Re: Review of Mueckenheims book.

In article <1174657899.144362.102630@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

On 22 Mrz., 17:09, William Hughes <wpihug...@xxxxxxxxxxx> wrote:


M: Every paths of T_U has an end.

Which of the following statements that you previously
agreed to do you now retract?

14. A path in a tree can only end at the last level

15. There is no last level in T_U

Both are correct. Nevertheless there is no infinite path in U(T(n)).

Then U(T(n)) is not a tree, or at least is not a CIBT.

That is an antinomy of infinity.

No, it is only a failure of WM's models to represent infinite trees.



Remember: The union of finite nodes (nodes with finite index n) makes
an infinite path.
But simultaneously the union of all finite paths makes the infinite
tree.

But how is it that WM's infinite tree has no paths at all?
Note: a set of paths is not a path.

Obviously we have too much elements of one sort, namely the infinite
paths. If U(T(n)) = T(oo), the complete tree, then we simply don't
need the infinite paths.

Then clearly WM's U(T(n)) is NOT equal to T(oo).

There are lots of models for a T(oo) which contain paths which are
infinite, and in all of them there re uncountably many such paths.
And there are no models for T(oo) without paths (maximal chains in which
each node connects via an edge to exactly one child node).

For example: Every partition of N into two subsets, one of the nodes to
be followed by a left branch and the other by nodes to be followed by a
right branch, determines a tree in an complete infinite binary tree
which has a root node and in which each node is followed by one left
child node and one right child node.

As the number of such partitions of N bijects with P(N), the number of
paths so defined is uncountable.

Such a tree is no more complicated to understand than N itself, so that
if WM does not understand it, he does not understand N.
.

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