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

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Virgil

Posts: 9,012
Registered: 1/6/11
Re: Matheology � 253
Posted: Apr 20, 2013 4:12 PM
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In article
<fd865b02-8eb3-45b9-bb71-3710cbe9ed97@y14g2000vbk.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.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.
--





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