Re: Cantor and the binary tree

Virgil said:
> In article <1117432107.613540.297700@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> mueckenh@xxxxxxxxxxxxxxxxx wrote:
>
> > *** T. Winter wrote:
> > > In article <1117381085.891784.283560@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
> > > mueckenh@xxxxxxxxxxxxxxxxx writes:
> > > > Examples.
> > > > Path 0,1000...is mapped on the node on level n = 0.
> > > > Path 0,01000... is mapped on the left node on level n = 1.
> > > > In this way all numbers (except 0 = 0.000...) which differ from all
> > > > other numbers by at least one digit are mapped on the nodes.
> > >
> > > Oh. On what node is 0,010101010... mapped? The numbers mapped on nodes
> > > at level n = 0 are of the form k/2, with k odd. The numbers mapped on
> > > nodes at level n = 1 are of the form k/4, with k odd. In general, the
> > > numbers mapped on the nodes of some level n are of the form k/(2^n),
> > > with k odd. On what level is there a node on which 1/3 is mapped?
> >
> > I don't know how many bits the number 1/3 has. But if it is a number
> > then it has a path in my tree. And if it has a path in my tree then it
> > has a node to be mapped on. You must know, there are infinitely many
> > nodes in my tree.
>
> Any path representing 1/3 will have to be "mapped on" infinitely
> infinitely many nodes, if "mapped on" means the path passes trough that
> node.
>
> In fact every infinite path corresponds uniquely to a subset of N as
> determined by which nodes it passes through.
>
Wrong. Every infinite path corresponds to ONE number with infinite digits, not
some subset of N. The branches in the path, right and left, correspond to the
bits in the number, 1 and 0. Wake up and know what you're talking about, for a
change.
--
Smiles,

Tony
.

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