Re: Cantor and the binary tree
- From: mueckenh@xxxxxxxxxxxxxxxxx
- Date: 15 Jun 2005 06:06:38 -0700
Virgil wrote:
> In a maximal binary tree, there is no path that "has an edge"(by which I
> presume WM means what we call a branch) to itself, since for any branch
> in any path, there are infinitely many other paths sharing that branch.
>
> No maximal path ever becomes 'separate' from ALL other maximal paths
> simultaneously, but any two such paths will eventually "separate".
That is uninteresting. Any path of an irrational number is always (in
the whole tree) accompanied by infinitely many paths which are due to
terminating rational numbers. These form a bunch which does never
separate. The due numbers are one and the same number. ==> Irrationals
do not exist.
Regards, WM
.
- Follow-Ups:
- Re: Cantor and the binary tree
- From: Virgil
- Re: Cantor and the binary tree
- References:
- Re: Cantor and the binary tree
- From: mueckenh
- Re: Cantor and the binary tree
- From: Robert Kolker
- Re: Cantor and the binary tree
- From: mueckenh
- Re: Cantor and the binary tree
- From: Robert Kolker
- Re: Cantor and the binary tree
- From: mueckenh
- Re: Cantor and the binary tree
- From: Robert Kolker
- Re: Cantor and the binary tree
- From: mueckenh
- Re: Cantor and the binary tree
- From: Robert Kolker
- Re: Cantor and the binary tree
- From: mueckenh
- Re: Cantor and the binary tree
- From: Virgil
- Re: Cantor and the binary tree
- From: mueckenh
- Re: Cantor and the binary tree
- From: Robert Kolker
- Re: Cantor and the binary tree
- From: mueckenh
- Re: Cantor and the binary tree
- From: Virgil
- Re: Cantor and the binary tree
- From: mueckenh
- Re: Cantor and the binary tree
- From: Virgil
- Re: Cantor and the binary tree
- Prev by Date: Re: what's wrong with this argument?
- Next by Date: Re: Cantor and the binary tree
- Previous by thread: Re: Cantor and the binary tree
- Next by thread: Re: Cantor and the binary tree
- Index(es):