Re: Cantor Confusion

mueckenh@xxxxxxxxxxxxxxxxx wrote:
cbrown@xxxxxxxxxxxxxxxxx schrieb:

mueckenh@xxxxxxxxxxxxxxxxx wrote:
*** T. Winter schrieb:

In article <1165923504.410525.226600@xxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
> *** T. Winter schrieb:
...
> > So you can not show a bijection (in your opinion), but nevertheless you
> > state that you have given a surjection. Do you not think you are
> > contradicting yourself a bit?
>
> I give a surjection on all existing paths.

But not from the edges. But from parts of edges.

> > > No? Even if there are enough parts to gather more than a whole edge to
> > > be mapped on every path?
> >
> > No. Each part of an edge may map to a single path, that does *not* give
> > a map from each edge to a single path.
>
> Since infinity = infinity became an allowed equation, 1/2 + 1/2 = 1 can
> no longer be true?

So if the first half of ab edge maps to, say, 1/2 and the other half to,
say, 3/4. To what single path maps the edge itself?

Why isn't it sufficient to collect the shares of two edegs for every
path? That shows that there are not more paths than edges. Can you
explain your objection to factions?


I previously asked you to "show" how you perform this mapping, given
that we agree to the following:

(1) Edges are countable.
(2) Paths can be bijected with the reals.
(3) B(e) is the (uncountable) set of all paths to which edge e belongs.
(4) Shares (of edges in the original diagram) are denoted (p,e) for
some path p and some edge e where e belongs to p.
(5) S(e) is the (uncountable) set of all shares of edge e.
(6) C(p) is the (countable) set of all shares which belong to path p.
(7) 1 + 1/2 + 1/4 + ... +1/2^n + ... = 2

Your claim (which you seem unable to completely articulate)

Don't mix up lacking understanding of he matter with lacking precision
in stating it.

But that is precisely the problem here: you are mixing up two different
definitions of "part of an edge", because you have not been precise in
stating what you mean by "part of an edge".


seems to be
that there are two "full edges" whose "shares" are exactly the shares
of edges composing a particular path p. But a path p is composed of a
countable number of shares; so a "full edge" can have at most a
countable number of shares.

You are aware of the number 2^omega being a countable one? If so, why
then do you speak of an uncountable set of shares?


Here's what I am "aware" of for (1)-(7): there is a bijection between
paths and reals. There is a bijection between shares of edges and
paths. Therefore, there is a bijection between reals and shares of
edges of any particular edge.

By "set X is uncountable" I mean there is a bijection between the
elements of X and the reals. By "set Y is countable" I mean there is a
bijection between Y and the naturals.

You claim that you can /prove/, using the 7 statements above upon which
we have agreement, that there is a bijection between the reals and and
the naturals; and that therefore countable = uncountable.

When you say "You are aware of the number 2^omega being a countable
one?", I would respond : that's what you're trying to /prove/. You
can't just assert it that it's true for other reasons, and claim that
therefore you have proven it using your "rational relation".
..
Do you think that lim {n-->oo}1/2^n * 2^n consists of uncountably many
parts?


lim {n->oo}(2^-n*2^n) = 1. If your question is "does the real number 1
consist of uncountably many parts", I must ask: what do you mean by a
"part" of the real number 1? There are many interpretations of the
term, each yielding a /different answer/.

Meanwhile, an edge in the original diagram
consists of an uncountable number of shares. So full edges can only be
said to be "the same as" edges in the original diagram if you /assume/
that countable is the same as uncountable.

But that is what you are trying to prove - you can't simply assume it
in a proof of it.

How can 1 + 1/2 + 1/4 + ... +1/2^n + ... yield the value 2 if we add
only countably many shares which are uncountably small?


Why would I think that the real number 1 in the above series is a
"share of an edge" in the /same sense/ that you have defined "shares of
an edge" as being equinumerous with the real numbers?

A "share of an edge" is associated with a particular edge and a
particular path. A path is associated with a real number. But the sum
of the real numbers that correspond to the paths to which edge e
belongs is not 1; it is instead undefined (the sum does not converge).

Would you mind to think about that topic again?


Sure. I think that because you fail to make clear definitions, you
don't even realize that you are using the term "share of an edge" to
mean two completely different things; and this is the source of your
mathematical error.

Now let's argue with aleph0...

Your other arguments are not so interesting to me.

Cheers - Chas

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