Virgil
Posts:
6,991
Registered:
1/6/11


Re: Matheology � 253
Posted:
Apr 20, 2013 4:12 PM


In article <fd865b028eb345b9bb713710cbe9ed97@y14g2000vbk.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 19 Apr., 22:50, Virgil <vir...@ligriv.com> wrote: > > > > There exist m, n, a b in N, > > > such that m is in FISON(a) and not in FISON(b) and n is in FISON(b) > > > and not in FISON(a). > > > > For each natural n in N, FISON(n) is defined by WM to be the set of all > > naturals less than or equal to the natural number n. > > Thus for natural numbers m ands n, > > n is in FISON(m) if and only if n <= m, and > > similarly n is NOT in FISON(m) if and only if m < n. > > Thus if m is in FISON(a) and not in FISON(b), then b < m <= a > > and if n is in FISON(b) and not in FISON(a), then a < n <= b > > > > Thus in WM's world one must have b < a and a < b simultaneously. > > It is not my world. It the world of those who insist that the sequence > 1 > 1, 2 > 1, 2, 3 > ... > contains all naturals, i.e., no one is missing, but that in no line > there are all naturals. That means, in no line all are together, which > are together in the union of all lines. Since when unioning, nothing > is added, that not has been previously in at least one of the lines, > the union must contain at least two numbers, call them m and n, which > were in the list, but not in the same line.
Nonsense!
Either m > n or n > m or m = n, but only one of these.
So either m > n and both are in every line that m is in, or n > m and both are in every line that n is in, or both are in every line the either is in.
So again WM is off in his dreamworld of Wolkenmuekenheim where anything he wishes for, except reality, occurs. 

