Re: Cantor and the binary tree

On Thu, 23 Jun 2005 10:55:08 -0400, Tony Orlow (aeo6)
<aeo6@xxxxxxxxxxx> wrote:

>Martin Shobe said:
>> On Wed, 22 Jun 2005 09:39:20 -0400, Tony Orlow (aeo6)
>> <aeo6@xxxxxxxxxxx> wrote:
>>
>> >Martin Shobe said:
>> >> Ok. Now why is this relevant? There is still no node, n, such that
>> >> f(n) = p.
>> >Of course there is, if you allow for infinite paths, which are a necessity of
>> >an infinite tree with finite branchings, such as a binary tree. Each unique
>> >infinite path is associated with the last branch of divergence, or last 1 in
>> >the series of bits, and of course, one can associate that path with the node at
>> >the end of that branch.
>>
>> Most paths do not have a "last 1 in the series of bits".
>>
>> > In cases where there is an unending string of bits
>> >which aren't all 1's or 0's, that branch/node is a conceptual branch at
>> >infinity, as defined by the pattern of bits.
>>
>> What is a "conceptual branch" and how does it differ from a branch?
>It is infinitely far from the root.

Then there aren't any in the tree in question.

>> > I don't see that, when dealing
>> >with an infinite tree, it makes any sense to discount the branchings that
>> >happen infinitely far from the root node, and there is no way one can have
>> >infinite sets/numbers of branches, nodes, or paths without infinitely long
>> >paths, mathematically.
>>
>> In the tree in question, all branches occure at a finite distance from
>> the root. It is part of the definition of that tree.

>Not in an infinite tree, unless you can somehow show that for finite x, 2^x is
>infinite (x is the number of levels in the tree).

Why would I need to show that an infinite tree is finite?

>> >> >> I've seen your child-sibling tree now. Since at no point on that tree
>> >> >> does any path diverge from all others, your argument fails.
>> >> >At the point of the sibling branch, a new path is formed. All child branches
>> >> >are continuations of the path to which the parent belongs. All sibling branches
>> >> >are the divergence of that parent's path into a new path. Therfore, every other
>> >> >branch is associated with a unique infinite path.
>> >>
>> >> So? The point was that not every unique infinite path has been
>> >> associated with a branch.
>> >That is precisely what I did. Each infinite path is associated with its last
>> >sibling branch, even if it infinitely far from the root.
>>
>> There are still paths you haven't accounted for, those that have no
>> last sibling branch.
>They are accounted for conceptually. An path like 0.01010101...., which is
>equivalent to 1/3, exists. Pretend the number is 0.010101.....010101. Poof!
>There is your last sibling node, which denotes the path. You are going to have
>a root path, then an additional path at level 1, 2 more at level 2, 4 more at
>level 3, etc, so at level N you have 2^(n-1) new paths created. Of course most
>paths are at infinite levels of depth. So what? It's an infinite tree.

The path .010101...010101 is not 1/3, as it terminates while 1/3 does
not.

Martin

.

Relevant Pages

  • Re: Cantor Confusion
    ... The completed infinite tree has no maximum level ... trees does not necessarily hold for the infinite tree. ... infinite set of paths, not a single path. ... is *no* edge that separates the path leading to 1/3 from ...
    (sci.math)
  • Re: An uncountable countable set
    ... members in any power set that in the set itself. ... insists that a property of finite trees must hold for infinite trees. ... What happens in an any part of the infinite tree is precisely described ...
    (sci.math)
  • Re: Cantor / Abbildung N auf R
    ... Zwar wird keine ... Zitat aus Beitrag in NG "Aspects of the infinite", ... extending to the complete infinite tree T, ... besitzt als IN. ...
    (de.sci.mathematik)
  • Re: Cantor Confusion
    ... In standard infinite trees, an edge only separates the uncountable set ... As WM has not explained how a path in an infinite tree can be "separated ... individual entity does not exist as an individual entity. ...
    (sci.math)
  • Re: Cantor Confusion
    ... far away from the root. ... infinite tree, because that contains edges that are *not* finitely far ... difference between the union of all finite trees and the infinite tree. ... The situation with the rationals is quite different, ...
    (sci.math)