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Topic: A BLATENT FLAW in Cantor's diag proof
Replies: 111   Last Post: Jun 14, 2010 6:15 PM

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 William Hughes Posts: 4,277 Registered: 12/6/04
Re: A BLATENT FLAW in Cantor's diag proof
Posted: Jun 7, 2010 11:04 PM

On Jun 7, 11:51 pm, "|-|ercules" <radgray...@yahoo.com> wrote:
> "William Hughes" <wpihug...@hotmail.com> wrote
>
>
>

> > On Jun 7, 11:41 pm, "|-|ercules" <radgray...@yahoo.com> wrote:
> >> "William Hughes" <wpihug...@hotmail.com> wrote
>
> >> > On Jun 7, 11:27 pm, "|-|ercules" <radgray...@yahoo.com> wrote:
> >> >> Here is an example of diagonalization
>
> >> >> 123
> >> >> 456
> >> >> 789

>
> >> >> Diag = 159
>
> >> >> AntiDiag = 260   <<<<<<<NEW SEQUENCE NOT ON THE LIST!
>
> >> >> YOU ALL THINK THIS WORKS ON THE LIST OF COMPUTABLE REALS!
>
> >> >> DON'T YOU!!!
>
> >> >> Gee it works for 159, must work in the infinite case too, who cares if there's
> >> >> no new digit sequence that can be formed.

>
> >> >> You're all DIM!  How can you form a new digit sequence when they're all
> >> >> computed up to infinite length?

>
> >> > You can't.  So you have a contradiction.  The assumption
> >> > that there is a list of all real numbers is wrong.

>
> >> >                - William Hughes
>
> >> You can't find a new sequence using diagonalization?
>
> > Not if you start with a list that does not exist.
>
> >                      - William Hughes
>
> I can compute the list of all computable reals.
> There's just some numbers that show
> up blank.
>
> It's trivial to compute a list that covers every digit sequence to all (infinite) finite lengths.
>

Indeed, however, although such a list
contains an infinite number
of sequences with a last digit,
it does not contain a sequence that
does not have a last digit. Since there are
computable sequences that do not have a last
digit, this is not a list of all computable
sequences.

-William Hughes

Date Subject Author
6/7/10 |-|ercules
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