SPQR
Posts:
411
Registered:
8/12/11


Re: Who's up for a friendly round of debating CANTORS PROOF?
Posted:
Aug 19, 2011 3:44 PM


In article <e0fd64110ef04e57809dd5b957625a3a@r12g2000vbe.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 19 Aug., 16:38, hagman <goo...@voneitzen.de> wrote: > > On 18 Aug., 14:37, WM <mueck...@rz.fhaugsburg.de> wrote: > > > > > > > > 0. > > > > > / \ > > > > > 0 1 > > > > > / \ / \ > > > > > 0 1 0 1 > > > > > / > > > > > 0 ... > > > > > > 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). > > > > With a proper definition of "complete binary tree" (let's call it > > B_oo), e.g. the follwoing: > > How can one write so much about that innocent Binary Tree? > But if you understand it better that way, it is ok. I understand it > better my way. > > > So, yes, this statement can be considered agreeable. > > Fine. > > > > > > 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.) > > > > > So in this sense, yes, B_oo is the limit of the countably many B_k. > > Fine. > > > > > 3) There is no B_(k+1) that contains two or more infinite paths that > > > are not contained in B_k. > > > > It is an easy exercise, that no B_k (or B_{k+1}) contains *any* > > infinite path at all. > > Yes. > > > This may be due to the fact that I used my definition of B_k, which > > has the advab´ntage > > of being a definition at all. > > It is obvious in any representation. > > > > As a consequence, the set of infinite paths in B_oo (i.e. in the > > "union" of all B_k) > > is not the union of the sets of infinite paths in the B_k. > > Nothing could be easier. > > > B_oo contains all infinite paths as subsets. That is important. > > > > > My argument shows that no infinite path is constructed when the > > > complete Binary Tree is constructed.
It does in my construction:
*************************************************************** * 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. * ***************************************************************
So I have shown, at least for > > > those who believe in infinite paths, that Cantor's argument is wrong. > > > The limit of an infinite sequence (here (B_k)) can contain entities > > > that are not in any of the terms of the sequence. It contains no elements that are not in some B_k, but since each B_k is finite and every path is infinite, no path is a subset of any B_k ,or of any other finite set, for that matter, and more than the entire tree is a subset of any B_k.
> > > > So what? > > The limit sqrt(2) of the sequence 1.4, 1.41, 1.414, 1.4142, > > 1.41421, ... > > has the property that its square is an integer, > > If the limit has a decimal representation, then it must have that > property. But does it?
It is not necessary for the existence of the limit, L, for that limit to have a decimal representation, it is enough that the difference between the alleged limit and the terms of the sequence from some point onwards be sufficiently small.
> Correct. And the limit of the complete Cantorlist may have the > property to contain its diagonal number.
A complete Cantor list is most unlikely to have a limit.
And every such a list of binary sequences can always be shown, by easy extensions of Cantors own diagonal argument, to have omitted as many such sequences as it contains.
> But the Binary Tree contains > *uncountably* many subsets that are not in any of the finite initial > segments B_k. That should trigger the process of wondering about > actual infinity.
Since the number of subsets of finite set increases so much more rapidly that the number of its members, why should one suddenly assume them to become equal in the limit? > > > What you observe is thet the operations > > "take the limit" (of trees or path sets) and "consider the set of > > infinite paths" (of trees) > > do not commute. Noone claims that they should. > > There are many examples of operations that do not commute, how about > > this one > > or this one: The limit, i.e., the complete Cantorlist contains every > real number including its antidiagonal whereas every finite initial > segment of the Cantor list contains only a finite number of reals > excluding its antidiagonal.
WM is off his rocker again.
A "Cantor list" is any mapping from N to 2^N, and none of them contain any of their own antidiagonals.

