In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 20 Apr., 23:36, Virgil <vir...@ligriv.com> wrote: > > > > > 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. > > Then for all j: there is an n not in s_j.
> Does the union contain more naturals than the complete sequence?
There is an order preserving bijection between the set of all naturals, j, and the set of all FISONs, s_j. > > > > > > 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 > > if all naturals are in the terms s_i of S but not all are in one s_i.
Claimed but not proved, and, as usual, not provable without making assumptions that do not hold outside of WMytheology.
> Yes, that's the contradiction.
Claims are not proofs, and WM, while great at making claims, is really bad at proving them, particularly the ones that are false, like that one, at least outside of his wild weird world of WMytheology. > > Otherwise there are more naturals in the union than in the terms of > the sequence.
Since one new natural is added with each one new FISON added to the union, at what point in adding FISONs to that union does this alleged miracle of an additional FISON failing to add a correspondingly additional natural take place?