Re: Cantor Confusion

In article <1166092128.058778.127950@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

Dik T. Winter schrieb:

In article <1166035958.305158.282870@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
mueckenh@xxxxxxxxxxxxxxxxx writes:
> Dik T. Winter schrieb:
> > > > > That one which goes left from the root is not engaged, because
> > > > > we need
> > > > > only half of the set of edges.
> > > >
> > > > So your construction is not a surjection.
> > >
> > > Of course it is! It is a surjection from the set of edges onto the
> > > set
> > > of paths.
> >
> > Than again. To what number maps the edge that goes left from the
> > root?
> > If it is a surjection, there should be one.
> >
> A surjection onto the paths covers all elements of the *range* = every
> edge. It is not necessary that every edge is mapped on a path, to show
> that there are not less edges than paths. It is only necessary that
> there is no path without edge mapped on it.

Yes. But you stated that you had constructed a surjection. And if you can
not state to what number a particular edge maps you have *not* constructed
a sufjection. So you now state that you did *not* construct a surjection?

I proved the possibility of a surjection.

No such proof is possible without making an assumption contrary to any
standard set theory, such as ZFC or NBG or NF.

But again your statement "it is only necessary that there is no path
without
edge mapped to it" is blatantly wrong. Consider the infinite set of
decimal
numbers. Map a digit (from 0 to 9) to a number when it has somewhere in
its expansion that digit. If there are more numbers a digit maps to, use
shares. By this reasoning (which is your reasoning) there is a surjection
of the set of digits 0 to 9 to the decimal numbers.

You have not understood, deplorably. The shares of every edge I use add
to 1 edge. And every share is mapped on one path only. And every path
gets as many shares to restate two full edges on its own.

Every path requires an infinite sequence of edges in WM's construction,
so it is not a mapping from individual edges but from infinite sequences
of edges, which ruins WM's claims.


No I simply state that the diagonal proof fails in case of the tree.

AS the diagonal proof is not allied to trees, so what?



We can simply count the edges. They are countable. So we an count the
biginnings of separated parts of paths (because every edge is a
beginnng of the separated part of a path, notwithstanding which it may
be).


Every finite beginning of a set of paths, however long, is the beginning
of as many paths as the set of all paths.

So there are as many paths following any such beginning as ther are
paths altogether.

Therefore the beginnings of separated parts of the paths are
countable.


It is the endings that are not countable.
Cut off as large a finite leading segment of any path as you like , and
the number of paths continuing from that point onward is the same as the
set of ALL PATHS.


It seems there is a gap
between 2^1 and 2^2. Why is that not a problem?

There is no gap in looking at the beginnings of separated parts of
paths. If always counting from the left side, we come to the result 3
edges

WM chooses only to look where the things of interest do not occur.


Look at the edges.

Look at the number of paths sprouting out of any edge.

Do you deny that every edge is the beginning of that
part of a path where it is separated from other paths?

I deny that it has any relevance to the "number" of paths.

Do you believe that paths beginn to run separated wihout any edge being
inolved?

I deny that it has any relevance to the "number" of paths.
.

Relevant Pages

  • Re: Cantor Confusion
    ... But you stated that you had constructed a surjection. ... Map a digit to a number when it ... The shares of every edge I use add ... No I simply state that the diagonal proof fails in case of the tree. ...
    (sci.math)
  • Re: Cantor Confusion
    ... It is a surjection from the set of edges onto the set ... its expansion that digit. ... The shares of every edge I use add ... >>> which is not in the tree. ...
    (sci.math)
  • Re: Cantor Confusion
    ... >> later on recombine those shares. ... Wolfgang claimed a surjection from the edges to the ... infinite path in the tree. ... If we have a contradiction then ZFC is inconsistent. ...
    (sci.math)
  • Re: Cantor Confusion
    ... >> later on recombine those shares. ... Wolfgang claimed a surjection from the edges to the ... Each node can be represented uniquely by a finite string of left/right ... If we have a contradiction then ZFC is inconsistent. ...
    (sci.math)
  • Re: Cantor Confusion
    ... I can't so you can not show a bijection. ... You said you had constructed a surjection from the ... To prove that show us a single edge that maps to ... Each part of an edge may map to a single path, ...
    (sci.math)