In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> 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.
And since for EVERY such s, max(s) + 1 is NOT a member of s, there cannot be all natruals in any s.
> 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.
Once before claimed and also once beforE DEBUNKED.
NOTE that a e s_b iff a <= b and ~( a e s_b) iff a > b, so that m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k iff (m <= j) & (m > k) & (n >j) & ( n <= k) But if (m <= j) & (n >j) then m < n And if (m > k) & (n <= k) then n < m So it follows that m > m and n > n, at least in WM's world.
But OUTSIDE WM's world, NOT all naturals can be in an one FISON, s_n > > 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.
Maybe no such evidence in Wolkenmuekenheim, but there is enough outside Wolkenmuekenheim, to satisfy everyone other than WM.
For every i in |N, max(s_i) = i and i+1 is not in s_i.
Which is sufficient evidence OUTSIDE WOLKENMUEKENHEIM to show that not all naturals are in any one of the s_i, at least not OUTSIDE WOLKENMUEKENHEIM . --