Date: Apr 20, 2013 4:16 AM
Subject: Matheology § 255
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
There is no term s_n of S that contains all natural numbers. This
(2) exist j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k.
(2) is in contradiction with (1).
A better readable version is in § 255 of