
Re: Matheology § 201
Posted:
Jan 27, 2013 12:05 PM


On 27 Jan., 17:49, William Hughes <wpihug...@gmail.com> wrote: > On Jan 27, 2:33 pm, WM <mueck...@rz.fhaugsburg.de> wrote: > > > > > > > On 27 Jan., 14:12, William Hughes <wpihug...@gmail.com> wrote: > > > > On Jan 27, 10:40 am, WM <mueck...@rz.fhaugsburg.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. > > The latter, > > > > > however, is claimed to follow.
No. it is claimed to follow that the diagonal differs from "all lines". That is different, for infinite sets, from "each line".
Example: Each line is followed by infinitely many lines. All lines are followed by infinitely many lines.
Further: If the list contains all terminating decimals, then the diagonal cannot differ up to a d_n that belongs to a terminating decimal, from all lines. But since there is no other d_n, the diagonal cannot differ from all lines (it differs from every line, though).
Further if we work in the terminating decimal only, then there is no actually infinite diagonal admitted and possible. This can be proved by induction. (d_1) is a finite number of digits. If (d_1, ...,d_n) is a finite number of digits, then (d_1, ...,d_n), d_n+1) is a finite number of digits too.
Regards, WM

