On Jan 27, 5:56 pm, "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.
I would agree with you. However, WM might not. WM has agreed that induction can be used to demonstrate statements like
"there is no natural number that does not have a prime decompostion"
So, I invoke induction where it may not be necessary.