On 20 Apr., 17:37, fom <fomJ...@nyms.net> wrote: > On 4/20/2013 3:16 AM, WM 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. > > 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). > > > There is no term s_n of S that contains all natural numbers. > > Since the s_n are defined in terms of finite sets, this > is true. > > > This > > condition requires > > (2) exist j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k. > > Notice that (2) is not a consequence of standard mathematics.
No? And this is all counter-argument you have collected in five posts???
How do you remove the problem that all n are in S, that all s_i are in S, that nothing else is in S, but that not all n are in one s_i? In "standard mathematics" they usually say that every guy dances with a girl, but that no guy dances with all girls. That means there is Joe who doesn't dance with Nancy and there is Karl who does not dance with Mary. But that is impossible in case of Steve who simultaneously dances with all girls he ever has met. Any other proposals?