On 20 Apr., 17:15, dullr...@sprynet.com wrote: > On Sat, 20 Apr 2013 01:16:36 -0700 (PDT), WM > > <mueck...@rz.fh-augsburg.de> wrote: > >Matheology § 255 > > >Let S = (1), (1, 2), (1, 2, 3), ... be a sequence of all finite > >initial sets s_n = (1, 2, 3, ..., n) of natural numbers n. > > >Every natural number is in some term of S. > > Yes... > > >U s_n = |N > >forall n exists i : n e s_i. > > >S is constructed by adding s_(i+1) after s_(i). So we have > >(1) forall n, forall i : (n < i <==> ~(n e s_i)) & (n >= i <==> n e > >s_i). > > No, that's backwards. n <= i if and only if n is in s_i.
You are right. The correct and better readable version (from which this has been copied with some errors) is in http://www.hs-augsburg.de/~mueckenh/KB/Matheology.pdf > > > > >There is no term s_n of S that contains all natural numbers. > > Obviously not. > > So what?
Small wonder, since nowhere "all" n can be found. If they could be found, what should them hinder to be in S after infinitely many attempts to collect them. You think there are never infinitely many attempts? Why then do you think there would be infinitely many natural numbers?