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Topic: Matheology § 201
Replies: 32   Last Post: Jan 28, 2013 2:26 PM

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 mueckenh@rz.fh-augsburg.de Posts: 18,076 Registered: 1/29/05
Re: Matheology S 201
Posted: Jan 27, 2013 12:40 PM

On 27 Jan., 17:56, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> William Hughes <wpihug...@gmail.com> writes:
> > On Jan 27, 2:33 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >> On 27 Jan., 14:12, William Hughes <wpihug...@gmail.com> wrote:
>
> >> > On Jan 27, 10:40 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >> >  <snip>

>
> >> > <... |N contains more than all (finite initial segments)
>
> >> > Piffle. |N does not contain more that all FISONS
> >> > nor has anyone claimed this.

>
> >> > the correct statement is
>
> >> >   For every initial seqment f, there is an element of |N, n(f)
> >> >   such that n(f) is not in f.  (note that n(f) may change if you
> >> > change
> >> >   f)

>
> >> > However,
>
> >> >   for ever n(f) there is an initial segment g
> >> >   such that g contains n(f)

>
> >> Of course. But these all are relative definitions. As  I said, it is
> >> impossible to define infinity absolutely. Same is valid for Cantor's
> >> diagonal: For every line n, there is an initial segment of the
> >> diagonal that differs from the first n lines. But that does not imply
> >> that there is a diagonal that differs from all lines.

>
> > It does imply that there is a diagonal that differs
> > from each line.   We start by noting that if a list L has a finite
> > definition, so does the antidiagonal of L, call it l.  By induction we
> > show that l differs from each line of L.

>
> Frankly, it seems to me that induction is utterly unnecessary.
>
> Once we define the anti-diagonal, it is a triviality to show that it
> differs from each line of L.
>
> A minor point, perhaps.

It the list is complete with respect to all terminating decimals, what
is possible, then the anti-diagonal cannot differ from all lines at
finite places.

Regards, WM