Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Matheology § 293
Replies: 44   Last Post: Jun 27, 2013 2:25 PM

 Messages: [ Previous | Next ]
 mueckenh@rz.fh-augsburg.de Posts: 18,076 Registered: 1/29/05
Re: WMytheology § 293
Posted: Jun 21, 2013 4:33 PM

muec...@rz.fh-augsburg.de wrote:

> Re: WMytheology § 293
> In article
> WM <muec...@rz.fh-augsburg.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 inclusion-monotonic. 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