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Re: Distinguishability argument x Cantor's arguments?
Posted:
Jan 8, 2013 4:19 PM
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On 8 Jan., 17:27, George Greene <gree...@email.unc.edu> wrote:
> On Jan 3, 9:19 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > because
> > no infinite diagonal of a Cantor list can be defined
>
> THERE IS *NO*SUCH*THING*AS* "a Cantor list"
> IN THIS CONTEXT! Cantor is REFUTING the list!
> Cantor is REBUTTING the list! The list comes FROM YOU!
In this context, a Cantor-list is a sequence of real numbers of the
unit interval, in general without repetitions. But that would not
disturb the argument.
> IT'S YOUR list! YOU allege that it is well-defined and that
> it contains all the reals!
No, I don't, in particular because there is no such thing as "all the
reals". Why should I claim such a nonsense?
> But you also allege that there are only countably many reals and
> that they are all definable!
I do not allege but can prove that not more than countably many
objects can exist a individuals in mathematics and elesewhere.
> In that case, a definable list of all the definable reals IS YOURS
> *AND*NOT*
> Cantor's!
It seems to be yours. It is neither mine nor Cantor's.
> If THAT list exists, then THE INFINITE DIAGONAL OF IT
> *CAN*BE*AND*IS*
> *TRIVIALLY* defined! It is just "the real whose nth place is the nth
> place of the nth-defined real ON YOUR, NOT CANTOR'S, list."
> It's YOUR list and THAT'S *YOUR* definition of its diagonal!
But I can define a list of all finite strings of digits, that is a
subset of all rational numbers of the unit interval, where most appear
more than once. Here it is:
0.0
0.1
0.00
0.01
0.10
0.11
0.000
...
> And if that diagonal can be defined then THE ANTI-diagonal CAN ALSO be
> defined!
> For ANY DEFINED bit-string, its complement IS WELL-defined!
> The nth digit of the complement is 1 - <the nth digit of the original
> string>.
> If the base is 10 then the nth digit of the anti-diagonal is just 9 -
> the nth digit of the diagonal!
> THESE ARE TRIVIAL, SIMPLE, STRAIGHTFORWARD definitions!
> ANYthing defined this way CAN BE *AND*IS*WELL*- defined!
For instance this list is so well defined, that it cannot be objected
that every finite string of the anti-diagonal is an entry of that
list. By the way this could also be constructed in decimal. But you
may be intelligent enough to understand that?
>
> However, if the list of all definable reals were ITSELF definable,
> then its anti-diagonal
> would be definable
Let that strawman rest where it is burried. I am not interested in the
question whether suchga list was definable.
> and WE WOULD HAVE A CONTRADICTION that the this
> definition
> of "the anti-diagonal-of-the-list-of-all-definable-reals" both WAS
> definable (since the
> above is a definition) AND WAS NOT definable (since it differs from
> every row of the
> list of definable reals).
This contradiction happens already in the above list of all finite
strings which is certainly definable. The anti-diagonal never deviates
from every entry (at a finite place - but others do not exist). This
contracition shows that it is nonsense to talk about completed
infinity.
> Conclusion: the list of all definable reals
> IS NOT ITSELF
> DEFINABLE! And it most certainly is NOT a CANTOR list!
As I said, definability is not interesting.
>
> IT'S
>
> * YOUR *
>
> DELUSION
>
You are in error. It is your delusion. I do not believe in completed
infinity. Not even in the existence of the list defined above.
> THAT *YOUR* list of all definable reals is definable, and yoru much
> deeper
> delusion that your list of all definable reals lists all the reals!
Do you join me when I say that the list containing all finite strings
of 0's and 1's given above does not exist? Do you join me when I say
that the notion of countability is nonsense?
Regards, WM
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