On 20 Apr., 22:32, Virgil <vir...@ligriv.com> wrote:
> > Nevertheless, please explain according to your knowledge of > > mathematics how all n can be in the union of all s_i without all n > > being in one and the same s_i. > > Simple! At least everywhere outside of Wolkenmuekenheim: > > For all i in |N, s_(i+1) \ s_i = i+1, > and > For all i in |N, i is a member of s_1.
No. > > Thus for every s_i there is i+1 in |N not in s_i, but Union(S) = |N.
That is trivial and requires no explanation.
U s_i = |N.
But either there are all naturals in one s or not. That is simple TND. If *not* all naturals in one s, then exist j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k.
If only ~(n e s_j) & n e s_k, then also m e s_k is possible, and then n and m are in s_k. No evidence that not all naturals in one s.