Re: Paths

On 11 Mai, 18:17, William Hughes <wpihug...@xxxxxxxxxxx> wrote:
On May 11, 11:43 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:





On 11 Mai, 16:49, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On May 11, 10:35 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

On 11 Mai, 14:33, Franziska Neugebauer <Franziska-

Neugeba...@xxxxxxxxxxxxxxxxxxx> wrote:

There is no p* which is "with" a given path p and
which is _different_ from p.

But, as I said already, it can also be proven that for no set of nodes
of p the company must differ.

Please do so.

Assume the set P* of paths p* =/= p which share some nodes with p is
finite and at every node at least one such path p* leaves p. Then the
set of common nodes CK = {K | K(p, n) = K(p*, n)} has a last node K
with respect to n. Then p is single at some node K(p, n+1). This is
wrong. Hence the set CK is infinite.

And more, it is easy to see that CK consists of exactly the nodes
of p. (CK is a set of common nodes of path p and a set of paths P*.
Thus
each node in CK is a node belonging to p.)

Can be seen with the naked eye. But by definition CK consists only of
nodes which belong to paths p* =/= p

This does not contradict the fact that every node in CK belongs to p
as well.

Of course not. Every node of CK belongs to p. But every node of CK
does also belong to a path p* which is different from p. That is the
consequence of infinity.
.

It is easy to prove that if a
node K(p, n) of CK belongs to a path p*, then all nodes K(p, m < n)
belong to p* too. Where lies the error?

In thinking that the fact that the nodes in CK belong to other paths
as
well as to p has any significance.

That's a bit too easy. The nodes in CK form one single path. Every
node belongs to a path which is not p. Therefore every node of CK
belongs to a path p* which is formed by these nodes.

Regards, WM

.

Quantcast