
Re: WMytheology § 293
Posted:
Jun 21, 2013 4:33 PM


muec...@rz.fhaugsburg.de wrote:
> Re: WMytheology § 293 > In article > <6a564a31b1354066ad04018b2a86260e@t8g2000vbh.googlegroups.com>, > WM <muec...@rz.fhaugsburg.de> wrote: > > > On 19 Jun., 23:38, Virgil <vir...@ligriv.com> wrote:
>>> {1}, {1,2}, {1,2,3}, ... is a proper subset of N, unioning all of them >>> cannot give any more than N. > >> Certainly not more, but a lot less.
>So, if, according to WM, the union of all FISONs does not include all of N, then WM must be claiming that there are natural numbers not in any FISON.
You should understand that logical proof is meaningless in case of a contradictory theory.
Every natural number that anybody has used or ever will use belongs to a FISON, such that aleph_0 numbers must follow which are never used and cannot even be addressed.
Therefore it is a contradiction to accept a set of natural numbers that has more than any finite number of elements.
> >> The sequence of sets is inclusionmonotonic. This means there are never two >> or more sets the union of which surpasses each of the sets.
> That only holds for unions of finitely many FISONs, but there are infinitely many FISONs.
That holds for inclusion monotonic sets. The number of elements does not play a role.
The infinite set (1/2^n) is strictly monotonic. Your arguing would allow that its limit is 100 and its sum is 1000.
> Any and every union of infinitely many FISONs includes all of N,
>> The sequence contains only sets each of which lack aleph_0 natural numbers.
> The sequence of FISONs contains aleph_0 FISONs, and any and every union of aleph_0 FISONs contains all of N.
>> Lacking infinitely many natural numbers cannot be compensated by unioning >> infinitely many of such deficient sets.
> It can, and does,
That would require that a union after infinitely many unions yields infinitely many more numbers than infinitely many unions before. That is not acceptable.
Regards, WM

