> 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.
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