P_a. Therefore
uncountably many P_a must exist in the tree without being defined by
nodes (because all nodes are constructed during the construction).

This means: There can be elements P_a or D in a union or limit of a
sequence (B_k) or (L_k) that are not in any of its terms.

Hence Cantor's diagonal argument fails.

Regrets, WM

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