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mueckenh@xxxxxxxxxxxxxxxxx wrote:
> cbrown@xxxxxxxxxxxxxxxxx schrieb:
> > mueckenh@xxxxxxxxxxxxxxxxx wrote:
> > > As those parts of an edge which are mapped on a path are not mapped on
> > > any other path, there is obviously a bijection, though not from
> > > undivided edges but from the shares of divided edges onto paths.
> > >
> >
> > Your original argument was: "Edges are countable. Paths are
> > uncountable. There exists a surjection T of edges onto paths; therefore
> > countable >= uncountable; contradiction".
> >
> > Now you say that T is /not/ a surjection of edges onto paths, but some
> > other thing. I have two questions:
> >
> > (1) What is the range and domain of the bijection T you claim to have
> > provided? More exactly, when you say T maps "shares of divided edges"
> > one-to-one onto "paths", how do you characterize the set of "shares of
> > divided edges"? What exactly are the elements of this set?
>
> The domain is all edges.
So now you return to claiming that your mapping T /is/ a surjection
from edges onto paths. Perhaps there is some confusion regarding "what
is the domain/range of T?".
To clarify, let e be any edge; say the first branch to the left in your
original diagram. According to your above statement, e is in the domain
of T. Which path is T(e)?
> >
> > (2) How does the existence of T then lead to a contradiction?
>
> The number of full edges mapped on a path is larger than 1.
Let p be the path representing the number 1/2. p is in the range of T
because T is a surjection. Which 2 edges in the domain "the set of all
edges" are mapped to p by T? (i.e., what is the pre-image T^-1(p)?)
<snip>
> > > No. The necessity of as many edges as path is so obvious that this fact
> > > is impossible to overlook - once one has discovered it.
> > >
> >
> > If it is so obvious, you should, as a professor, also be able to prove
> > it; otherwise, it is simply your firmly held conviction.
>
> I have proved it by rational relatin and by a random mapping.
You haven't proved it until you provide a surjective function T :
(edges -> paths). (And I am only interested in your "rational relation"
argument).
> >
> > > I add an appendix to one of my papers, where this is underlined I (here
> > > the arguing is based on nodes instead of edges, but that doesn't matter
> > 4> much):
> > >
> >
> > Your appendix fails to address the key question: what is the domain of
> > the function T? If e is an edge, what is the set of "shares of the
> > divided edge e"?
>
> What is your problem? The complete set of shares of one edge is the
> full edge.
So you have a bijection S : (edges -> complete sets of shares of an
edge). But given an edge e, what are the /elements/ of the set of
shares, S(e)? How many elements does S(e) have? 1? 32? A countable
number? An uncountable number?
Cheers - Chas
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