Re: Cantor and the binary tree

In article <1117020437.377622.305540@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

> Robert Kolker wrote:
> > Tony Orlow (aeo6) wrote:
> > >
> > > That's all very well and good, if you specify f anf g and figure those
> > > functions into your comparison. It's a mistake to ignore them.
> >
> > Are you capable of following a proof? Even a three line proof?
>
> Are you capable to follow a five lines proof without referring to "Big
> Brother" Cantor? Consistency of set theory is questioned, hence I do
> not accept Cantor's proof as an argument.
>
> line number n
> 0 0.
> 1 0 1
> 2 0 1 0 1
> ... ..................
>
>
> 1) Each real number of (0,1) is given by a path stretching over
> infinitely many nodes (bits).
> 2) All nodes (bits) of the tree belong to a countable set.
> 3) A node can only exist within a path.
> 4) Any node increases the number of paths by 1 from 1 coming in, to 2
> going out. 2 - 1 = 1.
> 5) Any node increases the number of nodes by 1.
>
> Please point out which step is wrong.

WM's "proof" disproved"

WM conflates bounded paths, having terminal or leaf nodes with unbounded
unending paths which have no terminal or leaf nodes, but contain
infinitely many intermediate nodes.

1) Each number of (0,1) is given by an UNENDING path stretching over
infinitely many nodes (bits).

2) All nodes (bits) of the tree belong to a countable set.

3) A node can only exist within a path.

4) Any node increases the number of ENDING paths, having terminal or
leaf nodes, by 1 from 1 coming in, to 2> going out. 2 - 1 = 1.

5) Any node increases the number of nodes by 1, but have absolutely
nothing to do with the number of unending paths.


All unending paths in an unending binary tree contain infinitely many
nodes.

The number of leaf nodes exactly equals the number of ending or finite
paths in any finite binary tree (in which all paths end).

Considering the binary tree whose root is "." and each branch is
indicated by a "0" or a "1", each leaf node, and therefore each path, is
represented by a terminating binary fraction , but each unending path is
represented by a non-terminating binary fraction.

There are moreof the non-terminating than of the terminating.

So WM is wrong yet again.
.

Relevant Pages

  • Re: Cantor and the binary tree
    ... >> unending paths which have no terminal or leaf nodes. ... >> paths in any finite binary tree. ... >> There are more of the non-terminating than of the terminating. ... > and if one is infinite, ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... > There are no finite paths in my tree. ... nothing to do with the number of unending paths. ... All unending paths in an unending binary tree contain infinitely many ... represented by a terminating binary fraction, ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... > nothing to do with the number of unending paths. ... > paths in any finite binary tree. ... infinite in he tree. ... > There are moreof the non-terminating than of the terminating. ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... If you are talking about an infinite tree it may be ... having terminal or leaf nodes with unbounded ... nothing to do with the number of unending paths. ... All unending paths in an unending binary tree contain infinitely many ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... > equivalent, but infinite or non-terminating paths have no leaf nodes, ... > 2) All nodes of the tree belong to a countable set. ... > nothing to do with the number of unending paths. ... > represented by a terminating binary fraction, ...
    (sci.math)