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 <c1d951ca5133458d9b2ec0ad6c9e6c63@e7g2000vbw.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.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 leftchild node and 2*n+1 * be its rightchild 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?

