Re: Cantor Confusion

mueckenh@xxxxxxxxxxxxxxxxx wrote:
cbrown@xxxxxxxxxxxxxxxxx schrieb:


I think, nobody would oppose to dividing the edges merely in two halves
each. If the series 1 + 1/2 + 1/4 + ... yields 2, then we can extend
this knowledge to bijections too.

If I say that the sets {a} and {p,q} have the same cardinality,

I do not follow this "path". Only for 1/2 edge together with another
1/2 edge together I assert to have 1 edge with cardinal number 1. And
that is correct.

Furthermore, many sets of things (for example, the set of all finite
simple abelian groups) have elements which cannot be "divided" in the
way you seem to imply. What is "half" of the group of order 7?

But a unit can be divided. You know what 1/2 is. And you know what 1/4
is. You know tat 1/2 + 1/2 = 1.


I know that 1/2 of the length of a unit edge + 1/2 of the length of a
unit = 1 unit edge in length; but I don't know that 1/2 of an edge +
1/2 of an edge = 1 edge; because I don't know what "1/2 of" an edge is.

Some things (unit lengths) can be divided in a sensible manner; and
some things cannot. Likewise, some things, such as /lengths/, can be
added so that 1/2 + 1/2 = 1; but 1/2 of the group of order 7 + 1/2 of
the group of order 7 is not the group of order 7; because "1/2 of" is
meaningless here.

Your argument seems to say something about lengths of paths and lengths
of edges; but not about number of paths and number of edges (where by
"number" I mean "cardinality").

If you dislike the fractions only...

I don't dislike fractions; some of my best friends are fractions.

I simply don't see that you have produced, from the functions f, g, and
h, a surjective function T (edges -> paths). You appear to have no
response to the request that you produce one; and instead change the
problem.

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
there 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?

(2) How does the existence of T then lead to a contradiction?

(Prediction: you are somehow conflating the fact that lim n->oo sum
(over all edges e) g(p,e,n) = 2 with the (false) assertion that sum
(over all edges e) lim n->oo g(p,e,n) = 2.)

So I take it you agree that your original argument is flawed?

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 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"?

<snip repetition of the undisputed definition of g: (paths X edges X N)
-> R with lim n->oo sum (over all edges e) g(p,e,n) = 2>

Cheers - Chas

.

Relevant Pages

  • Re: Cantor Confusion
    ... undivided edges but from the shares of divided edges onto paths. ... Now you say that T is /not/ a surjection of edges onto paths, ... What is the range and domain of the bijection T you claim to have ...
    (sci.math)
  • Re: Cantor Confusion
    ... I know that 1/2 of the length of a unit edge + 1/2 of the length of a ... undivided edges but from the shares of divided edges onto paths. ... Now you say that T is /not/ a surjection of edges onto paths, ...
    (sci.math)