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Topic: FAILURE OF THE DISTINGUISHABILITY ARGUMENT. THE TRIUMPH OF CANTOR:
THE REALS ARE UNCOUNTABLE!

Replies: 47   Last Post: Jan 12, 2013 11:33 AM

 Messages: [ Previous | Next ]
 ross.finlayson@gmail.com Posts: 2,720 Registered: 2/15/09
Re: FAILURE OF THE DISTINGUISHABILITY ARGUMENT. THE TRIUMPH OF
CANTOR: THE REALS ARE UNCOUNTABLE!

Posted: Jan 12, 2013 11:33 AM

On Jan 11, 5:57 pm, Virgil <vir...@ligriv.com> wrote:
> In article
>
>  WM <mueck...@rz.fh-augsburg.de> wrote:

> > On 11 Jan., 01:35, Virgil <vir...@ligriv.com> wrote:
>
> > Do you think that the union of the set of finite initial segments of
> > paths
> > 0.1
> > 0.11
> > 0.111
> > ...
> > contains the path 0.111...
> > or not?

>
> Since I have no idea what the "union" of even "0.1" and 0.11" would be,
> since neither of them appears to refer to any set, I cannot answer until
> WM explains what sets he is referring to.
>

> > If not, what node is missing? None.
>
> What nodes are present, unless you first tell us that you are working
> within a tree?
>
> Rephrase you  questions less ambiguously!
> --

What he asks is what is the union of the finite sets.

Does induction precede the axiomatization of infinity? Because, it
can be built from unbounded finite cases.

Then, does the unbounded imply the existence of the infinite? (Yes,
else it would be bounded.)

Not saying there's enough paper to write it out, saying paper isn't

For the path defined by a set of naturals, with the presence of a
natural in the set indicating a 1 else 0 for branch, that an n-set
describe .111 ... {n-many 1's} , what is the union of the n-sets'
paths?

Then, where the union of the n-sets is N, does induction not require
axiomatization of infinity? Because, a union of m<n-sets would define
induction through n, and there exists n > m for each m.

Then, wouldn't the set of finite sets, be Russell's that all don't
contain themselves? Now, that's a result.

Then, casually about one or zero being the first, ordering the n-sets
with all the sets containing one following all those not, and all
those containing two, and one, after all those not containing two, and
containing one, and all those containing two, and not one, after all
those containing neither two nor one, etcetera ad infinitum, the
ordering of the breadth-first traversal so defined sees the
antidiagonal result not follow.
1^1 + 2^2 + 3^3 + 4^4 ... = sum_i=1^n n^n rules define breadth first
ordering of n-sets as paths

The breadth-first traversal of the C.I.B.T. sees the antidiagonal
result not follow. And: rays through countably many ordinal points
are dense in the paths.

Regards,

Ross Finlayson