Re: Review of Mueckenheims book.
- From: mueckenh@xxxxxxxxxxxxxxxxx
- Date: 1 Apr 2007 00:43:35 -0700
On 29 Mrz., 20:53, "William Hughes" <wpihug...@xxxxxxxxxxx> wrote:
On Mar 29, 1:29 pm, mueck...@xxxxxxxxxxxxxxxxx wrote:
On 29 Mrz., 17:56, "William Hughes" <wpihug...@xxxxxxxxxxx> wrote:
On Mar 29, 9:38 am, mueck...@xxxxxxxxxxxxxxxxx wrote:
On 29 Mrz., 14:07, "William Hughes" <wpihug...@xxxxxxxxxxx> wrote:
On Mar 29, 6:45 am, mueck...@xxxxxxxxxxxxxxxxx wrote:
On 28 Mrz., 23:33, Virgil <vir...@xxxxxxxxxxx> wrote:
Do you need an infinite natural number to have the complete set of
finite natural numbers?
No. But I do need an infinite set of naturals to have a complete set.
Yes.
And, as paths are sets of nodes, one needs a sequentially complete set
of nodes to form a path in any tree,
And as finite paths are isomorphic natural numbers, one needs no
infinite path to have an infinite set.
However, such set is not a path in T_U, i.e. is not an element of P_U.
Do you agree with
It is useless to talk about any imaginary infinite paths unless with
have considered and answered the questions of my last post.
Done
Do you agree with
24. Every path in T_U contains an infinite set of nodes.
T_U is the Waft Maximum of the union of all finite trees. It is not
an actually infinite tree.
No. By agreed definition (check the March 19 post)
T_U is the union of all finite trees, U(T(n)).
[The Waft Maximum of U(T(n))] is not U(T(n)). Therefore
T_U is not "the Waft Maximum of the union of all finite trees".
The Waft Maximum of a linear set is the union of its initial segments.
The Waft Maximum of the finite trees is the union of all finite
trees.
No. That would mean the union of all finite initial segments of
Do you agree with
24. Every path in T_U contains an infinite set of nodes.
natural indexes like {1,2,3,...n} is an infinite segment. This would
mean that the Waft Maximum of these segments is N.
Regards, WM
.
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