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Topic: Matheology § 253
Replies: 30   Last Post: Apr 22, 2013 2:44 PM

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mueckenh@rz.fh-augsburg.de

Posts: 15,340
Registered: 1/29/05
Re: Matheology § 253
Posted: Apr 20, 2013 4:33 AM
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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. Otherwise, if you cannot
find such pair, then all are in one and the same line. Or there is no
"all". Simple as that.

Regards, WM




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