
Re: Distinguishability argument x Cantor's arguments?
Posted:
Jan 8, 2013 4:19 PM


On 8 Jan., 17:27, George Greene <gree...@email.unc.edu> wrote: > On Jan 3, 9:19 am, WM <mueck...@rz.fhaugsburg.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 Cantorlist 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 welldefined 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 nthdefined 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 ANTIdiagonal CAN ALSO be > defined! > For ANY DEFINED bitstring, its complement IS WELLdefined! > 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 antidiagonal 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 antidiagonal 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 antidiagonal > 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 antidiagonalofthelistofalldefinablereals" 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 antidiagonal 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

