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Topic: Who's up for a friendly round of debating CANTORS PROOF?
Replies: 134   Last Post: Aug 29, 2011 7:17 PM

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 SPQR Posts: 411 Registered: 8/12/11
Re: Who's up for a friendly round of debating CANTORS PROOF?
Posted: Aug 19, 2011 3:09 PM

In article
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 19 Aug., 02:23, at1with0 <at1wi...@gmail.com> wrote:
>

> > I disagree with the statement, "there is a set x and a function whose
> > domain is x which is onto the powerset of x."  This is a consequence
> > of the statement, "R is countable."

>
> I do not claim that |R is countable but I claim that there is no
> actual infinity, hence no complete set |R.

If |R is to exist at all, it must have more than any finite number of
members, so that WM blows hot and cold with one breath.
>
> > Since your conclusion is wrong, there is at least one step in your
> > argument which is wrong.

>
> Or set theory with its claim of finished infinity is inconsistent.

Nope, WM is the one who is wrong.
>
> The Binary Tree is a simple representation of all infinite binary
> sequences, where identical initial segments are represented only once.
> Nothing more, nothing less. A mathematician should be able to grasp
> it.

***************************************************************
* A Simple Model for a Complete Infinite Binary Tree:
*
* Let N be the set of positive natural numbers, {1,2,3,...} as the
* set of nodes, with 1 as the root node.
*
* For any node n in N, let 2*n be its left-child node and 2*n+1
* be its right-child node.
*
* It is easily seen that every node has two and only two child nodes,
* and so this construction forms a complete infinite binary tree
* from N.
*
* A path, P, in this tree is any minimal subset of N such that
* (1) 1 is in P, and
* (2) if n is in P then either 2*n, its left child, or 2*n+1,
* its right child, is in P, but not both
*
* Note that each child node can also be distinguished
* from its opposite child by whether it is an even natural and left
* child denoted by 0, or an odd natural and right child, denoted
* by 1, so that any infinite sequence, 1xxxxxx..., where each x is
* either a 0 or a 1, also designates a unique infinite path.
*
* Note that every path is necessarily an infinite subset of N.
*
* E.g., {1,2,4,8,16,...} (or 1000...)
* and {1,3,7,15,31,...} (or 1111...) are the paths with
* only left children and only right children, respectively.
*
* It has been proved that the set of all such paths bijects
* with the set of infinite binary sequences, and as the latter
* is well known to be uncountably infinite, so is the former.
*
***************************************************************

>
> The following three statements should be easy to check:
> Please note the first statement that you do not agree with.
>
> 1) The complete Binary Tree contains all infinite paths that
> represent
> real numbers of the unit interval [0, 1] (some of them even in
> duplicate, but that does not matter).
>
> 2) The complete Binary Tree is the limit of the sequence (B_k) of its
> finite initial segments. (A sequence consists of a countable number
> of
> terms.)
>
> 3) There is no B_(k+1) that contains two or more infinite paths that
> are not contained in B_k.

There is, in fact, no B_k or B_(k+1) which contains any paths since no
finite set can contain as a subset an infinite set.

However the 3 statements by WM prove absolutely nothing about complete
infinite binary trees.
>
>

> > I have never considered the diagonal argument as it is usually
> > presented to be rigorous.

>
> So you are one of very few.

> >
> > I would like you to provide a function from N onto [0,1].  Can you do
> > that (or at least prove that such a function exists)?-

f: N -> [0,1] : n -> 0 is such a function.
g: N -> [0,1] : n -> 1 is such a function.
h: N -> [0,1] : n -> 1/2 is such a function.
How many of them do you need?

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