george
Posts:
800
Registered:
8/5/08


Re: Distinguishability argument x Cantor's arguments?
Posted:
Jan 8, 2013 11:27 AM


On Jan 3, 9:19 am, WM <mueck...@rz.fhaugsburg.de> wrote: > because > no infinite diagonal of a Cantor list can be defined
DANG, you're stupid. 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! IT'S YOUR list! YOU allege that it is welldefined and that it contains all the reals! But you also allege that there are only countably many reals and that they are all definable! In that case, a definable list of all the definable reals IS YOURS *AND*NOT* 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! 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!
However, if the list of all definable reals were ITSELF definable, then its antidiagonal would be 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). Conclusion: the list of all definable reals IS NOT ITSELF DEFINABLE! And it most certainly is NOT a CANTOR list!
IT'S
* YOUR *
DELUSION
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!
that YOUR list

