Re: Cantor and the binary tree

On 25 May 2005 04:49:46 -0700, mueckenh@xxxxxxxxxxxxxxxxx wrote:

>
>
>Martin Shobe wrote:
>
>> >If you prefer "is", you may use it. That does not matter. It is
>> >obviously impossible that the set of paths is uncountable when the set
>> >of nodes is countable, because every pair of paths springs off from one
>> >node, while one path leads to that node. Try to find an error n the
>> >arguing, not in the result.
>>
>> Sure thing. The error is that "It is simply impossible to assume that
>> one of these numbers becomes uncountably infinite while the other
>> remains countably infinite" does not follow from the previous
>> statements in your proof.
>
>Then point out, please, which step is wrong.
>
>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.

The tree is static. The number of paths does not increase or
decrease.

>5) Any node increases the number of nodes by 1.

Again the tree is static. The number of nodes does not increase or
decrease.

> The relationships between nodes and paths
>> you supplied require that the tree be finite.
>
>If you assert that, you should give a number n as an upper bound.

An upper bound of what? The only tree that has been specifically
pointed out is the infinite one. I certainly can't give you an upper
bound for that.

I will admit that I was wrong when I said that the relationship
requires that the tree be finite. I should have said that the proofs
of that relationship requires the tree to be finite.

Would
>you assert that
>0.
>1
>1
>1
>...
>
>must be finite?

No. But the proof of the relationships between nodes and branches
does not hold for this tree.

>Would you assert that
> 0.
>0 1
>0 1
>0 1
>...
>
>must be finite?

As above.

>Which branching must terminate the tree?

If we allow infinite trees, they don't have to terminate.


> How many nodes are admitted?

Are they sick?


> Attempting to use those
>> relationships on infinite trees (without first proving that the
>> relationships hold) is simply a fallicy.
>
>To prove that a branching is a node is easy, because it is defined so.
>More is not necessary.

But what you need to prove isn't "a branching is a node". You need to
prove that for a complete infinite binary tree, there is a bijection
betweeen the paths and the nodes.

Martin

.

Relevant Pages

  • Re: Cantor and the binary tree
    ... >Martin Shobe wrote: ... Each path "contains" an infinite number of edges. ... >Not those in the tree. ... If a separation does not occur in this tree ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... >>Martin Shobe wrote: ... > from every other path by an infinite number of edges. ... >>Not those in the tree. ... If a separation does not occur in this tree ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... >>Martin Shobe wrote: ... There is no point in the tree where any given path seperates ... there are not uncountably many infinite paths any more than there ... Without gauging the relative infinites of paths and ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... > one of these numbers becomes uncountably infinite while the other ... All nodes of the tree belong to a countable set. ... If you assert that, you should give a number n as an upper bound. ... To prove that a branching is a node is easy, ...
    (sci.math)
  • Re: Cantor Confusion
    ... And as 10^-k is never zero, that sum is never an irrational number. ... The infinite binary tree contains this limit. ... Therefore we can calculate the union of all ...
    (sci.math)
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