Re: Answer to OJ

In article
<1b27b853-cc43-47e9-8bd9-7fc5420d7de2@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
WM <mueckenh@xxxxxxxxxxxxxxxxx> wrote:

On 1 Apr., 22:23, Virgil <Vir...@xxxxxxxxx> wrote:


The infinite binary tree (except for the root node) consists of
elements only that are identical with that one shown above.

Each of these elements shows us that one line goes in, two lines go
out, and the difference is due to the node.

One edge goes in (to all but the root node) and two edges come out.
Lines are just edges.

Lines, other than as edges, are irrelevant.

Lines are infinite sequences (of nodes or, alternatively, of edges)
that can be distinguished.

Having sequence of nodes or edges does not require that there be any
path containing that sequence without some appropriate sequencing rule,
which has not been stated.


There are only two lines that can be
distinguished between the root node and the first level.

Path === binary representation of a real of the unit uinterval?

This won't work as there are reals in that interval with two binary
representations. E.g., 0.0111... and 0.1000... both represent 1/2.

Some reals have double binary representation. Those have also two
paths representing them. Therefore:
Path === binary representation of a real of the unit uinterval!

*Any* iterative construction of paths will omit paths with some
property, even if it demonstrably visits infinitely many different
nodes in the tree, or every node in the tree, at some point.

That is important: Every node is constructed. And there is no path
that exists and remains unconstructed.

WRONG! One can have at least one path through every node in a COMPLETE
infinite tree without needing every path.

Classical example: in a binary tree, at least one path with only
finitely many right branchings passes through each node, but no path
with infinitely many right branchings need occur in order to cover all
those nodes

None of them *can* occur if all nodes have been occupied by paths with
drive towards LHS (and if mathematics of numbers consists of digits
and not of superstition, as should not be necessary to mention).

None of WHICH can occur? WM seems unable to eliminate or even control
his many superstitions about infiniteness.

Cantor's binary sequence proof still rules.

WM's superstitions about it still fail.
.

Relevant Pages

  • Re: The complete infinite binary tree has only countably many infinite paths.
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  • Re: The complete infinite binary tree has only countably many infinite paths, says WM.
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  • Re: Binary Tree and Pairs of Nodes
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  • Re: Cantor Confusion
    ... > The domain of a sequence is a natural number. ... The domain of an infinite ... > Therefore there is no uncuntable set of path in the union tree. ...
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  • Re: Review of Mueckenheims book.
    ... ....2222, they are actually infinitely distant elements of a sequence, ... smallest positive number, on the infinite scale. ... That's the ordering used by Cantor to prove their countability. ... If a set S is countable, then there is a total order < of S such that ...
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